🧠:[<think>Alright, let's try to tackle this problem step by step. It's a two-part question involving calculus and inequalities. Let me start with part (1).Problem (1): Find the minimum value of h(x) = (x - a)e^x + a for x in [-1, 1].Okay, so to find the minimum value of a function on a closed interval, I remember that we need to check the critical points inside the interval and also the endpoints. The critical points are where the derivative is zero or undefined. Since h(x) is a combination of polynomials and exponentials, which are differentiable everywhere, the critical points will just be where the derivative is zero.First, let's compute the derivative of h(x). h(x) = (x - a)e^x + aTaking the derivative using the product rule:h'(x) = d/dx [ (x - a)e^x ] + d/dx [a]The derivative of a constant term a is zero. For the first term, product rule says derivative of u*v is u'v + uv', where u = (x - a) and v = e^x.So:u' = derivative of (x - a) = 1v' = derivative of e^x = e^xTherefore:h'(x) = (1)(e^x) + (x - a)(e^x) = e^x [1 + (x - a)]Simplify that:h'(x) = e^x (x - a + 1)To find critical points, set h'(x) = 0:e^x (x - a + 1) = 0But e^x is always positive, so the equation reduces to:x - a + 1 = 0Therefore, x = a - 1So the critical point is at x = a - 1. Now, we need to check whether this critical point lies within the interval [-1, 1]. So, depending on the value of a, x = a - 1 might be inside or outside the interval.Therefore, the minimum value of h(x) on [-1, 1] will occur either at x = -1, x = 1, or at x = a - 1 if a - 1 is in [-1, 1].Therefore, we need to evaluate h(x) at these points and compare them.Let's first write expressions for h(-1), h(1), and h(a - 1):1. h(-1) = (-1 - a)e^{-1} + a = [ - (1 + a) ] e^{-1} + a2. h(1) = (1 - a)e^{1} + a = (1 - a)e + a3. If x = a - 1 is in [-1, 1], then h(a - 1) = ( (a - 1) - a ) e^{a - 1} + a = (-1) e^{a - 1} + a = a - e^{a - 1}So, depending on the value of a, the critical point x = a - 1 may or may not be in the interval. Let's analyze the possible cases.Case 1: a - 1 is in [-1, 1]Which implies:-1 ≤ a - 1 ≤ 1Adding 1 to all parts:0 ≤ a ≤ 2Therefore, if a is between 0 and 2, the critical point x = a - 1 is inside the interval. Otherwise, the critical point is outside, and the minimum occurs at one of the endpoints.So, if a is in [0,2], then we need to compare h(-1), h(1), and h(a - 1). If a < 0 or a > 2, then the minimum is at the endpoints.Therefore, let's structure our solution as follows:First, compute h(-1), h(1), and h(a - 1) (when applicable), then determine which is the minimum based on the value of a.But the problem says "If x ∈ [-1,1], find the minimum value of the function h(x)". It doesn't specify a particular a, so the answer will depend on a.Wait, but part (1) is phrased as: "(1) If x ∈ [-1,1], find the minimum value of the function h(x);"Wait, is the function h(x) given as (x - a)e^x + a. So, in part (1), do we need to find the minimum value as a function of a? Or is a a fixed parameter? The problem doesn't specify a value for a in part (1), so I think the answer will have to be expressed in terms of a.Therefore, we need to find the minimum value of h(x) over x in [-1,1], which depends on a. So, depending on the value of a, the minimum could be at different points.Therefore, the answer will be a piecewise function depending on a.So, summarizing:- For a in [0, 2], the critical point x = a - 1 is in [-1,1]. Therefore, the