🧠:[<think>Okay, let's tackle this problem step by step. There are three parts, so I need to handle each one carefully.Part 1: Find the value of ( m ) and determine the explicit formula for ( f(x) ).First, the function given is a power function: ( f(x) = (m^2 + m - 1)x^{-2m^2 + m + 3} ). We're told that it's increasing on the interval ( (0, +infty) ). To find ( m ), I need to recall that for a power function ( x^k ), the function is increasing if its derivative is positive on the interval. Since the derivative of ( x^k ) is ( kx^{k-1} ), and since ( x > 0 ), the sign of the derivative depends on the exponent ( k ). Therefore, the function is increasing if ( k > 0 ).But wait, also, the coefficient of the power function is ( m^2 + m - 1 ). Since power functions are defined such that their coefficients are non-zero (otherwise, it's just a constant zero function), so we need ( m^2 + m - 1 neq 0 ). But more importantly, for the function to be increasing, both the coefficient and the exponent must satisfy certain conditions. Wait, actually, the coefficient's sign combined with the exponent's behavior determines if the function is increasing or decreasing.Let me clarify. Suppose ( f(x) = a x^k ). Then the derivative is ( f'(x) = a k x^{k - 1} ). For the derivative to be positive on ( (0, +infty) ), both ( a k ) must be positive. Since ( x^{k - 1} ) is always positive for ( x > 0 ), the sign of the derivative depends entirely on ( a k ). Therefore, ( a k > 0 ).Therefore, for ( f(x) ) to be increasing on ( (0, +infty) ), we must have:1. The coefficient ( m^2 + m - 1 ) and the exponent ( -2m^2 + m + 3 ) must multiply to a positive number. So:( (m^2 + m - 1)(-2m^2 + m + 3) > 0 )Additionally, since it's a power function, the exponent ( -2m^2 + m + 3 ) can be any real number, but the coefficient must be non-zero, as mentioned. So, ( m^2 + m - 1 neq 0 ).So, first, solve the inequality ( (m^2 + m - 1)(-2m^2 + m + 3) > 0 ).Let me factor both quadratics if possible.First quadratic: ( m^2 + m - 1 ). Discriminant ( D = 1 + 4 = 5 ), roots at ( m = [-1 pm sqrt{5}]/2 ).Second quadratic: ( -2m^2 + m + 3 ). Let's factor this. Let me write it as ( -2m^2 + m + 3 = -(2m^2 - m - 3) ). Let's factor ( 2m^2 - m - 3 ). Looking for two numbers a and b such that ( a * b = 2*(-3) = -6 ) and ( a + b = -1 ). Hmm, factors of -6 that add to -1. Let's see: 2 and -3. 2 + (-3) = -1. So, split the middle term:( 2m^2 - m - 3 = 2m^2 + 2m - 3m - 3 = 2m(m + 1) -3(m + 1) = (2m - 3)(m + 1) ). Therefore, ( -2m^2 + m + 3 = - (2m - 3)(m + 1) = (3 - 2m)(m + 1) ).Therefore, the second quadratic factors as ( (3 - 2m)(m + 1) ).So now, the inequality becomes:( (m^2 + m - 1)(3 - 2m)(m + 1) > 0 )But ( m^2 + m -1 ) has roots at ( m = [ -1 pm sqrt{5} ] / 2 ). Let's approximate these roots numerically to understand the intervals:( sqrt{5} approx 2.236 ), so:Root 1: ( (-1 + 2.236)/2 ≈ 1.236/2 ≈ 0.618 )Root 2: ( (-1 - 2.236)/2 ≈ -3.236/2 ≈ -1.618 )Therefore, the first quadratic ( m^2 + m -1 ) can be written as ( (m - 0.618)(m + 1.618) ), approximately.The second quadratic factors: ( (3 - 2m)(m + 1) ). So roots at ( m = 3/2 = 1.5 ) and ( m = -1 ).So now, we need to analyze the sign of the product ( (m^2 + m -1)(3 - 2m)(m + 1) ).To do this, let's note the critical points (roots) in order:From left to right on the number line:- ( m = -1.618 ) (root of first quadratic)- ( m = -1 ) (root from the second quadratic)- ( m = 0.618 ) (root of first quadratic)- ( m = 1.5 ) (root of the second quadratic)So the intervals to consider are:1. ( (-infty, -1.618) )2. ( (-1.618, -1) )3. ( (-1, 0.618) )4. ( (0.618, 1.5) )5. ( (1.5, infty) )In each interval, we can test the sign of each factor.Let's create a table:| Interval | m^2 + m -1 | 3 - 2m | m + 1 | Product ||------------------|------------|--------|-------|---------|| (-∞, -1.618) | + (since m^2 dominates) | + (3 - 2m when m is very negative, 3 - 2m is large positive) | - (m +1 < -1.618 +1 = -0.618, so negative) | + * + * - = - || (-1.618, -1) | - (between its roots) | + (3 - 2m: m is between -1.618 and -1, so 3 - 2m > 3 - 2*(-1) = 5, positive) | - (m +1 < -1 +1 = 0, so negative) | - * + * - = + || (-1, 0.618) | - (still between its roots) | + (3 - 2m: m between -1 and 0.618, 3 - 2m > 3 - 2*(0.618) ≈ 3 - 1.236 = 1.764, positive) | + (m +1 > -1 +1 = 0) | - * + * + = - || (0.618, 1.5) | + (outside its roots) | + (3 - 2m: m < 1.5, so 3 - 2m > 0) | + | + * + * + = + || (1.5, ∞) | + | - (3 - 2m becomes negative) | + | + * - * + = - |So the product is positive in intervals (-1.618, -1), (0.618, 1.5). Therefore, the inequality ( (m^2 + m -1)(3 - 2m)(m + 1) > 0 ) holds when ( m in (-1.618, -1) cup (0.618, 1.5) ).But we need to remember that ( m ) must also satisfy that ( f(x) ) is a power function, so the coefficient ( m^2 + m -1 neq 0 ). The roots of ( m^2 + m -1 = 0 ) are approximately ( m ≈ 0.618 ) and ( m ≈ -1.618 ), so we must exclude those exact values. Since the intervals we found are open around these roots, we're okay.Therefore, the possible values of ( m ) are ( m in (-1.618, -1) cup (0.618, 1.5) ). But since the problem states "Find the value of ( m )", implying a unique solution, maybe there's an integer or a simple fraction in these intervals? Wait, perhaps we need to check if ( m ) is an integer, but the problem doesn't specify. Hmm.Wait, maybe I made a mistake here. The problem says "the power function ( f(x) = (m^2 + m -1)x^{-2m^2 + m + 3} ) is increasing on ( (0, +infty) )". To be increasing, the derivative must be positive. So as we discussed earlier, the product ( (m^2 + m -1)(-2m^2 + m +3) > 0 ). But in our analysis, we broke down the product into ( (m^2 + m -1)(3 - 2m)(m +1) ), which is correct. Then we found the intervals where this product is positive. However, since ( m ) is a real number, unless there are additional constraints, the solution is those intervals. But the problem asks to "Find the value of ( m )", which suggests that there is a unique value. Therefore, perhaps there is a miscalculation here.Wait, maybe I need to reconsider. Since the problem is part 1 of a three-part question, maybe there is a unique integer solution for m. Let's check the intervals again:The intervals where the product is positive are (-1.618, -1) and (0.618, 1.5). Let's check if there are integer values in these intervals.First interval: between approximately -1.618 and -1. The integers in this interval would be none, since -1 is excluded.Second interval: between approximately 0.618 and 1.5. The integers here would be m = 1. So m = 1 is within (0.618, 1.5). Let's check if m = 1 satisfies the original inequality.If m = 1:Coefficient: ( 1 + 1 -1 = 1 ).Exponent: ( -2(1)^2 +1 +3 = -2 +1 +3 = 2.So f(x) = 1 * x^2, which is a parabola opening upwards. However, on the interval ( (0, +infty) ), x^2 is increasing? Wait, no. Wait, x^2 is increasing on ( (0, +infty) ) only if the derivative is positive. The derivative of x^2 is 2x, which is positive for x > 0. So yes, x^2 is indeed increasing on ( (0, +infty) ). So m = 1 is a valid solution.Wait, but in the interval (0.618, 1.5), m can be any real number there. But the problem says "Find the value of m", which suggests there's a unique solution. Hmm. Maybe there is a mistake in assuming multiple solutions? Wait, let's check if m = 1 is the only integer solution. Let's see.Wait, m must be real. But maybe the problem is designed so that m is an integer. Let me check m = 1:As above, m = 1 gives f(x) = x^2, which is indeed increasing on (0, ∞). What about m = 0. Let's check m = 0.5, which is in (0.618, 1.5)? Wait, 0.5 is less than 0.618, so not in the interval. How about m = 1. Let's check m = 1.5. Wait, m = 1.5 is the upper bound of the interval (0.618, 1.5), but it's excluded since the interval is open. So if m = 1.5, the exponent becomes:Exponent: -2*(1.5)^2 +1.5 +3 = -2*(2.25) +1.5 +3 = -4.5 +4.5 = 0. So exponent is 0, which would make f(x) = (m^2 + m -1)x^0 = (m^2 + m -1). But m =1.5:Coefficient: (2.25 + 1.5 -1) = 2.75. So f(x) = 2.75, a constant function. The derivative is zero, so it's not increasing. Hence, m cannot be 1.5.So m must be in (0.618, 1.5). If m =1, that's valid. Let's check another value in the interval, say m =1. Then that works. Let's check m = 0.618 approximately. Let's take m =0.618, which is the root. But m=0.618 is excluded since the coefficient would be zero. So in the interval (0.618,1.5), m=1 is an integer there, which works.But the problem says "Find the value of m", so maybe there is a unique solution? But according to the inequality, there's an interval of solutions. Hmm. Maybe there's a mistake in my approach. Let me re-express the derivative.Wait, f(x) = (m² + m -1)x^{-2m² + m +3}. The derivative f’(x) = (m² + m -1)(-2m² + m +3)x^{-2m² + m +2}. For f(x) to be increasing on (0, +infty), f’(x) > 0 for all x >0. Since x^{-2m² + m +2} is positive for all x >0, regardless of the exponent. Therefore, the sign of f’(x) is determined by the product (m² + m -1)(-2m² + m +3). Therefore, this product must be positive. So yes, that's correct.But the original problem says "Find the value of m". So if the solution is an interval, but the problem says "the value", maybe I need to check if there is a specific m that makes the exponent an integer or something. Wait, or maybe m must be integer? The problem doesn't specify. Wait, the problem says "real number t" in part 2, so maybe m is also a real number. Therefore, the answer for part 1 is the interval we found: ( m in (-(1+sqrt{5})/2, -1) cup ( (sqrt{5}-1)/2, 3/2 ) ). But the problem says "Find the value of m", so perhaps there is a specific m. Maybe there's an error in my previous thought.Wait, maybe there is a miscalculation when solving the inequality. Let me check again.Wait, original inequality is (m² + m -1)(-2m² + m +3) > 0. Which factors into (m² + m -1)(3 - 2m)(m +1) > 0.The roots are at m ≈ -1.618, -1, 0.618, 1.5. So sign chart as before. But perhaps the problem requires m to be an integer. Let me check m =1, which is in (0.618,1.5). For m=1:Coefficient: 1 +1 -1 =1.Exponent: -2(1) +1 +3 =2. So f(x) =x², which is indeed increasing on (0, ∞) since derivative is 2x >0. So m=1 is valid.Is there another integer? m=0: in (-1.618, -1)? But m=0 is not in that interval. m=-2: in (-∞, -1.618). The interval (-1.618, -1) is between approximately -1.618 and -1. So m=-1.5 is in that interval. Let's check m=-1.5:Coefficient: (-1.5)^2 + (-1.5) -1 =2.25 -1.5 -1= -0.25.Exponent: -2*(-1.5)^2 + (-1.5) +3 = -2*(2.25) -1.5 +3= -4.5 -1.5 +3= -3. So exponent is -3.Thus, f(x)= -0.25 x^{-3}. The derivative is -0.25*(-3)x^{-4}=0.75x^{-4}, which is positive for x>0. Therefore, the function is increasing. So m=-1.5 is also a valid solution.But m=-1.5 is a real number. Therefore, the problem might accept any m in those intervals. But the problem says "Find the value of m", which is confusing. Maybe there's a restriction I missed.Wait, looking back at the problem statement: "Given the power function f(x)=(m² +m−1)x^{-2m² +m+3} is increasing on the interval (0, +∞)". The problem says "the power function", which is a term that sometimes refers specifically to functions of the form f(x) = kx^n where k ≠0. So the coefficient (m² +m−1) must not be zero, which we already considered. But maybe the problem is from a textbook where m is an integer? If so, m=1 is the only integer solution. But the problem didn't specify. Hmm.Wait, checking again the problem statement in Chinese might clarify, but since it's translated here, maybe there's a standard assumption. Alternatively, maybe I need to check if there's a unique m such that the exponent is even or something. Wait, but exponent is -2m² +m +3. If exponent is 2 as in m=1 case, that's a parabola. If exponent is another number, like with m=-1.5, exponent is -3. So the function x^{-3} is decreasing, but multiplied by a negative coefficient, so f(x)= -0.25x^{-3} = -0.25/x³. The derivative is 0.75/x⁴, which is positive. So it's increasing. So that's a valid case.Therefore, unless there is more constraints, there are infinitely many solutions for m. However, the problem says "Find the value of m", implying a unique solution. So maybe I made a mistake in the derivative.Wait, wait. Let's recast the problem. For f(x) to be increasing, the exponent must satisfy that the function is increasing. For a power function x^k, if k >0, then x^k is increasing for x>0 only if k >0 and the coefficient is positive? Wait, no. Wait, f(x) = Cx^k. The derivative is Ck x^{k-1}. For this to be positive for all x>0, need Ck >0. So either C>0 and k>0, or C<0 and k<0.But in our case, f(x) is given as a power function, which is increasing on (0, ∞). Therefore, either:Case 1: Coefficient (m² +m−1) >0 and exponent (-2m² +m +3) >0, orCase 2: Coefficient (m² +m−1) <0 and exponent (-2m² +m +3) <0.Therefore, we need to solve these two cases.Let's consider Case 1:1. m² +m−1 >0 and -2m² +m +3 >0.First inequality: m² +m−1 >0. As before, roots at m = [ -1 ± sqrt(5) ] /2. Therefore, m < [ -1 - sqrt(5) ] /2 ≈ -1.618 or m > [ -1 + sqrt(5) ] /2 ≈0.618.Second inequality: -2m² +m +3 >0. This is equivalent to 2m² -m -3 <0. Factorizing as before: (2m -3)(m +1) <0. So roots at m=1.5 and m=-1. Therefore, the inequality holds for m ∈ (-1,1.5).Therefore, Case 1 intersection: (m < -1.618 or m >0.618) intersect (-1 <m <1.5). Therefore:For m < -1.618 intersect (-1 <m <1.5): empty set.For m >0.618 intersect (-1 <m <1.5): (0.618,1.5).So Case 1 solutions: m ∈ (0.618,1.5).Case 2:Coefficient <0 and exponent <0.Coefficient <0: m² +m -1 <0 ⇒ m ∈ (-1.618,0.618).Exponent <0: -2m² +m +3 <0 ⇒2m² -m -3 >0 ⇒ m ∈ (-∞, -1) ∪ (1.5, ∞).Therefore, Case 2 intersection: (-1.618,0.618) intersect [ (-∞, -1) ∪ (1.5, ∞) ] ⇒ (-1.618, -1).Thus, Case 2 solutions: m ∈ (-1.618, -1).Therefore, combining Case 1 and Case 2, solutions are m ∈ (-1.618, -1) ∪ (0.618,1.5), as before.But again, since the problem says "Find the value of m", perhaps there's an integer in these intervals. In Case 2 interval (-1.618, -1), there are no integers. In Case 1 interval (0.618,1.5), the only integer is m=1. Therefore, m=1 is the only integer solution, and likely the answer expected here.Therefore, m=1. Then, substituting m=1 into f(x):Coefficient: 1 +1 -1=1.Exponent: -2(1)^2 +1 +3= -2+1+3=2.Thus, f(x)=x².So part 1 answer: m=1, f(x)=x².Part 2: For any ( x in [1, 2] ), there exist ( x_1, x_2 in [1, 2] ) such that ( f(x) leq f(x_1) ), ( g(x) leq g(x_2) ), if ( f(x_1) = g(x_2) ), find the value of the real number ( t ).First, we need to parse the problem statement."For any ( x in [1, 2] ), there exist ( x_1, x_2 in [1, 2] ) such that ( f(x) leq f(x_1) ), ( g(x) leq g(x_2) ), if ( f(x_1) = g(x_2) ), find the value of the real number ( t )."So, for every x in [1,2], we can find some x1 and x2 in [1,2] such that f(x) is less than or equal to f(x1), and g(x) is less than or equal to g(x2). Then, given that f(x1) = g(x2), find t.Wait, but f(x) is given as x² (from part 1), and g(x) is given as -x² +2|x| +t. Since x is in [1,2], |x|=x, so g(x) simplifies to -x² +2x +t.First, note that for any x in [1,2], f(x) =x². The maximum value of f(x) on [1,2] is at x=2, which is 4. The minimum is at x=1, which is 1. Similarly, for g(x) = -x² +2x +t. Let's analyze g(x):g(x) = -x² +2x +t. It's a quadratic function opening downward. The vertex is at x = -b/(2a) = -2/(2*(-1))=1. So at x=1, g(x) reaches its maximum. Then, on [1,2], g(x) is decreasing. Therefore, the maximum of g on [1,2] is at x=1: g(1) = -1 +2 +t =1 + t. The minimum is at x=2: g(2)= -4 +4 +t = t.The problem states that for any x in [1,2], there exist x1, x2 in [1,2] such that f(x) ≤ f(x1) and g(x) ≤ g(x2). Since f(x) ≤ f(x1) for some x1, and since f(x1) is maximized at x1=2 (where f(x1)=4), then for any x, f(x) ≤4. Similarly, for g(x) ≤g(x2), since g(x2) is maximized at x2=1 (g(x2)=1 + t). Therefore, for any x in [1,2], f(x) ≤4 and g(x) ≤1 +t. But the problem says "there exist x1, x2" such that these inequalities hold. However, since for any x in [1,2], f(x) ≤4 (since 4 is the maximum of f on [1,2]), then x1=2 will work. Similarly, for g(x) ≤1 +t, since 1 +t is the maximum of g on [1,2], then x2=1 will work.But then, the condition is that f(x1) = g(x2). Since x1 is 2 and x2 is 1, then f(2)=4 and g(1)=1 + t. Therefore, setting f(x1) = g(x2) gives 4 =1 + t, so t=3.Wait, but let me check again. The problem says "For any x ∈ [1,2], there exist x1, x2 ∈ [1,2] such that f(x) ≤ f(x1) and g(x) ≤ g(x2), if f(x1) = g(x2), find t".So, for every x in [1,2], there's some x1 and x2 (possibly depending on x) in [1,2] such that f(x) ≤ f(x1) and g(x) ≤ g(x2), and also f(x1) = g(x2). So the equality f(x1) = g(x2) must hold for these x1 and x2 that work for each x.But how is this possible? Because for each x, x1 and x2 could vary. However, perhaps the key is that the maximum of f and the maximum of g must be equal, so that for any x, we can take x1=2 and x2=1, leading to f(x1)=4 and g(x2)=1 +t. So setting 4=1 +t gives t=3. Alternatively, maybe for some x, we need different x1 and x2?Wait, but if we take x1 and x2 depending on x, but the problem says "for any x ∈ [1,2], there exist x1, x2 ∈ [1,2] such that f(x) ≤ f(x1), g(x) ≤g(x2), and f(x1)=g(x2)". So for each x, there's some x1 and x2 (possibly depending on x) such that those inequalities hold and f(x1)=g(x2).But then, for each x, f(x) ≤ f(x1) and g(x) ≤g(x2), with f(x1)=g(x2). So for each x, f(x) ≤ some value, and g(x) ≤ the same value. Therefore, the maximum of f(x) and g(x) on [1,2] must be equal, and t is chosen so that the maximum of f equals the maximum of g.Wait, but f(x) reaches maximum 4, and g(x) reaches maximum 1 + t. So setting 4 =1 +t gives t=3. Then, for any x, f(x) ≤4 and g(x) ≤4. So for each x, we can take x1=2 and x2=1 (since g(1)=4 when t=3), so that f(x1)=4 and g(x2)=4, and f(x) ≤4 and g(x) ≤4 for all x.Alternatively, even if x varies, since f(x) ≤4 and g(x) ≤4 (if t=3), then for any x, choosing x1=2 and x2=1 gives the required inequalities. Therefore, t=3 is the answer.But let's verify with t=3. Then g(x)= -x² +2x +3. The maximum at x=1 is -1 +2 +3=4, and the minimum at x=2 is -4 +4 +3=3. So for any x in [1,2], g(x) ≤4. Then, for any x in [1,2], f(x) ≤4 and g(x) ≤4. So we can take x1=2 and x2=1, which gives f(x1)=4 and g(x2)=4, hence f(x1)=g(x2). Therefore, for all x, f(x) ≤4 and g(x) ≤4, so the conditions are satisfied. Therefore, t=3 is correct.Part 3: If ( 2^x h(2x) + lambda h(x) geq 0 ) holds for all ( x in [1, 2] ), find the range of the real number ( lambda ).Given ( h(x) = 2^x - 2^{-x} ). Let's first compute h(2x):h(2x) = 2^{2x} - 2^{-2x} = (2^x)^2 - (2^{-x})^2.Therefore, the expression becomes:2^x [ (2^x)^2 - (2^{-x})^2 ] + lambda (2^x - 2^{-x}) ≥0.Simplify:2^x*(2^{2x} - 2^{-2x}) + λ*(2^x -2^{-x}) = 2^{3x} - 2^{-x} + λ*2^x - λ*2^{-x}.Let me write all terms in terms of 2^x and 2^{-x}. Let’s denote y = 2^x. Since x ∈ [1,2], then y ∈ [2,4]. Therefore, 2^{-x} =1/y.So substituting:2^{3x} = y^3,-2^{-x} = -1/y,λ*2^x = λ y,-λ*2^{-x} = -λ/y.Therefore, the expression becomes:y^3 - (1/y) + λ y - (λ / y) ≥0.Factor terms:y^3 + λ y - (1/y + λ / y) ≥0.Factor out y from the first two terms and -1/y from the last two:y(y^2 + λ) - (1 + λ)/y ≥0.Alternatively, combine all terms over a common denominator y:[ y^4 + λ y^2 -1 -λ ] / y ≥0.Since y >0 (as y=2^x >0), the denominator y is positive, so the inequality simplifies to:y^4 + λ y^2 -1 -λ ≥0 for all y ∈ [2,4].Let’s denote z = y^2. Since y ∈ [2,4], z ∈ [4,16].Then the inequality becomes:z^2 + λ z -1 -λ ≥0 for all z ∈ [4,16].Let’s rearrange:z^2 -1 + λ(z -1) ≥0.Factor:(z^2 -1) + λ(z -1) = (z -1)(z +1) + λ(z -1) = (z -1)(z +1 + λ) ≥0.So the inequality is (z -1)(z +1 + λ) ≥0 for all z ∈ [4,16].Since z ∈ [4,16], z -1 ≥3 >0. Therefore, the inequality reduces to (z +1 + λ) ≥0 for all z ∈ [4,16].Because (z -1) is positive, so the product is non-negative if and only if (z +1 + λ) ≥0.Therefore, z +1 + λ ≥0 for all z ∈ [4,16]. The minimum value of z is 4, so 4 +1 +λ ≥0 ⇒5 +λ ≥0 ⇒λ ≥-5. However, since we need the inequality to hold for all z up to 16, but z +1 +λ is increasing in z. Therefore, the minimal value occurs at z=4. So if 4 +1 +λ ≥0, then for all higher z (up to16), z +1 +λ is larger, so automatically non-negative. Therefore, the condition is λ ≥-5.But wait, let me confirm.The inequality (z -1)(z +1 + λ) ≥0. Since z -1 >0 for z ≥4, then the entire expression is non-negative if and only if z +1 +λ ≥0. Therefore, z +1 +λ ≥0 for all z ∈[4,16]. The minimal z is 4, so 4 +1 +λ ≥0 ⇒λ ≥-5. Hence, the range of λ is λ ≥-5.But let me check with z=4:At z=4: (4 -1)(4 +1 +λ) =3*(5 +λ) ≥0. Since 3>0, need 5 +λ ≥0 ⇒λ ≥-5.For z=16: (16 -1)(16 +1 +λ)=15*(17 +λ) ≥0. Since 15>0, 17 +λ ≥0 ⇒λ ≥-17. But since we need this to hold for all z ∈[4,16], the most restrictive condition is λ ≥-5. Because when z increases, z +1 +lambda increases, so the minimum is at z=4. Thus, λ ≥-5.Therefore, the range of λ is [-5, ∞).But let's test λ=-5:Plug into the original inequality:(z -1)(z +1 -5) = (z -1)(z -4). For z ∈ [4,16], when z=4, (4-1)(4-4)=3*0=0, which is okay. For z>4, (z-1)(z -4) ≥0 since z-4 ≥0, z-1 >0. So the expression is non-negative. Therefore, λ=-5 is acceptable. If λ < -5, then at z=4, (4 -1)(5 +λ) =3*(5 +λ) <0, which violates the inequality. Therefore, λ must be ≥-5.Hence, the range of λ is [-5, +∞).Final Answer(1) ( m = boxed{1} ), ( f(x) = boxed{x^2} );(2) ( t = boxed{3} );(3) The range of ( lambda ) is ( boxed{[-5, +infty)} ).</think>Given the power function ( f(x) = (m^2 + m - 1)x^{-2m^2 + m + 3} ) is increasing on the interval ( (0, +infty) ), we need to find the value of ( m ) and determine the explicit formula for ( f(x) ).1. For ( f(x) ) to be increasing on ( (0, +infty) ), the product of the coefficient and the exponent must be positive. Solving the inequality ( (m^2 + m - 1)(-2m^2 + m + 3) > 0 ), we find the intervals for ( m ). The valid integer solution within these intervals is ( m = 1 ). Substituting ( m = 1 ) into ( f(x) ), we get ( f(x) = x^2 ).2. For any ( x in [1, 2] ), there exist ( x_1, x_2 in [1, 2] ) such that ( f(x) leq f(x_1) ) and ( g(x) leq g(x_2) ). Given ( f(x_1) = g(x_2) ), we need to find ( t ). The maximum values of ( f(x) ) and ( g(x) ) on ([1, 2]) are 4 and ( 1 + t ) respectively. Setting ( 4 = 1 + t ), we find ( t = 3 ).3. For the inequality ( 2^x h(2x) + lambda h(x) geq 0 ) to hold for all ( x in [1, 2] ), we substitute ( h(x) = 2^x - 2^{-x} ) and simplify. Converting to ( z = y^2 ) where ( y = 2^x ), we find the inequality reduces to ( (z - 1)(z + 1 + lambda) geq 0 ). The minimum value of ( z ) is 4, leading to ( lambda geq -5 ).Final answers:1. ( m = boxed{1} ), ( f(x) = boxed{x^2} );2. ( t = boxed{3} );3. The range of ( lambda ) is ( boxed{[-5, +infty)} ).,<think>Okay, so I need to solve this three-part problem. Let's start with part (1). We have a power function f(x) = (m² + m -1)x^{-2m² + m +3}, and it's given that this function is increasing on the interval (0, +infty). I need to find the value of m and determine the explicit formula for f(x).First, recall that a power function is generally of the form f(x) = kx^n, where k is a constant and n is the exponent. For f(x) to be increasing on (0, ∞), its derivative must be positive on that interval.So let's compute the derivative of f(x). The derivative of x^n is n x^{n - 1}. Therefore, f'(x) = (m² + m -1) * (-2m² + m +3) x^{-2m² + m +3 -1}.Simplify the exponent: -2m² + m +3 -1 = -2m² + m +2.So f'(x) = (m² + m -1)(-2m² + m +2) x^{-2m² + m +2}.For f(x) to be increasing on (0, ∞), the derivative f'(x) must be positive for all x > 0. Since x^{-2m² + m +2} is positive for all x > 0 (because any positive number raised to a real exponent is positive), the sign of f'(x) depends on the product (m² + m -1)(-2m² + m +2).Therefore, we need (m² + m -1)(-2m² + m +2) > 0.So our inequality is (m² + m -1)(-2m² + m +2) > 0.Let me solve this inequality. To do this, I need to find the values of m where this product is positive. First, let's factor both quadratics if possible, or find their roots.First quadratic: m² + m -1. The discriminant is 1 + 4 = 5. So roots are [-1 ± sqrt(5)]/2. Let me compute them:Root1: [-1 + sqrt(5)]/2 ≈ (-1 + 2.236)/2 ≈ 1.236/2 ≈ 0.618Root2: [-1 - sqrt(5)]/2 ≈ (-1 -2.236)/2 ≈ -3.236/2 ≈ -1.618Second quadratic: -2m² + m +2. Let's write it as -2m² + m +2. Multiply by -1 to make it easier: 2m² - m -2. The discriminant is 1 + 16 = 17. So roots are [1 ± sqrt(17)]/(4). Wait, sqrt(17) ≈4.123, so:Root1: [1 + sqrt(17)]/4 ≈ (1 +4.123)/4 ≈5.123/4≈1.281Root2: [1 - sqrt(17)]/4≈(1 -4.123)/4≈-3.123/4≈-0.781But since the original quadratic is -2m² + m +2, the roots are the same as for 2m² -m -2=0, which are [1 ± sqrt(17)]/4. Wait, let me check:Quadratic equation: ax² + bx +c =0. The roots are [-b ± sqrt(b² -4ac)]/(2a). So for 2m² -m -2=0, the roots are [1 ± sqrt(1 +16)]/(4) = [1 ± sqrt(17)]/4. So that's correct.But since the original quadratic is -2m² + m +2, its roots are the same as 2m² -m -2=0, so m = [1 ± sqrt(17)]/4. Therefore, the quadratic -2m² +m +2 can be written as -2(m - [1 + sqrt(17)]/4)(m - [1 - sqrt(17)]/4).But maybe instead of factoring, it's better to analyze the sign of each quadratic.So, let's analyze the sign charts for each quadratic:First quadratic: m² + m -1. As we saw, roots at m ≈0.618 and m≈-1.618.This is a parabola opening upwards (since coefficient of m² is positive). Therefore, it is positive outside the roots and negative between the roots.So, m² + m -1 >0 when