minimum is the minimum among h(-1), h(1), and h(a - 1).Wait, actually, in the interval, the minimum could be at x = a -1, but we need to check whether that critical point is a minimum. Since h'(x) = e^x (x - a + 1). Let's check the second derivative or analyze the sign of h'(x) around x = a -1 to confirm whether it's a minima or maxima.Alternatively, since the critical point is x = a -1, and the derivative changes sign around that point. Let's check:For x < a -1, h'(x) = e^x (x - a +1). Since x - a +1 < 0 (because x < a -1), so h'(x) is negative.For x > a -1, x - a +1 >0, so h'(x) is positive. Therefore, the critical point x = a -1 is a local minimum.Therefore, if the critical point is in the interval, it's a local minimum, so the minimum of h(x) on [-1,1] will be at x = a -1. But we need to compare with the endpoints as well, because sometimes endpoints can be lower.Wait, but in general, even if there's a local minimum, the absolute minimum could be at an endpoint. Wait, but since the function is differentiable and the critical point is a local minimum, if it's inside the interval, then the absolute minimum is either at that critical point or at the endpoints. So we need to evaluate h(a -1), h(-1), h(1), and see which is the smallest.But maybe, depending on a, one of them is the smallest.Alternatively, maybe h(a -1) is always the minimum when a is in [0,2], but not sure. Let's test with specific values of a.Let me take a =1. Then the critical point is x=0. Let's compute h(0) = (0 -1)e^0 +1 = (-1)(1) +1 = 0.h(-1) = (-1 -1)e^{-1} +1 = (-2)/e +1 ≈ -0.7358 +1 ≈ 0.2642h(1) = (1 -1)e^1 +1 =0 +1 =1So here, h(0) =0 is the minimum. So in this case, critical point gives the minimum.Another example: a=0. Then critical point x = -1. Since a=0, x= a -1 = -1, which is the left endpoint. Then h(-1) = (-1 -0)e^{-1} +0 = -e^{-1} ≈ -0.3679. h(1) = (1 -0)e +0 = e ≈2.718. So the minimum is at x=-1, which is the critical point here. But since the critical point is at x=-1, which is the endpoint. So if a=0, the critical point is at the left endpoint. So in this case, the minimum is at x=-1.Similarly, if a=2, the critical point is x=1. Then h(1) = (1 -2)e +2 = (-1)e +2 ≈ -2.718 +2 ≈ -0.718. h(-1) = (-1 -2)/e +2 = (-3)/e +2 ≈ -1.1036 +2 ≈0.8964. So the minimum is at x=1. So in this case, the critical point is at the right endpoint.So for a in [0,2], the critical point is in the interval, and depending on a, it could be a local minimum. But since we saw when a=1, the critical point is at x=0, which gives a lower value than the endpoints. Therefore, when the critical point is in the interior of the interval, it gives the absolute minimum. However, when the critical point is at the endpoints (a=0 or a=2), then the minimum is at that endpoint.Therefore, for a in [0,2], the minimum is at x=a-1. For a outside [0,2], the minimum is at x=-1 if a <0, and at x=1 if a >2.Wait, but wait. Let me check for a=3. If a=3, which is greater than 2. Then critical point x =3 -1 =2, which is outside the interval [-1,1]. Therefore, the minimum is at one of the endpoints. Let's compute h(-1) and h(1).h(-1)= (-1 -3)e^{-1} +3 = (-4)/e +3 ≈ -1.4715 +3 ≈1.5285h(1)= (1 -3)e +3 = (-2)e +3 ≈-5.436 +3 ≈-2.436So here, the minimum is at x=1.Similarly, for a=-1, which is less than 0. Then critical point x=-1 -1= -2, outside the interval. So compute h(-1) and h(1):h(-1)= (-1 -(-1))e^{-1} + (-1)= (0)e^{-1} -1= -1h(1)= (1 -(-1))e + (-1)=2e -1≈5.436 -1≈4.436Thus, the minimum is at x=-1.Therefore, the conclusion is:- If a <0, the minimum is at x=-1: h(-1) = [ - (1 + a) ] e^{-1} + a- If 0 ≤ a ≤2, the minimum is at x=a-1: h(a-1) =a - e^{a -1}- If a >2, the minimum is at x=1: h(1) = (1 -a)e +aTherefore, the minimum value is a piecewise function:Minimum value = { [ - (1 + a) ] e^{-1} + a, if a <0; a - e^{a -1}, if 0 ≤ a ≤2; (1 -a)e +a, if a >2 }But let me check if for a=0, the first and second cases give the same result. For a=0:First case: a <0: h(-1)= [ - (1 +0) ] e^{-1} +0= -1/e≈-0.3679Second case: 0 ≤a ≤2: h(a -1)=0 - e^{-1}≈-0.3679. So same result. Similarly, for a=2:Second case: h(1)=2 - e^{1}≈2 -2.718≈-0.718Third case: (1 -2)e +2= -e +2≈-2.718 +2≈-0.718. So same result. Therefore, the piecewise function is continuous at the boundaries.Hence, the answer to part (1) is this piecewise function.But the problem says "find the minimum value of the function h(x)". So I need to present it as such. Maybe we can write it more compactly, but I think writing the three cases is acceptable.Wait, but let's check if the expressions can be simplified.For a <0: h(-1) = [ - (1 + a) ] e^{-1} + a = - (1 + a)/e + a = a(1 - 1/e ) - 1/eSimilarly, for a >2: h(1) = (1 -a)e +a = a(1 - e) + eBut maybe they prefer it written as it is. Let me check with a=1, which is in [0,2], the expression is a - e^{a -1}, which is 1 - e^{0}=1 -1=0, which matches our previous calculation. Correct.Therefore, part (1) is answered with the piecewise function.Problem (2): When a=3, if for all x₁ ∈ [-1,1], there exists an x₂ ∈ [1,2] such that h(x₁) ≥ x₂² - 2b x₂ - a e + e +15/2 holds, find the range of b.Alright, this part is more complicated. Let's parse the problem.Given a=3, so h(x) becomes (x -3)e^x +3.We need to ensure that for all x₁ in [-1,1], there exists an x₂ in [1,2] such that h(x₁) ≥ [x₂² - 2b x₂ - 3e + e +15/2].Simplify the right-hand side: x₂² - 2b x₂ -3e +e +15/2 = x₂² -2b x₂ -2e +15/2.So the inequality becomes h(x₁) ≥ x₂² -2b x₂ -2e +15/2.We need to find the range of b such that for every x₁ in [-1,1], we can find some x₂ in [1,2] making that inequality true.In other words, for each x₁ in [-1,1], the value h(x₁) must be greater than or equal to the minimum value (over x₂ ∈ [1,2]) of x₂² -2b x₂ -2e +15/2. Wait, no. Because the problem states "for all x₁, exists x₂ such that...". So for each x₁, there exists an x₂ (which can depend on x₁) such that the inequality holds.But actually, the way to approach this is that the inequality must hold for each x₁, with some x₂ (possibly depending on x₁). Therefore, the condition is equivalent to: for every x₁ ∈ [-1,1], the value h(x₁) is ≥ the minimum of (x₂² -2b x₂ -2e +15/2) over x₂ ∈ [1,2]. Because if h(x₁) is greater than or equal to the minimal value of the right-hand side, then there exists at least one x₂ (the one that achieves the minimum) such that the inequality holds. Conversely, if h(x₁) is less than the minimal value, then no x₂ would work.Wait, actually, that's a good point. Let me think.Suppose that for some x₁, the minimal value of the right-hand side (RHS) over x₂ ∈ [1,2] is m. Then if h(x₁) ≥ m, there exists at least one x₂ (the one that gives m) such that h(x₁) ≥ RHS. Conversely, if h(x₁) < m, then for all x₂, RHS ≥ m (since m is the minimum), so h(x₁) < m ≤ RHS, which would imply h(x₁) < RHS for all x₂, which would mean no such x₂ exists. Therefore, the