m < (-1 -sqrt(5))/2 or m > (-1 + sqrt(5))/2.Similarly, the second quadratic: -2m² +m +2. Since the coefficient of m² is negative, it's a downward opening parabola. The roots are at m≈1.281 and m≈-0.781. Therefore, the quadratic is positive between its roots and negative outside.So, -2m² +m +2 >0 when -0.781 < m <1.281.Now, our inequality is (m² + m -1)(-2m² +m +2) >0. We need to find m where the product is positive.To solve this, we can consider the intervals determined by the critical points of both quadratics: m≈-1.618, m≈-0.781, m≈0.618, m≈1.281.So intervals are:1. (-infty, -1.618)2. (-1.618, -0.781)3. (-0.781, 0.618)4. (0.618, 1.281)5. (1.281, +infty)In each interval, we can determine the sign of each quadratic and then the product.Let's proceed interval by interval.1. Interval (-infty, -1.618):First quadratic (m² + m -1): here, m is less than -1.618. Let's pick m = -2. Then m² +m -1 = 4 -2 -1 =1 >0.Second quadratic (-2m² +m +2): m = -2. Compute -2*(4) + (-2) +2 = -8 -2 +2 = -8 <0.Product: positive * negative = negative. So the product is negative here. Doesn't satisfy the inequality.2. Interval (-1.618, -0.781):Take m = -1.First quadratic: (-1)^2 + (-1) -1 =1 -1 -1= -1 <0.Second quadratic: -2*(-1)^2 + (-1) +2 = -2*1 -1 +2 = -2 -1 +2= -1 <0.Product: negative * negative = positive. So product positive here. This interval satisfies the inequality.3. Interval (-0.781, 0.618):Take m=0.First quadratic: 0 +0 -1= -1 <0.Second quadratic: -0 +0 +2=2>0.Product: negative * positive = negative. Doesn't satisfy.4. Interval (0.618, 1.281):Take m=1.First quadratic:1 +1 -1=1>0.Second quadratic: -2*1 +1 +2= -2 +3=1>0.Product: positive * positive = positive. Satisfies the inequality.5. Interval (1.281, +infty):Take m=2.First quadratic:4 +2 -1=5>0.Second quadratic: -2*4 +2 +2= -8 +4= -4<0.Product: positive * negative = negative. Doesn't satisfy.So the solutions are the intervals (-1.618, -0.781) and (0.618, 1.281). However, m needs to be such that the exponent in f(x) is valid? Wait, the problem states it's a power function. A power function is defined for x >0, and the exponent can be any real number. So as long as the exponent is a real number, which it is here. So we don't have restrictions on m other than the ones from the derivative being positive.But the problem is likely expecting integer values for m? Wait, not necessarily. But maybe m is an integer? Wait, the problem says "Find the value of m", so likely a specific value. Wait, but the intervals are ranges. So maybe m is an integer within those intervals? Let's check.Wait, the intervals are approximately (-1.618, -0.781) and (0.618,1.281). If m is an integer, possible m values in the first interval would be m=-1, but -1.618 to -0.781 is between approximately -1.6 to -0.78. So integers in that interval: only m=-1. Let's check m=-1.Check m=-1:First quadratic: (-1)^2 + (-1) -1=1-1-1=-1 <0.Second quadratic: -2*(-1)^2 + (-1) +3= -2*1 -1 +3= -2 -1 +3=0.Wait, original exponent is -2m² +m +3. Wait, in the derivative we had the exponent as -2m² +m +2. Wait, actually, let me check again.Wait, the original exponent in f(x) is -2m² +m +3. Then when taking the derivative, we subtract 1, so exponent becomes -2m² +m +3 -1 = -2m² +m +2. But in the derivative, the coefficient is (m² + m -1)*(-2m² +m +2). Wait, but when m=-1, let's compute the two quadratics.First quadratic: m² +m -1 =1 -1 -1=-1.Second quadratic: -2m² +m +2= -2*1 +(-1)+2= -2 -1 +2= -1.So their product is (-1)*(-1)=1>0. So m=-1 would satisfy the inequality. But when m=-1, the original exponent in f(x) is -2*(-1)^2 + (-1) +3= -2*1 -1 +3= -2 -1 +3=0. So f(x)= (m² +m -1)x^0= (1 -1 -1)x^0= (-1)*1= -1. So f(x)=-1. Then f(x) is a constant function. But the derivative of a constant function is zero. So the problem states that f is increasing. However, if f is constant, it's not increasing. So there's a contradiction here.Wait, so m=-1 gives f(x)=-1, which is a constant function. But the problem says it's increasing, so m=-1 is invalid. Therefore, even though m=-1 is in the interval where the product is positive, the function is not increasing but constant. Hence, m=-1 is not acceptable.So we need to check if for other integer m's in those intervals.First interval (-1.618, -0.781): possible integer m=-1, but m=-1 gives a constant function, which is invalid. So no integers here.Second interval (0.618,1.281). Possible integer m=1. Let's check m=1.First quadratic: m=1, 1 +1 -1=1>0.Second quadratic: -2*1 +1 +3= -2 +1 +3=2>0. Wait, but the second quadratic in the derivative was -2m² +m +2. Wait, hold on.Wait, when m=1, the second quadratic is -2(1)^2 +1 +2= -2 +1 +2=1>0.So product is 1*1=1>0. Then exponent in f(x) is -2*1 +1 +3= -2 +1 +3=2. So f(x)=(1 +1 -1)x^2= (1)x^2. Therefore, f(x)=x². The derivative is 2x, which is positive for x>0. So f(x)=x² is indeed increasing on (0, +infty). Wait, but hold on: f(x)=x² is decreasing on (-infty,0) and increasing on (0, +infty). Wait, but the problem states that f is increasing on (0, +infty), which is true. So m=1 is acceptable.Wait, but let me check the derivative again. For m=1, f(x)=x², derivative is 2x, which is positive for x>0. So indeed, increasing on (0, ∞). So m=1 is a solution.Wait, but the interval (0.618,1.281) includes m=1. Is there another integer here? m=0.618 to 1.281, so m=1 is the only integer. So m=1 is a solution.But wait, in the interval (0.618,1.281), m can take real values. But the problem says "Find the value of m". If they expect a specific value, but the inequality is satisfied for all m in that interval. However, since the problem is likely structured to have integer solutions, maybe m=1 is the answer. Let me check.But let's see if there are other possible m's. For example, suppose m=0. Let's check m=0.But m=0 is not in the interval (0.618,1.281). So m=0 would be in the interval (-0.781,0.618). But m=0 is not in the solution intervals. So m=1 is the only integer solution. Wait, but what if m is not an integer? For example, if m=1.2, which is in the interval (0.618,1.281). Then, compute f(x):First, coefficient m² +m -1: (1.44) +1.2 -1=1.64>0.Second, exponent: -2*(1.44) +1.2 +3= -2.88 +4.2=1.32. So f(x)=1.64x^{1.32}. The derivative is 1.64*1.32x^{0.32}, which is positive since all terms are positive. So any m in (0.618,1.281) would work. However, the problem says "Find the value of m", implying a unique solution. But m is in an interval. That seems conflicting.Wait, maybe the exponent must be an integer? Because power functions with integer exponents are more standard. But the problem doesn't specify that. Hmm. Let's check the problem statement again.It says "Given the power function f(x)=(m² +m -1)x^{-2m² +m +3} is increasing on the interval (0, +infty)...". So it's a power function, which can have any real exponent. Therefore, m can be any real number in the intervals where the derivative is positive. However, the problem says "Find the value of m", which suggests that m is uniquely determined. So maybe there's an error in my previous reasoning.Wait, let's check again. The problem says "Find the value of m". So likely, there is only one m. But according to our analysis, there's an interval of m's. Wait, but maybe there are constraints from the coefficient m² + m -1. Wait, because f(x) is a power function, but the coefficient must not be zero, since a power function is usually of the form kx^n where k ≠0. Therefore, m² +m -1 ≠0. So m ≠ [ -1 ± sqrt(5) ] /2 ≈0.618 and ≈-1.618. So m cannot be those roots, but those are already the critical points from the first quadratic. So in our previous intervals, m is not equal to 0.618 or -1.618, but the intervals exclude those points.But still, the problem says "Find the value of m". If it's expecting a unique solution, perhaps there is a mistake in my analysis.Wait, maybe I need to check the original function. The function f(x) is a power function, which is increasing on (0, ∞). For a power function x^n, it's increasing on (0, ∞) if n >0. Wait, but here the coefficient is (m² +m -1). So f(x) = (m² +m -1) x^{exponent}.But if the coefficient is negative, even if the exponent is positive, the function would be decreasing. For example, if coefficient is negative and exponent is positive, then as x increases, x^exponent increases, but multiplied by a negative coefficient, making f(x) decrease.Similarly, if the coefficient is positive and exponent positive, the function increases. If coefficient is positive and exponent negative, the function decreases. If coefficient is negative and exponent negative, the function increases (since x^{-n} is decreasing, multiplied by negative coefficient becomes increasing). Wait, let's see.Suppose f(x) = k x^n. Then:- If k >0 and n >0: increasing on (0, ∞) when n >0? No, actually, the function x^n with n >0 is increasing for x >0 if n >0, yes. Wait, no, wait. Wait, x^n with n >0 is increasing for x >0 only if n >1. Wait, no. Let me check.Wait, the derivative of x^n is n x^{n-1}. So if n >0, the derivative is positive for x >0, so the function is increasing on (0, ∞) if n >0. If n=0, it's a constant. If n <0, the derivative is negative, so decreasing. Wait, that's right. For example, x^2 is increasing on (0, ∞) since derivative 2x >0. x^{1} is also increasing. x^{0.5} (sqrt(x)) has derivative 0.5x^{-0.5} >0, so increasing. Similarly, x^{-1} has derivative -x^{-2} <0, decreasing. So yes, if n >0, the power function x^n is increasing on (0, ∞). If n <0, it's decreasing. But here, f(x) is (m² +m -1) x^{-2m² +m +3}. So the exponent is -2m² +m +3. For f(x) to be increasing, two things must hold:1. The coefficient (m² +m -1) must be positive, because if the coefficient is positive and the exponent is positive, then the function is increasing. If the coefficient is negative and the exponent is negative, then the function is increasing (since negative times negative exponent gives positive). Wait, let me formalize:If f(x) = k x^n. Then:- If k >0 and n >0: f is increasing on (0, ∞) because derivative k n x^{n-1} >0.- If k >0 and n <0: f is decreasing.- If k <0 and n >0: f is decreasing (since negative times positive derivative).- If k <0 and n <0: f is increasing (negative times negative derivative is positive).Therefore, for f(x) to be increasing, either:1. k >0 and n >0, or2. k <0 and n <0.Therefore, we have two cases:Case 1: (m² +m -1) >0 and (-2m² +m +3) >0.Case 2: (m² +m -1) <0 and (-2m² +m +3) <0.Therefore, our previous inequality (m² +m -1)(-2m² +m +2) >0 was from the derivative. Wait, but in the derivative, the exponent after differentiation is -2m² +m +2. Wait, but actually, the derivative is:f'(x) = (m² +m -1) * (-2m² +m +3) x^{-2m² +m +2}.So the exponent in the derivative is -2m² +m +2, but the coefficient is (m² +m -1)*(-2m² +m +3).Wait, but for the function to be increasing, the derivative must be positive. So:(m² +m -1)*(-2m² +m +3) >0.But the original exponent in f(x) is -2m² +m +3. However, in the derivative, the exponent becomes -2m² +m +3 -1= -2m² +m +2.But regardless, the critical part is the sign of the coefficient (m² +m -1)*(-2m² +m +3). So as analyzed before.But going back to the two cases for the function to be increasing:Either:1. Coefficient (m² +m -1) >0 and exponent (-2m² +m +3) >0, leading to f increasing because positive coefficient and positive exponent.OR2. Coefficient (m² +m -1) <0 and exponent (-2m² +m +3) <0, leading to f increasing because negative coefficient and negative exponent (so product is positive).Therefore, the two cases. Let's solve these two cases.Case 1:(m² +m -1 >0) and (-2m² +m +3 >0).Case 2:(m² +m -1 <0) and (-2m² +m +3 <0).Let's solve Case 1 first.Case 1: m² +m -1 >0 and -2m² +m +3 >0.We already found that m² +m -1 >0 when m < (-1 -sqrt(5))/2 ≈-1.618 or m > (-1 +sqrt(5))/2 ≈0.618.-2m² +m +3 >0. Let's solve -2m² +m +3 >0.Multiply both sides by -1 (inequality sign reverses): 2m² -m -3 <0.Find roots of 2m² -m -3=0.Discriminant: 1 +24=25. So roots are [1 ±5]/4.So m=(1 +5)/4=6/4=1.5 or m=(1 -5)/4=-4/4=-1.Therefore, the quadratic 2m² -m -3 factors as (2m -3)(m +1). So 2m² -m -3 <0 when -1 <m <1.5.Therefore, original inequality -2m² +m +3 >0 when -1 <m <1.5.Therefore, in Case 1, we need m < -1.618 or m >0.618 (from first inequality) and -1 <m <1.5 (from second inequality). Therefore, intersecting these intervals:For m < -1.618 and -1 <m <1.5: no overlap.For m >0.618 and -1 <m <1.5: intersection is (0.618,1.5).Therefore, Case 1 solutions: (0.618,1.5).Case 2: m² +m -1 <0 and -2m² +m +3 <0.m² +m -1 <0 when -1.618 <m <0.618.-2m² +m +3 <0 when m < -1 or m >1.5.Intersection of -1.618 <m <0.618 and m < -1 or m >1.5: The overlap is (-1.618 <m < -1).Therefore, Case 2 solutions: (-1.618, -1).Therefore, overall solutions for m are (-1.618, -1) and (0.618,1.5).But wait, in the previous analysis, we had considered the product (m² +m -1)(-2m² +m +2) >0. Wait, but now according to the two cases, the solutions are different intervals. Wait, but there's a discrepancy here. Which is correct?Wait, perhaps I made a mistake earlier when considering the derivative. Let's clarify.The derivative is f'(x) = (m² +m -1)*(-2m² +m +3) x^{-2m² +m +2}.We need f'(x) >0 for all x >0. Since x^{-2m² +m +2} >0 for all x >0, the sign of f'(x) is determined by (m² +m -1)*(-2m² +m +3).Therefore, the condition is (m² +m -1)*(-2m² +m +3) >0.This is equivalent to the two cases:Either both factors positive or both negative.So, either:1. m² +m -1 >0 and -2m² +m +3 >0, or2. m² +m -1 <0 and -2m² +m +3 <0.Which is exactly what we did in Case1 and Case2.So solutions are m ∈ (-1.618, -1) ∪ (0.618,1.5).But earlier when considering the intervals, we had (-1.618, -0.781) and (0.618,1.281). Wait, but that was for the product (m² +m -1)(-2m² +m +2). Wait, but there's a discrepancy here. Wait, no, the derivative coefficient is (m² +m -1)*(-2m² +m +3). Not (-2m² +m +2). Earlier, I might have made a mistake here.Wait, let's check the exponent in the derivative. The original exponent in f(x) is -2m² +m +3. Then the derivative of x^{n} is n x^{n -1}, so the exponent becomes -2m² +m +3 -1= -2m² +m +2. Therefore, the derivative is:f'(x) = (m² +m -1)*(-2m² +m +3) x^{-2m² +m +2}.Wait, the coefficient is (m² +m -1)*(-2m² +m +3). Therefore, the previous analysis where we considered (m² +m -1)(-2m² +m +3) >0 is correct.But when I first analyzed the product (m² +m -1)(-2m² +m +2) >0, that was a mistake. The correct coefficient in the derivative is (m² +m -1)*(-2m² +m +3). Therefore, I had an error earlier. So the correct inequality is (m² +m -1)*(-2m² +m +3) >0.That explains the confusion. Therefore, we need to redo the analysis with the correct quadratic.So, correct quadratics:First quadratic: m² +m -1, roots at m = [-1 ± sqrt(5)]/2 ≈-1.618 and 0.618.Second quadratic: -2m² +m +3. Let's solve for roots.Quadratic equation: -2m² +m +3=0. Multiply by -1: 2m² -m -3=0. As before, discriminant 1 +24=25. Roots: [1 ±5]/4. Therefore, m= (6)/4=1.5 and m=(-4)/4=-1. So roots at m=-1 and m=1.5. Since the quadratic is -2m² +m +3, which opens downward (coefficient -2), the quadratic is positive between its roots: -1 <m <1.5.Therefore, the inequality (m² +m -1)(-2m² +m +3) >0.Let's analyze the intervals determined by the roots: m=-1.618, m=-1, m=0.618, m=1.5.So the intervals are:1. (-infty, -1.618)2. (-1.618, -1)3. (-1, 0.618)4. (0.618,1.5)5. (1.5, +infty)Now, let's check the sign of each quadratic in these intervals and their product.1. Interval (-infty, -1.618):m² +m -1: let m=-2. 4 -2 -1=1>0.-2m² +m +3: m=-2. -8 +(-2)+3=-7<0.Product: positive*negative=negative.2. Interval (-1.618, -1):Take m=-1.5.First quadratic: (-1.5)^2 + (-1.5) -1=2.25 -1.5 -1=-0.25<0.Second quadratic: -2*(-1.5)^2 + (-1.5) +3= -2*(2.25) -1.5 +3= -4.5 -1.5 +3= -3<0.Product: negative*negative=positive.3. Interval (-1,0.618):Take m=0.First quadratic:0 +0 -1=-1<0.Second quadratic: -0 +0 +3=3>0.Product: negative*positive=negative.4. Interval (0.618,1.5):Take m=1.First quadratic:1 +1 -1=1>0.Second quadratic:-2 +1 +3=2>0.Product: positive*positive=positive.5. Interval (1.5, +infty):Take m=2.First quadratic:4 +2 -1=5>0.Second quadratic:-8 +2 +3=-3<0.Product: positive*negative=negative.Therefore, the product is positive in intervals (-1.618,-1) and (0.618,1.5).So the solutions are m ∈ (-1.618, -1) ∪ (0.618,1.5).But the problem asks to "Find the value of m". This suggests a specific value. Therefore, perhaps m must be an integer. Let's check integers in these intervals.First interval (-1.618, -1): possible integer is m=-1. But check m=-1.Check m=-1 in the original function:f(x)=(m² +m -1)x^{-2m² +m +3}= (1 -1 -1)x^{-2(1) + (-1) +3}= (-1)x^{-2 -1 +3}= (-1)x^{0}= -1. So f(x)=-1, which is a constant function. Therefore, derivative is zero, not positive. Hence, m=-1 is invalid.Second interval (0.618,1.5): possible integers m=1.Check m=1.f(x)=(1 +1 -1)x^{-2(1) +1 +3}=1*x^{2}=x². The derivative is 2x, which is positive for x>0. So m=1 is valid.Next possible integer in (0.618,1.5) is m=1. Let me check m=1.5, but 1.5 is not an integer, and the interval is up to 1.5. So no other integers. So m=1 is the only integer solution.But wait, what if m is a real number? Then there are infinitely many solutions. However, the problem asks for "the value of m", which is phrased in singular. So likely m=1 is the intended answer. Additionally, in part (2) and (3), m is used to define f(x), so it must be a specific value. Therefore, I think the problem expects m=1.Therefore, the answer to part (1) is m=1 and f(x)=x².Now, moving to part (2).For any x ∈ [1,2], there exist x1, x2 ∈ [1,2] such that f(x) ≤ f(x1), g(x) ≤g(x2), and if f(x1)=g(x2), find real number t.First, let's parse the problem.Given x in [1,2], there exists x1, x2 in [1,2] such that f(x) ≤ f(x1) and g(x) ≤g(x2). This means that for each x, f(x) is bounded above by the maximum of f on [1,2], which is f(x1), and similarly for g. So x1 is a maximizer of f on [1,2], and x2 is a maximizer of g on [1,2]. Then, given that f(x1)=g(x2), find t.Therefore, we need to find t such that the maximum value of f on [1,2] equals the maximum value of g on [1,2].First, since we found m=1, f(x)=x². Therefore, on [1,2], f(x)=x² is increasing, so its maximum is at x=2: f(2)=4.Now, g(x)=-x² +2|x| +t. On the interval [1,2], x is positive, so |x|=x. Therefore, g(x)=-x² +2x +t.We need to find the maximum of g(x) on [1,2]. Let's analyze g(x).g(x)= -x² +2x +t. This is a quadratic function opening downward (coefficient of x² is -1). Its vertex is at x = -b/(2a) = -2/(2*(-1)) = 1. So the vertex is at x=1. Since it's a downward opening parabola, the maximum occurs at the vertex x=1. Therefore, on [1,2], g(x) is decreasing from x=1 to x=2. Therefore, maximum of g on [1,2] is at x=1: g(1)= -1 +2 +t=1 +t. The minimum is at x=2: g(2)= -4 +4 +t=0 +t=t.But the problem states that for any x ∈ [1,2], there exist x1, x2 ∈ [1,2] such that f(x) ≤ f(x1) and g(x) ≤g(x2). So here, x1 would be the maximizer of f (which is x1=2) and x2 would be the maximizer of g (which is x2=1). Then, given that f(x1)=g(x2), which is f(2)=g(1)=4=1 +t. Therefore, 1 +t=4, so t=3.Wait, that seems straightforward. Let me check.Since f(x)=x² on [1,2], which is increasing, so maximum at x=2: f(2)=4.g(x)= -x² +2x +t. As mentioned, it's a downward parabola with vertex at x=1, so maximum at x=1: g(1)= -1 +2 +t=1 +t.Given that f(x1)=g(x2), and x1=2, x2=1, so 4=1 +t => t=3.Therefore, the value of t is 3.But let me verify this.If t=3, then g(x)= -x² +2x +3. Then on [1,2], the maximum is at x=1: g(1)= -1 +2 +3=4, and f(2)=4. Therefore, equality holds. So yes, t=3.Therefore, part (2) answer is t=3.Now, part (3): If 2^x h(2x) + λ h(x) ≥0 holds for all x ∈ [1,2], find the range of real number λ.Given that h(x)=2^x -2^{-x}.First, let's write the inequality: 2^x h(2x) + λ h(x) ≥0 for all x ∈ [1,2].Compute h(2x): h(2x)=2^{2x} -2^{-2x}.Therefore, substitute into the inequality:2^x (2^{2x} -2^{-2x}) + λ(2^x -2^{-x}) ≥0.Simplify each term:First term: 2^x *2^{2x}=2^{3x}, 2^x *(-2^{-2x})=-2^{x -2x}=-2^{-x}.So the inequality becomes:2^{3x} -2^{-x} + λ(2^x -2^{-x}) ≥0.Let me write this as:2^{3x} -2^{-x} + λ2^x -λ2^{-x} ≥0.Combine like terms:2^{3x} + λ2^x - (1 + λ)2^{-x} ≥0.Perhaps factor terms with 2^x and 2^{-x}:Let me denote y=2^x. Since x ∈ [1,2], y ∈ [2,4].Then 2^{3x}=y^3, 2^x=y, 2^{-x}=1/y.Therefore, substitute into the inequality:y^3 + λ y - (1 + λ)(1/y) ≥0.Multiply both sides by y (since y >0, this doesn't change inequality direction):y^4 + λ y^2 - (1 + λ) ≥0.So the inequality becomes y^4 + λ y^2 - (1 + λ) ≥0 for all y ∈ [2,4].Let me denote the left-hand side as Q(y) = y^4 + λ y^2 - (1 + λ).We need Q(y) ≥0 for all y ∈ [2,4].So Q(y) = y^4 + λ(y^2 -1) -1.We need to find λ such that Q(y) ≥0 for all y ∈ [2,4].Let's rearrange Q(y):Q(y) = y^4 -1 + λ(y^2 -1).Factor y^4 -1: (y²)^2 -1 = (y² -1)(y² +1).Therefore, Q(y) = (y² -1)(y² +1) + λ(y² -1) = (y² -1)(y² +1 +λ).So Q(y) = (y² -1)(y² +1 +λ).Therefore, the inequality is (y² -1)(y² +1 +λ) ≥0 for all y ∈ [2,4].Note that for y ∈ [2,4], y² -1 ≥4 -1=3>0. Therefore, y² -1 is positive. So the sign of Q(y) is determined by (y² +1 +λ).Since y² -1 >0, the inequality Q(y) ≥0 simplifies to (y² +1 +λ) ≥0.But since y² +1 +λ ≥0 must hold for all y ∈ [2,4]. The minimal value of y² +1 +λ occurs at the minimal y, which is y=2. Therefore:At y=2: 4 +1 +λ ≥0 =>5 +λ ≥0 =>λ ≥-5.But wait, but y ranges from 2 to4. So y² +1 ranges from 5 to 17. So y² +1 +λ ≥0 for all y in [2,4] is equivalent to the minimal value of y² +1 +λ being ≥0, which is at y=2: 5 +λ ≥0. Hence, λ ≥-5.But this seems too straightforward, but let me check.Wait, the original inequality Q(y)=(y² -1)(y² +1 +λ) ≥0.Since y² -1 ≥3 >0, the entire expression is ≥0 if and only if y² +1 +λ ≥0. Therefore, y² +1 +λ ≥0 for all y ∈ [2,4]. The minimal value of y² +1 is at y=2, which is 5. Therefore, 5 +λ ≥0 => λ ≥-5.Thus, the range of λ is [-5, ∞).But wait, the problem states "find the range of the real number λ". However, this seems too easy. Let me verify with y=4:At y=4, y² +1 +λ=16 +1 +λ=17 +λ. If λ ≥-5, then 17 +λ ≥12 >0. So for all y in [2,4], y² +1 +λ ≥5 +λ ≥0.Therefore, the inequality holds for λ ≥-5.Wait, but maybe there's a mistake in the factorization.Wait, let's redo the factorization step.Original Q(y):Q(y) = y^4 + λ y² - (1 +λ).But I factored as (y² -1)(y² +1) +λ(y² -1) = (y² -1)(y² +1 +λ). Let's verify:(y² -1)(y² +1) = y^4 -1.Then add λ(y² -1): y^4 -1 +λ y² -λ= y^4 +λ y² - (1 +λ). Yes, that matches Q(y). Therefore, correct.Therefore, Q(y) = (y² -1)(y² +1 +λ). Since y² -1 >0 for y≥2, then Q(y) ≥0 ⇨ y² +1 +λ ≥0 ⇨ λ ≥ -y² -1.But we need this to hold for all y ∈ [2,4]. Therefore, λ must be ≥ the maximum of (-y² -1) over y ∈ [2,4].But since -y² -1 is decreasing in y, the maximum occurs at y=2: -4 -1=-5. Therefore, λ ≥-5.Hence, the range is λ ∈ [-5, ∞).But wait, is this correct? Let me plug λ=-5 into the original inequality.If λ=-5, then Q(y) = (y² -1)(y² +1 -5) = (y² -1)(y² -4).So for y=2: (4 -1)(4 -4)=3*0=0. So equality holds.For y=3: (9 -1)(9 -4)=8*5=40>0.For y=4: (16 -1)(16 -4)=15*12=180>0.For y between 2 and4: y² -1 is positive, y² -4 is ≥0 when y ≥2 (since y starts at 2). Therefore, y² -4 ≥0, so Q(y)=(y² -1)(y² -4) ≥0.Therefore, for λ=-5, the inequality holds with equality at y=2 and strictly positive elsewhere in [2,4].If λ < -5, say λ=-6, then at y=2: Q(2)=(4 -1)(4 +1 -6)=3*(-1)=-3 <0, which violates the inequality. Therefore, λ must be ≥-5.Therefore, the range of λ is [-5, ∞). So the answer is λ ≥-5.Final Answer(1) The value of ( m ) is (boxed{1}) and the explicit formula for ( f(x) ) is (boxed{x^2}).(2) The value of ( t ) is (boxed{3}).(3) The range of ( lambda ) is (boxed{[-5, +infty)}).</think>Given the power function ( f(x) = (m^2 + m - 1)x^{-2m^2 + m + 3} ) is increasing on the interval ( (0, +infty) ), we need to find the value of ( m ) and determine the explicit formula for ( f(x) ). 1. Finding the value of ( m ): - For ( f(x) ) to be increasing, its derivative ( f'(x) ) must be positive for all ( x > 0 ). - The derivative is ( f'(x) = (m^2 + m - 1)(-2m^2 + m + 3)x^{-2m^2 + m + 2} ). - The sign of ( f'(x) ) depends on the product ( (m^2 + m - 1)(-2m^2 + m + 3) ). - Solving the inequality ( (m^2 + m - 1)(-2m^2 + m + 3) > 0 ), we find the intervals where the product is positive. - After analyzing the intervals and checking possible integer values, we find ( m = 1 ) is the valid solution. - Substituting ( m = 1 ) into ( f(x) ), we get ( f(x) = x^2 ).2. Determining the value of ( t ): - Given ( f(x_1) = g(x_2) ) where ( x_1 ) and ( x_2 ) are the maximum points of ( f ) and ( g ) on ([1, 2]). - The maximum value of ( f(x) = x^2 ) on ([1, 2]) is ( f(2) = 4 ). - The function ( g(x) = -x^2 + 2|x| + t ) simplifies to ( g(x) = -x^2 + 2x + t ) on ([1, 2]), with maximum value at ( x = 1 ), giving ( g(1) = 1 + t ). - Setting ( f(2) = g(1) ), we solve ( 4 = 1 + t ) to find ( t = 3 ).3. Finding the range of ( lambda ): - Given the inequality ( 2^x h(2x) + lambda h(x) geq 0 ) for ( x in [1, 2] ). - Substituting ( h(x) = 2^x - 2^{-x} ) and simplifying, we rewrite the inequality in terms of ( y = 2^x ). - The inequality becomes ( (y^2 - 1)(y^2 + 1 + lambda) geq 0 ) for ( y in [2, 4] ). - Since ( y^2 - 1 > 0 ), we need ( y^2 + 1 + lambda geq 0 ). The minimum value occurs at ( y = 2 ), leading to ( lambda geq -5 ). Final Answers1. The value of ( m ) is (boxed{1}) and the explicit formula for ( f(x) ) is (boxed{x^2}).2. The value of ( t ) is (boxed{3}).3. The range of ( lambda ) is (boxed{[-5, +infty)}).]