condition is equivalent to h(x₁) ≥ m for all x₁ ∈ [-1,1], where m is the minimum of RHS over x₂ ∈ [1,2].Therefore, the problem reduces to: Find the range of b such that the minimum of h(x₁) over x₁ ∈ [-1,1] is ≥ the minimum of (x₂² -2b x₂ -2e +15/2) over x₂ ∈ [1,2].Wait, but this is only true if the minimum of RHS is achieved. So actually, the original condition is equivalent to: min_{x₁ ∈ [-1,1]} h(x₁) ≥ min_{x₂ ∈ [1,2]} [x₂² -2b x₂ -2e +15/2]But let me verify that. Suppose that for every x₁, there exists x₂ such that h(x₁) ≥ RHS(x₂). Then, in particular, for each x₁, h(x₁) ≥ min_{x₂} RHS(x₂). Therefore, the minimal h(x₁) must be ≥ min RHS(x₂). Conversely, if min h(x₁) ≥ min RHS(x₂), then for every x₁, h(x₁) ≥ min RHS, so there exists x₂ (the one that gives min RHS) such that h(x₁) ≥ RHS(x₂). Therefore, yes, the condition is equivalent to min h(x₁) ≥ min RHS(x₂).Therefore, we can compute the minimum of h(x₁) over [-1,1] when a=3, then compute the minimum of the quadratic in x₂ over [1,2], and set the former ≥ the latter, then solve for b.So let's first compute the minimum of h(x₁) when a=3.From part (1), since a=3, which is greater than 2, the minimum of h(x) on [-1,1] is at x=1, as per our previous analysis. Let's confirm.h(x) = (x -3)e^x +3.From part (1), for a >2, the minimum is at x=1. Let's compute h(1):h(1) = (1 -3)e +3 = (-2)e +3 ≈ -5.436 +3 ≈ -2.436.But let's also check h(-1):h(-1) = (-1 -3)e^{-1} +3 = (-4)/e +3 ≈ -1.4715 +3 ≈1.5285.h(1) is indeed lower. Also, critical point for a=3 is x=3 -1=2, which is outside the interval. Therefore, the minimum is at x=1, so min h(x₁) = -2e +3.Wait, let's compute exactly:h(1) = (1 -3)e +3 = (-2)e +3 = 3 - 2e.So the minimum value is 3 - 2e.Now, we need to compute the minimum of the quadratic function in x₂: f(x₂) = x₂² -2b x₂ -2e +15/2.Quadratic in x₂: f(x₂) = x₂² -2b x₂ + ( -2e +15/2 )This is a quadratic function opening upwards (since coefficient of x₂² is 1 >0), so its minimum is at the vertex. The vertex is at x = -B/(2A) = (2b)/2 = b.So the vertex is at x₂ = b. However, the domain of x₂ is [1,2]. Therefore, the minimum of f(x₂) over [1,2] is:- If b ≤1, then the minimum is at x₂=1.- If 1 < b <2, then the minimum is at x₂ =b.- If b ≥2, then the minimum is at x₂=2.Therefore, the minimum value is:min f(x₂) = { f(1), if b ≤1; f(b), if 1 <b <2; f(2), if b ≥2 }Compute each case:1. If b ≤1: f(1) =1 -2b(1) -2e +15/2 =1 -2b -2e +7.5=8.5 -2b -2e2. If 1 <b <2: f(b)=b² -2b*b -2e +15/2 = -b² -2e +15/23. If b ≥2: f(2)=4 -4b -2e +15/2=4 +7.5 -4b -2e=11.5 -4b -2eSo, the minimum of f(x₂) is:min f = 8.5 -2b -2e (for b ≤1), -b² -2e +7.5 (for 1 <b <2), 11.5 -4b -2e (for b ≥2)We need min h(x₁) ≥ min f(x₂)We have min h(x₁) =3 -2e.So:3 -2e ≥ min f(x₂)Therefore, depending on the value of b, we have different expressions for min f(x₂). Let's consider each interval of b.Case 1: b ≤1Then, min f(x₂) =8.5 -2b -2eSo inequality becomes:3 -2e ≥8.5 -2b -2eSubtract -2e from both sides:3 ≥8.5 -2bThen:-2b ≤3 -8.5= -5.5Multiply both sides by (-1) (inequality flips):2b ≥5.5So:b ≥5.5/2=2.75But in this case, we assumed b ≤1. However, b ≥2.75 and b ≤1 is impossible. Therefore, in this case, there are no solutions.Case 2:1 <b <2Then, min f(x₂)= -b² -2e +7.5So inequality becomes:3 -2e ≥ -b² -2e +7.5Add 2e to both sides:3 ≥ -b² +7.5Then:-b² +7.5 ≤3Subtract 7.5:-b² ≤ -4.5Multiply by (-1), inequality flips:b² ≥4.5Therefore:b ≥ sqrt(4.5) or b ≤ -sqrt(4.5)But