🔑:(This question is worth 10 points) Solution: (1) According to the definition of a power function: m^2+m-1=1, which simplifies to m^2+m-2=0, Solving this gives: m=-2 or m=1, Since f(x) is increasing on (0, +infty), we have -2m^2+m+3>0, solving this inequality gives -10, we have lambda geq -(2^{2x}+1) Let k(x)=-(2^{2x}+1), for x in [1, 2], next we find the maximum value of k(x); Since x in [1, 2], -(2^{2x}+1) in [-17, -5], thus k_{text{max}}(x)=-5 Hence, the range of lambda is boxed{[-5, +infty)}…(12 points)

🧠:[<think>Okay, so I need to find the range of the function ( f(x) = frac{sin x - 1}{sqrt{3 - 2 cos x - 2 sin x}} ) where ( 0 leq x leq 2pi ). Hmm, let's start by understanding what this function looks like and how to approach finding its range.First, the numerator is ( sin x - 1 ). Since the sine function ranges between -1 and 1, subtracting 1 means the numerator will range between -2 and 0. So the numerator is always non-positive. That might be useful later.The denominator is ( sqrt{3 - 2 cos x - 2 sin x} ). Since it's a square root, the expression inside must be non-negative. So we have to ensure that ( 3 - 2 cos x - 2 sin x geq 0 ). Let me check if that's true for all x in [0, 2π]. Maybe I should analyze the expression inside the square root first.Let me denote the expression inside the square root as ( D(x) = 3 - 2 cos x - 2 sin x ). To find when D(x) is non-negative:( D(x) = 3 - 2(cos x + sin x) geq 0 )So, ( 3 geq 2(cos x + sin x) )Divide both sides by 2:( frac{3}{2} geq cos x + sin x )But the maximum value of ( cos x + sin x ) is ( sqrt{2} ) (since ( cos x + sin x = sqrt{2} sin(x + frac{pi}{4}) ), which has amplitude ( sqrt{2} )). Since ( sqrt{2} approx 1.414 ), and ( 3/2 = 1.5 ), which is greater than ( sqrt{2} ), so ( cos x + sin x ) never exceeds ( sqrt{2} ), which is less than 1.5. Therefore, ( 3 - 2(cos x + sin x) ) is always positive. So the denominator is always real and positive. That's good; the function is defined for all x in [0, 2π].So the denominator is always positive, and the numerator is non-positive. Therefore, the function ( f(x) ) is always non-positive. The range will be from some minimum value up to 0.But we need to find exactly what the minimum is. So I need to find the minimum value of ( f(x) ), which is the most negative value it can attain.One approach is to let ( y = f(x) ), and then try to express this equation in terms of trigonometric identities or manipulate it to find the possible values of y.Let me write:( y = frac{sin x - 1}{sqrt{3 - 2 cos x - 2 sin x}} )Since y is non-positive, let's square both sides to eliminate the square root, but we have to remember that squaring might introduce extraneous solutions. However, since we know y is non-positive, the squared equation will give us the square of the absolute value, and we can later take the negative square roots to find the original y.So:( y^2 = frac{(sin x - 1)^2}{3 - 2 cos x - 2 sin x} )Let me compute the numerator and denominator:Numerator: ( (sin x - 1)^2 = sin^2 x - 2 sin x + 1 )Denominator: ( 3 - 2 cos x - 2 sin x )So,( y^2 = frac{sin^2 x - 2 sin x + 1}{3 - 2 cos x - 2 sin x} )Hmm, maybe we can manipulate this equation to express it in terms of a single trigonometric function. Let's see. Let's try to express both numerator and denominator in terms of sin x and cos x, perhaps use some identity.Alternatively, maybe we can let t = sin x, and then express cos x in terms of t. But since cos x = sqrt(1 - t^2), but that introduces square roots, which might complicate things.Alternatively, maybe express both numerator and denominator in terms of sin x and cos x, then use substitution variables.Alternatively, note that both numerator and denominator can be written in terms of sin x and cos x. Let me try to see:First, numerator is ( sin^2 x - 2 sin x + 1 ). That is similar to ( (sin x - 1)^2 ), which it is. The denominator is 3 - 2 cos x - 2 sin x.Another idea: Maybe use substitution variables. Let me set u = sin x and v = cos x, with the constraint that ( u^2 + v^2 = 1 ). Then, the equation becomes:( y^2 = frac{(u - 1)^2}{3 - 2v - 2u} )So, we can think of this as an equation in u and v with the constraint ( u^2 + v^2 = 1 ). Maybe we can use Lagrange multipliers or something to find the extrema? Hmm, that might be a bit involved, but possible.Alternatively, note that the denominator is 3 - 2(u + v). So, denominator is 3 - 2(u + v).So the expression for y^2 is ( frac{(u - 1)^2}{3 - 2(u + v)} )But since u^2 + v^2 = 1, perhaps we can express v in terms of u or vice versa.Let me solve for v in terms of u: v = sqrt(1 - u^2), but with a ± sign. However, that might complicate things. Alternatively, perhaps express u + v in terms of a single variable.Wait, another approach: Let's consider the denominator 3 - 2(u + v). Let me denote S = u + v. Then, denominator is 3 - 2S. The numerator is (u - 1)^2.So, y^2 = (u - 1)^2 / (3 - 2S)But S = u + v, and u^2 + v^2 = 1. So, perhaps express S^2 = u^2 + 2uv + v^2 = 1 + 2uv. Therefore, uv = (S^2 - 1)/2. Hmm, but I don't know if that helps.Alternatively, express S = u + v. Then, the maximum value of S is sqrt(2) as we saw earlier. The minimum is -sqrt(2). So S ranges between -sqrt(2) and sqrt(2). Therefore, denominator 3 - 2S ranges between 3 - 2*sqrt(2) and 3 + 2*sqrt(2). Since sqrt(2) is approx 1.414, so 3 - 2*sqrt(2) is approx 3 - 2.828 = 0.172, and 3 + 2*sqrt(2) is approx 5.828. So denominator is always positive as we already knew.But I need to relate S and the numerator (u - 1)^2. Hmm.Alternatively, perhaps express (u - 1)^2 = u^2 - 2u + 1. Since u^2 = 1 - v^2, so substitute:(u - 1)^2 = (1 - v^2) - 2u + 1 = 2 - v^2 - 2uBut not sure if that helps. Alternatively, maybe express both numerator and denominator in terms of S and something else.Alternatively, let's consider that the function y is given in terms of sin x and cos x. Maybe parametrize x using a parameter θ such that we can write sin x and cos x in terms of θ. For example, set t = tan(x/2), but that might be messy.Alternatively, note that the denominator is similar to the numerator if we manipulate it. Let me see:Denominator: 3 - 2 cos x - 2 sin x = 3 - 2(cos x + sin x). Let's see if we can relate this to the numerator.Wait, the numerator is (sin x - 1). Maybe if I consider the denominator as 3 - 2(cos x + sin x), perhaps write 3 as 2 + 1, so 2 + 1 - 2(cos x + sin x) = 2(1 - (cos x + sin x)) + 1. Not sure.Alternatively, maybe complete the square for the denominator. Let's see:3 - 2 cos x - 2 sin x = 3 - 2(cos x + sin x). Let me think of cos x + sin x as sqrt(2) sin(x + π/4). So:cos x + sin x = sqrt(2) sin(x + π/4). Therefore, denominator is 3 - 2*sqrt(2) sin(x + π/4). Then, since sin(x + π/4) ranges between -1 and 1, denominator ranges between 3 - 2*sqrt(2) and 3 + 2*sqrt(2), which matches our previous calculation.But how does this help? Hmm.Alternatively, maybe write both numerator and denominator in terms of sin(x + something). Let's see:Numerator: sin x - 1. Maybe write as sin x - sin(π/2). Using the identity sin A - sin B = 2 cos((A+B)/2) sin((A - B)/2). So:sin x - sin(π/2) = 2 cos((x + π/2)/2) sin((x - π/2)/2). Not sure if this helps, but let's compute:= 2 cos(x/2 + π/4) sin(x/2 - π/4)Similarly, the denominator: 3 - 2(cos x + sin x). As before, cos x + sin x = sqrt(2) sin(x + π/4). So denominator is 3 - 2*sqrt(2) sin(x + π/4). Hmm.Alternatively, maybe express the entire function f(x) as a single trigonometric function. Let me think.Alternatively, maybe set t = x + π/4 or some other angle to align the sine and cosine terms.Wait, another approach: Let's consider substituting variables to simplify the expression. Let me set θ = x - π/2. Then, sin x = sin(θ + π/2) = cos θ, and cos x = cos(θ + π/2) = -sin θ. Let's try that substitution.So, let θ = x - π/2, so x = θ + π/2.Then,sin x = sin(θ + π/2) = cos θcos x = cos(θ + π/2) = -sin θSo, substitute into f(x):Numerator: sin x - 1 = cos θ - 1Denominator: sqrt(3 - 2 cos x - 2 sin x) = sqrt(3 - 2*(-sin θ) - 2 cos θ) = sqrt(3 + 2 sin θ - 2 cos θ)So, f(x) becomes:f(θ) = (cos θ - 1)/sqrt(3 + 2 sin θ - 2 cos θ)Hmm, does this make it simpler? Let's check:Numerator: cos θ - 1 = -2 sin²(θ/2) (using the identity 1 - cos θ = 2 sin²(θ/2))Denominator: sqrt(3 + 2 sin θ - 2 cos θ)Maybe try to simplify denominator as well.Let me see:3 + 2 sin θ - 2 cos θ= 3 - 2 cos θ + 2 sin θHmm, perhaps factor differently. Let me group constants and trigonometric terms:= 3 - 2 cos θ + 2 sin θHmm, maybe express this as 1 + 2(1 - cos θ) + 2 sin θ. Because 3 = 1 + 2, so:= 1 + 2(1 - cos θ) + 2 sin θBut 1 - cos θ = 2 sin²(θ/2), and sin θ = 2 sin(θ/2) cos(θ/2). Let me try substituting:= 1 + 2*(2 sin²(θ/2)) + 2*(2 sin(θ/2) cos(θ/2))= 1 + 4 sin²(θ/2) + 4 sin(θ/2) cos(θ/2)Hmm, not sure if that helps. Alternatively, maybe write the entire expression in terms of sin(θ/2) and cos(θ/2). Let me set t = θ/2. Then θ = 2t, so:Denominator:sqrt(3 + 2 sin 2t - 2 cos 2t)Let me compute 3 + 2 sin 2t - 2 cos 2t:= 3 + 2 sin 2t - 2 cos 2tUsing double-angle identities:= 3 + 2*(2 sin t cos t) - 2*(1 - 2 sin²t)= 3 + 4 sin t cos t - 2 + 4 sin²t= 1 + 4 sin t cos t + 4 sin²tHmm, maybe factor this:= 1 + 4 sin²t + 4 sin t cos tHmm, 4 sin²t + 4 sin t cos t + 1Not sure. Alternatively, let me try to write this as a square:Suppose it's (a sin t + b cos t + c)^2. Let me see:(a sin t + b cos t + c)^2 = a² sin² t + b² cos² t + c² + 2ab sin t cos t + 2ac sin t + 2bc cos tBut our expression is 4 sin²t + 4 sin t cos t + 1. Let's compare:If I set a² = 4, so a = 2Then 2ab = 4, so with a=2, 2*2*b=4 => 4b=4 => b=1Then the cos² t term is b² = 1, but in our expression, there's no cos² t term. Wait, the original expression is 4 sin²t + 4 sin t cos t + 1. So, maybe not a perfect square. Alternatively, maybe factor as (2 sin t + cos t)^2 + ... ?Let me compute (2 sin t + cos t)^2 = 4 sin²t + 4 sin t cos t + cos²t. Which is 4 sin²t + 4 sin t cos t + cos²t. Comparing to our expression 4 sin²t + 4 sin t cos t + 1, the difference is cos²t vs 1. Since cos²t = 1 - sin²t, so:(2 sin t + cos t)^2 = 4 sin²t + 4 sin t cos t + 1 - sin²t = 3 sin²t + 4 sin t cos t + 1But that's not matching. Hmm.Alternatively, maybe use another identity. Let me think.Alternatively, perhaps write the denominator expression in terms of sin(t + φ). Let me see:3 + 2 sin θ - 2 cos θ = 3 + 2 sin θ - 2 cos θWait, if I consider 2 sin θ - 2 cos θ, that's equal to 2(sin θ - cos θ) = 2√2 sin(θ - π/4). Because sin θ - cos θ = √2 sin(θ - π/4). Let's verify:sin θ - cos θ = √2 sin(θ - π/4)Yes, because sin(θ - π/4) = sin θ cos π/4 - cos θ sin π/4 = (sin θ - cos θ)/√2. So multiplying by √2 gives sin θ - cos θ.Therefore, 2(sin θ - cos θ) = 2√2 sin(θ - π/4)So, denominator expression:3 + 2 sin θ - 2 cos θ = 3 + 2√2 sin(θ - π/4)Therefore, denominator is sqrt(3 + 2√2 sin(θ - π/4))So, putting it all together, after substitution θ = x - π/2, we have:f(θ) = (cos θ - 1)/sqrt(3 + 2√2 sin(θ - π/4))But I'm not sure if this helps yet. Let's think.Alternatively, since we have f(θ) expressed in terms of θ, which is x shifted by π/2, but θ ranges from -π/2 to 3π/2 (since x is from 0 to 2π). Maybe this substitution complicates the limits. Maybe another substitution is better.Alternatively, let's consider the original function f(x) = (sin x - 1)/sqrt(3 - 2 cos x - 2 sin x). Let's consider making a substitution to simplify the denominator.Let me denote the denominator as sqrt(3 - 2 cos x - 2 sin x). Let me set this denominator to be some variable, say, D = sqrt(3 - 2 cos x - 2 sin x). Then, perhaps express both numerator and denominator in terms of D.But that might not help. Alternatively, maybe consider squaring the function and then using trigonometric identities.We had earlier:y^2 = (sin x - 1)^2 / (3 - 2 cos x - 2 sin x)Let me compute both numerator and denominator:Numerator: sin²x - 2 sin x + 1Denominator: 3 - 2 cos x - 2 sin xNote that sin²x = 1 - cos²x, so substitute:Numerator: (1 - cos²x) - 2 sin x + 1 = 2 - cos²x - 2 sin xSo,y^2 = (2 - cos²x - 2 sin x)/(3 - 2 cos x - 2 sin x)Hmm, this seems a bit complicated, but maybe we can rearrange terms.Let me write the denominator as 3 - 2 sin x - 2 cos x. Let me rearrange numerator:2 - cos²x - 2 sin x = 2 - 2 sin x - cos²xSo,y^2 = (2 - 2 sin x - cos²x)/(3 - 2 sin x - 2 cos x)Hmm, perhaps factor numerator and denominator?Alternatively, note that cos²x = 1 - sin²x, so substitute into numerator:2 - 2 sin x - (1 - sin²x) = 2 - 2 sin x -1 + sin²x = 1 - 2 sin x + sin²x = (1 - sin x)^2So, numerator simplifies to (1 - sin x)^2, denominator is 3 - 2 sin x - 2 cos xTherefore, y^2 = (1 - sin x)^2 / (3 - 2 sin x - 2 cos x)Wait, that's a nice simplification! So, the numerator is (1 - sin x)^2 and the denominator is 3 - 2 sin x - 2 cos x.Therefore, y^2 = (1 - sin x)^2 / (3 - 2 sin x - 2 cos x)But since y is non-positive, we can write y = - (1 - sin x) / sqrt(3 - 2 sin x - 2 cos x). Because (sin x - 1) = - (1 - sin x), so original function is y = - (1 - sin x)/sqrt(denominator). Therefore, y is equal to negative of (1 - sin x)/sqrt(denominator). So squaring gives y^2 = (1 - sin x)^2 / denominator, which is the same as above.So, perhaps this form is easier to handle. Let's write:Let me denote t = sin x. Then, since sin x ranges from -1 to 1, t ∈ [-1, 1]. But in our case, the numerator (1 - sin x) is 1 - t, which ranges from 0 to 2. Since in the original function y, we have y = - (1 - t)/sqrt(3 - 2t - 2 cos x). However, we still have cos x in the denominator. So perhaps express cos x in terms of t.Since cos x = sqrt(1 - t^2), but with a ± sign. However, since x is between 0 and 2π, cos x can be positive or negative. But in this case, since denominator is sqrt(3 - 2t - 2 cos x), and we have to ensure that it's real, which we already did.But this might complicate things. Alternatively, maybe express the denominator in terms of t.Let me try:Denominator: 3 - 2t - 2 cos xBut cos x = sqrt(1 - t^2) or -sqrt(1 - t^2). Hmm, but this might lead to different cases. Let's see.Wait, but for a given t = sin x, cos x can be either positive or negative. So perhaps we have to consider both cases. That complicates things. Maybe there's a better approach.Alternatively, use substitution variables u = sin x and v = cos x, with u^2 + v^2 = 1, and then express y^2 in terms of u and v.We have y^2 = (1 - u)^2 / (3 - 2u - 2v)So, let me set up the equation:(1 - u)^2 = y^2 (3 - 2u - 2v)But since u^2 + v^2 = 1, we can write v = ±sqrt(1 - u^2). However, dealing with the square root might not be helpful. Alternatively, consider this as an optimization problem: maximize and minimize y given the constraint u^2 + v^2 = 1.So, treating y as a function of u and v:y = - (1 - u)/sqrt(3 - 2u - 2v)Subject to u^2 + v^2 = 1.To find the extrema of y, we can use Lagrange multipliers. Let's set up the Lagrangian:L(u, v, λ) = - (1 - u)/sqrt(3 - 2u - 2v) + λ(u^2 + v^2 - 1)Take partial derivatives with respect to u, v, and λ, set them to zero.But this seems quite involved. Let's compute the derivative of L with respect to u:First, let me denote D = 3 - 2u - 2v, so sqrt(D) is the denominator.Then, y = - (1 - u)/sqrt(D)So, derivative of L with respect to u:dL/du = derivative of [- (1 - u)/sqrt(D)] + derivative of [λ(u^2 + v^2 -1)]First term:d/du [ - (1 - u)/sqrt(D) ] = - [ (-1) * sqrt(D) - (1 - u)*(1/(2 sqrt(D)))*(-2) ] / DWait, actually, let's use quotient rule or product rule.Let me write it as - (1 - u) * D^(-1/2)So, derivative with respect to u:- [ d/du (1 - u) * D^(-1/2) + (1 - u) * d/du (D^(-1/2)) ]= - [ (-1) * D^(-1/2) + (1 - u) * (-1/2) D^(-3/2) * (-2) ) ]Wait, step by step:First, derivative of (1 - u) is -1.Derivative of D^(-1/2) with respect to u is (-1/2) D^(-3/2) * (-2) = D^(-3/2)Wait, D = 3 - 2u - 2v, so dD/du = -2 - 2 dv/du. Wait, but in partial derivatives, when taking dL/du, we treat v as independent variable? Wait, no, in the Lagrangian multiplier method, u and v are variables subject to the constraint u^2 + v^2 = 1, so their derivatives are related through the constraint.Wait, perhaps this is getting too complicated. Maybe another approach.Alternatively, since we have y expressed in terms of u and v, with u^2 + v^2 = 1, we can parametrize u and v as u = cos θ, v = sin θ, but θ here is a different parameter, not the original x. Wait, that might be confusing, but let's try.Let me set u = cos φ, v = sin φ, where φ ∈ [0, 2π). Then, our expression becomes:y = - (1 - cos φ)/sqrt(3 - 2 cos φ - 2 sin φ)But φ is a parameter here, not related to the original x. Wait, but in this substitution, we have u = cos φ, v = sin φ, so φ is just a parameter to represent u and v on the unit circle. So, we can think of y as a function of φ:y(φ) = - (1 - cos φ)/sqrt(3 - 2 cos φ - 2 sin φ)Our goal is to find the range of y(φ) as φ ranges from 0 to 2π.This seems manageable. So now, the problem reduces to finding the minimum and maximum of y(φ) over φ ∈ [0, 2π). But since y is non-positive, we only need to find its minimum value (most negative) and confirm that its maximum is 0.But when does y(φ) reach 0? When the numerator is 0, i.e., when 1 - cos φ = 0 ⇒ cos φ = 1 ⇒ φ = 0. So, y(0) = 0. So the maximum value is 0.Therefore, the range is [m, 0], where m is the minimum value of y(φ).So now, we need to find the minimum of y(φ). To find the minimum, since y is non-positive, it's equivalent to finding where y(φ) is most negative, which corresponds to where (1 - cos φ)/sqrt(3 - 2 cos φ - 2 sin φ) is maximized (since y is negative of that).So, we can instead find the maximum value of (1 - cos φ)/sqrt(3 - 2 cos φ - 2 sin φ), and then the minimum of y will be the negative of that maximum.Let me denote z(φ) = (1 - cos φ)/sqrt(3 - 2 cos φ - 2 sin φ). Then, we need to find the maximum of z(φ).To find the maximum of z(φ), we can take the derivative of z(φ) with respect to φ, set it to zero, and solve for φ. However, this might be complicated, but let's try.First, write z(φ) = (1 - cos φ) / sqrt(3 - 2 cos φ - 2 sin φ)Let me denote D(φ) = 3 - 2 cos φ - 2 sin φ, so z = (1 - cos φ)/sqrt(D)Compute derivative dz/dφ:dz/dφ = [ (sin φ) * sqrt(D) - (1 - cos φ) * (1/(2 sqrt(D))) * (2 sin φ - 2 cos φ) ] / DWait, let me do it step by step.Using the quotient rule:If z = numerator / denominator, where numerator = 1 - cos φ, denominator = sqrt(D)Then, dz/dφ = [ (d/dφ numerator) * denominator - numerator * (d/dφ denominator) ] / (denominator)^2First, compute d/dφ numerator:d/dφ (1 - cos φ) = sin φd/dφ denominator = d/dφ [D^(1/2)] = (1/2) D^(-1/2) * dD/dφCompute dD/dφ:D = 3 - 2 cos φ - 2 sin φdD/dφ = 2 sin φ - 2 cos φTherefore, d/dφ denominator = (1/2) D^(-1/2) * (2 sin φ - 2 cos φ) = (sin φ - cos φ)/sqrt(D)Putting it all together:dz/dφ = [ sin φ * sqrt(D) - (1 - cos φ) * (sin φ - cos φ)/sqrt(D) ] / DMultiply numerator and denominator by sqrt(D):= [ sin φ * D - (1 - cos φ)(sin φ - cos φ) ] / D^(3/2)So, critical points occur when numerator is zero:sin φ * D - (1 - cos φ)(sin φ - cos φ) = 0Substitute D = 3 - 2 cos φ - 2 sin φ:sin φ*(3 - 2 cos φ - 2 sin φ) - (1 - cos φ)(sin φ - cos φ) = 0Let me expand both terms:First term: 3 sin φ - 2 sin φ cos φ - 2 sin² φSecond term: - (1 - cos φ)(sin φ - cos φ) = - [ sin φ - cos φ - sin φ cos φ + cos² φ ]= - sin φ + cos φ + sin φ cos φ - cos² φSo, total equation:3 sin φ - 2 sin φ cos φ - 2 sin² φ - sin φ + cos φ + sin φ cos φ - cos² φ = 0Combine like terms:(3 sin φ - sin φ) + (-2 sin φ cos φ + sin φ cos φ) + (-2 sin² φ) + (cos φ) + (- cos² φ) = 0Which is:2 sin φ - sin φ cos φ - 2 sin² φ + cos φ - cos² φ = 0Let me write all terms:2 sin φ - sin φ cos φ - 2 sin² φ + cos φ - cos² φ = 0Hmm, this seems complicated. Maybe factor terms where possible.Group terms with sin φ:2 sin φ - sin φ cos φ - 2 sin² φ = sin φ (2 - cos φ - 2 sin φ)Terms with cos φ:cos φ - cos² φ = cos φ (1 - cos φ)So, the equation becomes:sin φ (2 - cos φ - 2 sin φ) + cos φ (1 - cos φ) = 0Let me check if this is correct:Yes:sin φ*(2 - cos φ - 2 sin φ) + cos φ*(1 - cos φ)= 2 sin φ - sin φ cos φ - 2 sin² φ + cos φ - cos² φWhich matches.Now, let's denote this equation as:sin φ (2 - cos φ - 2 sin φ) + cos φ (1 - cos φ) = 0Let me see if we can factor this or find common terms.Let me expand sin φ*(2 - cos φ - 2 sin φ):= 2 sin φ - sin φ cos φ - 2 sin² φThen, adding cos φ (1 - cos φ):= 2 sin φ - sin φ cos φ - 2 sin² φ + cos φ - cos² φAnother approach: Let's express everything in terms of sin φ and cos φ, and try to write as a single equation.Let me collect terms:-2 sin² φ - cos² φ - sin φ cos φ + 2 sin φ + cos φ = 0Hmm. Maybe use Pythagorean identities. For example, replace sin² φ with 1 - cos² φ.Wait, let's try that:Replace -2 sin² φ with -2(1 - cos² φ) = -2 + 2 cos² φSo, equation becomes:(-2 + 2 cos² φ) - cos² φ - sin φ cos φ + 2 sin φ + cos φ = 0Simplify:-2 + 2 cos² φ - cos² φ - sin φ cos φ + 2 sin φ + cos φ = 0Which is:-2 + cos² φ - sin φ cos φ + 2 sin φ + cos φ = 0Rearranged:cos² φ + cos φ - sin φ cos φ + 2 sin φ - 2 = 0Hmm, still complicated. Maybe factor terms with cos φ:cos² φ + cos φ - sin φ cos φ = cos φ (cos φ + 1 - sin φ)So, equation becomes:cos φ (cos φ + 1 - sin φ) + 2 sin φ - 2 = 0Let me write:cos φ (cos φ + 1 - sin φ) + 2(sin φ - 1) = 0Hmm, notice that sin φ - 1 is similar to the term in the original function's numerator. Let me factor:= cos φ (cos φ + 1 - sin φ) - 2(1 - sin φ) = 0Factor out (1 - sin φ):= (1 - sin φ)( - cos φ ( (cos φ + 1)/ (1 - sin φ) ) - 2 ) = 0Wait, this might not be straightforward. Alternatively, set t = sin φ - 1, but not sure.Alternatively, maybe consider specific angles where the equation could be satisfied.Let me assume possible values for φ where the equation holds.For example, try φ = π/2. Then sin φ = 1, cos φ = 0.Plug into equation:cos φ (cos φ + 1 - sin φ) + 2 sin φ - 2 = 0*(0 + 1 -1) + 2*1 - 2 = 0 + 2 -2 = 0. So φ=π/2 is a solution.Similarly, check φ=0: sin φ=0, cos φ=1Equation becomes:1*(1 +1 -0) +0 -2= 2 +0 -2=0. So φ=0 is a solution.But φ=0 and φ=π/2 are critical points. But we need to check if these correspond to maxima or minima.Wait, but when φ=0, z(φ)=(1 -1)/sqrt(3 -2*1 -2*0)=0/sqrt(1)=0. So that's the maximum value of z(φ)=0, which corresponds to y=0.When φ=π/2, z(π/2)=(1 -0)/sqrt(3 -0 -2*1)=1/sqrt(1)=1. So z=1, which would mean y=-1. So that's a candidate for the minimum.Wait, but is φ=π/2 the only critical point? Let's check another angle, say φ=π.At φ=π, sin φ=0, cos φ=-1Equation becomes:(-1)(-1 +1 -0 ) +0 -2= (-1)(0) +0 -2= -2 ≠0. So not a solution.How about φ=3π/2, sin φ=-1, cos φ=0:Equation becomes 0*(0 +1 - (-1)) +2*(-1) -2=0 + (-2) -2=-4≠0.Another angle: φ=π/4. sin φ=√2/2, cos φ=√2/2.Compute:cos φ (cos φ +1 - sin φ) +2 sin φ -2= √2/2 ( √2/2 +1 - √2/2 ) + 2*(√2/2) -2= √2/2 (1) + √2 -2 ≈ 0.707 +1.414 -2 ≈ 0.121≠0So not zero.Another angle: Let's assume that 1 - sin φ ≠0, so we can factor it out.We had:cos φ (cos φ +1 - sin φ) -2(1 - sin φ)=0Factor (1 - sin φ):But it's not straightforward. Let's rearrange:cos φ (cos φ +1 - sin φ) = 2(1 - sin φ)Divide both sides by (1 - sin φ):cos φ [ (cos φ +1 - sin φ)/(1 - sin φ) ] = 2Simplify the fraction:(cos φ +1 - sin φ)/(1 - sin φ) = [ (cos φ +1) - sin φ ]/(1 - sin φ)Let me split the numerator:= [ (cos φ +1) - sin φ ] / (1 - sin φ) = [ (1 + cos φ) - sin φ ] / (1 - sin φ)Can we write this as [ (1 + cos φ) - sin φ ] / (1 - sin φ) = A?Alternatively, multiply numerator and denominator by (1 + sin φ):[ (1 + cos φ) - sin φ ](1 + sin φ) / [ (1 - sin φ)(1 + sin φ) ]Denominator becomes 1 - sin²φ = cos²φNumerator:(1 + cos φ)(1 + sin φ) - sin φ(1 + sin φ)= (1 + sin φ + cos φ + sin φ cos φ) - (sin φ + sin²φ)= 1 + sin φ + cos φ + sin φ cos φ - sin φ - sin²φSimplify:1 + cos φ + sin φ cos φ - sin²φ= 1 + cos φ + sin φ cos φ - (1 - cos²φ)= 1 + cos φ + sin φ cos φ -1 + cos²φ= cos φ + sin φ cos φ + cos²φ= cos φ(1 + sin φ + cos φ)So, overall:[ numerator ] / [ denominator ] = [ cos φ(1 + sin φ + cos φ) ] / cos²φ = (1 + sin φ + cos φ)/cos φTherefore, the left-hand side of the equation becomes:cos φ * [ (1 + sin φ + cos φ)/cos φ ] = 1 + sin φ + cos φSo, the equation is:1 + sin φ + cos φ = 2Thus,1 + sin φ + cos φ = 2 ⇒ sin φ + cos φ = 1Ah, this is a simpler equation!So, after simplifying, we find that the critical points occur when either:1. 1 - sin φ = 0 ⇒ sin φ = 1 ⇒ φ = π/2 (which we already considered), or2. sin φ + cos φ = 1Therefore, the critical points are φ=0, φ=π/2, and solutions to sin φ + cos φ =1.Wait, but wait, when we divided by (1 - sin φ), we assumed that 1 - sin φ ≠0. So the equation sin φ + cos φ =1 comes from the case when 1 - sin φ ≠0. So in total, the critical points are:- φ=0 (from 1 - sin φ =0)- φ=π/2 (from earlier solution)- φ such that sin φ + cos φ =1So, let's solve sin φ + cos φ =1.This equation can be written as:sin φ + cos φ =1Divide both sides by sqrt(2):(1/sqrt(2)) sin φ + (1/sqrt(2)) cos φ = 1/sqrt(2)Which is:sin φ cos π/4 + cos φ sin π/4 = sin(φ + π/4) = 1/sqrt(2)Therefore,sin(φ + π/4) = 1/sqrt(2)Solutions are:φ + π/4 = π/4 + 2πk ⇒ φ = 0 + 2πkorφ + π/4 = 3π/4 + 2πk ⇒ φ = π/2 + 2πkBut since φ ∈ [0, 2π), the solutions are φ=0 and φ=π/2. However, these are already considered. Wait, but sin(φ + π/4) = 1/sqrt(2) also has solutions:φ + π/4 = π/4 ⇒ φ=0φ + π/4 = 3π/4 ⇒ φ=π/2But also, φ + π/4 = 9π/4 ⇒ φ=2π, which is equivalent to φ=0, and φ + π/4= 7π/4 ⇒ φ=3π/2 - π/4=3π/2 - π/4=5π/4. Wait, no, wait:Wait, solving sin α = 1/sqrt(2), where α = φ + π/4.Solutions are α = π/4 + 2πk or α=3π/4 + 2πk.Therefore, φ = α - π/4:φ = π/4 - π/4 + 2πk = 0 + 2πkφ = 3π/4 - π/4 + 2πk = π/2 + 2πkSo within [0, 2π), the solutions are φ=0 and φ=π/2. Which we already have. Therefore, the critical points are φ=0 and φ=π/2.Wait, but this seems contradictory, because when we divided by (1 - sin φ), we obtained sin φ + cos φ =1, whose solutions are φ=0 and φ=π/2, but φ=0 makes 1 - sin φ=0, which was our initial case. So this suggests that the only critical points are φ=0 and φ=π/2. But when we tested φ=π/2, we got z=1, and at φ=0, z=0. So are these the only critical points?Wait, but earlier when we did the substitution with θ, φ, etc., we might have lost some solutions. Let me verify.Wait, the original equation after simplifying was:sin φ + cos φ =1Which only has solutions φ=0 and φ=π/2 in [0, 2π). Let me check φ=π/2:sin(π/2) + cos(π/2) =1 +0=1 ✔️φ=0:sin(0)+cos(0)=0 +1=1 ✔️Other angles? For example, φ=π/2 + π=3π/2:sin(3π/2)+cos(3π/2)= -1 +0= -1≠1So indeed, only φ=0 and φ=π/2 satisfy sin φ + cos φ=1.Therefore, the critical points are φ=0 and φ=π/2. But φ=0 gives z=0 and φ=π/2 gives z=1. But wait, is that the only critical points?Wait, in the earlier derivative calculation, we considered two cases: when 1 - sin φ=0 and when the other factor is zero. But after factoring, we saw that the critical points are φ=0 and φ=π/2. But this seems too few. Perhaps there are other critical points where the derivative is undefined? The denominator sqrt(D) is always positive, so no. Or perhaps my mistake came during the algebra.Alternatively, maybe there are other critical points. Let's check another angle where z(φ) might attain a maximum. Let's try φ=2π/3.At φ=2π/3, sin φ=√3/2 ≈0.866, cos φ=-1/2.z(φ)=(1 - (-1/2))/sqrt(3 -2*(-1/2) -2*(√3/2))=(3/2)/sqrt(3 +1 -√3)= (3/2)/sqrt(4 -√3)Compute denominator: sqrt(4 -√3)≈sqrt(4 -1.732)=sqrt(2.268)≈1.506So z≈ (1.5)/1.506≈0.996, which is close to 1. So z≈1, but slightly less.Wait, but at φ=π/2, z=1. So maybe z=1 is the maximum. But let's check another angle.At φ=π/3, sin φ=√3/2≈0.866, cos φ=1/2.z(φ)=(1 -1/2)/sqrt(3 -2*(1/2) -2*(√3/2))= (1/2)/sqrt(3 -1 -√3)= (0.5)/sqrt(2 -√3)≈0.5 / sqrt(2 -1.732)=0.5/sqrt(0.268)=0.5/0.517≈0.968, which is less than 1.Hmm. At φ=5π/6, sin φ=1/2, cos φ=-√3/2.z=(1 -(-√3/2))/sqrt(3 -2*(-√3/2) -2*(1/2))=(1 +√3/2)/sqrt(3 +√3 -1)=(1 +0.866)/sqrt(2 +1.732)=1.866/sqrt(3.732)≈1.866/1.932≈0.966, again less than 1.At φ=π/6, sin φ=0.5, cos φ=√3/2.z=(1 -√3/2)/sqrt(3 -2*(√3/2) -2*(0.5))=(1 -0.866)/sqrt(3 -√3 -1)=0.134/sqrt(2 -1.732)=0.134/0.517≈0.259.So it seems like z=1 at φ=π/2 is the maximum. So then, the minimum value of y is -1, achieved when φ=π/2, which corresponds to x=φ + π/2=π/2 + π/2=π. Wait, no, wait, we had earlier substituted φ =x - π/2. So φ=π/2 corresponds to x=φ + π/2=π/2 + π/2=π. Wait, no:Wait, in the substitution earlier, we set φ = x - π/2. So when φ=π/2, x=φ + π/2=π/2 + π/2=π. So x=π. Let me verify:At x=π, sin x=0, cos x=-1. So original function:f(x)=(0 -1)/sqrt(3 -2*(-1) -2*0)=(-1)/sqrt(3 +2)=(-1)/sqrt(5). But sqrt(5)≈2.236, so f(x)≈-0.447. But according to our substitution, when φ=π/2, z=1, so y=-1. But this contradicts with the direct calculation. Hmm, there must be a mistake here.Wait, no, when we substituted φ=x - π/2, then when x=π, φ=π - π/2=π/2. Then in terms of the substitution:y(φ)= (cos φ -1)/sqrt(3 +2 sin φ -2 cos φ)At φ=π/2, cos φ=0, sin φ=1:y(π/2)= (0 -1)/sqrt(3 +2*1 -2*0)= (-1)/sqrt(5)=≈-0.447.But earlier, when we calculated z(φ)= (1 - cos φ)/sqrt(3 -2 sin φ -2 cos φ), but in the substitution, after replacing x with φ= x - π/2, the denominator became sqrt(3 +2 sin φ -2 cos φ). So z(phi) as per substitution is (1 - cos phi)/sqrt(3 +2 sin phi -2 cos phi). Wait, but in the previous analysis, we had z(phi)= (1 - cos phi)/sqrt(3 -2 cos phi -2 sin phi). But after substitution, the denominator became 3 + 2 sin phi - 2 cos phi. So there was a sign change. Therefore, in the previous analysis, the z(phi) after substitution was different.But this seems to have created confusion. Let me clarify.Original function f(x) = (sin x -1)/sqrt(3 -2 cos x -2 sin x). After substitution phi =x - pi/2, we have:f(phi) = (cos phi -1)/sqrt(3 +2 sin phi -2 cos phi)So, z(phi) = (1 - cos phi)/sqrt(3 +2 sin phi -2 cos phi)But when we did the derivative and found critical points at phi=0 and phi=pi/2, which correspond to x=pi/2 and x=pi.At phi=0: x=pi/2, sin x=1, cos x=0: f(x)=(1 -1)/sqrt(3 -0 -2*1)=0/sqrt(1)=0.At phi=pi/2: x=pi, sin pi=0, cos pi=-1: f(x)=(0 -1)/sqrt(3 -2*(-1) -2*0)=(-1)/sqrt(5)= -1/sqrt(5).But according to our previous analysis, when we computed z(phi)=1 at phi=pi/2, but in reality, z(pi/2)= (1 - cos(pi/2))/sqrt(3 +2 sin(pi/2)-2 cos(pi/2))= (1 -0)/sqrt(3 +2*1 -2*0)=1/sqrt(5). Therefore, y(pi/2)= -1/sqrt(5), which is approximately -0.447.But earlier, when solving for critical points, we found that at phi=pi/2, z=1/sqrt(5), so y=-1/sqrt(5). But according to the derivative, that was a critical point. However, when I computed z(phi=2π/3), I got z≈0.996, which is close to 1, but in reality, when substituting into the actual function, z(phi) can't exceed 1/sqrt(5)≈0.447. Wait, there's a contradiction here.Wait, no. Wait, there's confusion between variables. Let me start over.Original function f(x) = (sin x -1)/sqrt(3 -2 cos x -2 sin x). We want to find its range.After substituting phi = x - pi/2, we get f(phi) = (cos phi -1)/sqrt(3 +2 sin phi -2 cos phi). Let's call this function g(phi) = (cos phi -1)/sqrt(3 +2 sin phi -2 cos phi)We found that the critical points are phi=0 and phi=pi/2. At phi=0: g(0)= (1 -1)/sqrt(3 +0 -2*1)=0/sqrt(1)=0. At phi=pi/2: g(pi/2)= (0 -1)/sqrt(3 +2*1 -2*0)= (-1)/sqrt(5)= -1/sqrt(5).But when we tried plugging in other angles, like phi=2pi/3, we found that z(phi)= (1 - cos phi)/sqrt(denominator). Wait, but no, in the substitution, the function becomes g(phi)= (cos phi -1)/sqrt(denominator). So z(phi)= -g(phi)= (1 - cos phi)/sqrt(denominator). So when I calculated z(2pi/3)≈0.996, that was in the substituted function. But in reality, after substitution, the denominator is different.This seems too convoluted. Perhaps a better approach is to parametrize the original function using t = tan(x/2), which sometimes helps in rationalizing trigonometric functions.Let me use the Weierstrass substitution:Let t = tan(x/2). Then, sin x = 2t/(1 + t²), cos x = (1 - t²)/(1 + t²). Then, dx = 2 dt/(1 + t²). But since we're dealing with a function of x, maybe express f(x) in terms of t.So, substituting:Numerator: sin x -1 = (2t)/(1 + t²) -1 = (2t -1 - t²)/(1 + t²)Denominator: sqrt(3 -2 cos x -2 sin x) = sqrt(3 -2*( (1 - t²)/(1 + t²) ) -2*(2t/(1 + t²)) )Simplify denominator:= sqrt(3 - [2(1 - t²) + 4t]/(1 + t²))= sqrt( [3(1 + t²) - 2(1 - t²) -4t]/(1 + t²) )Compute numerator inside sqrt:3(1 + t²) -2(1 - t²) -4t =3 +3t² -2 +2t² -4t=1 +5t² -4tTherefore, denominator= sqrt( (1 +5t² -4t)/(1 + t²) )= sqrt( (5t² -4t +1)/(1 + t²) )So, f(x) becomes:f(t)= [ (2t -1 -t²)/(1 + t²) ] / sqrt( (5t² -4t +1)/(1 + t²) )Simplify:= [ ( -t² +2t -1 ) / (1 + t²) ] / [ sqrt(5t² -4t +1)/sqrt(1 + t²) ) ]= [ (- (t² -2t +1) ) / (1 + t²) ] * [ sqrt(1 + t²)/sqrt(5t² -4t +1) )= [ - (t -1)^2 / (1 + t²) ] * [ sqrt(1 + t²) / sqrt(5t² -4t +1) )= - (t -1)^2 / sqrt( (1 + t²)(5t² -4t +1) )So, f(t)= - (t -1)^2 / sqrt( (1 + t²)(5t² -4t +1) )This looks complicated, but maybe we can analyze it further.We need to find the range of f(t) as t ranges over real numbers (since x ∈ [0, 2π), t = tan(x/2) ranges from t=0 to t approaching ±∞, but repeating periodically). However, since x is from 0 to 2π, t goes from 0 to ∞, then from -∞ to 0 as x goes from π to 2π. So t ∈ ℝ.But since the function f(t) is periodic with period 2π, but expressed in terms of t, which covers all real numbers.However, analyzing this function might still be tough. Let's consider squaring the function:f(t)^2 = ( (t -1)^4 ) / ( (1 + t²)(5t² -4t +1) )But since f(t) is non-positive, the range of f(t) is determined by the maximum of |f(t)|, which corresponds to the maximum of sqrt(f(t)^2). So, we can find the maximum of f(t)^2, take its square root, and then the minimum of f(t) is the negative of that.So, let's define h(t) = f(t)^2 = (t -1)^4 / [ (1 + t²)(5t² -4t +1) ]We need to find the maximum value of h(t) over t ∈ ℝ.To find the maximum of h(t), take derivative h'(t), set to zero.But this would be quite involved. Alternatively, notice that h(t) is a rational function, and we can look for critical points by setting derivative to zero.Let me compute h'(t):Let h(t) = N(t)/D(t), where N(t)=(t -1)^4, D(t)=(1 + t²)(5t² -4t +1)Then, h'(t) = [N'(t)D(t) - N(t)D'(t)] / D(t)^2Compute N'(t)=4(t-1)^3Compute D(t)= (1 + t²)(5t² -4t +1)Compute D'(t)= derivative of first * second + first * derivative of second= 2t*(5t² -4t +1) + (1 + t²)*(10t -4)= 10t^3 -8t² +2t +10t^3 -4t² +10t -4= 20t^3 -12t² +12t -4Therefore, h'(t)= [4(t-1)^3*(1 + t²)(5t² -4t +1) - (t -1)^4*(20t^3 -12t² +12t -4)] / [ (1 + t²)^2(5t² -4t +1)^2 ]Set numerator to zero:4(t-1)^3*(1 + t²)(5t² -4t +1) - (t -1)^4*(20t^3 -12t² +12t -4) =0Factor out (t -1)^3:(t -1)^3 [4*(1 + t²)(5t² -4t +1) - (t -1)*(20t^3 -12t² +12t -4) ] =0Solutions are t=1 (with multiplicity 3) and solutions from the bracketed term.Set the bracketed term to zero:4*(1 + t²)(5t² -4t +1) - (t -1)*(20t^3 -12t² +12t -4)=0Expand first term:4*(5t² -4t +1 +5t^4 -4t^3 +t²)=4*(5t^4 -4t^3 +6t² -4t +1)=20t^4 -16t^3 +24t² -16t +4Second term:-(t -1)*(20t^3 -12t² +12t -4)= -[20t^4 -12t^3 +12t^2 -4t -20t^3 +12t^2 -12t +4]= -[20t^4 -32t^3 +24t^2 -16t +4]= -20t^4 +32t^3 -24t^2 +16t -4Add both terms:20t^4 -16t^3 +24t² -16t +4 -20t^4 +32t^3 -24t^2 +16t -4= (20t^4 -20t^4) + (-16t^3 +32t^3) + (24t² -24t²) + (-16t +16t) + (4 -4)=16t^3=0Thus, 16t^3=0 ⇒t=0Therefore, critical points are t=1 (multiplicity 3) and t=0.Therefore, the critical points are t=0 and t=1.Now, we need to evaluate h(t) at these points.At t=1:h(1)= (1 -1)^4 / [ (1 +1)(5*1 -4*1 +1) ]=0 / [2*2]=0At t=0:h(0)= (0 -1)^4 / [ (1 +0)(0 -0 +1) ]=1 / (1*1)=1Therefore, the maximum of h(t) is 1 at t=0, and another critical point at t=1 gives h(t)=0.But wait, we need to check the behavior as t approaches infinity.As t→∞, h(t)= (t^4)/[t²*(5t²)]=t^4/(5t^4)=1/5. So h(t)→1/5 as t→∞. Therefore, the maximum value of h(t) is 1, achieved at t=0, and another local maximum at infinity approaching 1/5.Therefore, the maximum of h(t)=1, so the maximum of |f(t)|=sqrt(1)=1, so the minimum value of f(t) is -1.But wait, when t=0, which corresponds to x=0 (since t=tan(x/2)=0 ⇒x=0). At x=0, sin x=0, cos x=1, so f(0)=(0 -1)/sqrt(3 -2*1 -2*0)=(-1)/sqrt(1)=-1.But earlier, when we checked x=pi, we got f(pi)= -1/sqrt(5)≈-0.447. But according to this analysis, the minimum value is -1, achieved at x=0. But wait, at x=0, the numerator is sin0 -1= -1, denominator sqrt(3 -2*1 -0)=sqrt(1)=1. So f(0)= -1/1= -1.However, when we checked phi=pi/2 (x=pi), we got f(pi)= -1/sqrt(5). So the minimum is -1 at x=0, and the function approaches -1/sqrt(5) at x=pi. But according to the critical points analysis, the maximum of z(t)=1, achieved at t=0 (x=0). So the minimum of y is -1.But wait, the function at x=0 is -1, at x=pi/2 is 0, at x=pi is -1/sqrt(5), and at x=3pi/2 is 0/something (which is 0). Wait, no, at x=3pi/2, sin x=-1, so numerator is -1 -1=-2, denominator is sqrt(3 -2*0 -2*(-1))=sqrt(3 +2)=sqrt(5), so f(3pi/2)= (-2)/sqrt(5)≈-0.894. This is more negative than -1/sqrt(5). But according to our analysis using the Weierstrass substitution, the minimum is -1.There's a contradiction here. Let's verify with x=3pi/2.At x=3pi/2, sin x= -1, cos x=0.So, f(x)= (-1 -1)/sqrt(3 -2*0 -2*(-1))= (-2)/sqrt(3 +2)= (-2)/sqrt(5)≈-0.894.But according to our previous analysis, the minimum is -1 at x=0. However, here, we have a more negative value at x=3pi/2. This suggests that our analysis using the Weierstrass substitution missed some critical points.Wait, in the Weierstrass substitution analysis, we found critical points at t=0 and t=1. t=0 corresponds to x=0, and t=1 corresponds to x=pi/2. However, x=3pi/2 corresponds to t=tan(3pi/4)= -1. So t=-1. Let's compute h(t) at t=-1.h(-1)= (-1 -1)^4 / [ (1 +1)(5*1 -4*(-1) +1) ]= (16)/[2*(5 +4 +1)]=16/(2*10)=16/20=0.8Therefore, h(-1)=0.8, which is 4/5. Therefore, |f(-1)|=sqrt(4/5)=2/sqrt(5)=≈0.894, so f(-1)= -2/sqrt(5)≈-0.894. Therefore, this is a local maximum of h(t) but not the global maximum. The global maximum of h(t) is 1 at t=0.So, the minimum value of f(x) is -1, and another critical point at t=-1 gives f(x)= -2/sqrt(5). But according to the analysis, t=0 gives the global maximum of h(t)=1, thus f(t)=-1.But how to reconcile this with the fact that at x=3pi/2, the function is -2/sqrt(5)≈-0.894, which is greater than -1.Wait, no, -0.894 is greater than -1. So the minimum value is -1, and other values are less negative (i.e., higher). So the range is [-1, 0].But wait, at x=3pi/2, the function is -2/sqrt(5)≈-0.894, which is higher than -1, so the minimum is indeed -1 at x=0.But why does the substitution method give h(t)=1 at t=0 (x=0), and another critical point at t=-1 with h(t)=0.8, but we also have x=3pi/2 corresponding to t=-1 giving f(x)= -2/sqrt(5). But this is less than the minimum of -1? No, wait, -2/sqrt(5)≈-0.894 is greater than -1.Wait, but in reality, -1 is less than -0.894, so the minimum value is -1.But when x=3pi/2, the function is approximately -0.894, which is higher (less negative) than -1. Therefore, the true minimum is -1 at x=0.But this seems to contradict the earlier critical point analysis where solving via Lagrange multipliers and substitution only found critical points at x=0 and x=pi, but missed x=3pi/2. But in reality, when we used the Weierstrass substitution, we found another critical point at t=-1 (x=3pi/2), but its corresponding h(t)=0.8, which is less than the maximum of 1 at t=0. Therefore, the minimum value of f(x) is -1, and the other critical points have higher values.But why didn't the derivative method with phi find this?Wait, when we did the substitution phi = x - pi/2, we transformed the function and found critical points at phi=0 and phi=pi/2, corresponding to x=pi/2 and x=pi. But the critical point at x=3pi/2 (phi=3pi/2 - pi/2=pi) was already considered? No, wait, phi ranges from -pi/2 to 3pi/2 when x ranges from 0 to 2pi. So phi=pi corresponds to x=pi + pi/2=3pi/2. Wait, no:Wait, phi = x - pi/2. So when x=3pi/2, phi=3pi/2 - pi/2=pi. So phi=pi. But in our analysis of critical points for the substituted function g(phi), we found critical points at phi=0 and phi=pi/2, but not at phi=pi. But when phi=pi, which corresponds to x=3pi/2, we have:g(pi)= (cos pi -1)/sqrt(3 +2 sin pi -2 cos pi)= (-1 -1)/sqrt(3 +0 -2*(-1))= (-2)/sqrt(5)= -2/sqrt(5). Which is the same as x=3pi/2.But in our critical point analysis for the substituted function, we only found critical points at phi=0 and phi=pi/2. Why didn't we get phi=pi?Because when we computed the derivative for the substituted function, we found critical points at phi=0 and phi=pi/2, which correspond to x=pi/2 and x=pi. However, the critical point at x=3pi/2 (phi=pi) wasn't found because when we solved the equation, we only got phi=0 and phi=pi/2.This suggests that there might be an error in the critical point analysis. However, when using the Weierstrass substitution, we found critical points at t=0, t=1, and t=-1 (the latter corresponding to phi=pi), but in the substituted function, t=-1 corresponds to x=3pi/2. However, the maximum of h(t) at t=0 gives the minimum value of f(x) as -1, and other critical points have higher values.Therefore, the range of f(x) is [-1, 0].But to confirm, let's check x=0: f(0)= -1/sqrt(1)= -1.At x=3pi/2: f(3pi/2)= -2/sqrt(5)≈-0.894.At x=pi: f(pi)= -1/sqrt(5)≈-0.447.At x=pi/2: f(pi/2)=0.At x=2pi: same as x=0, f(2pi)= -1.Therefore, the function attains its minimum value of -1 at x=0 and x=2pi, and other critical points have higher values. Therefore, the range is [-1, 0].Final AnswerThe range of the function is boxed{[-1, 0]}.</think>To find the range of the function ( f(x) = frac{sin x - 1}{sqrt{3 - 2 cos x - 2 sin x}} ) for ( 0 leq x leq 2pi ):1. Numerator and Denominator Analysis: - The numerator (sin x - 1) ranges from -2 to 0. - The denominator (sqrt{3 - 2 cos x - 2 sin x}) is always positive because (3 - 2(cos x + sin x) geq 0) for all (x).2. Substitution and Simplification: - Substitute (phi = x - pi/2), transforming the function into ( f(phi) = frac{cos phi - 1}{sqrt{3 + 2 sin phi - 2 cos phi}} ). - This substitution helps in simplifying the expression and analyzing critical points.3. Critical Points and Extrema: - Using the Weierstrass substitution ( t = tan(x/2) ), the function is transformed into a rational function, leading to the analysis of critical points. - Critical points are found at ( t = 0 ) and ( t = 1 ), corresponding to ( x = 0 ) and ( x = pi/2 ).4. Evaluation at Critical Points: - At ( x = 0 ), the function value is (-1). - At ( x = pi/2 ), the function value is (0). - Other critical points such as ( x = 3pi/2 ) yield values less than (-1) but not the minimum.5. Conclusion: - The minimum value of the function is (-1) at ( x = 0 ) and ( x = 2pi ). - The maximum value of the function is (0) at ( x = pi/2 ).Thus, the range of the function is (boxed{[-1, 0]}).,<think>Alright, so I need to find the range of the function f(x) = (sin x - 1)/sqrt(3 - 2 cos x - 2 sin x) where x is between 0 and 2π. Hmm, okay. Let me try to break this down step by step. First, let's understand the function. The numerator is sin x - 1, and the denominator is the square root of 3 - 2 cos x - 2 sin x. Since there's a square root in the denominator, the expression inside the square root must be positive. So, I should first figure out the domain where 3 - 2 cos x - 2 sin x > 0. But the problem already specifies 0 ≤ x ≤ 2π, so maybe the entire interval is valid? Or maybe not? Let me check.Let me compute 3 - 2 cos x - 2 sin x. Let's see if this is always positive. To check that, maybe I can write the expression as 3 - 2(cos x + sin x). Then, since cos x + sin x has a maximum value of sqrt(1^2 + 1^2) = sqrt(2) from the identity a cos x + b sin x = sqrt(a² + b²) cos(x - φ). So, cos x + sin x has a maximum of sqrt(2) and minimum of -sqrt(2). Therefore, 3 - 2(cos x + sin x) would range from 3 - 2*sqrt(2) ≈ 3 - 2.828 ≈ 0.172 to 3 + 2*sqrt(2) ≈ 5.828. So, the expression inside the square root is always positive since even the minimum is approximately 0.172, which is positive. Therefore, the denominator is always defined and positive for all x in [0, 2π]. So, the domain is indeed 0 ≤ x ≤ 2π.Okay, so next, I need to find the range of f(x). The numerator is sin x - 1. Since sin x ranges from -1 to 1, sin x -1 ranges from -2 to 0. Therefore, the numerator is always non-positive, ranging from -2 to 0. The denominator is always positive as we saw. Therefore, f(x) is a non-positive function, so its range will be from some negative value up to 0.To find the exact range, maybe we can express the function in terms of a single trigonometric function or find a substitution. Let me see. Let's denote t = x or maybe use some substitution to simplify the expression.Alternatively, maybe square the function to make it easier, but since the numerator is negative and the denominator is positive, squaring will give us the square of the function, which is (sin x - 1)^2 / (3 - 2 cos x - 2 sin x). Then, perhaps analyze this expression and then take the square root at the end, considering the sign. But since we know f(x) is non-positive, the range will just be from -sqrt(max of squared function) to 0.Alternatively, maybe express both numerator and denominator in terms of sin x and cos x, and try to find a relation. Let me try to write the denominator as sqrt(3 - 2 cos x - 2 sin x). Maybe we can write 3 - 2 cos x - 2 sin x as something squared? Let me check.Alternatively, notice that 3 - 2 cos x - 2 sin x can be rewritten as 3 - 2(cos x + sin x). Let me recall that cos x + sin x can be written as sqrt(2) sin(x + π/4). Therefore, 3 - 2*sqrt(2) sin(x + π/4). So, 3 - 2*sqrt(2) sin(x + π/4). The denominator then becomes sqrt(3 - 2*sqrt(2) sin(x + π/4)). Hmm, not sure if that helps.Alternatively, let's consider the denominator expression 3 - 2 cos x - 2 sin x. Maybe complete the square or relate it to the numerator. The numerator is sin x - 1.Alternatively, let's set t = x - π/2. Then, sin x = sin(t + π/2) = cos t, and cos x = cos(t + π/2) = -sin t. Maybe substitution?Wait, perhaps we can write the denominator expression 3 - 2 cos x - 2 sin x in terms of sin x and cos x. Let me see:3 - 2 cos x - 2 sin x = 3 - 2(cos x + sin x). Let me set u = cos x + sin x. Then, we can write the denominator as 3 - 2u.Also, note that sin x - 1 can be written as -(1 - sin x). So, f(x) = -(1 - sin x)/sqrt(3 - 2u). Let's see if we can relate u and 1 - sin x.We have u = cos x + sin x. Let's compute (1 - sin x). Maybe express u in terms of sin x.Alternatively, let's square u: u² = cos²x + 2 sin x cos x + sin²x = 1 + sin 2x. So, sin 2x = u² - 1. Not sure if that helps.Alternatively, let's try to express both numerator and denominator in terms of sin x or cos x. Let me try to express the denominator in terms of sin x. Let's see:Denominator inside sqrt: 3 - 2 cos x - 2 sin x. Let me express cos x in terms of sin x. Since cos x = sqrt(1 - sin²x), but that might complicate things. Alternatively, perhaps use a substitution variable. Let me let y = sin x. Then cos x = sqrt(1 - y²), but that introduces square roots and signs. Since x is between 0 and 2π, we have to consider the sign of cos x. Maybe split the interval based on the quadrants, but that might not be efficient.Alternatively, perhaps use a substitution like t = tan(x/2). Then, using the Weierstrass substitution:sin x = 2t/(1 + t²)cos x = (1 - t²)/(1 + t²)But that might lead to a complicated expression, but let's try.So, substituting into numerator and denominator:Numerator: sin x - 1 = (2t)/(1 + t²) - 1 = (2t - (1 + t²))/(1 + t²) = (-t² + 2t - 1)/(1 + t²) = -(t² - 2t + 1)/(1 + t²) = -(t - 1)^2 / (1 + t²)Denominator: sqrt(3 - 2 cos x - 2 sin x) = sqrt(3 - 2*(1 - t²)/(1 + t²) - 2*(2t)/(1 + t²))Let's compute inside the square root:3 - 2*(1 - t²)/(1 + t²) - 4t/(1 + t²)= 3 - [2(1 - t²) + 4t]/(1 + t²)= 3 - [2 - 2t² + 4t]/(1 + t²)= [3(1 + t²) - 2 + 2t² - 4t]/(1 + t²)Compute numerator of the denominator:3 + 3t² - 2 + 2t² - 4t = (3 - 2) + (3t² + 2t²) -4t = 1 + 5t² -4tTherefore, denominator becomes sqrt( (1 + 5t² -4t)/(1 + t²) )So, putting it all together, f(x) becomes:[-(t - 1)^2 / (1 + t²)] / sqrt( (1 + 5t² -4t)/(1 + t²) )Simplify the division:= [-(t - 1)^2 / (1 + t²)] * [sqrt(1 + t²) / sqrt(1 + 5t² -4t)]= - (t - 1)^2 / [ (1 + t²) * sqrt( (1 + 5t² -4t)/(1 + t²) ) ]= - (t - 1)^2 / [ sqrt( (1 + t²)(1 + 5t² -4t) ) ]Hmm, this seems complicated. Maybe not the best approach. Let's try another way.Alternatively, note that both numerator and denominator can be related to some trigonometric identity. For instance, the denominator is 3 - 2 cos x - 2 sin x. Let me see if this can be written in terms of (1 - sin x) or (1 - cos x). Let's compute:Let me compute 3 - 2 cos x - 2 sin x. Let's group terms:3 = 1 + 1 + 1, maybe. Not sure. Alternatively, think of 3 - 2 cos x - 2 sin x as (1 - 2 sin x) + (2 - 2 cos x). Not sure. Alternatively, 3 - 2(cos x + sin x). Maybe consider the expression (sin x -1) in the numerator. Let's see if there is a relationship between (sin x -1) and the denominator.Alternatively, let's consider writing the denominator as sqrt(3 - 2 cos x - 2 sin x). Let me square the function f(x) to get rid of the square root:f(x)^2 = (sin x - 1)^2 / (3 - 2 cos x - 2 sin x)So, if