since in this case, 1 <b <2, we have to check if sqrt(4.5) is less than 2.sqrt(4.5)=sqrt(9/2)= (3)/sqrt(2)≈2.1213Which is greater than 2, so within the interval 1 <b <2, b² ≥4.5 would imply b ≥sqrt(4.5)≈2.1213, but since b <2, there's no solution in this interval as well.Case 3: b ≥2min f(x₂)=11.5 -4b -2eInequality:3 -2e ≥11.5 -4b -2eAdd 2e to both sides:3 ≥11.5 -4bSubtract 11.5:-8.5 ≥ -4bMultiply by (-1), flip inequality:8.5 ≤4bTherefore:b ≥8.5/4=2.125Since in this case, b ≥2, the inequality reduces to b ≥2.125. Therefore, in this case, the solutions are b ≥2.125.But wait, the problem states "find the range of b" such that for all x₁ ∈ [-1,1], there exists x₂ ∈ [1,2] such that h(x₁) ≥ RHS.But according to the above, in case 3 (b ≥2), the inequality 3 -2e ≥11.5 -4b -2e holds when b ≥2.125. But 2.125 is 17/8 or 2.125. So b ≥17/8.But 17/8 is 2.125. Therefore, combining all cases:Case1: no solutionCase2: no solutionCase3: b≥17/8Hence, the range of b is [17/8, ∞)But let's check with b=2.125.Compute RHS min f(x₂):If b=17/8=2.125, then since b ≥2, min f(x₂)=11.5 -4*(17/8) -2e=11.5 -8.5 -2e=3 -2e, which is equal to the left-hand side. Therefore, equality holds.If b>17/8, then min f(x₂)=11.5 -4b -2e. Since 11.5 -4b -2e <3 -2e if 11.5 -4b <3 => -4b < -8.5 => b >8.5/4=2.125=17/8. Wait, wait. Wait, the inequality is 3 -2e ≥11.5 -4b -2e. So that simplifies to 3 ≥11.5 -4b => -8.5 ≥-4b => multiplying by (-1):8.5 ≤4b => b≥8.5/4=2.125=17/8.Therefore, for b≥17/8, 11.5 -4b -2e ≤3 -2e, which is exactly the required condition.But wait, if b increases beyond 17/8, 11.5 -4b -2e decreases, so the inequality 3 -2e ≥11.5 -4b -2e is maintained for all b≥17/8.Therefore, the range of b is b≥17/8.But let me confirm with a specific value. Let's take b=3. Then min f(x₂)=11.5 -4*3 -2e=11.5 -12 -2e≈-0.5 -5.436≈-5.936. Then min h(x₁)=3 -2e≈3 -5.436≈-2.436. So -2.436≥-5.936, which is true.If b=17/8=2.125, min f(x₂)=11.5 -4*(17/8)=11.5 -8.5=3. Then 3 -2e≈-2.436. Wait, 3 -2e≈3 -5.436≈-2.436. So 3 -2e≥3 -2e, which is equality. So equality holds.If b is exactly 17/8, then min f(x₂)=3 -2e, which is equal to the left-hand side, so the inequality holds.Therefore, the range is b≥17/8, which is 17/8=2.125=17/8=2.125.Expressed as a fraction, 17/8 is 2 and 1/8.Therefore, the range of b is [17/8, ∞).But let's check if there are any other intervals. For example, if b is between 1 and 2, even though in case 2 we saw no solution, but maybe I made a mistake.Wait, let's take b=2. Then since b=2 is in the case b≥2, so min f(x₂)=11.5 -4*2 -2e=11.5 -8 -2e=3.5 -2e≈3.5 -5.436≈-1.936. Then min h(x₁)=3 -2e≈-2.436. So the inequality is -2.436≥-1.936, which is false. Therefore, when b=2, the inequality does not hold. Therefore, our previous conclusion that b must be ≥17/8≈2.125 is correct.Therefore, the answer is b≥17/8, which is boxed{[dfrac{17}{8}, +infty)}.Final Answer(1) The minimum value of ( h(x) ) is (boxed{ begin{cases} -dfrac{1 + a}{e} + a & text{if } a < 0, a - e^{a - 1} & text{if } 0 leq a leq 2, (1 - a)e + a & text{if } a > 2.end{cases} })(2) The range of ( b ) is (boxed{[dfrac{17}{8}, +infty)}).