I can find the range of f(x)^2, then take square roots and adjust the sign. Since f(x) is non-positive, the range of f(x) will be from -sqrt(max(f(x)^2)) to 0.So, let's compute f(x)^2:Let me denote y = sin x. Then cos x = sqrt(1 - y²) with appropriate sign. But since x is between 0 and 2π, cos x can be positive or negative. However, this might complicate things. Alternatively, maybe use a substitution variable t = sin x + cos x or something else.Alternatively, express both the numerator and denominator in terms of sin x -1. Let's see:Numerator is sin x -1. Let's denote u = sin x -1. Then, sin x = u + 1. Then, cos x = sqrt(1 - (u +1)^2) but that's messy because (u +1)^2 would be sin²x, so cos x = sqrt(1 - sin²x) = sqrt( (1 - sin x)(1 + sin x) ) = sqrt( (-u)(2 + u) ). But this may not be helpful.Wait, perhaps we can consider writing the denominator in terms of (sin x -1). Let's compute denominator:3 - 2 cos x - 2 sin x = 3 - 2 sin x - 2 cos x. Let's see if we can relate this to (sin x -1). Maybe expand (sin x -1)^2:(sin x -1)^2 = sin²x - 2 sin x +1. Not directly related. The denominator is linear in sin x and cos x, while the numerator is linear in sin x.Alternatively, maybe parametrize the function using a parameter t. Let me set t = x. Then, we can think of f(x) as a function of t, but that's not helpful. Alternatively, use substitution for theta = x + α, such that we can simplify the denominator. For example, if we write 3 - 2 cos x - 2 sin x as R - 2 cos(x + φ), but not sure.Wait, earlier I mentioned that 3 - 2(cos x + sin x) can be written as 3 - 2sqrt(2) sin(x + π/4). Let's verify:cos x + sin x = sqrt(2) sin(x + π/4). Yes, because:sqrt(2) sin(x + π/4) = sqrt(2)(sin x cos π/4 + cos x sin π/4) = sqrt(2)( (sin x + cos x)/sqrt(2) ) = sin x + cos x. So, that's correct.Therefore, denominator is sqrt(3 - 2sqrt(2) sin(x + π/4)). Let's denote θ = x + π/4. Then, x = θ - π/4. Then, sin x = sin(θ - π/4) = sin θ cos π/4 - cos θ sin π/4 = (sin θ - cos θ)/sqrt(2). Therefore, sin x -1 = (sin θ - cos θ)/sqrt(2) -1. Hmm, maybe not helpful. Alternatively, maybe write f(x) in terms of θ.Wait, f(x) = [sin x -1]/sqrt(3 - 2sqrt(2) sin θ), where θ = x + π/4. But θ ranges from π/4 to 2π + π/4, so sin θ ranges between -1 and 1. So, denominator is sqrt(3 - 2sqrt(2) sin θ). So, the denominator is sqrt(3 - 2sqrt(2) sin θ). Hmm, but I don't see an immediate relationship between numerator and θ.Alternatively, since both numerator and denominator are expressed in terms of sin x and sin(theta), maybe relate them.Alternatively, perhaps use calculus. Since f(x) is continuous on [0, 2π], its range will be an interval [m, 0], where m is the minimum value. To find m, we can take the derivative of f(x) and find critical points.So, let's try that. Let me compute f'(x) and set it to zero to find critical points.First, f(x) = (sin x -1)/sqrt(3 - 2 cos x - 2 sin x). Let's denote denominator as D(x) = sqrt(3 - 2 cos x - 2 sin x). Then, f(x) = (sin x -1)/D(x). Then, f'(x) = [cos x * D(x) - (sin x -1) * (1/(2 D(x))) * (2 sin x - 2 cos x) ] / D(x)^2.Wait, let me compute this step by step. Using the quotient rule: if f = numerator/denominator, then f’ = (num’ * denom - num * denom’)/denom².First, compute numerator’:Numerator = sin x -1, so derivative is cos x.Denominator = sqrt(3 - 2 cos x - 2 sin x). Let me denote inside the sqrt as g(x) = 3 - 2 cos x - 2 sin x. So, denominator = sqrt(g(x)). Then, the derivative of denominator is (1/(2 sqrt(g(x)))) * g’(x). Compute g’(x):g’(x) = 0 - 2*(-sin x) - 2*cos x = 2 sin x - 2 cos x.Therefore, denominator’ = (2 sin x - 2 cos x)/(2 sqrt(g(x))) = (sin x - cos x)/sqrt(g(x)).Therefore, f’(x) = [cos x * sqrt(g(x)) - (sin x -1) * (sin x - cos x)/sqrt(g(x))] / g(x)Multiply numerator and denominator by sqrt(g(x)) to simplify:= [cos x * g(x) - (sin x -1)(sin x - cos x)] / g(x)^(3/2)So, the critical points occur when the numerator is zero:cos x * g(x) - (sin x -1)(sin x - cos x) = 0Compute each term:First term: cos x * (3 - 2 cos x - 2 sin x)Second term: -(sin x -1)(sin x - cos x)Let me expand the second term:-(sin x -1)(sin x - cos x) = -[sin²x - sin x cos x - sin x + cos x] = -sin²x + sin x cos x + sin x - cos xSo, the equation becomes:cos x * (3 - 2 cos x - 2 sin x) - sin²x + sin x cos x + sin x - cos x = 0Let me expand the first term:3 cos x - 2 cos²x - 2 sin x cos x - sin²x + sin x cos x + sin x - cos x = 0Combine like terms:3 cos x - cos x = 2 cos x-2 cos²x - sin²x-2 sin x cos x + sin x cos x = - sin x cos x+ sin xSo, combining:2 cos x -2 cos²x - sin²x - sin x cos x + sin x = 0Hmm, this seems complicated. Let me try to simplify further. Let's note that sin²x = 1 - cos²x. Substitute that in:2 cos x -2 cos²x - (1 - cos²x) - sin x cos x + sin x = 0Simplify:2 cos x -2 cos²x -1 + cos²x - sin x cos x + sin x = 0Combine like terms:2 cos x - cos²x -1 - sin x cos x + sin x = 0Let me group terms:(-cos²x) + (2 cos x - sin x cos x) + (-1 + sin x) = 0Factor where possible:= -cos²x + cos x (2 - sin x) + (-1 + sin x) = 0Hmm, not obvious. Let me rearrange:-cos²x -1 + 2 cos x - sin x cos x + sin x = 0Maybe factor terms with sin x:= -cos²x -1 + 2 cos x + sin x (1 - cos x) = 0So, we have:-cos²x -1 + 2 cos x + sin x (1 - cos x) = 0Let me write cos²x as 1 - sin²x:= -(1 - sin²x) -1 + 2 cos x + sin x (1 - cos x)= -1 + sin²x -1 + 2 cos x + sin x (1 - cos x)= sin²x - 2 + 2 cos x + sin x (1 - cos x)Hmm, still complicated. Let me see if substituting t = cos x or t = sin x helps. Let me set t = sin x. Then cos x = sqrt(1 - t²) but with possible sign. Alternatively, express everything in terms of sin x and cos x.Alternatively, let me try to factor this equation. Let me see:Starting from:-cos²x -1 + 2 cos x + sin x (1 - cos x) = 0Let me factor out (1 - cos x) from the last two terms:-cos²x -1 + 2 cos x + (1 - cos x)(sin x) = 0But I don't see a clear factor. Alternatively, perhaps consider specific values of x where this equation holds. For instance, maybe x = π/2 where sin x =1, but then sin x -1 =0, which would be the maximum of the numerator. Let's check x = π/2.At x = π/2:Check if it's a critical point. Let's compute f'(π/2):First, cos(π/2) =0, sin(π/2)=1.Original function f(π/2) = (1 -1)/sqrt(3 -0 -2*1) = 0/sqrt(1) =0. So, f(x)=0 here.But when computing f'(π/2), let's plug into the derivative expression. From before:Numerator of f’(x) is:cos x * g(x) - (sin x -1)(sin x - cos x)At x=π/2:cos x=0, sin x=1.So, first term: 0 * g(x)=0Second term: (1 -1)(1 -0)=0*1=0Therefore, numerator is 0 -0=0. So, x=π/2 is a critical point. But since f(π/2)=0, which is the upper bound of the range.Similarly, check x=3π/2, sin x=-1, cos x=0.At x=3π/2, numerator is -1 -1 = -2, denominator is sqrt(3 -0 -2*(-1))=sqrt(3 +2)=sqrt(5). So, f(3π/2)= -2/sqrt(5). Maybe this is the minimum?But is this a critical point? Let's check f'(3π/2):cos x=0, sin x=-1Numerator of f’(x) is:0 * g(x) - (-1 -1)(-1 -0) = 0 - (-2)(-1) =0 -2= -2 ≠0. So, not a critical point. So, maybe not.Alternatively, maybe other points. Let's think. Let me try x=0. At x=0, sin x=0, cos x=1. f(0)= (0 -1)/sqrt(3 -2*1 -2*0)= (-1)/sqrt(1)= -1. So, f(0)=-1. Let's check derivative at x=0:Numerator:cos0 * g(0) - (sin0 -1)(sin0 -cos0)=1*(3 -2*1 -0) - (-1)(0 -1)=1*(1) - (-1)(-1)=1 -1=0. Therefore, x=0 is a critical point.So, f'(0)=0. Therefore, x=0 is a critical point.Similarly, check x=π. At x=π, sin x=0, cos x=-1.Numerator of f’(x):cos π * g(π) - (sin π -1)(sin π - cos π)= (-1)*(3 -2*(-1)-2*0) - (-1)(0 - (-1))= (-1)*(3 +2) - (-1)(1)= (-5) - (-1)= -5 +1= -4 ≠0. So, not a critical point.So, critical points at x=0, x=π/2, and maybe others. Let's check another point. Let's take x=π/4. Let me compute f’(π/4):First, compute cos(π/4)=sqrt(2)/2, sin(π/4)=sqrt(2)/2.Compute g(π/4)=3 -2*(sqrt(2)/2) -2*(sqrt(2)/2)=3 -2*sqrt(2).Numerator:cos(π/4)*g(π/4) - (sin(π/4)-1)(sin(π/4)-cos(π/4)) = (sqrt(2)/2)(3 -2*sqrt(2)) - (sqrt(2)/2 -1)(0) = (sqrt(2)/2)(3 -2*sqrt(2)) -0. So, not zero. Therefore, x=π/4 is not a critical point.Alternatively, perhaps there's symmetry here. Let me think.Alternatively, let's try to find critical points by solving the equation:-cos²x -1 + 2 cos x + sin x (1 - cos x) = 0This seems complicated. Maybe instead, we can parametrize the problem using t = sin x + cos x as before.Let me denote u = sin x + cos x. Then, u² = sin²x + 2 sin x cos x + cos²x = 1 + sin 2x. So, sin 2x = u² -1. Also, we know that the denominator is 3 - 2u, so D = sqrt(3 - 2u).The numerator is sin x -1. Let me express sin x -1 in terms of u. Since u = sin x + cos x, perhaps express sin x as u - cos x. But that might not help. Alternatively, write sin x -1 = (u - cos x) -1 = u - cos x -1. Hmm, not helpful.Alternatively, perhaps square both sides. Let me consider y = f(x) = (sin x -1)/sqrt(3 - 2u). Then, y = (sin x -1)/sqrt(3 - 2u). Let me square both sides:y² = (sin x -1)² / (3 - 2u)But we can also express (sin x -1)² = sin²x -2 sin x +1 = (1 - cos²x) -2 sin x +1 = 2 -2 sin x -cos²x.Hmm, but how does that relate to u? Since u = sin x + cos x, perhaps express cos x = u - sin x.Then, substitute cos x = u - sin x into cos²x:cos²x = (u - sin x)^2 = u² - 2 u sin x + sin²xTherefore, (sin x -1)^2 = 2 -2 sin x - (u² -2 u sin x + sin²x) = 2 -2 sin x -u² +2 u sin x - sin²xBut this seems complicated. Maybe not the way to go.Alternatively, let's consider that u = sin x + cos x. Then, the denominator is 3 - 2u. The numerator is sin x -1. Let me express sin x as u - cos x. Then, numerator is u - cos x -1. Not sure.Alternatively, let me set v = sin x -1. Then, sin x = v +1. Then, cos x = sqrt(1 - (v +1)^2). Wait, but (v +1)^2 = sin²x, so 1 - sin²x = cos²x, but that would be sqrt( -v(v +2) ). Since v = sin x -1, which ranges from -2 to 0, so v(v +2) = (sin x -1)(sin x +1) = sin²x -1 = -cos²x. Therefore, sqrt(-v(v +2)) = |cos x|. So, cos x = ±sqrt(-v(v +2)). But this might complicate things.Alternatively, maybe use substitution t = sin x. Then, cos x = sqrt(1 - t²), but again, with sign considerations. The denominator becomes sqrt(3 - 2 sqrt(1 - t²) - 2 t). Hmm, complicated.Alternatively, perhaps use Lagrange multipliers, treating this as an optimization problem with trigonometric functions. But maybe overkill.Alternatively, parametrize x as an angle and consider f(x) as a ratio involving coordinates on the unit circle. Let me think.Since x is an angle, we can represent sin x and cos x as coordinates on the unit circle. So, point P(cos x, sin x) is on the unit circle. Then, the numerator is sin x -1, which is the y-coordinate minus 1. The denominator is sqrt(3 - 2 cos x - 2 sin x). Let's see what 3 - 2 cos x -2 sin x represents. If we write it as 3 - 2(cos x + sin x). The term (cos x + sin x) is the sum of coordinates. So, 3 - 2(cos x + sin x). Hmm, maybe this is the equation of a line in terms of coordinates? Not exactly, since we are evaluating it at a point on the unit circle.Alternatively, perhaps consider 3 - 2 cos x -2 sin x as the distance from some point. Wait, 3 - 2 cos x -2 sin x can be rewritten as 3 - 2(cos x + sin x). Let me think: The expression a cos x + b sin x can be written as C cos(x - φ). In this case, we have cos x + sin x, which is sqrt(2) cos(x - π/4). Therefore, 3 - 2 sqrt(2) cos(x - π/4). Therefore, 3 - 2 sqrt(2) cos θ where θ = x - π/4. So, the denominator becomes sqrt(3 - 2 sqrt(2) cos θ). Then, the numerator is sin x -1 = sin(θ + π/4) -1. Hmm, maybe this substitution helps.Express sin(θ + π/4) using angle addition formula:sin(θ + π/4) = sin θ cos π/4 + cos θ sin π/4 = (sin θ + cos θ)/sqrt(2)Therefore, numerator becomes (sin θ + cos θ)/sqrt(2) -1So, f(x) = [ (sin θ + cos θ)/sqrt(2) -1 ] / sqrt(3 - 2 sqrt(2) cos θ )Hmm, this might not directly help, but let's see. Let me denote phi = theta for simplicity:Let phi = theta = x - π/4. Then, f(x) = [ (sin(phi) + cos(phi))/sqrt(2) -1 ] / sqrt(3 - 2 sqrt(2) cos(phi) )Simplify the numerator:(sin phi + cos phi)/sqrt(2) -1 = sin(phi + π/4) -1, since sin(phi + π/4) = (sin phi + cos phi)/sqrt(2). Therefore, numerator is sin(phi + π/4) -1.Therefore, f(x) = [ sin(phi + π/4) -1 ] / sqrt(3 - 2 sqrt(2) cos phi )But phi = x - π/4, so phi ranges from -π/4 to 2π - π/4 = 7π/4. Not sure if this helps.Alternatively, perhaps consider another substitution. Let me set phi = phi.Alternatively, consider writing 3 - 2 sqrt(2) cos phi as something squared. For example, 3 - 2 sqrt(2) cos phi. Let me see if this can be written as (sqrt(2) - cos phi)^2 or something similar.Expand (sqrt(2) - cos phi)^2 = 2 - 2 sqrt(2) cos phi + cos² phi. Not quite 3 - 2 sqrt(2) cos phi. Hmm.Alternatively, (a cos phi + b)^2 = a² cos² phi + 2ab cos phi + b². Compare with 3 - 2 sqrt(2) cos phi. If we could match coefficients:Let me suppose that 3 - 2 sqrt(2) cos phi = (A cos phi + B)^2 + C.But not sure. Alternatively, maybe use the identity A cos phi + B sin phi + C. Not sure.Alternatively, maybe set t = cos phi. Then, we can write 3 - 2 sqrt(2) t. But the numerator is sin(phi + π/4) -1. Not helpful.Alternatively, since this seems too complicated, let's return to calculus approach. We found critical points at x=0 and x=π/2. Let's check the value of f(x) at these points.At x=0: f(0) = (0 -1)/sqrt(3 -2*1 -0) = (-1)/sqrt(1) = -1At x=π/2: f(π/2) =0 as before.At x=3π/2: f(3π/2) = (-1 -1)/sqrt(3 -0 -2*(-1))= (-2)/sqrt(5) ≈ -0.894So, f(3π/2)≈-0.894 which is higher than -1.At x=π: f(π)= (0 -1)/sqrt(3 - (-2) -0)= (-1)/sqrt(5)≈-0.447So, so far, the lowest value is -1 at x=0, but wait, at x=3π/2, it's -2/sqrt(5)≈-0.894, but x=0 gives -1. So, -1 is lower. Wait, but earlier when I calculated f(3π/2), I had:Denominator: sqrt(3 -2 cos x -2 sin x)= sqrt(3 -2*0 -2*(-1))=sqrt(3 +2)=sqrt(5). Numerator: sin x -1 = -1 -1= -2. Therefore, f(3π/2)= -2/sqrt(5)≈-0.894. So, that's higher than -1.So, the critical point at x=0 gives f(0)=-1, which might be a local minimum. But let's check another critical point. We found that x=0 is a critical point. Maybe there's another critical point where the function attains its minimum.Wait, when we set the derivative to zero, we had:-cos²x -1 + 2 cos x + sin x (1 - cos x) =0Let me try to solve this equation.Rearrange the terms:sin x (1 - cos x) = cos²x +1 -2 cos xNote that the RHS is cos²x -2 cos x +1 = (cos x -1)^2So, sin x (1 - cos x) = (cos x -1)^2Multiply both sides by -1:sin x (cos x -1) = (cos x -1)^2If cos x ≠1, we can divide both sides by (cos x -1):sin x = cos x -1But if cos x =1, then x=0 or 2π, which we already considered.So, assuming cos x ≠1, we get sin x = cos x -1So, let's solve sin x = cos x -1Let me write this as sin x - cos x = -1Square both sides:(sin x - cos x)^2 =1sin²x -2 sin x cos x + cos²x=11 - sin 2x =1Therefore, -sin 2x=0 => sin 2x=0 => 2x =nπ =>x= nπ/2So, possible solutions in [0,2π] are x=0, π/2, π, 3π/2, 2π.Check which of these satisfy sin x = cos x -1.At x=0: sin0=0, cos0 -1=1-1=0. So, 0=0, valid. So x=0 is a solution.At x=π/2: sin π/2=1, cos π/2 -1=0 -1=-1. 1≠-1. Not valid.At x=π: sin π=0, cos π -1=-1 -1=-2. 0≠-2. Not valid.At x=3π/2: sin3π/2=-1, cos3π/2 -1=0 -1=-1. So, -1=-1. Valid.At x=2π: sin2π=0, cos2π -1=1 -1=0. 0=0. Valid.Therefore, the solutions are x=0, 3π/2, 2π. But x=2π is same as x=0.So, critical points at x=0, 3π/2, and 2π.But we already checked x=0 and x=3π/2.At x=0, f(x)=-1At x=3π/2, f(x)= -2/sqrt(5)≈-0.894At x=2π, same as x=0: f(2π)= (0 -1)/sqrt(3 -2*1 -0)= -1/sqrt(1)= -1.Therefore, the critical points are x=0, 2π with f(x)=-1, and x=3π/2 with f(x)= -2/sqrt(5).Additionally, we had x=π/2 as a critical point where f(x)=0.Therefore, the function attains its minimum at x=0 and x=2π with value -1, and another critical point at x=3π/2 with value -2/sqrt(5). Also, x=π/2 is a critical point with value 0.But wait, at x=π/2, the function is 0. Is this a maximum? Since f(x) is non-positive, 0 is the upper bound.So, now, we need to check if these critical points are minima or maxima.For x=0 and x=2π: f(x)=-1. Let's check around x=0.Take x approaching 0 from the right: f(x) approaches -1. Take a small ε>0, f(ε)= (sin ε -1)/sqrt(3 -2 cos ε -2 sin ε). Since sin ε ≈ε, cos ε≈1 -ε²/2. So numerator≈ -1 + ε, denominator≈sqrt(3 -2(1 -ε²/2) -2ε)=sqrt(3 -2 + ε² -2ε)=sqrt(1 -2ε +ε²)=sqrt((1 -ε)^2)=1 -ε. Therefore, f(ε)≈(-1 +ε)/(1 -ε)≈-1. So, near x=0, f(x) is approximately -1. Since x=0 is a critical point, and the function is continuous, it's likely a local minimum.At x=3π/2, f(x)= -2/sqrt(5). Let's check around x=3π/2. Take x=3π/2 + ε. For small ε.sin(3π/2 + ε)= -cos ε≈-1 + ε²/2cos(3π/2 + ε)= sin ε≈εSo, numerator≈ (-1 + ε²/2 -1)= -2 + ε²/2Denominator≈sqrt(3 -2*(ε) -2*(-1 + ε²/2 ))=sqrt(3 -2ε +2 - ε²)=sqrt(5 -2ε -ε²)≈sqrt(5)(1 - (2ε +ε²)/10 )Therefore, f(x)≈(-2)/sqrt(5). So, the value is approximately -2/sqrt(5) near x=3π/2, so likely a local minimum? Wait, but when I perturb x around 3π/2, does the function increase or decrease?Wait, take x slightly less than 3π/2, say x=3π/2 - ε.sin(3π/2 - ε)= -cos ε≈-1 + ε²/2cos(3π/2 - ε)= -sin ε≈-εNumerator≈ (-1 + ε²/2 -1)= -2 + ε²/2Denominator≈sqrt(3 -2*(-ε) -2*(-1 + ε²/2 ))=sqrt(3 +2ε +2 - ε²)=sqrt(5 +2ε -ε²)≈sqrt(5)(1 + (2ε)/10 )Therefore, f(x)≈(-2)/sqrt(5)*(1/(1 + ε/5 ))≈-2/sqrt(5)*(1 -ε/5). So, slightly more than -2/sqrt(5). Similarly, for x=3π/2 + ε, f(x)≈-2/sqrt(5)*(1 -ε/5). So, in both cases, moving away from 3π/2, the function increases. Therefore, x=3π/2 is a local maximum.But wait, since the function is negative, a local maximum would mean it's less negative, so higher value. Therefore, x=3π/2 is a local maximum (i.e., the function has a "hill" there in the negative range). Whereas x=0 is a local minimum.But we already saw that f(0)=-1 is lower than f(3π/2)=≈-0.894. So, x=0 is the global minimum, and x=3π/2 is a local maximum.Therefore, the range of f(x) is from -1 to 0. But wait, need to verify.Wait, but let's check another point. For example, at x=7π/6, which is 210 degrees. sin x= -1/2, cos x= -sqrt(3)/2.Compute f(x)= (-1/2 -1)/sqrt(3 -2*(-sqrt(3)/2) -2*(-1/2))= (-3/2)/sqrt(3 + sqrt(3) +1)= (-3/2)/sqrt(4 + sqrt(3))≈(-1.5)/sqrt(4 +1.732)= (-1.5)/sqrt(5.732)≈(-1.5)/2.394≈-0.626. Which is higher than -1 and higher than -0.894.So, seems like the function ranges from -1 (at x=0, 2π) up to 0 (at x=π/2). The other critical point at x=3π/2 gives a local maximum of -2/sqrt(5)≈-0.894, which is higher than -1, so the function reaches its lowest point at -1, then goes up to 0 at π/2, and has other local maxima and minima in between.But wait, maybe there are other critical points? Earlier, when solving sin x = cos x -1, we found solutions at x=0, 3π/2, and 2π. But perhaps there are more critical points where cos x=1. When cos x=1, then denominator expression becomes 3 -2*1 -2 sin x=1 -2 sin x. But when cos x=1, x=0 or 2π, which we already considered.Alternatively, maybe I missed something in the derivative calculation.Wait, when solving the derivative equation, we arrived at two cases: cos x=1 or sin x=cos x -1.But when cos x=1, the equation sin x=cos x -1 becomes sin x=0, which is true when x=0 or 2π. So, those points are already included.Therefore, the critical points are x=0,2π (minima), x=3π/2 (local maximum), and x=π/2 (upper bound zero). Therefore, the function reaches its minimum at x=0 and x=2π with f(x)=-1, and its maximum at x=π/2 with f(x)=0. However, there's also a local maximum at x=3π/2 with f(x)≈-0.894. So, the overall maximum is 0 and the overall minimum is -1. Therefore, the range is [-1, 0].But wait, wait a second. Let me check another point. For example, take x=π/6. sin x=0.5, cos x=sqrt(3)/2≈0.866.Compute f(x)= (0.5 -1)/sqrt(3 -2*sqrt(3)/2 -2*0.5)= (-0.5)/sqrt(3 -sqrt(3) -1)= (-0.5)/sqrt(2 -sqrt(3))≈ (-0.5)/sqrt(2 -1.732)= (-0.5)/sqrt(0.268)≈ (-0.5)/0.517≈-0.967. Which is lower than -0.894 but higher than -1. So, this is between -1 and -0.894. Therefore, the function does attain values between -1 and -0.894. So, this suggests that the function's minimum is -1, but there are other values lower than the local maximum at x=3π/2. Therefore, the range might actually be [-1,0].Wait, but in that case, the function at x=π/6 is≈-0.967, which is between -1 and -0.894. So, lower than x=3π/2. So, the function decreases from x=0 to some point, reaches a minimum at x=0, but wait, no. Wait, at x=0, it's -1, which is the lowest. Then, moving from x=0 to x=π/2, the function increases from -1 to 0. Then, from x=π/2 to x=π, it decreases again? Wait, but at x=π, f(x)= (0 -1)/sqrt(3 - (-2) -0)= (-1)/sqrt(5)≈-0.447. Then, from x=π to x=3π/2, it decreases further to≈-0.894? Wait, but at x=3π/2, it's -2/sqrt(5)≈-0.894. Then, from x=3π/2 to x=2π, it increases back to -1.Wait, but this seems contradictory. How can the function go from x=0 (-1) to x=π/2 (0), then to x=π (-0.447), then to x=3π/2 (-0.894), then back to x=2π (-1). So, it's like the function has a minimum at x=0 and x=2π, peaks at x=π/2 (0), and has another trough at x=3π/2 (-0.894). Wait, but this can't be. Because between x=π and x=3π/2, the function goes from -0.447 to -0.894, which is decreasing, then from x=3π/2 to x=2π, it increases back to -1. So, the function is periodic, but over the interval [0,2π], it has multiple extrema.But given that we found the critical points at x=0, x=3π/2, x=π/2, and x=2π, and the function is continuous, then the minimum value is -1, and the maximum value is 0. So, the range is [-1,0]. But in the calculation at x=π/6, we had≈-0.967, which is greater than -1, so between -1 and 0. Therefore, even though there are local minima and maxima, the absolute minimum is -1, and the absolute maximum is 0. Therefore, the range is [-1,0].Wait, but to confirm, let's check another point. Let's take x approaching π/2 from the left. For example, x=π/2 - ε. Then, sin x=cos ε≈1 -ε²/2, cos x=sin ε≈ε. So, numerator≈ (1 -ε²/2 -1)= -ε²/2. Denominator≈sqrt(3 -2*ε -2*(1 -ε²/2))=sqrt(3 -2ε -2 +ε²)=sqrt(1 -2ε +ε²)=sqrt((1 -ε)^2)=1 -ε. Therefore, f(x)≈(-ε²/2)/(1 -ε)≈0 as ε approaches 0. So, near x=π/2, the function approaches 0 from below (since numerator is negative and denominator is positive). Therefore, the function reaches 0 exactly at x=π/2.Therefore, the function f(x) ranges from -1 to 0. So, the range is [-1, 0].But wait, let me check with x=7π/4. At x=7π/4, sin x=-sqrt(2)/2≈-0.707, cos x=sqrt(2)/2≈0.707.Compute f(x)= (-sqrt(2)/2 -1)/sqrt(3 -2*(sqrt(2)/2) -2*(-sqrt(2)/2))= (-(sqrt(2)/2 +1))/sqrt(3 -sqrt(2) +sqrt(2))= (-(sqrt(2)/2 +1))/sqrt(3)= (-(sqrt(2)/2 +1))/sqrt(3)≈ (-(0.707 +1))/1.732≈ (-1.707)/1.732≈-0.986. This is close to -1, but still greater than -1. So, this supports that the minimum is -1.Therefore, after all this analysis, the range of the function is from -1 to 0. Therefore, the answer is [-1, 0], which in boxed form is boxed{[-1, 0]}.Final AnswerThe range of the function is boxed{[-1, 0]}.</think>To find the range of the function ( f(x) = frac{sin x - 1}{sqrt{3 - 2 cos x - 2 sin x}} ) for ( 0 leqslant x leqslant 2pi ):1. Domain Analysis: - The denominator ( 3 - 2 cos x - 2 sin x ) must be positive. - Using trigonometric identities, we determine that ( 3 - 2(cos x + sin x) ) is always positive because the minimum value of ( cos x + sin x ) is (-sqrt{2}), leading to ( 3 - 2(-sqrt{2}) approx 5.828 ), and the maximum value is (sqrt{2}), leading to ( 3 - 2sqrt{2} approx 0.172 ).2. Numerator and Denominator Behavior: - The numerator ( sin x - 1 ) ranges from (-2) to (0). - The denominator is always positive, so ( f(x) ) is non-positive.3. Critical Points and Extrema: - Using calculus, we find the critical points by setting the derivative ( f'(x) ) to zero. - Solving the derivative equation, we find critical points at ( x = 0, 2pi ) (minimum values), ( x = pi/2 ) (maximum value), and ( x = 3pi/2 ) (local maximum).4. Evaluating Function at Critical Points: - At ( x = 0 ) and ( x = 2pi ), ( f(x) = -1 ). - At ( x = pi/2 ), ( f(x) = 0 ). - At ( x = 3pi/2 ), ( f(x) = -2/sqrt{5} approx -0.894 ).5. Range Determination: - The minimum value of ( f(x) ) is (-1) at ( x = 0 ) and ( x = 2pi ). - The maximum value of ( f(x) ) is (0) at ( x = pi/2 ).Thus, the range of the function ( f(x) ) is (boxed{[-1, 0]}).]

🔑:When x neq frac{pi}{2}, f(x)=frac{sin x-1}{sqrt{(sin x-1)^{2}+(cos x-1)^{2}}}=-frac{1}{sqrt{1+left(frac{cos x-1}{sin x-1}right)^{2}}},Since k=frac{cos x-1}{sin x-1} represents the slope of the line connecting a point on the unit circle x^{2}+y^{2}=1 to (1,1), i.e., k in[0,+infty) Rightarrow f(x) in[-1,0); when x=frac{pi}{2}, f(x)=0.Therefore, the range of the function f(x)=frac{sin x-1}{sqrt{3-2 cos x-2 sin x}}(0 leqslant x leqslant 2 pi) is [-1,0].

🧠:[<think>Okay, let's see. So we have a problem where in a school, each desk has exactly two people. Some percentage of boys have boys as their desk partners, and some percentage of girls have girls as their desk partners. We need to find how many times more boys there are than girls. Hmm. Let me try to break this down.First, let's define variables. Let's say the number of boys is B and the number of girls is G. We need to find B/G, right? Because the question is asking how many times more boys there are than girls, so that ratio.Now, according to the problem, exactly 70% of the boys have a boy as a desk partner. That means 70% of the boys are sitting with another boy. Similarly, 40% of the girls have a girl as a desk partner. So 40% of the girls are sitting with another girl. Wait, but each desk has two people. So when they say 70% of the boys have a boy as a desk partner, those boys are paired with another boy. So each such pair would account for two boys. Similarly, for the girls, 40% of the girls are paired with another girl. So each of those pairs would have two girls.But then, the remaining boys (30%) must be sitting with girls, and the remaining girls (60%) must be sitting with boys. Hmm. But wait, if a boy is sitting with a girl, then the girl he's sitting with must also be counted in the girls who are sitting with boys. So the number of boys sitting with girls should equal the number of girls sitting with boys. That seems important.Let me write that down. Let's denote:- The number of boys sitting with boys: 70% of B = 0.7B. Since each of these boys is part of a pair, the number of boy-boy desks would be (0.7B)/2. Because each desk has two boys. Similarly, the number of girls sitting with girls: 40% of G = 0.4G. So the number of girl-girl desks is (0.4G)/2.Now, the remaining boys are 30% of B = 0.3B, and they must be sitting with girls. Similarly, the remaining girls are 60% of G = 0.6G, and they must be sitting with boys. But the number of boys sitting with girls must equal the number of girls sitting with boys because each boy-girl desk has one boy and one girl. So the total number of such mixed desks would be equal to both 0.3B and 0.6G.Therefore, 0.3B = 0.6G. Let me check that. If 30% of boys are sitting with girls, each contributing one boy per desk, then the number of mixed desks is 0.3B. Similarly, 60% of girls are sitting with boys, each contributing one girl per desk, so the number of mixed desks is 0.6G. Since these must be equal, we have 0.3B = 0.6G. Solving for B/G, we divide both sides by G and by 0.3:B/G = 0.6 / 0.3 = 2. So B is twice as much as G. Wait, that seems straightforward. But let me verify this with another approach to make sure.Alternative approach: Let's consider the total number of desks. Each desk is either boy-boy, girl-girl, or boy-girl. Let's compute the number of each type.Number of boy-boy desks: (0.7B)/2. Since 70% of boys are in boy-boy pairs, each desk has two boys, so divide by 2.Number of girl-girl desks: (0.4G)/2. Similarly, 40% of girls in girl-girl pairs, divided by 2.Number of mixed desks: Let's see, the remaining boys are 0.3B and the remaining girls are 0.6G. Since each mixed desk has one boy and one girl, the number of mixed desks should be equal to both 0.3B and 0.6G, as each mixed desk uses one boy and one girl. Therefore, 0.3B = 0.6G. So again, B/G = 0.6 / 0.3 = 2.But wait, is there a way that this could be inconsistent? Let's check the total number of boys and girls.Total boys: B = number in boy-boy pairs + number in mixed pairs. That would be (0.7B) + (0.3B) = B. Correct.Total girls: G = number in girl-girl pairs + number in mixed pairs. That would be (0.4G) + (0.6G) = G. Correct.So everything adds up. Therefore, the ratio B/G is 2. Therefore, there are twice as many boys as girls. So the answer is 2. But let me check once more because sometimes these problems can have a twist.Wait, but when you have boy-boy pairs, each pair has two boys, right? So the number of boys in boy-boy pairs is 2*(number of boy-boy desks). Similarly, the number of girls in girl-girl pairs is 2*(number of girl-girl desks). But in the problem statement, it says "exactly 70% of the boys have a boy as a desk partner". So that 70% of the boys are in boy-boy pairs. Each boy in a boy-boy pair is paired with another boy, so the total number of boys in boy-boy pairs is 0.7B, which is equal to 2*(number of boy-boy desks). Therefore, number of boy-boy desks is 0.7B / 2. Similarly, the number of girl-girl desks is 0.4G / 2. Then the number of mixed desks is the remaining desks.But then, the total number of desks can also be calculated as (Total students)/2 = (B + G)/2. Therefore, number of boy-boy desks + girl-girl desks + mixed desks = (B + G)/2.But let's write that out:Number of boy-boy desks: 0.7B / 2Number of girl-girl desks: 0.4G / 2Number of mixed desks: 0.3B (since 0.3B boys are in mixed desks, and each desk has 1 boy, so 0.3B desks). Alternatively, 0.6G girls are in mixed desks, so 0.6G desks. So, 0.3B = 0.6G => B = 2G, as before.Therefore, total desks:0.7B/2 + 0.4G/2 + 0.3B = (0.35B + 0.2G + 0.3B) = 0.65B + 0.2GBut total desks should also be (B + G)/2.Therefore,0.65B + 0.2G = (B + G)/2Multiply both sides by 2:1.3B + 0.4G = B + GSubtract B and G from both sides:0.3B - 0.6G = 0Which gives 0.3B = 0.6G => B = 2G. So same result. So that checks out. Therefore, the ratio is 2. So the answer is 2 times more boys than girls.I think that's solid. Let me check once again with actual numbers. Suppose there are 2 boys and 1 girl. Wait, but according to B = 2G, if G is 10, B is 20.Let's take G = 10, B = 20.Then, 70% of boys (14 boys) are sitting with boys. So number of boy-boy desks is 14 / 2 = 7 desks.40% of girls (4 girls) are sitting with girls. So girl-girl desks: 4 / 2 = 2 desks.Remaining boys: 20 - 14 = 6 boys. These 6 boys are sitting with girls. So they need 6 girls. But total girls are 10. Girls already sitting with girls: 4. So remaining girls: 10 - 4 = 6. Perfect, 6 girls to pair with 6 boys. So mixed desks: 6 desks.Total desks: 7 + 2 + 6 = 15.Total students: 20 + 10 = 30. 30 / 2 = 15. Correct.So numbers check out. So ratio B/G = 20/10 = 2. So indeed, the answer is 2. So the boys are twice as many as the girls. Therefore, the answer is 2.Final Answerboxed{2}</think>Let ( B ) be the number of boys and ( G ) be the number of girls. We need to find the ratio ( frac{B}{G} ).1. Boys Analysis: - 70% of the boys have a boy as a desk partner. This means ( 0.7B ) boys are paired with another boy, forming ( frac{0.7B}{2} ) boy-boy desks. - The remaining 30% of the boys (( 0.3B )) must be paired with girls.2. Girls Analysis: - 40% of the girls have a girl as a desk partner. This means ( 0.4G ) girls are paired with another girl, forming ( frac{0.4G}{2} ) girl-girl desks. - The remaining 60% of the girls (( 0.6G )) must be paired with boys.3. Mixed Desks: - The number of boys paired with girls (( 0.3B )) must equal the number of girls paired with boys (( 0.6G )). Therefore, we have the equation: [ 0.3B = 0.6G ] - Solving for ( frac{B}{G} ): [ frac{B}{G} = frac{0.6}{0.3} = 2 ]4. Verification: - Total desks calculation confirms the consistency: - Boy-boy desks: ( frac{0.7B}{2} ) - Girl-girl desks: ( frac{0.4G}{2} ) - Mixed desks: ( 0.3B ) - Total desks: ( frac{B + G}{2} ) - Substituting ( B = 2G ) into the total desks equation confirms the solution is correct.Thus, the number of boys is twice the number of girls. The final answer is (boxed{2}).,<think>Okay, let's see. I need to figure out how many times more boys there are than girls in this school. The problem says that at each desk, there are exactly 2 people. So, all desks are occupied by two students. It also mentions that exactly 70% of the boys have a boy as a desk partner, and exactly 40% of the girls have a girl as a desk partner. Hmm. So, let me break this down.First, let's denote the number of boys as B and the number of girls as G. The question is asking for the ratio B/G, or how many times more boys there are than girls.Now, let's think about the desk partnerships. Since each desk has two people, the total number of desks is (B + G)/2. But maybe that's not directly necessary. Instead, let's consider the partnerships from the perspective of boys and girls.70% of the boys have a boy as a partner. That means 70% of B boys are paired with another boy. So, the number of boys paired with boys is 0.7B. Similarly, 40% of the girls have a girl as a partner, so the number of girls paired with girls is 0.4G.But when boys are paired with boys, that's a desk with two boys, and similarly for girls. On the other hand, the remaining boys must be paired with girls, and the remaining girls must be paired with boys.Wait a minute, but when a boy is paired with a girl, that counts as one boy and one girl. So, the number of boys paired with girls should be equal to the number of girls paired with boys. Let's verify that.If 70% of boys are paired with boys, then 30% of boys are paired with girls. Similarly, 40% of girls are paired with girls, so 60% of girls are paired with boys. The number of boys paired with girls is 0.3B, and the number of girls paired with boys is 0.6G. Since these must be equal (because each boy-girl pair has one boy and one girl), we can set up the equation:0.3B = 0.6GSolving for B/G:0.3B = 0.6G Divide both sides by 0.3G: B/G = 0.6 / 0.3 = 2So, the ratio of boys to girls is 2, meaning there are twice as many boys as girls.But let me check this again because it seems straightforward, but maybe I made a mistake in setting up the equations.So, the total number of boy-boy desks is 0.7B boys, but since each desk has two boys, the number of boy-boy desks is (0.7B)/2. Similarly, the number of girl-girl desks is (0.4G)/2.Then, the number of boy-girl desks would be the remaining desks. Let's compute the total number of students:Total students = B + G = 2*(number of desks)But maybe another approach is better.Let me think in terms of partnerships. Each boy has one desk partner, so each boy is either paired with a boy or a girl.Similarly, each girl is either paired with a girl or a boy.So, the number of boy-boy partnerships is 70% of the boys, but since each partnership consists of two boys, the total number of boys in boy-boy pairs is 2 * (number of boy-boy desks). Wait, but 70% of the boys are in boy-boy pairs. So, 0.7B = 2 * (number of boy-boy desks). Therefore, the number of boy-boy desks is 0.7B / 2.Similarly, the number of girl-girl desks is 0.4G / 2.Then, the number of mixed desks (boy-girl) would be the total desks minus the boy-boy desks minus the girl-girl desks.But total desks are (B + G)/2.Alternatively, the number of mixed desks can also be calculated from the boys' perspective: the remaining 30% of boys (0.3B) are paired with girls. Each mixed desk has one boy and one girl, so the number of mixed desks is 0.3B. Similarly, from the girls' perspective, 60% of girls (0.6G) are paired with boys, so the number of mixed desks is 0.6G. Since these are the same, 0.3B = 0.6G, which gives B = 2G. So, the same answer as before.Therefore, the ratio of boys to girls is 2:1, so there are twice as many boys as girls. Therefore, the answer should be 2.But wait, let me confirm once more. Let me use variables.Let B be the number of boys, G the number of girls.Number of boys with boys: 0.7B. Each of these requires pairing, so the number of boy-boy pairs is 0.7B / 2.Number of girls with girls: 0.4G. Similarly, girl-girl pairs: 0.4G / 2.Number of boys with girls: 0.3B (remaining boys). Number of girls with boys: 0.6G (remaining girls). Since each boy-girl pair consists of one boy and one girl, the number of such pairs is 0.3B = 0.6G.So, 0.3B = 0.6G => B = 2G. So, B/G = 2.Yes, that seems consistent.Alternatively, if I consider total number of partnerships:Each boy is in one partnership. 70% of the boys are in boy-boy partnerships. Each boy-boy partnership has two boys, so the total number of boy-boy partnerships is (0.7B)/2.Similarly, the number of girl-girl partnerships is (0.4G)/2.The number of boy-girl partnerships is 0.3B (from boys' side) or 0.6G (from girls' side), which must be equal.But total partnerships should also be (B + G)/2.Let me check:Total partnerships = boy-boy + girl-girl + boy-girl = (0.7B)/2 + (0.4G)/2 + 0.3B But since 0.3B = 0.6G, let's substitute:= (0.7B)/2 + (0.4G)/2 + 0.6GBut B = 2G, so substitute B = 2G:= (0.7*2G)/2 + (0.4G)/2 + 0.6G = (1.4G)/2 + 0.2G + 0.6G = 0.7G + 0.2G + 0.6G = 1.5GTotal students: B + G = 2G + G = 3G, so total partnerships should be 3G/2 = 1.5G, which matches. So the equations are consistent.Therefore, the ratio is 2:1. So the answer is 2.Final Answerboxed{2}</think>Let ( B ) be the number of boys and ( G ) be the number of girls. We need to determine the ratio ( frac{B}{G} ).Given:- 70% of the boys have a boy as a desk partner.- 40% of the girls have a girl as a desk partner.From the boys' perspective:- 70% of boys are paired with boys, so ( 0.7B ) boys are in boy-boy pairs.- The remaining 30% of boys are paired with girls, so ( 0.3B ) boys are in boy-girl pairs.From the girls' perspective:- 40% of girls are paired with girls, so ( 0.4G ) girls are in girl-girl pairs.- The remaining 60% of girls are paired with boys, so ( 0.6G ) girls are in boy-girl pairs.Since the number of boys paired with girls must equal the number of girls paired with boys, we set up the equation:[ 0.3B = 0.6G ]Solving for ( frac{B}{G} ):[ 0.3B = 0.6G ][ frac{B}{G} = frac{0.6}{0.3} = 2 ]Thus, the number of boys is twice the number of girls. The final answer is:[boxed{2}]]

🔑:Answer: 2 times.Solution. Let the number of boys be x, and the number of girls be y. Note that 30% of the boys sit at desks with girls and 60% of the girls sit at desks with boys. Since exactly 2 people sit at each desk, then 0.3 x = 0.6 y, from which x = 2 y. Thus, there are 2 times more boys than girls.## Criteria## 7 p. Any complete solution of the problem.## 2 p. A particular case is provided.1 p. Only the correct answer.

🧠:[<think>Okay, so I need to figure out the range of the parameter ( a ) such that the function ( f(x) = log_{frac{1}{2}}(x^{2} - a x + 3) ) is monotonically increasing on the interval ( (-infty, 1) ). The answer choices are given, so maybe after working through the problem, I can match my conclusion with the options provided. Let's start breaking this down step by step.First, I know that logarithmic functions have certain properties depending on their base. The base here is ( frac{1}{2} ), which is between 0 and 1. I remember that when the base of a logarithm is between 0 and 1, the logarithmic function is decreasing. So, ( log_{frac{1}{2}}(u) ) is a decreasing function in terms of ( u ). Therefore, for ( f(x) ) to be monotonically increasing, the argument inside the logarithm, which is ( u(x) = x^{2} - a x + 3 ), must be decreasing. Because if the outer function is decreasing, then for the composition ( f(x) = log_{frac{1}{2}}(u(x)) ) to be increasing, ( u(x) ) must be decreasing. That makes sense because a decreasing function of a decreasing function is increasing. Wait, let me verify that.Suppose ( h(x) = g(u(x)) ). If ( g ) is decreasing and ( u(x) ) is decreasing, then ( h(x) ) would be increasing. For example, if ( u(x) ) decreases as ( x ) increases, then ( g(u(x)) ) would increase as ( u(x) ) decreases (since ( g ) is decreasing). So yes, the composition would be increasing if both the outer function is decreasing and the inner function is decreasing. Therefore, the key here is that ( u(x) = x^2 - a x + 3 ) must be decreasing on ( (-infty, 1) ).So, the first step is to ensure that ( u(x) = x^2 - a x + 3 ) is decreasing on ( (-infty, 1) ). To check where a quadratic function is decreasing, I need to look at its derivative. The derivative of ( u(x) ) is ( u'(x) = 2x - a ). A quadratic function is decreasing where its derivative is negative. So, we need ( u'(x) = 2x - a < 0 ) for all ( x in (-infty, 1) ).Let me solve the inequality ( 2x - a < 0 ). That simplifies to ( x < frac{a}{2} ). So, the quadratic function ( u(x) ) is decreasing on the interval ( (-infty, frac{a}{2}) ). Therefore, to make sure that ( u(x) ) is decreasing on ( (-infty, 1) ), the interval ( (-infty, 1) ) must be a subset of ( (-infty, frac{a}{2}) ). In other words, ( frac{a}{2} geq 1 ), which implies ( a geq 2 ). So that's one condition: ( a geq 2 ).But there's another important consideration when dealing with logarithmic functions: the argument inside the logarithm must be positive for all ( x ) in the domain where the function is defined. The original function ( f(x) = log_{frac{1}{2}}(x^{2} - a x + 3) ) is only defined when ( x^2 - a x + 3 > 0 ). Therefore, we must ensure that ( x^2 - a x + 3 > 0 ) for all ( x in (-infty, 1) ). If this quadratic is positive on that interval, then the function is defined there. Otherwise, if the quadratic becomes zero or negative at some point in ( (-infty, 1) ), the logarithm would be undefined there, which would complicate things. So, the second condition is that ( x^2 - a x + 3 > 0 ) for all ( x < 1 ).Therefore, we need to analyze the quadratic ( x^2 - a x + 3 ) and ensure it's positive for all ( x < 1 ). Let's denote ( q(x) = x^2 - a x + 3 ). For ( q(x) > 0 ) for all ( x < 1 ), we need to consider the behavior of this quadratic. Since the coefficient of ( x^2 ) is positive (1), the parabola opens upwards. Therefore, if the minimum of the quadratic occurs at some point, and if the minimum value is positive, then the quadratic is always positive. Alternatively, if the quadratic has real roots, we need to ensure that both roots are to the right of 1, so that for all ( x < 1 ), the quadratic is positive.So, let's find the vertex of the quadratic. The x-coordinate of the vertex of a quadratic ( ax^2 + bx + c ) is at ( -frac{b}{2a} ). Here, the quadratic is ( x^2 - a x + 3 ), so ( a_{text{coeff}} = 1 ), ( b_{text{coeff}} = -a ). Therefore, the x-coordinate of the vertex is at ( -frac{-a}{2*1} = frac{a}{2} ). So, the vertex is at ( x = frac{a}{2} ).Since the parabola opens upwards, the minimum value of the quadratic is at ( x = frac{a}{2} ). Therefore, the minimum value of ( q(x) ) is ( qleft( frac{a}{2} right) ). Let's compute that:( qleft( frac{a}{2} right) = left( frac{a}{2} right)^2 - a cdot frac{a}{2} + 3 = frac{a^2}{4} - frac{a^2}{2} + 3 = -frac{a^2}{4} + 3 ).Therefore, the minimum value of the quadratic is ( 3 - frac{a^2}{4} ). For the quadratic to be positive for all ( x ), the minimum value must be positive. That is, ( 3 - frac{a^2}{4} > 0 implies frac{a^2}{4} < 3 implies a^2 < 12 implies |a| < 2sqrt{3} approx 3.464 ). However, this condition ensures that the quadratic is always positive for all real numbers ( x ). But in our problem, we only need the quadratic to be positive for ( x < 1 ). So, maybe the quadratic can have real roots as long as both roots are greater than or equal to 1. Then, for ( x < 1 ), the quadratic would be positive. Let's explore both possibilities.Case 1: The quadratic has no real roots. Then, it is always positive. The discriminant is ( D = (-a)^2 - 4 * 1 * 3 = a^2 - 12 ). So, if ( D < 0 implies a^2 - 12 < 0 implies a^2 < 12 implies |a| < 2sqrt{3} approx 3.464 ). So, if ( |a| < 2sqrt{3} ), then the quadratic is always positive, which would satisfy the condition for ( x < 1 ).Case 2: The quadratic has real roots (i.e., ( a^2 geq 12 )), but both roots are greater than or equal to 1. In that case, for ( x < 1 ), the quadratic would be positive because it's opening upwards and doesn't cross the x-axis until 1 or beyond. Therefore, even if the quadratic has real roots, as long as both roots are greater than or equal to 1, ( q(x) > 0 ) for all ( x < 1 ).So, we need to consider these two cases. Let's first analyze the roots of the quadratic. The roots are given by:( x = frac{a pm sqrt{a^2 - 12}}{2} ).So, if the quadratic has real roots (when ( a geq 2sqrt{3} ) or ( a leq -2sqrt{3} )), we need both roots to be greater than or equal to 1. Since the quadratic is ( x^2 - a x + 3 ), if ( a ) is positive, then the roots are both positive (since the product of the roots is 3, which is positive, and the sum of the roots is ( a ), which is positive). If ( a ) is negative, the roots would have a negative sum and positive product, so they would both be negative. But since we are looking for roots greater than or equal to 1, and if ( a ) is negative, the roots would be negative, which would not interfere with ( x < 1 ). Wait, but if ( a ) is negative, the quadratic becomes ( x^2 - a x + 3 ), and since ( a ) is negative, the linear term is positive. Let me check that again.Wait, the quadratic is ( x^2 - a x + 3 ). If ( a ) is negative, then the linear term becomes ( -a x ), which is positive since ( a ) is negative. So, the quadratic becomes ( x^2 + |a| x + 3 ). Then, the roots would be:( x = frac{ -|a| pm sqrt{a^2 - 12} }{2} ).But since ( a ) is negative, the roots would have sum ( a ), which is negative, and product 3, which is positive. Therefore, both roots would be negative. Therefore, if ( a ) is negative, the quadratic has two negative roots (if any real roots exist). Therefore, for ( x < 1 ), since 1 is positive, and all x in ( (-infty, 1) ) include all the negative numbers up to 1. However, if the quadratic has roots at negative values, then between those roots (which are both negative) the quadratic would be negative. Wait, but the quadratic opens upwards, so it is positive outside the roots. Since the roots are both negative, then for ( x > text{larger root} ), the quadratic is positive, but since the larger root is negative, for all ( x > text{larger root} ), which includes all x in ( (-infty, 1) ) except between the two roots, but wait, no.Wait, if the quadratic has two roots at, say, ( x_1 ) and ( x_2 ), both negative, then the quadratic is positive when ( x < x_1 ) or ( x > x_2 ). But since ( x_2 ) is the larger root (still negative), for all ( x > x_2 ), which includes ( x ) approaching positive infinity, the quadratic is positive. But in between ( x_1 ) and ( x_2 ), it's negative. However, if ( x_2 ) is negative, then between ( x_1 ) and ( x_2 ), which are both negative, the quadratic is negative. But the interval ( (-infty, 1) ) includes all x less than 1, so between the roots (negative) and beyond. Wait, if the roots are both negative, then for ( x ) values less than the smaller root, the quadratic is positive (since parabola opens upwards). For ( x ) between the two roots, it's negative, and for ( x ) greater than the larger root (which is still negative), it's positive. So, if the interval ( (-infty, 1) ) includes the region between the larger root and 1, which is positive. Wait, this is getting confusing. Let's take an example.Suppose ( a = -4 ). Then the quadratic becomes ( x^2 - (-4)x + 3 = x^2 + 4x + 3 ). The roots of this quadratic are ( x = frac{-4 pm sqrt{16 - 12}}{2} = frac{-4 pm 2}{2} ), so ( x = -1 ) and ( x = -3 ). Therefore, the quadratic is positive when ( x < -3 ) or ( x > -1 ), and negative between ( -3 ) and ( -1 ). So, on the interval ( (-infty, 1) ), the quadratic is positive except between ( -3 ) and ( -1 ). Therefore, the quadratic is not positive for all ( x < 1 ); it's negative between ( -3 ) and ( -1 ). Therefore, for ( a = -4 ), the logarithm would be undefined in that interval, so ( f(x) ) is not defined on all of ( (-infty, 1) ). Therefore, even though the quadratic has roots at negative values, the quadratic is not positive for all ( x < 1 ). Therefore, if ( a ) is negative, even if ( a geq 2 ), but wait, ( a = -4 ) is not in our consideration since earlier we found that ( a geq 2 ). Wait, the first condition was ( a geq 2 ), so we don't need to consider negative ( a ).Wait, hold on. Earlier, we found that for ( u(x) ) to be decreasing on ( (-infty, 1) ), we needed ( a geq 2 ). Therefore, the possible values of ( a ) we need to consider are ( a geq 2 ). So, even if negative ( a ) could somehow satisfy the positivity condition, they would fail the first condition of ( a geq 2 ). Therefore, we can restrict our analysis to ( a geq 2 ).So, for ( a geq 2 ), we need to ensure that ( x^2 - a x + 3 > 0 ) for all ( x < 1 ). Let's consider two scenarios for ( a geq 2 ):1. When ( a geq 2 ) and the quadratic has no real roots (i.e., ( a < 2sqrt{3} approx 3.464 )), so the quadratic is always positive. Therefore, in this case, the quadratic is positive for all ( x ), so certainly for all ( x < 1 ).2. When ( a geq 2sqrt{3} ), the quadratic has real roots. In this case, we need to ensure that both roots are greater than or equal to 1, so that ( x < 1 ) lies entirely in the region where the quadratic is positive (since parabola opens upwards, and if both roots are ≥1, then for x <1, it's positive).Therefore, let's find the conditions for the roots to be greater than or equal to 1. Let's denote the roots as ( x_1 ) and ( x_2 ), with ( x_1 leq x_2 ). We need both roots to be ≥1. To ensure this, the smaller root ( x_1 ) must be ≥1. So, let's compute the smaller root:The roots are ( frac{a pm sqrt{a^2 - 12}}{2} ). The smaller root is ( frac{a - sqrt{a^2 - 12}}{2} ). We need this smaller root to be ≥1.Set up the inequality:( frac{a - sqrt{a^2 - 12}}{2} geq 1 )Multiply both sides by 2:( a - sqrt{a^2 - 12} geq 2 )Subtract ( a ) from both sides:( - sqrt{a^2 - 12} geq 2 - a )Multiply both sides by -1 (which reverses the inequality):( sqrt{a^2 - 12} leq a - 2 )Now, since the left side ( sqrt{a^2 - 12} ) is non-negative, the right side must also be non-negative. Therefore:( a - 2 geq 0 implies a geq 2 ), which is already our case.Now, square both sides to eliminate the square root:( a^2 - 12 leq (a - 2)^2 )Expand the right side:( a^2 - 12 leq a^2 - 4a + 4 )Subtract ( a^2 ) from both sides:( -12 leq -4a + 4 )Add ( 4a ) to both sides:( 4a - 12 leq 4 )Add 12 to both sides:( 4a leq 16 )Divide by 4:( a leq 4 )Therefore, combining this with the previous conditions:We have ( a geq 2sqrt{3} ) (to have real roots) and ( a leq 4 ). But ( 2sqrt{3} approx 3.464 ), so the range here is ( 3.464 leq a leq 4 ).But let's verify if squaring both sides introduced any extraneous solutions. Let's check when ( a = 4 ):The smaller root is ( frac{4 - sqrt{16 - 12}}{2} = frac{4 - 2}{2} = 1 ). So, when ( a = 4 ), the smaller root is 1, which satisfies the condition. For ( a = 2sqrt{3} approx 3.464 ), the smaller root is:( frac{2sqrt{3} - sqrt{(2sqrt{3})^2 - 12}}{2} = frac{2sqrt{3} - sqrt{12 - 12}}{2} = frac{2sqrt{3}}{2} = sqrt{3} approx 1.732 ), which is greater than 1. Therefore, in this case, the roots are both greater than 1, so for ( a geq 2sqrt{3} ), the roots start at around 1.732 and increase as ( a ) increases. Wait, but when ( a ) increases, how do the roots behave?Wait, actually, the roots are:( x = frac{a pm sqrt{a^2 - 12}}{2} ). As ( a ) increases, both the numerator terms increase. Let's see for a large ( a ), the roots would be approximately ( frac{a pm a}{2} ), so one root approaches ( a ) and the other approaches 0. Wait, that's not correct. Let me analyze:For large ( a ), ( sqrt{a^2 - 12} approx a - frac{6}{a} ). So, the roots are approximately:( frac{a pm (a - frac{6}{a})}{2} ). Therefore, the larger root is approximately ( frac{a + a - frac{6}{a}}{2} = frac{2a - frac{6}{a}}{2} = a - frac{3}{a} ), which approaches ( a ) as ( a ) increases. The smaller root is approximately ( frac{a - (a - frac{6}{a})}{2} = frac{frac{6}{a}}{2} = frac{3}{a} ), which approaches 0 as ( a ) increases. Therefore, as ( a ) increases beyond ( 2sqrt{3} ), the smaller root decreases towards 0. But this contradicts our previous conclusion. Wait, maybe my approximation is wrong.Wait, let's take ( a = 5 ). Then, the roots are:( frac{5 pm sqrt{25 - 12}}{2} = frac{5 pm sqrt{13}}{2} approx frac{5 pm 3.605}{2} ). So, the roots are approximately ( (5 + 3.605)/2 approx 4.3025 ) and ( (5 - 3.605)/2 approx 0.6975 ). Therefore, when ( a = 5 ), the smaller root is approximately 0.6975, which is less than 1. Therefore, the quadratic would be negative between 0.6975 and 4.3025. Thus, on the interval ( (-infty, 1) ), the quadratic is positive when ( x < 0.6975 ) and negative when ( 0.6975 < x < 1 ). Therefore, the quadratic is not positive for all ( x < 1 ), which violates our requirement.Therefore, our earlier conclusion that when ( a leq 4 ), the smaller root is greater than or equal to 1 is incorrect. Wait, but when we solved the inequality:We had ( sqrt{a^2 - 12} leq a - 2 ), leading to ( a leq 4 ). However, when ( a = 4 ), the smaller root is exactly 1, as computed earlier. Let's confirm:For ( a = 4 ), quadratic is ( x^2 - 4x + 3 ). The roots are ( x = 1 ) and ( x = 3 ). So, the quadratic is positive when ( x < 1 ) or ( x > 3 ). Therefore, on ( (-infty, 1) ), the quadratic is positive. Therefore, when ( a = 4 ), it's okay. But when ( a > 4 ), like ( a = 5 ), the smaller root is less than 1, so the quadratic is negative between the roots, which includes part of ( (-infty, 1) ).Therefore, the condition that the smaller root is ≥1 is only satisfied when ( a leq 4 ). Therefore, combining the two cases:1. When ( 2 leq a < 2sqrt{3} approx 3.464 ), the quadratic has no real roots and is always positive, so it's positive for all ( x < 1 ).2. When ( 2sqrt{3} leq a leq 4 ), the quadratic has real roots, but the smaller root is ≥1 (exactly 1 when ( a = 4 )), so the quadratic is positive for all ( x < 1 ).Therefore, combining these two intervals, the values of ( a ) that satisfy ( x^2 - a x + 3 > 0 ) for all ( x < 1 ) are ( 2 leq a leq 4 ).But we also need to remember the first condition, which was ( a geq 2 ) for the quadratic ( u(x) ) to be decreasing on ( (-infty, 1) ). Therefore, combining both conditions:- ( a geq 2 ) (to ensure ( u(x) ) is decreasing on ( (-infty, 1) ))- ( a leq 4 ) (to ensure ( x^2 - a x + 3 > 0 ) for all ( x < 1 ))Therefore, the range of ( a ) is ( [2, 4] ). But wait, need to check if 4 is included or not. When ( a = 4 ), the quadratic is ( x^2 - 4x + 3 ), which factors as ( (x - 1)(x - 3) ). Therefore, at ( x = 1 ), the quadratic is 0, but our domain is ( (-infty, 1) ), not including 1. Therefore, since ( x = 1 ) is not included, and for ( x < 1 ), the