</think>Given the function ( h(x) = (x - a)e^x + a ):(1) To find the minimum value of ( h(x) ) for ( x in [-1, 1] ):1. Compute the derivative: ( h'(x) = e^x (x - a + 1) ).2. Find the critical point: ( x = a - 1 ).3. Determine if the critical point ( x = a - 1 ) lies within the interval ([-1, 1]): - If ( 0 leq a leq 2 ), the critical point ( x = a - 1 ) is within the interval. - If ( a < 0 ), the minimum occurs at ( x = -1 ). - If ( a > 2 ), the minimum occurs at ( x = 1 ).Evaluating ( h(x) ) at the critical point and endpoints:- For ( a < 0 ): ( h(-1) = -frac{1 + a}{e} + a )- For ( 0 leq a leq 2 ): ( h(a - 1) = a - e^{a - 1} )- For ( a > 2 ): ( h(1) = (1 - a)e + a )The minimum value of ( h(x) ) is:[boxed{ begin{cases} -dfrac{1 + a}{e} + a & text{if } a < 0, a - e^{a - 1} & text{if } 0 leq a leq 2, (1 - a)e + a & text{if } a > 2.end{cases} }](2) When ( a = 3 ), we need to find the range of ( b ) such that for all ( x_1 in [-1, 1] ), there exists ( x_2 in [1, 2] ) such that ( h(x_1) geq x_2^2 - 2bx_2 - 3e + e + frac{15}{2} ).1. The minimum value of ( h(x_1) ) when ( a = 3 ) is at ( x = 1 ): ( h(1) = 3 - 2e ).2. The right-hand side expression simplifies to ( x_2^2 - 2bx_2 - 2e + frac{15}{2} ).3. Find the minimum value of the quadratic function ( f(x_2) = x_2^2 - 2bx_2 - 2e + frac{15}{2} ) over ( x_2 in [1, 2] ): - For ( b leq 1 ): minimum at ( x_2 = 1 ) - For ( 1 < b < 2 ): minimum at ( x_2 = b ) - For ( b geq 2 ): minimum at ( x_2 = 2 )4. Compare the minimum values with ( 3 - 2e ): - For ( b geq 2 ), the minimum value is ( 11.5 - 4b - 2e ). Solving ( 3 - 2e geq 11.5 - 4b - 2e ) gives ( b geq frac{17}{8} ).The range of ( b ) is:[boxed{[dfrac{17}{8}, +infty)}],<think>Okay, let's tackle this problem step by step. First, part (1) asks for the minimum value of the function ( h(x) = (x - a)e^x + a ) when ( x in [-1, 1] ). Then part (2) is about finding the range of ( b ) given some conditions when ( a = 3 ). Let's start with part (1).Problem (1): Find the minimum of ( h(x) ) on [-1,1].To find the minimum value of a function on a closed interval, I remember that we need to check critical points inside the interval and also the endpoints. So first, I need to find the derivative of ( h(x) ) to locate critical points.Given ( h(x) = (x - a)e^x + a ). Let's compute ( h'(x) ).Using the product rule for differentiation on the term ( (x - a)e^x ):The derivative of ( (x - a) ) is 1, and the derivative of ( e^x ) is ( e^x ). So,( h'(x) = (1)e^x + (x - a)e^x = e^x (1 + x - a) ).Simplify that:( h'(x) = e^x (x - a + 1) ).To find critical points, set ( h'(x) = 0 ):( e^x (x - a + 1) = 0 ).But ( e^x ) is never zero for any real x. So the critical point occurs when:( x - a + 1 = 0 ) ⇒ ( x = a - 1 ).Therefore, the critical point is at ( x = a - 1 ). However, this critical point is only relevant if it lies within the interval [-1, 1]. So we need to check whether ( a - 1 ) is between -1 and 1.So:Case 1: If ( a - 1 in [-1, 1] ), i.e., if ( a in [0, 2] ), then the critical point is within the interval. Then the minimum could be at x = a - 1 or at the endpoints x = -1 or x = 1.Case 2: If ( a - 1 < -1 ), i.e., ( a < 0 ), then the critical point is to the left of the interval, so the minimum would be at one of the endpoints.Case 3: If ( a - 1 > 1 ), i.e., ( a > 2 ), then the critical point is to the right of the interval, so again the minimum is at the endpoints.Therefore, depending on the value of ( a ), the location of the critical point changes, and we have different scenarios.Now, let's analyze each case.Case 1: ( a in [0, 2] ).In this case, the critical point ( x = a - 1 ) is in [-1, 1]. Therefore, the minimum could be at x = a - 1, x = -1, or x = 1. So we need to evaluate ( h(x) ) at all three points