quadratic is positive (since the root at 1 is not included, and the other root is at 3). Therefore, at ( a = 4 ), the quadratic is positive for all ( x < 1 ), so ( a = 4 ) is allowed. Therefore, the upper bound is inclusive.Similarly, when ( a = 2 ), the quadratic is ( x^2 - 2x + 3 ). The discriminant is ( 4 - 12 = -8 ), so no real roots. Therefore, it's always positive. And the derivative of ( u(x) = x^2 - 2x + 3 ) is ( 2x - 2 ), which is negative when ( x < 1 ), so ( u(x) ) is decreasing on ( (-infty, 1) ). Therefore, ( a = 2 ) is allowed. Therefore, the interval is ( [2, 4] ).But wait, let's check for ( a = 4 ). We have the quadratic ( x^2 - 4x + 3 ), which is positive for ( x < 1 ) and ( x > 3 ). So on ( (-infty, 1) ), it's positive. However, we must also check that the logarithmic function is defined at the limit as ( x ) approaches 1 from the left. As ( x ) approaches 1 from the left, ( x^2 - 4x + 3 ) approaches ( 1 - 4 + 3 = 0 ). Therefore, ( log_{frac{1}{2}}(x^2 - 4x + 3) ) approaches ( log_{frac{1}{2}}(0^+) ), which tends to ( +infty ). But since 1 is not included in the interval ( (-infty, 1) ), the function approaches infinity at the boundary, but it's still defined on the entire interval ( (-infty, 1) ). Therefore, ( a = 4 ) is acceptable.Wait, but the problem says the function is monotonically increasing on ( (-infty, 1) ). However, if at ( a = 4 ), as ( x ) approaches 1 from the left, the function tends to infinity, but it's still increasing throughout the interval. Let me confirm the monotonicity at ( a = 4 ).Since ( u(x) = x^2 -4x +3 ), derivative ( u'(x) = 2x -4 ). For ( x < 1 ), ( 2x -4 < 2*1 -4 = -2 < 0 ). Therefore, ( u(x) ) is decreasing on ( (-infty, 1) ), hence ( f(x) = log_{frac{1}{2}}(u(x)) ) is increasing. So, yes, even at ( a =4 ), the function is monotonically increasing on ( (-infty,1) ). Therefore, 4 is included.Therefore, the range of ( a ) is ( [2,4] ), which is option C. However, looking back at the answer choices: A: ( (2,+infty) ), B: ( [2,+infty) ), C: ( [2,4] ), D: ( [2,4) ). So, the correct answer is C.But wait, hold on. Let me cross-verify. If ( a =4 ), then the quadratic is zero at x=1. But since x=1 is not included in the interval, the logarithm is defined on ( (-infty,1) ). However, even though the quadratic approaches zero as x approaches 1, the logarithm will approach infinity, but the function is still increasing on the entire interval. Therefore, the answer is C.However, sometimes in such problems, even though the function approaches infinity, if the domain is open, endpoints are not considered. But in this case, the function is defined and increasing on the entire open interval ( (-infty,1) ), even as it approaches infinity. Therefore, 4 is allowed. Hence, the answer is C: [2,4].But wait, let me check another edge case. Suppose ( a = 4 ), then the quadratic is ( x^2 -4x +3 ), which is zero at x=1 and x=3. But for all x <1, the quadratic is positive. So, the function ( f(x) ) is defined and increasing on ( (-infty,1) ). Therefore, a=4 is acceptable.Another check: Take a=3, which is between 2√3 (~3.464) and 2. Let's see. For a=3, the quadratic is ( x^2 -3x +3 ). The discriminant is 9 -12 = -3 <0, so no real roots. Therefore, it's always positive. The derivative of u(x) is 2x -3. For x <1, 2x -3 <2*1 -3= -1 <0. So, u(x) is decreasing, hence f(x) is increasing. So a=3 is acceptable.Take a=2√3 (~3.464). Quadratic is ( x^2 - 2√3 x +3 ). Discriminant: (2√3)^2 -12=12-12=0. Therefore, the quadratic has a double root at x= (2√3)/2=√3 ~1.732. Therefore, the quadratic is positive for all x except x=√3. Since √3 ~1.732 >1, then for x <1, the quadratic is positive. Therefore, the function f(x) is defined on (-infty,1), and u(x) is decreasing there. Therefore, a=2√3 is allowed, and since 2√3 ~3.464 is within [2,4], that's okay.Wait, but when a=2√3 (~3.464), the quadratic has a double root at ~1.732, which is greater than 1. Therefore, for x <1, quadratic is positive, so function is defined. The derivative of u(x) is 2x -2√3. For x <1, 2x -2√3 <2*1 -2√3≈2 -3.464≈-1.464 <0, so u(x) is decreasing, hence f(x) is increasing. Therefore, a=2√3 is acceptable. Therefore, the upper limit is indeed 4.Hence, combining all these cases, the range of a is [2,4], which is option C.But wait, the given answer options include D: [2,4). Maybe there is a detail I missed. Let me check when a=4. Is the function f(x) actually defined on all of (-infty,1)? As a=4, quadratic is x^2 -4x +3 = (x-1)(x-3). For x <1, quadratic is positive, because both factors (x-1) and (x-3) are negative, so their product is positive. Therefore, yes, x^2 -4x +3 >0 for x <1. Therefore, the logarithm is defined. Therefore, a=4 is acceptable. So the upper limit is inclusive, so the answer should be C: [2,4].But maybe there's another consideration. For example, even though the quadratic is positive for x <1 when a=4, perhaps the function f(x) must be monotonically increasing on the entire interval (-infty,1), but as x approaches 1 from the left, f(x) approaches infinity. However, in calculus, a function can be monotonically increasing on an interval even if it tends to infinity at the endpoint. Monotonicity doesn't require the function to be bounded, just that it never decreases. Therefore, even though it tends to infinity, it's still increasing. Therefore, a=4 is allowed.Therefore, the answer should be C: [2,4]. But the options given are C: [2,4], D: [2,4). Let me check again the problem statement.Wait, the problem says "monotonically increasing on (−∞,1)". So as long as it's increasing on that interval, regardless of behavior at the endpoint. Therefore, even if at a=4, the function tends to infinity as x approaches 1, it's still increasing on the entire interval. Therefore, 4 should be included. So the correct answer is C.However, sometimes in such problems, the upper limit might be excluded if the function ceases to be defined at some point. But in this case, at a=4, the function is still defined on the entire interval. Therefore, I think the correct answer is C: [2,4].But let me check once more. If a=4, the quadratic is zero at x=1 and x=3. Since x=1 is not included in the interval, the quadratic is positive on (−∞,1). So, the logarithm is defined there, and since the quadratic is decreasing, the logarithm is increasing. Therefore, a=4 is valid. Hence, the answer is C.Final Answerboxed{C}</think>To determine the range of ( a ) such that the function ( f(x) = log_{frac{1}{2}}(x^2 - ax + 3) ) is monotonically increasing on ( (-infty, 1) ), we need to consider two main conditions:1. The argument of the logarithm, ( u(x) = x^2 - ax + 3 ), must be positive for all ( x in (-infty, 1) ).2. The quadratic function ( u(x) ) must be decreasing on ( (-infty, 1) ). Step-by-Step Analysis:1. Monotonicity of ( u(x) ): - The derivative of ( u(x) ) is ( u'(x) = 2x - a ). - For ( u(x) ) to be decreasing on ( (-infty, 1) ), ( u'(x) < 0 ) for all ( x in (-infty, 1) ). - This requires ( 2x - a < 0 ) for all ( x < 1 ), leading to ( a geq 2 ).2. Positivity of ( u(x) ): - The quadratic ( u(x) = x^2 - ax + 3 ) must be positive for all ( x < 1 ). - The vertex of the quadratic is at ( x = frac{a}{2} ). The minimum value of ( u(x) ) at the vertex is ( 3 - frac{a^2}{4} ). - For ( u(x) ) to be positive for all ( x ), the discriminant ( a^2 - 12 ) must be negative (i.e., ( a < 2sqrt{3} approx 3.464 )) or the roots of the quadratic must be greater than or equal to 1.3. Roots Analysis: - The roots of the quadratic are ( frac{a pm sqrt{a^2 - 12}}{2} ). - For the quadratic to be positive on ( (-infty, 1) ), the smaller root must be greater than or equal to 1. This leads to the condition ( frac{a - sqrt{a^2 - 12}}{2} geq 1 ), which simplifies to ( a leq 4 ). Combining Conditions:- The interval for ( a ) must satisfy both ( a geq 2 ) (for monotonicity) and ( a leq 4 ) (for positivity). - Checking the boundary values: - At ( a = 2 ), the quadratic ( x^2 - 2x + 3 ) has no real roots and is always positive. - At ( a = 4 ), the quadratic ( x^2 - 4x + 3 ) has roots at 1 and 3, but is positive for all ( x < 1 ).Thus, the range of ( a ) is ( [2, 4] ).[boxed{C}],<think>Okay, let's see. I need to determine the range of the parameter 'a' such that the function f(x) = log base 1/2 of (x² - a x + 3) is monotonically increasing on the interval (-∞, 1). The answer choices are given, so hopefully, after working through the problem, I can match my result to one of them. Let me start by recalling some logarithmic function properties.First, since the logarithm's base is 1/2, which is between 0 and 1, the logarithmic function itself is a decreasing function. That means if the argument inside the log increases, the log function will decrease, and vice versa. So, for f(x) to be increasing, the argument of the log, which is the quadratic expression x² - a x + 3, must be decreasing. Because the outer function (log base 1/2) is decreasing, so the composition f(x) will be increasing when the inner function is decreasing. Is that right?Yes, because if the inner function is decreasing, then as x increases, the inner function's output decreases, but since the log is also decreasing, a decrease in the inner function's output would result in an increase in the log's output. So, the composition would be increasing. That makes sense. So, the key is to ensure that the quadratic expression x² - a x + 3 is decreasing on (-∞, 1).Now, I need to check when the quadratic function g(x) = x² - a x + 3 is decreasing on (-∞, 1). Quadratics are typically decreasing to the left of their vertex and increasing to the right if the coefficient of x² is positive. Here, the coefficient is 1, which is positive. So, the parabola opens upward. The vertex of this quadratic is at x = -b/(2a) where the quadratic is in the form ax² + bx + c. Wait, in this case, the quadratic is x² - a x + 3, so a coefficient here is 1, and the linear coefficient is -a. So, the vertex is at x = -(-a)/(2*1) = a/2.Since the parabola opens upward, the function is decreasing on (-∞, a/2) and increasing on (a/2, ∞). So, for the quadratic to be decreasing on (-∞, 1), the interval (-∞,1) must be a subset of (-∞, a/2). That is, the vertex at a/2 must be to the right of 1. Therefore, a/2 ≥ 1, which implies a ≥ 2. So, a must be at least 2. But wait, is that all?But we also need to make sure that the quadratic expression is positive for all x in (-∞,1) because the logarithm is only defined when its argument is positive. So, x² - a x + 3 > 0 for all x in (-∞,1). Therefore, we have two conditions here:1. The quadratic is decreasing on (-∞,1), which requires a ≥ 2.2. The quadratic is positive for all x < 1.So, I need to ensure that x² - a x + 3 > 0 for all x in (-∞,1). Let's handle this second condition.For the quadratic to be positive on (-∞,1), the quadratic must not cross the x-axis to the left of x = 1. That can be achieved in two ways:Either the quadratic has no real roots at all, meaning its discriminant is negative.Or, if it has real roots, both roots are greater than or equal to 1. But since the parabola opens upward, if there are roots, the quadratic would be negative between the roots. Therefore, to have the quadratic positive on (-∞,1), either there are no real roots, or the interval (-∞,1) is entirely to the left of the smaller root.Wait, let's think again. If the quadratic has real roots, say r1 and r2, with r1 ≤ r2, then the quadratic is positive outside the interval [r1, r2]. So, to have x² - a x + 3 > 0 for all x in (-∞,1), either there are no real roots (discriminant < 0) or the interval (-∞,1) is entirely to the left of r1. That is, r1 ≥ 1.So, two cases:Case 1: Discriminant < 0.Discriminant D = a² - 12 < 0 ⇒ a² < 12 ⇒ -2√3 < a < 2√3. Since √3 ≈ 1.732, so 2√3 ≈ 3.464. But from the first condition, a ≥ 2. So, combining these, 2 ≤ a < 2√3 ≈ 3.464.But 2√3 is approximately 3.464, which is less than 4. However, the answer choices given are integers. Wait, the answer options are [2,4), etc. So, maybe 2√3 is approximately 3.464, so perhaps in exact terms, we need another approach.Case 2: If discriminant ≥ 0, then the roots must satisfy r1 ≥ 1. Let's compute the roots.The roots of x² - a x + 3 = 0 are [a ± sqrt(a² - 12)] / 2.The smaller root r1 is [a - sqrt(a² - 12)] / 2.We need r1 ≥ 1.So, [a - sqrt(a² - 12)] / 2 ≥ 1.Multiply both sides by 2:a - sqrt(a² - 12) ≥ 2.Then, move sqrt(a² -12) to the other side:a - 2 ≥ sqrt(a² - 12).Now, since sqrt(a² -12) is non-negative, the left side must also be non-negative. So, a - 2 ≥ 0 ⇒ a ≥ 2, which is already satisfied from the first condition.Now, square both sides to eliminate the square root:(a - 2)² ≥ a² - 12Expand left side:a² -4a +4 ≥ a² -12Subtract a² from both sides:-4a +4 ≥ -12Subtract 4:-4a ≥ -16Divide by -4 (inequality sign reverses):a ≤ 4.Therefore, in the case where discriminant ≥ 0 (i.e., a² ≥ 12 ⇒ a ≥ 2√3 ≈ 3.464), we have that a ≤ 4. Therefore, combining these, in case 2, we have 3.464... ≤ a ≤ 4.Therefore, combining both cases:Case 1: 2 ≤ a < 2√3 (≈3.464)Case 2: 2√3 ≤ a ≤4Therefore, the total range is 2 ≤ a ≤4.But wait, in case 1, discriminant is negative, so quadratic is always positive. So, as long as the quadratic is always positive (a² <12), which for a ≥2, is 2 ≤a <2√3. Then, in case 2, when quadratic is not always positive, but the smaller root is ≥1, which requires a between 2√3 and 4. Therefore, combining both, the total range is [2,4]. Wait, but in the problem, the answer options include [2,4] as option C and [2,4) as D. Hmm.Wait, need to check if a=4 is allowed. Let's check when a=4.If a=4, then the quadratic is x² -4x +3. Let's factor that: (x-1)(x-3). So, the roots are 1 and 3. Therefore, the quadratic is positive on (-∞,1) ∪ (3, ∞). So, on (-∞,1), the quadratic is positive. So, the logarithm is defined. Also, since the quadratic is decreasing on (-∞, a/2) = (-∞, 2). Since we are looking at the interval (-∞,1), which is a subset of (-∞,2). Therefore, the quadratic is decreasing on (-∞,1), so the log is increasing. Therefore, a=4 is allowed. So, the upper limit is 4 inclusive.But wait, when a approaches 4 from below? For example, a=3.5, the roots would be?For a=3.5, discriminant is 12.25 -12 =0.25, which is positive. Then roots are [3.5 ±0.5]/2, so [3.5 -0.5]/2=3/2=1.5 and [3.5 +0.5]/2=4/2=2. So, roots at 1.5 and 2. Then, quadratic is positive on (-∞,1.5) ∪ (2, ∞). So, on (-∞,1), it's positive. So, even when a=3.5, which is less than 4, the quadratic is positive on (-∞,1). So, as long as a ≤4, then in the case of real roots, the smaller root is greater than or equal to 1.Wait, when a=4, the smaller root is 1. So, the quadratic is positive on (-∞,1) union (3, ∞). So, at x=1, the quadratic is zero, but since our interval is (-∞,1), which is open, x=1 is not included. Therefore, even when a=4, the quadratic is positive on (-∞,1). So, a=4 is acceptable.But what about a >4? Let's take a=5. Then quadratic is x² -5x +3. The discriminant is 25 -12=13. Roots are [5 ±sqrt(13)]/2. sqrt(13)≈3.606. So, roots ≈ (5 -3.606)/2≈0.697 and (5 +3.606)/2≈4.303. Therefore, the quadratic is positive on (-∞,0.697) ∪ (4.303, ∞). But our interval is (-∞,1). So, between 0.697 and1, the quadratic would be negative. Therefore, for a=5, which is greater than 4, the quadratic is negative in part of (-∞,1), which would make the logarithm undefined there. Therefore, a cannot be greater than 4.Similarly, if a=4, the quadratic is x² -4x +3=(x-1)(x-3), so it's positive on (-∞,1) and (3, ∞). So, on (-∞,1), it's positive. Hence, a=4 is okay.But if a is slightly more than 4, like 4.1, the roots would be less than 1 and more than 3? Wait, let's compute. For a=4.1, discriminant is (4.1)^2 -12=16.81 -12=4.81. sqrt(4.81)=2.193. So, roots are (4.1 ±2.193)/2. So, (4.1 -2.193)/2≈1.907/2≈0.9535, and (4.1 +2.193)/2≈6.293/2≈3.1465. Therefore, the quadratic is positive on (-∞,0.9535) and (3.1465, ∞). But our interval is (-∞,1). So, between 0.9535 and1, the quadratic is negative, which is part of the interval. Therefore, the logarithm is not defined there. Thus, a=4.1 is invalid. Therefore, a must be ≤4.Therefore, the allowable values for a are from 2 to4, inclusive.So, the range of a is [2,4]. So, answer C.Wait, but let's verify with a=2. When a=2, the quadratic is x² -2x +3. The discriminant is 4 -12 = -8 <0, so quadratic is always positive. The vertex is at a/2=1. Since the quadratic is decreasing on (-∞,1), and since it's always positive, then the logarithm is defined and increasing on (-∞,1). So, a=2 is okay.When a=3, which is in [2,4], the quadratic is x² -3x +3. Discriminant 9-12=-3 <0, so always positive. Decreasing on (-∞,1.5), so on (-∞,1), it's decreasing, so log is increasing. So, okay.When a=4, as discussed, quadratic is (x-1)(x-3), positive on (-∞,1), decreasing on (-∞,2), so on (-∞,1), decreasing. Therefore, log is increasing. So, a=4 is okay.If a is between 2 and4, say a=3. Then it's okay. If a is greater than4, like a=5, as discussed, quadratic becomes negative on part of (-∞,1), which invalidates the logarithm's domain. If a is less than2, say a=1, then the vertex is at 0.5, so the quadratic is decreasing on (-∞,0.5) and increasing on (0.5, ∞). Therefore, on (-∞,1), the quadratic first decreases until x=0.5, then increases. Hence, the function f(x) would first increase (since log is decreasing, and the argument is decreasing) until x=0.5, then decrease. So, it's not monotonic on (-∞,1). Therefore, a must be ≥2.Therefore, combining all these, the range is [2,4], which is option C.But wait, the answer choices are:A: (2,+∞)B: [2,+∞)C: [2,4]D: [2,4)Hmm, so according to my reasoning, the correct answer is C. However, let me check once more for a=4. When a=4, the quadratic is x² -4x +3, which factors as (x-1)(x-3). So, the quadratic is positive when x <1 or x>3. Therefore, on (-∞,1), it's positive. The quadratic's vertex is at x=4/2=2. Since it's a parabola opening upwards, it's decreasing on (-∞,2). Therefore, on (-∞,1), it's decreasing. Therefore, the log function with base 1/2 is increasing there. So, a=4 is allowed. Therefore, the interval should include 4. So, answer C: [2,4].But let's check the answer options again. The options are C: [2,4] and D: [2,4). So, which one is correct?Wait, but when a=4, as shown, the quadratic is zero at x=1 and x=3. The original function f(x) is log base 1/2 of (x² -4x +3). At x approaching 1 from the left, the quadratic approaches 0 from the positive side (since x=1 is a root). So, log base 1/2 of a number approaching 0 from the positive side tends to +infinity. Therefore, at x approaching1 from the left, f(x) approaches +infinity. However, the function is defined on (-∞,1), but the problem states that f(x) is monotonically increasing on (-∞,1). So, even though at x=1, the function tends to infinity, the function is increasing on the entire interval (-∞,1). So, since the interval doesn't include x=1, the function is defined and increasing on (-∞,1) when a=4. So, a=4 is acceptable. Therefore, the correct answer should include 4, so option C.But wait, another thought: when a=4, the quadratic touches the x-axis at x=1. But since x=1 is not included in the domain (the interval is open at 1), the logarithm is still defined on (-∞,1). Therefore, even though the quadratic has a root at x=1, the function f(x) is still defined on (-∞,1) and is increasing there. Hence, a=4 is allowed.Therefore, the answer is C: [2,4].But let me check the answer choices again. Yes, C is [2,4], D is [2,4). So, the correct answer should be C. But wait, some sources might consider that at a=4, since the quadratic has a root at x=1, which is the endpoint of the interval, even though it's open. But in calculus terms, the function just needs to be increasing on the open interval (-∞,1), regardless of behavior at the endpoint. Therefore, as long as the quadratic is positive and decreasing on (-∞,1), even if it approaches zero at x=1, the function is still increasing on that interval. Therefore, a=4 is acceptable, so 4 is included. Therefore, the answer is C.But let me verify with another example. Take a=4. The quadratic is x² -4x +3. Its derivative is 2x -4. On (-∞,1), the derivative is 2x -4. For x <1, 2x -4 <2*1 -4= -2 <0. Therefore, the quadratic is decreasing on (-∞,1). Therefore, the log function is increasing. So, even though at x=1 the quadratic is zero, since x=1 is not in the interval, the function is increasing on (-∞,1). Therefore, a=4 is acceptable. Therefore, the answer is [2,4], so option C.Therefore, the correct answer is C.Final Answerboxed{C}</think>To determine the range of (a) such that the function (f(x) = log_{frac{1}{2}}(x^2 - ax + 3)) is monotonically increasing on ((-infty, 1)), we need to consider the following:1. Monotonicity of the Logarithmic Function: Since the base of the logarithm is (frac{1}{2}) (which is between 0 and 1), the logarithmic function is decreasing. Therefore, for (f(x)) to be increasing, the argument (g(x) = x^2 - ax + 3) must be decreasing on ((-infty, 1)).2. Quadratic Function Analysis: The quadratic function (g(x) = x^2 - ax + 3) is a parabola opening upwards (since the coefficient of (x^2) is positive). The vertex of this parabola is at (x = frac{a}{2}). For (g(x)) to be decreasing on ((-infty, 1)), the vertex must be at or to the right of 1, i.e., (frac{a}{2} geq 1 implies a geq 2).3. Positivity of the Quadratic Function: The quadratic (g(x)) must be positive for all (x in (-infty, 1)). This requires either the quadratic to have no real roots (discriminant negative) or the smaller root (if real) to be greater than or equal to 1. - Discriminant Condition: The discriminant of (g(x)) is (a^2 - 12). If (a^2 - 12 < 0), then (2 leq a < 2sqrt{3}) (approximately (3.464)). - Roots Condition: If the discriminant is non-negative ((a geq 2sqrt{3})), the smaller root (frac{a - sqrt{a^2 - 12}}{2}) must be greater than or equal to 1. Solving this gives (a leq 4).Combining these conditions, we find that (a) must satisfy (2 leq a leq 4).Thus, the range of (a) is ([2, 4]), which corresponds to option (boxed{C}).]

🔑:Let t=g(x)=x^{2}-ax+3, then y=log,_{frac {1}{2}}t is a decreasing function.If f(x)=log,_{frac {1}{2}}(x^{2}-ax+3) is monotonically increasing on (-infty,1),then t=g(x)=x^{2}-ax+3 is monotonically decreasing on (-infty,1), and g(1)geqslant 0,that is, -frac {-a}{2}=frac {a}{2}geqslant 1 and 1-a+3geqslant 0,then ageqslant 2 and aleqslant 4, i.e., 2leqslant aleqslant 4,therefore, the answer is boxed{C}.This problem is solved by using the substitution method in combination with the relationship between the monotonicity of composite functions. The key to solving this problem is to utilize the substitution method in conjunction with the properties of logarithmic functions and quadratic functions. This problem primarily tests the application of function monotonicity.

🧠:[<think>Okay, so I need to figure out what this sum is equal to. Let me write it down again to make sure I have it right:1/2 + (2¹)/(2²) + (2²)/(2³) + ... + (2²⁰⁰²)/(2²⁰⁰³) + (2²⁰⁰³)/(2²⁰⁰⁴)Hmm. Let's see. Each term seems to be of the form (2^(n-1))/(2^n) starting from n=1 up to n=2003? Wait, let me check the exponents.Wait, the first term is 1/2, which is 2⁰/2¹. Then the next term is 2¹/2², then 2²/2³, and so on, up until 2²⁰⁰³/2²⁰⁰⁴. So each term is (2^(k-1))/2^k where k starts at 1 and goes to 2004. But the last term given is (2²⁰⁰³)/(2²⁰⁰⁴), which would be when k=2004. So there are 2004 terms here? Because starting from k=1 (term 1) up to k=2004 (term 2004). Let me confirm:First term: k=1: 2^(1-1)/2^1 = 1/2Second term: k=2: 2^(2-1)/2^2 = 2^1/2^2 = 2/4 = 1/2Wait a second, is every term 1/2? Let me check the first few terms:1/2 is the first term.Then 2^1 / 2^2 = 2/4 = 1/2.Then 2^2 / 2^3 = 4/8 = 1/2.Wait, hold on! So each term simplifies to 1/2? Because 2^(n-1)/2^n = (2^(n-1))/(2*2^(n-1)) )= 1/2. So every single term here is 1/2. Therefore, the sum is just the number of terms multiplied by 1/2.How many terms are there? The first term is when exponent in the numerator is 0 (2^0/2^1) and the last term is when exponent in the numerator is 2003 (2^2003/2^2004). So the exponents in the numerator go from 0 to 2003, inclusive. That's 2004 terms (since 2003 - 0 + 1 = 2004). Therefore, the total sum is 2004 * (1/2) = 1002.But wait, the options are (A) 1002, (B) 501, (C) 1/2^2004, (D) 2004, (E) 2004/2^2004. So 1002 is option A. But let me verify once again to be absolutely sure.Each term: 2^k / 2^{k+1} = 1/2. So each term is indeed 1/2. Then the number of terms. The first term is k=0 (2^0/2^1) and the last term is k=2003 (2^2003/2^2004). So k ranges from 0 to 2003, which is 2004 terms. Therefore, 2004 * 1/2 = 1002. So answer A. That seems straightforward, but maybe there's a trick here?Wait, let me check the original problem again. It says:1/2 + 2¹/2² + 2²/2³ + ... + 2²⁰⁰²/2²⁰⁰³ + 2²⁰⁰³/2²⁰⁰⁴So yes, the first term is 1/2, which is 2⁰/2¹, then 2¹/2², ..., up to 2²⁰⁰³/2²⁰⁰⁴. So exponents in the numerator go from 0 to 2003. That's 2004 terms. Each term is 1/2, so total sum is 2004 * 1/2 = 1002. So answer is A, 1002.But why is option B 501? Maybe I made a mistake in the number of terms? Let's count again. Let's list a few terms:Term 1: exponent 0 (numerator) and 1 (denominator) => 2⁰/2¹ = 1/2Term 2: exponent 1 (numerator) and 2 (denominator) => 2¹/2² = 1/2...Last term: exponent 2003 (numerator) and 2004 (denominator) => 2²⁰⁰³/2²⁰⁰⁴ = 1/2So how many terms? From exponent 0 to 2003 in the numerator, that's 2004 terms. So 2004*(1/2) = 1002, which is A.Alternatively, maybe the problem is written in a confusing way. Let's check the indices again. If the first term is 1/2, then next term is 2^1 / 2^2, which is 1/2, then 2^2 / 2^3 = 1/2, ..., up to 2^2003 / 2^2004 = 1/2. So starting from exponent 0 in the numerator (first term) to exponent 2003 (last term). So exponents in the numerator: 0,1,2,...,2003. That's 2004 terms. Each term is 1/2, so sum is 2004*(1/2)=1002. So answer A. I think that's correct.But why is 501 an option? Maybe someone might think the number of terms is 2003? Let's see. If you subtract 2003 - 0 + 1? Wait, 2003 - 0 is 2003, plus 1 is 2004. So 2004 terms. So 2004/2=1002. So A is correct. Therefore, the answer must be A.Alternatively, if there was a miscalculation where someone thought the number of terms is 1002, then 1002*(1/2)=501, but that would be wrong. But in reality, the number of terms is 2004. Therefore, the answer is 1002. So A.Final Answerboxed{A}</think>The given sum is:[frac{1}{2} + frac{2^1}{2^2} + frac{2^2}{2^3} + cdots + frac{2^{2002}}{2^{2003}} + frac{2^{2003}}{2^{2004}}]Each term in the series can be simplified as follows:[frac{2^{k-1}}{2^k} = frac{1}{2} quad text{for} quad k = 1, 2, ldots, 2004]This is because:[frac{2^{k-1}}{2^k} = frac{2^{k-1}}{2 cdot 2^{k-1}} = frac{1}{2}]The series consists of 2004 terms, each equal to (frac{1}{2}). Therefore, the sum of the series is:[2004 times frac{1}{2} = 1002]Thus, the answer is (boxed{A}).,<think>Okay, so I have this problem here: a sum that starts with 1/2, then adds 2^1 over 2^2, then 2^2 over 2^3, and so on, up to 2^2003 over 2^2004. And I need to figure out which of the options A to E this sum equals. Let me first write down the expression properly to make sure I understand it.The sum is: 1/2 + (2^1)/(2^2) + (2^2)/(2^3) + ... + (2^2003)/(2^2004). Hmm, let me check the exponents. Each numerator is 2 raised to a power that starts at 1 and goes up to 2003, and each denominator is 2 raised to the next power. So the first term after 1/2 is 2^1/2^2, which simplifies to 2/4 = 1/2. Wait, is that right? Let me verify.Wait, 2^1 is 2, and 2^2 is 4, so 2/4 is indeed 1/2. Then the next term is 2^2/2^3, which is 4/8 = 1/2. Wait, hold on. Is every term after the first one also 1/2? Because 2^k over 2^(k+1) simplifies to 1/2 for any k. Let me check that.Let's take a general term in the sum: starting from the second term, each term is 2^{n}/2^{n+1} where n starts at 1 and goes up to 2003. So 2^{n}/2^{n+1} = 2^{n - (n + 1)} = 2^{-1} = 1/2. So every term from n=1 to n=2003 is 1/2. The first term is also 1/2. So how many terms are there in total?Wait, the first term is 1/2, then the next terms are from n=1 to n=2003, which is 2003 terms. But wait, let's see:Original sum:Term 1: 1/2 (maybe this is n=0? Let's check. If the first term is written as 2^0 / 2^1, which would be 1/2. Then the next term is 2^1 / 2^2, and so on up to 2^2003 / 2^2004. So if that's the case, then the sum is from n=0 to n=2003 of 2^n / 2^{n+1}.So each term is 2^n / 2^{n+1} = 1/2. So each term is 1/2. How many terms are there? From n=0 to n=2003, that's 2004 terms. So the sum is 2004 * (1/2) = 1002. Therefore, the answer should be A) 1002.Wait, but let me make sure. Let me recount the number of terms. If the first term is n=0, then n=0,1,2,...,2003. That's 2004 terms because you subtract 0 from 2003 and add 1. So 2003 - 0 + 1 = 2004 terms. Each term is 1/2, so total sum is 2004*(1/2) = 1002.So the answer is A) 1002. But let me check again to see if there's a trick here. Maybe the initial term is not part of the pattern? Let's look again.The problem states: 1/2 + 2^1/2^2 + 2^2/2^3 + ... + 2^2003/2^2004. So yes, the first term is 1/2, which is 2^0/2^1. Then the next term is 2^1/2^2, which is 1/2, and so on until 2^2003/2^2004. So each term is indeed 1/2, and the total number of terms is 2004. Therefore, 2004*(1/2) = 1002. So answer is A.But wait, the options include 1002, which is A. Let me check if the problem is written correctly. The original problem says: 1/2 + 2^1 / 2^2 + 2^2 / 2^3 + ... + 2^{2002}/2^{2003} + 2^{2003}/2^{2004}. Wait, hold on, in the problem statement, is the second last term 2^{2002}/2^{2003} and the last term is 2^{2003}/2^{2004}? Then the exponents go from 1 to 2003 in the numerators and 2 to 2004 in the denominators. Wait, but the first term is 1/2, which would be 2^0 / 2^1.Wait, let me parse the problem again:frac{1}{2}+frac{2^{1}}{2^{2}}+frac{2^{2}}{2^{3}}+cdots+frac{2^{2002}}{2^{2003}}+frac{2^{2003}}{2^{2004}}So the first term is 1/2, which is 2^0 / 2^1. Then the next term is 2^1 / 2^2, then 2^2 / 2^3, and so on, up to 2^{2003}/2^{2004}. So the exponents in the numerators start at 0 and go up to 2003. Thus, the number of terms is from n=0 to n=2003, which is 2004 terms. Each term is 1/2, so total sum is 2004*(1/2) = 1002. So answer A) 1002. That seems straightforward.But maybe the problem is written differently. Wait, in the problem statement, after the first term 1/2, the next term is 2^1 / 2^2, which is 1/2, then 2^2 / 2^3, which is 1/2, etc., up to 2^{2003}/2^{2004}, which is also 1/2. So all terms after the first are 1/2? Wait, no. Wait, the first term is already 1/2. Then the next terms are 1/2 each. So how many terms?Let me list them:Term 1: 1/2 (n=0)Term 2: 2^1 /2^2 =1/2 (n=1)Term 3: 2^2 /2^3 =1/2 (n=2)...Last term: 2^{2003}/2^{2004}=1/2 (n=2003)So total terms: n=0 to n=2003, which is 2004 terms. Each term is 1/2, so sum is 2004*(1/2)=1002. Therefore, answer is A.But wait, wait, wait. Let me check again. Let me count the number of terms. Let's see:The sequence starts at n=0 (term 1: 1/2) and goes up to n=2003 (term 2004). So yes, 2004 terms. Each term is 1/2, so sum is 2004*(1/2)=1002. Therefore, the answer is A) 1002.But the options are:(A) 1002(B) 501(C) 1/2^{2004}(D) 2004(E) 2004/2^{2004}So if my calculation is correct, answer is A. But let me check if there's an alternative way to compute this sum, perhaps a telescoping series or geometric series?Wait, but each term is 1/2. So sum is number of terms multiplied by 1/2. So unless there's a mistake in counting the number of terms, this should be correct.Alternatively, maybe the problem was written differently. Let me check the original problem again.Original problem: (1/2) + (2^1)/(2^2) + (2^2)/(2^3) + ... + (2^{2002})/(2^{2003}) + (2^{2003})/(2^{2004})Wait, hold on. The first term is (1)/2 = 1/2, which is 2^0 / 2^1. Then the next term is 2^1 / 2^2, which is 1/2, then 2^2 / 2^3 = 1/2, and so on, up until 2^{2003}/2^{2004} =1/2.But in the problem statement, is the second last term 2^{2002}/2^{2003}, and then the last term is 2^{2003}/2^{2004}? So if we count the exponents in the numerator: starting from 0 (for the first term) up to 2003. So the exponents in the numerator are 0,1,2,...,2003. So the number of terms is 2004. Therefore, each term is 1/2, sum is 2004*(1/2)=1002. So answer is A.Alternatively, maybe the problem is presented as starting from exponent 1 in the numerator? Wait, first term is 1/2 = 2^0 / 2^1. So exponent 0 in the numerator. Then exponent 1, 2,...,2003. So total terms are 2004. So 2004*(1/2)=1002.Alternatively, maybe if we think of the sum as starting from n=1, but let me check:If the first term is 1/2, which is n=0, then n ranges from 0 to 2003. So 2004 terms.Alternatively, if someone miscalculates and thinks the sum starts at n=1, then the number of terms would be 2003, leading to 2003*(1/2)=1001.5, which is not an integer, but the answer options are all integers or fractions with denominator 2^2004. So that can't be.Alternatively, maybe there's a different approach here. Let me try to write the sum as:Sum_{k=0}^{2003} (2^k)/(2^{k+1}) = Sum_{k=0}^{2003} 1/2 = 2004*(1/2) = 1002.Therefore, answer is A. So I think that is correct.Wait, but maybe the problem is written in a way that the first term is not 2^0 /2^1 but just 1/2, and the next term is 2^1/2^2, etc., so the total number of terms is 2004. Hence, sum is 2004*(1/2)=1002.Alternatively, if someone tries to interpret the sum as starting at n=1, but then the last term would be 2^{2003}/2^{2004}, so n=1 to n=2003: 2003 terms. Then sum would be 2003*(1/2)=1001.5, but that is not an answer choice. So that approach is invalid.Alternatively, maybe there's a different pattern here. Let's check the first few terms and see.First term: 1/2Second term: 2/4 =1/2Third term:4/8=1/2Fourth term:8/16=1/2So each term is 1/2. So if there are N terms, the sum is N*(1/2). Therefore, if we can find N, which is the number of terms.Given that the last term is 2^{2003}/2^{2004}, so the exponent in the numerator is 2003, which corresponds to n=2003. If the first term is n=0, then total number of terms is 2004. Hence, sum is 2004*(1/2)=1002. So answer is A.Therefore, I think the answer is A) 1002.Final Answerboxed{A}</think>The given problem is to evaluate the sum:[frac{1}{2} + frac{2^1}{2^2} + frac{2^2}{2^3} + cdots + frac{2^{2002}}{2^{2003}} + frac{2^{2003}}{2^{2004}}]First, we observe that each term in the sum can be simplified. For a general term (frac{2^k}{2^{k+1}}) (where (k) starts from 0), we have:[frac{2^k}{2^{k+1}} = frac{1}{2}]This simplification applies to each term in the series. Next, we count the number of terms. The series starts with (k = 0) (which is (frac{2^0}{2^1} = frac{1}{2})) and ends with (k = 2003) (which is (frac{2^{2003}}{2^{2004}} = frac{1}{2})). Therefore, the number of terms is (2004) (from (k = 0) to (k = 2003)).Each term is (frac{1}{2}), so the sum of all terms is:[2004 times frac{1}{2} = 1002]Thus, the answer is:[boxed{A}]]

🔑:Here we use the rule for manipulating exponents frac{2^{a}}{2^{b}}=2^{a-b}.Therefore, each of the 2003 fractions after the first fraction is equal to 2^{-1}=frac{1}{2}.This gives us 2004 copies of frac{1}{2} being added up, for a total of 1002 .Answer: (A)

🧠:[<think>Okay, so I need to find the minimum value of the product (1 - x₁)(1 - x₂)...(1 - xₙ) where each xᵢ is a non-negative real number and their sum is at most 1/2. Hmm, let's see. Since all the x's are non-negative and add up to no more than 1/2, each xᵢ must be between 0 and 1/2, right? Because if even one xᵢ were greater than 1/2, the sum would exceed 1/2 already. So each term (1 - xᵢ) is between 1 - 0 = 1 and 1 - 1/2 = 1/2. So the product is going to be a number between (1/2)^n and 1^n = 1. But the question is asking for the minimum value of this product. So we need to make the product as small as possible. Wait, but how does the product behave when each xᵢ is increased or decreased? Since each (1 - xᵢ) is a decreasing function as xᵢ increases, the product will decrease if we make each xᵢ as large as possible. But there's a constraint on the sum of the xᵢ. So maybe to minimize the product, we need to maximize each xᵢ as much as possible under the constraint that their sum is ≤ 1/2. But how exactly?If we have multiple variables, maybe distributing the "total allowed sum" unevenly would give a lower product? Wait, actually, I remember that for products, the maximum product under a sum constraint is achieved when all variables are equal, due to the AM-GM inequality. But here, since we are dealing with (1 - xᵢ), which is similar to a product of terms that are decreasing as xᵢ increases. So maybe to minimize the product, we need to make the terms as unequal as possible? Hmm, maybe the opposite of AM-GM?Wait, let's think. Suppose we have two variables, x and y, such that x + y ≤ 1/2. We need to minimize (1 - x)(1 - y). Let's test this. If x + y = 1/2, then we can consider x = 1/2 - y. Then the product becomes (1 - (1/2 - y))(1 - y) = (1/2 + y)(1 - y) = (1/2)(1 - y) + y(1 - y) = 1/2 - 1/2 y + y - y² = 1/2 + 1/2 y - y². To find the minimum, take derivative with respect to y: 1/2 - 2y. Set to zero: 1/2 - 2y = 0 => y = 1/4. Then x = 1/2 - 1/4 = 1/4. So the minimum is achieved when x = y = 1/4, which gives (1 - 1/4)(1 - 1/4) = (3/4)^2 = 9/16 ≈ 0.5625. Alternatively, if we set one variable to 1/2 and the other to 0, the product is (1 - 1/2)(1 - 0) = 1/2 * 1 = 1/2 = 0.5. Wait, that's smaller! But that contradicts the previous conclusion. Hmm, so which is correct?Wait, maybe I made a mistake here. Let's calculate again. If we set x = 1/2 and y = 0, then the sum is 1/2, which is allowed. The product is (1 - 1/2)(1 - 0) = 1/2 * 1 = 1/2. But when we set x = y = 1/4, the product is (3/4)^2 = 9/16 ≈ 0.5625, which is larger. So actually, making the variables unequal gives a smaller product. So in two variables, the minimum occurs when one variable is as large as possible (1/2) and the other is 0. But wait, is this the case?Wait, but if we set x = 1/2 and y = 0, the product is 1/2. If we set x = a and y = 1/2 - a, then the product is (1 - a)(1 - (1/2 - a)) = (1 - a)(1/2 + a). Let's compute this function for a between 0 and 1/2. Expanding: (1 - a)(1/2 + a) = 1*(1/2) + 1*a - a*(1/2) - a*a = 1/2 + a - (1/2)a - a² = 1/2 + (1/2)a - a². Taking derivative: 1/2 - 2a. Setting derivative to zero: 1/2 - 2a = 0 => a = 1/4. So the minimum is at a = 1/4? Wait, but when a = 1/4, the product is 1/2 + (1/2)(1/4) - (1/4)^2 = 1/2 + 1/8 - 1/16 = 8/16 + 2/16 - 1/16 = 9/16 ≈ 0.5625. But when a = 1/2, the product is (1 - 1/2)(1/2 + 1/2) = (1/2)(1) = 1/2. Wait, but how come the derivative suggested a minimum at a = 1/4? But at a = 1/4, the function is 9/16, which is higher than 1/2. So actually, the critical point at a = 1/4 is a maximum? Wait, maybe I messed up the derivative.Wait, if we have f(a) = (1 - a)(1/2 + a) = 1/2 + a - a² - (1/2)a. Wait, no, expanding again: (1 - a)(0.5 + a) = 1*0.5 + 1*a - a*0.5 - a*a = 0.5 + a - 0.5a - a² = 0.5 + 0.5a - a². Then derivative f’(a) = 0.5 - 2a. Setting to zero: 0.5 - 2a = 0 => a = 0.25. So at a = 0.25, the function is 0.5 + 0.5*(0.25) - (0.25)^2 = 0.5 + 0.125 - 0.0625 = 0.5625. Then, at the endpoints: when a = 0, f(0) = 0.5 + 0 - 0 = 0.5; when a = 0.5, f(0.5) = 0.5 + 0.5*(0.5) - (0.5)^2 = 0.5 + 0.25 - 0.25 = 0.5. Wait, so the function f(a) reaches a maximum at a = 0.25 and is 0.5 at both endpoints. So in this case, the minimum is 0.5, achieved at the endpoints when either a = 0.5 or a = 0. So that would mean that in two variables, the minimum product is 1/2, achieved when one variable is 1/2 and the other is 0. So in the two-variable case, distributing the total sum to one variable gives the minimum product. So perhaps in general, for n variables, to minimize the product (1 - x₁)(1 - x₂)...(1 - xₙ) under the constraint that the sum of xᵢ ≤ 1/2, we should set as many variables as possible to 0 and one variable to 1/2? Wait, but the sum only needs to be ≤ 1/2. So if we set one variable to 1/2 and the rest to 0, the sum is exactly 1/2, which is allowed, and the product is (1 - 1/2)(1 - 0)^{n - 1} = (1/2)(1)^{n - 1} = 1/2. But is this the minimum? Let's test with three variables.Suppose n = 3. If we set x₁ = 1/2, x₂ = x₃ = 0. Then the product is (1 - 1/2)(1 - 0)(1 - 0) = 1/2 * 1 * 1 = 1/2. Alternatively, if we set x₁ = x₂ = 1/4, x₃ = 0, sum is 1/2. Then the product is (1 - 1/4)(1 - 1/4)(1 - 0) = (3/4)^2 * 1 = 9/16 ≈ 0.5625, which is higher than 1/2. If we set all three variables to 1/6, sum is 3*(1/6) = 1/2. Then the product is (1 - 1/6)^3 = (5/6)^3 ≈ 0.5787, still higher than 1/2. So again, putting all the weight on one variable gives a lower product. So maybe in general, the minimal product is 1/2, achieved by setting one variable to 1/2 and the others to 0. But wait, is this the case for all n?Wait, let's check n = 1. If n = 1, then the constraint is x₁ ≤ 1/2. The product is (1 - x₁). To minimize this, we need to maximize x₁. So set x₁ = 1/2, then the product is 1 - 1/2 = 1/2. So for n = 1, the minimum is 1/2.For n = 2, we saw that it's also 1/2. For n = 3, same result. Wait, so no matter the value of n ≥ 1, the minimum is 1/2? But that seems counterintuitive. If n is very large, say n = 1000. Then, if we set one variable to 1/2 and the rest to 0, the product is (1 - 1/2) * (1 - 0)^999 = 1/2 * 1 = 1/2. But if we spread out the 1/2 over all variables, say each xᵢ = 1/(2*1000) = 1/2000. Then the product would be (1 - 1/2000)^1000 ≈ e^{-1/2} ≈ 0.6065, which is greater than 1/2. So even with very large n, putting all the sum into one variable gives 1/2, which is lower. But is there a case where distributing the sum more gives a lower product?Wait, maybe not. Let's think about the general case. Suppose we have two different allocations: one where one variable is 1/2 and the rest are 0, giving product 1/2. Another where two variables are 1/4 each and the rest 0, giving product (3/4)^2 ≈ 0.5625. Still higher. If we spread the sum equally among all variables, each xᵢ = 1/(2n). Then the product is (1 - 1/(2n))^n. As n increases, this approaches e^{-1/2} ≈ 0.6065, which is higher than 1/2. So indeed, the minimal product seems to always be 1/2, regardless of n.But wait, hold on. Let me check for n = 4. Suppose we set two variables to 1/4 each and the rest to 0. The product is (3/4)^2 * 1 * 1 = 9/16 ≈ 0.5625. If we set one variable to 1/2 and the rest to 0, the product is 1/2. So 1/2 is still lower. If we set three variables to 1/6 each, sum is 3*(1/6) = 1/2. The product is (5/6)^3 ≈ 0.5787. Still higher than 1/2. So no matter how we distribute the sum, the product is always higher than 1/2, except when we set one variable to 1/2 and others to 0. Therefore, the minimal value should be 1/2 for any n ≥ 1.But wait, let me test with n = 2 again. If n = 2 and sum is 1/2. If we set both variables to 1/4, the product is (3/4)^2 = 9/16 ≈ 0.5625. If we set one variable to 1/2 and the other to 0, the product is 1/2. So indeed, 1/2 is smaller. So this seems consistent. Therefore, the minimal value is 1/2.But wait, is there a mathematical principle that justifies this? The problem is to minimize the product (1 - x₁)(1 - x₂)...(1 - xₙ) given that the sum of xᵢ ≤ 1/2 and each xᵢ ≥ 0. To minimize the product, we need to maximize each (1 - xᵢ) term's deviation from 1, i.e., minimize each (1 - xᵢ). But since each term is between 1/2 and 1, the product is minimized when as many terms as possible are minimized. However, since we can only minimize a term by maximizing xᵢ, which requires using up the sum. So if we concentrate the sum into one variable, we minimize one term to 1 - 1/2 = 1/2, and keep all others at 1. If we spread the sum over multiple variables, each term is closer to 1, so their product is larger. Hence, concentrating the sum into one variable gives the minimal product.This seems similar to the concept that for products, the more uneven the distribution (under a sum constraint), the smaller the product. Which is the opposite of AM-GM, where the product is maximized when variables are equal. So here, to minimize the product, we make variables as unequal as possible.Therefore, the minimum value is 1/2, achieved when one variable is 1/2 and the rest are 0. Therefore, regardless of n, the minimum is 1/2.Final AnswerThe minimum value is boxed{dfrac{1}{2}}.</think>Given non-negative real numbers ( x_1, x_2, ldots, x_n ) that satisfy the inequality ( x_1 + x_2 + cdots + x_n leqslant frac{1}{2} ), we need to find the minimum value of the product ( (1 - x_1)(1 - x_2) cdots (1 - x_n) ).To minimize the product ( (1 - x_1)(1 - x_2) cdots (1 - x_n) ), we need to consider how the values of ( x_i ) affect the product. Each term ( (1 - x_i) ) is a decreasing function of ( x_i ), meaning that increasing ( x_i ) decreases ( (1 - x_i) ). However, the sum of all ( x_i ) is constrained by ( frac{1}{2} ).We analyze the product by considering different distributions of the sum ( frac{1}{2} ) among the variables ( x_i ). For two variables, we found that setting one variable to ( frac{1}{2} ) and the other to 0 results in the product ( frac{1}{2} ), which is smaller than distributing the sum equally. Extending this reasoning to ( n ) variables, we conclude that concentrating the entire sum ( frac{1}{2} ) into one variable (setting it to ( frac{1}{2} )) and setting the remaining variables to 0 will minimize the product.This approach is consistent for any ( n geq 1 ). By setting one ( x_i ) to ( frac{1}{2} ) and the rest to 0, the product becomes ( left(1 - frac{1}{2}right) times 1 times 1 times cdots times 1 = frac{1}{2} ).Thus, the minimum value of the product ( (1 - x_1)(1 - x_2) cdots (1 - x_n) ) is (boxed{dfrac{1}{2}}).,<think>Okay, let's see. The problem is to find the minimum value of the product (1 - x₁)(1 - x₂)...(1 - xₙ) given that the sum of the non-negative real numbers x₁, x₂, ..., xₙ is at most 1/2. Hmm. So we need to minimize this product under the constraint that their sum is ≤ 1/2. Alright.First, I remember that for optimization problems like this, especially with products and sums, inequalities like AM-GM might come into play. But here, it's a product of terms (1 - x_i), and we want to minimize that product. So maybe we need to maximize the individual x_i's? Wait, because if each (1 - x_i) is smaller, the product would be smaller. But since the x_i's are non-negative and their sum is bounded, there's a trade-off here.Let me think. If we want to minimize the product, we need to make as many terms as possible as small as possible. But each term (1 - x_i) can't be negative because x_i is non-negative and up to... Wait, x_i can be up to 1, but since the sum is at most 1/2, each x_i is at most 1/2? Wait, no. If all x_i's are non-negative and their sum is ≤1/2, then each x_i individually can be at most 1/2, but actually, if one x_i is 1/2, the rest must be zero. So, each x_i is in [0, 1/2]. Therefore, each (1 - x_i) is in [1/2, 1]. So the product is going to be between (1/2)^n and 1.But how do we find the minimum? Since the product is minimized when the terms are as small as possible, so when each x_i is as large as possible. But subject to the sum being at most 1/2. So to minimize the product, we need to maximize the x_i's. But since they are variables, how do we distribute the total sum among them to make the product as small as possible?I recall that for products, the product is minimized when the terms are as unequal as possible. Wait, but in some cases, like when you have a fixed sum and want to maximize a product, you use AM-GM which says that the product is maximized when all terms are equal. Conversely, if you want to minimize the product given a fixed sum, you might make the terms as unequal as possible. But in this case, we have the product of (1 - x_i) and the sum of x_i is fixed. So maybe to minimize the product, we need to maximize the sum of x_i? But the sum is bounded above by 1/2. So actually, the minimal product would occur when the sum is exactly 1/2, right? Because if the sum is less than 1/2, then maybe we can increase some x_i's to make the product smaller. So the minimal product occurs when the sum is exactly 1/2.Therefore, we can assume that x₁ + x₂ + ... + xₙ = 1/2. So now, the problem reduces to minimizing the product (1 - x₁)(1 - x₂)...(1 - xₙ) given that x₁ + x₂ + ... + xₙ = 1/2 and each x_i ≥ 0.Now, how to minimize this product. Let's think about the case when n=1 first. If n=1, then x₁ = 1/2, so the product is (1 - 1/2) = 1/2. So the minimum is 1/2. For n=2, we have x₁ + x₂ = 1/2. The product is (1 - x₁)(1 - x₂). To minimize this, we can set x₁ as large as possible and x₂ as small as possible? Wait, but with the sum fixed. For two variables, if the sum is fixed, the product of (1 - x)(1 - y) is minimized when one variable is as large as possible and the other as small as possible. Let's check.Suppose x₁ = 1/2, x₂ = 0. Then the product is (1 - 1/2)(1 - 0) = (1/2)(1) = 1/2. Alternatively, if both x₁ = x₂ = 1/4, then the product is (3/4)(3/4) = 9/16 ≈ 0.5625, which is larger. So indeed, the product is minimized when one variable is as large as possible, and the others as small as possible. So in the case of n=2, the minimum is 1/2.Similarly, for n=3, maybe setting one variable to 1/2 and the others to 0, but wait, sum would be 1/2. Then the product would be (1 - 1/2)(1 - 0)(1 - 0) = 1/2 * 1 * 1 = 1/2. Alternatively, distributing the 1/2 equally among all three variables: each x_i = 1/6. Then the product is (1 - 1/6)^3 = (5/6)^3 ≈ 0.5787, which is again larger. So the minimum is still 1/2.Wait, so for any n, if we set one variable to 1/2 and the rest to 0, then the product is 1/2 * 1^{n-1} = 1/2. But is this the minimum?Wait, let's check n=4. If one variable is 1/2, others 0, product is 1/2. If we split 1/2 equally among two variables: each x_i = 1/4. Then product is (1 - 1/4)^2 * (1 - 0)^2 = (3/4)^2 * 1 * 1 = 9/16 ≈ 0.5625, still higher than 1/2. If we split 1/2 among three variables: each x_i ≈ 1/6, product is (5/6)^3 * 1 ≈ 0.578, again higher. So even with more variables, the minimum seems to be 1/2.But wait, maybe there's another distribution where more variables are non-zero, but the product is lower? Let's see.Suppose for n=2, if we set both x₁ and x₂ to 1/2, but wait, sum is 1, which exceeds 1/2. So that's not allowed. So in order to have sum 1/2, if we have two variables, each can be up to 1/2. But if we set one to 1/2, the other must be 0.Alternatively, let's take n=3. Suppose we set two variables to 1/4 each, and the third to 0. Then sum is 1/4 + 1/4 = 1/2. The product is (1 - 1/4)^2 * (1 - 0) = (3/4)^2 * 1 = 9/16 ≈ 0.5625. If we set one variable to 1/2 and the rest to 0, product is 1/2. So 1/2 is lower. So even with distributing the sum among more variables, the product is higher. Therefore, in order to minimize the product, it's better to set as many variables as possible to 0, and concentrate the sum into one variable.But why is that? Let me think in terms of the function we're trying to minimize. The product is the product of (1 - x_i). If we spread the x_i's out, each term (1 - x_i) is closer to 1, but multiplied together. If we concentrate them into one term, then one term is much smaller, but the rest are 1. So perhaps the product is smaller in that case.Wait, for example, take n=2. If we have (1 - a)(1 - b). If a + b = 1/2. Then if a = 1/2, b=0, the product is (1 - 1/2)(1 - 0) = 1/2. If a = b = 1/4, then (3/4)(3/4) = 9/16 ≈ 0.5625. So 1/2 < 9/16, so indeed, the product is minimized when one variable is as large as possible.Similarly, in general, if we have (1 - x₁)...(1 - xₙ) with sum x_i = S. Then the product is minimized when one x_i is S and the others are 0, since the product becomes (1 - S) * 1^{n - 1} = 1 - S. If instead, we spread S over k variables, the product becomes product_{i=1}^k (1 - s/k) * 1^{n - k} } = (1 - s/k)^k. Then, we can compare (1 - s/k)^k with 1 - S.But when S is fixed, which is smaller? Let's take S = 1/2. Then (1 - 1/(2k))^k vs 1 - 1/2 = 1/2.We can compute (1 - 1/(2k))^k for different k. For k=1: (1 - 1/2)^1 = 1/2. For k=2: (1 - 1/4)^2 = (3/4)^2 = 9/16 ≈ 0.5625. For k=3: (1 - 1/6)^3 ≈ (5/6)^3 ≈ 0.5787. For k=4: (1 - 1/8)^4 ≈ (7/8)^4 ≈ 0.586. As k increases, (1 - 1/(2k))^k approaches e^{-1/2} ≈ 0.6065, which is larger than 1/2. So indeed, the minimal product is achieved when k=1, that is, when all the sum is concentrated in one variable.Therefore, for any n, the minimal product is 1 - 1/2 = 1/2. Wait, but in this problem, n can be any positive integer. But even for n=1, if x₁ = 1/2, then the product is 1 - 1/2 = 1/2, which is the same as for higher n.Wait, but if n is 1, then the sum x₁ ≤ 1/2, so the maximum x₁ is 1/2, so product is 1 - 1/2 = 1/2. If n=2, then to minimize (1 - x₁)(1 - x₂), set x₁=1/2, x₂=0, product is 1/2. Similarly, for any n, set one x_i=1/2 and the rest 0, the product is 1/2. But can we get a smaller product if n is larger?Wait, no. Because even if n is larger, the product is still (1 - 1/2) multiplied by 1's. So it's always 1/2. Therefore, the minimal value is 1/2, regardless of n?But wait, let me check with n=3. Suppose we set two variables to 1/2. Wait, but then the sum would be 1, which exceeds the constraint of 1/2. So we can't do that. So the maximum any single variable can be is 1/2. Therefore, the minimal product is 1/2.But wait, let me think again. Suppose we have n=2. If we set x₁ = 1/2, x₂=0, then the product is 1/2. If instead, we set x₁ = 1/2 - ε, and x₂=ε, then the product becomes (1 - (1/2 - ε))(1 - ε) = (1/2 + ε)(1 - ε) = 1/2(1 - ε) + ε(1 - ε) = 1/2 - ε/2 + ε - ε² = 1/2 + ε/2 - ε². For small ε, this is 1/2 + ε/2, which is larger than 1/2. So the minimal is indeed at ε=0, so x₁=1/2, x₂=0. So the product is minimized when we set as much as possible into one variable.Therefore, regardless of n ≥1, the minimal product is 1/2. But wait, is that possible?Wait, let me test n=3. Suppose we set two variables to 1/4 each, and the third to 0. Then sum is 1/2, product is (1 - 1/4)^2*(1 - 0) = (3/4)^2 = 9/16 ≈ 0.5625. If we set one variable to 1/2, others to 0, product is 1/2. So 1/2 is smaller. So yes, it's better to set one variable to 1/2. Similarly, for n=4, same thing. So for any n ≥1, the minimal product is 1/2.But the problem says "non-negative real numbers", so they can be zero. Therefore, the minimal value is 1/2. So the answer is 1/2.Wait, but wait. Let me check with n=3. Suppose we set two variables to 1/2 each, but that's not allowed because the sum would be 1. But the constraint is sum ≤1/2. So if we set one variable to 1/2, the rest have to be 0. So in that case, product is 1/2.Alternatively, what if we set three variables each to 1/6. Then sum is 3*(1/6)=1/2. The product is (1 - 1/6)^3 = (5/6)^3 ≈ 0.5787, which is larger than 1/2. So the minimal is still 1/2.Therefore, regardless of the number of variables, the minimal value is 1/2. Therefore, the answer is 1/2. So boxed, it's boxed{dfrac{1}{2}}.Wait, but let me confirm with another example. Suppose n=4. If we set two variables to 1/4 each and the other two to 0. Then product is (1 - 1/4)^2 * (1 - 0)^2 = (3/4)^2 ≈ 0.5625. If we set one variable to 1/2, the rest to 0, product is 1/2. So again, 1/2 is smaller. So regardless of n, as long as n ≥1, the minimal product is 1/2.Therefore, the answer is boxed{dfrac{1}{2}}.Final Answerboxed{dfrac{1}{2}}</think>Given non-negative real numbers (x_1, x_2, ldots, x_n) that satisfy the inequality (x_1 + x_2 + cdots + x_n leqslant frac{1}{2}), we need to find the minimum value of the product ((1 - x_1)(1 - x_2) cdots (1 - x_n)).To minimize the product ((1 - x_1)(1 - x_2) cdots (1 - x_n)), we need to consider the constraint (x_1 + x_2 + cdots + x_n leqslant frac{1}{2}). The product is minimized when the terms ((1 - x_i)) are as small as possible. This occurs when the values (x_i) are as large as possible, given the constraint.We can achieve the minimum value by setting one of the (x_i) to the maximum possible value under the constraint, which is (frac{1}{2}), and the remaining (x_j) (where (j neq i)) to 0. This configuration satisfies the constraint (x_1 + x_2 + cdots + x_n = frac{1}{2}).For example, if we set (x_1 = frac{1}{2}) and (x_2 = x_3 = cdots = x_n = 0), the product becomes:[(1 - x_1)(1 - x_2) cdots (1 - x_n) = left(1 - frac{1}{2}right)(1 - 0) cdots (1 - 0) = frac{1}{2} times 1 times cdots times 1 = frac{1}{2}.]Testing other distributions of the sum (frac{1}{2}) among multiple (x_i) values (e.g., distributing equally) results in a product larger than (frac{1}{2}). For instance, distributing (frac{1}{2}) equally among (k) variables results in each (x_i = frac{1}{2k}), leading to a product (left(1 - frac{1}{2k}right)^k), which is always greater than (frac{1}{2}) for (k geq 2).Thus, the minimum value of ((1 - x_1)(1 - x_2) cdots (1 - x_n)) is (boxed{dfrac{1}{2}}).]

🔑:We seek the extremum point, first conduct some experiments:Experiment 1: Take x_{1}=x_{2}=x_{3}=x_{4}=frac{1}{8}, x_{5}=x_{6}=cdots=x_{n}=0, at this point we havebegin{array}{c}x_{1}+x_{2}+cdots+x_{n}=frac{1}{2}, left(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right)=left(frac{7}{8}right)^{4}=frac{2401}{4096} ;end{array}Experiment 2: Take x_{1}=x_{2}=frac{1}{4}, x_{3}=x_{4}=cdots=x_{n}=0, at this point we havebegin{array}{c}x_{1}+x_{2}+cdots+x_{n}=frac{1}{2}, left(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right)=left(frac{3}{4}right)^{2}=frac{9}{16} ;end{array}Experiment 3: Take x_{1}=frac{1}{2}, x_{2}=x_{3}=cdots=x_{n}=0, at this point we havebegin{array}{c}x_{1}+x_{2}+cdots+x_{n}=frac{1}{2} left(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right)=frac{1}{2} .end{array}Comparing the above results, we find that the extremum point of this problem is different from the previous examples, and it is not when all variables x_{1}, x_{2}, cdots, x_{n} are equal that left(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right) achieves its minimum value. For example, in Experiment 1, four values are equal (all equal to left.frac{1}{8}right), and in Experiment 2, two values are equal left(right. both equal to left.frac{1}{4}right), and their corresponding left(1-x_{1}right)left(1-x_{2}right) cdots(1- left.x_{n}right) values are respectivelyfrac{2401}{4096}>frac{1}{2}, frac{9}{16}>frac{1}{2} .In Experiment 3, only x_{1}=frac{1}{2}, the rest are 0, at this point left(1-x_{1}right)left(1-x_{2}right) cdots(1- left.x_{n}right)=frac{1}{2}.We can conjecture that the extremum point of this problem isleft(frac{1}{2}, 0,0, cdots, 0right) .Therefore, our goal is to keep the sum x_{1}+x_{2}+cdots+x_{n} unchanged, and to eliminate x_{2}, x_{3}, cdots, x_{n}.For this, we setbegin{array}{c}x_{i}^{prime}=x_{i}, i=1,2, cdots, n-2, x_{n-1}^{prime}=x_{n-1}+x_{n}, x_{n}^{prime}=0 .end{array}At this point we havebegin{aligned}left(1-x_{n-1}^{prime}right)left(1-x_{n}^{prime}right) & =1-left(x_{n-1}^{prime}+x_{n}^{prime}right)+x_{n-1}^{prime} x_{n}^{prime} & =1-left(x_{n-1}+x_{n}right)+0 . left(1-x_{n-1}right)left(1-x_{n}right) & =1-left(x_{n-1}+x_{n}right)+x_{n-1} x_{n} .end{aligned}Thus,left(1-x_{n-1}^{prime}right)left(1-x_{n}^{prime}right) leqslantleft(1-x_{n-1}right)left(1-x_{n}right),i.e., left(1-x_{1}^{prime}right)left(1-x_{2}^{prime}right) cdotsleft(1-x_{n}^{prime}right) leqslantleft(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right).Further, we letbegin{array}{c}x_{i}^{prime prime}=x_{i}, i=1,2, cdots, n-3 . x_{n-2}^{prime prime}=x_{n-2}+x_{n-1}+x_{n}, x_{n-1}^{prime prime}=0, x_{n}^{prime prime}=0,end{array}From the above, we also haveleft(1-x_{1}^{prime prime}right)left(1-x_{2}^{prime prime}right) cdotsleft(1-x_{n}^{prime prime}right) leqslantleft(1-x_{1}^{prime}right)left(1-x_{2}^{prime}right) cdotsleft(1-x_{n}^{prime}right) .Thus, after at most n-1 transformations, we havebegin{array}{l}x_{1}^{0}=x_{1}+x_{2}+cdots+x_{n} leqslant frac{1}{2}, x_{2}^{0}=x_{3}^{0}=cdots=x_{n}^{0}=0 .end{array}Therefore,left(1-x_{1}^{0}right)left(1-x_{2}^{0}right) cdotsleft(1-x_{n}^{0}right) geqslant frac{1}{2} .That is, the minimum value of left(1-x_{1}right)left(1-x_{2}right) cdotsleft(1-x_{n}right) is frac{1}{2}.As previously noted, this problem is not smoothed when the variables are equal, so the equality of variables is not the essential characteristic of the smoothing transformation.