and compare.First, compute ( h(a - 1) ):( h(a - 1) = ( (a - 1) - a ) e^{a - 1} + a = (-1)e^{a - 1} + a = a - e^{a - 1} ).Then compute ( h(-1) ):( h(-1) = (-1 - a)e^{-1} + a = (- (1 + a))e^{-1} + a ).Similarly, ( h(1) = (1 - a)e^{1} + a ).So we need to compare these three values to find the minimum.But since we need the minimum value, we have to check which of these is the smallest.However, this seems like it might be complicated because it depends on the value of ( a ). So perhaps we need to analyze further.Alternatively, maybe we can consider the behavior of the function. Since ( h'(x) = e^x (x - a + 1) ), and ( e^x > 0 ), the sign of ( h'(x) ) depends on ( x - a + 1 ).So when ( x < a - 1 ), ( h'(x) < 0 ), and when ( x > a - 1 ), ( h'(x) > 0 ). Therefore, the function is decreasing before ( x = a - 1 ) and increasing after. Therefore, the critical point at ( x = a - 1 ) is a local minimum.Therefore, if ( x = a - 1 ) is in the interval, that's the point where the function attains its minimum. Otherwise, the minimum is at the left or right endpoint, depending on the critical point's position.Hence, summarizing:- If ( a in [0, 2] ), the minimum is at ( x = a - 1 ), value is ( a - e^{a - 1} ).- If ( a < 0 ), critical point is at x < -1, so the function is increasing on [-1,1], hence minimum at x = -1.- If ( a > 2 ), critical point is at x > 1, so the function is decreasing on [-1,1], hence minimum at x = 1.Wait, hold on. Wait, when ( a - 1 < -1 ), which means a < 0, then the critical point is at x = a - 1 < -1. So on the interval [-1,1], the derivative h'(x) = e^x (x - a + 1). Since x ∈ [-1,1], and a < 0, then x - a + 1 = x + (1 - a). Since a < 0, 1 - a > 1, so x + (1 - a) > x + 1. Since x ≥ -1, x + 1 ≥ 0, so x + (1 - a) ≥ 0 + (1 - a) > 1 - a. But a < 0, so 1 - a > 1. Therefore, x - a + 1 > 1 for all x in [-1,1], so h'(x) > 0, which means the function is increasing on the entire interval. Therefore, the minimum is at x = -1.Similarly, when a > 2, critical point x = a - 1 > 1. Then on [-1,1], x - a + 1 = x - (a - 1). Since a - 1 > 1, then x - (a -1 ) ≤ 1 - (a -1 ) = 2 - a. But a > 2, so 2 - a < 0. Therefore, x - a + 1 < 0 for all x ∈ [-1,1]. Therefore, h'(x) < 0 on entire interval, so the function is decreasing, and the minimum is at x = 1.Therefore, consolidating:- If a < 0: minimum at x = -1, value h(-1) = (-1 - a)e^{-1} + a- If 0 ≤ a ≤ 2: minimum at x = a - 1, value h(a -1 ) = a - e^{a -1 }- If a > 2: minimum at x = 1, value h(1) = (1 - a)e + aTherefore, the minimum value is a piecewise function depending on a.But the problem says "If x ∈ [-1,1], find the minimum value of the function h(x)". Wait, the problem didn't specify a range for a; it just says "given the function h(x) = ...". So we need to express the minimum value in terms of a. So the answer would be expressed piecewise.So part (1) is answered by:Minimum value =- ( (-1 - a)e^{-1} + a ) when ( a < 0 )- ( a - e^{a -1} ) when ( 0 leq a leq 2 )- ( (1 - a)e + a ) when ( a > 2 )But let's check if this is correct by verifying with specific a values.For example, take a = 1 (which is in [0,2]). Then the minimum is at x = 0. Compute h(0) = (0 -1)e^0 +1 = (-1)(1) +1 = 0. Alternatively, using the formula: a - e^{a -1} = 1 - e^{0} = 1 -1 = 0. Correct.Take a = -1 (a <0). Then h(-1) = (-1 - (-1))e^{-1} + (-1) = (0)e^{-1} -1 = -1. Wait, but according to our formula: (-1 - a)e^{-1} + a, when a = -1: (-1 - (-1))e^{-1} + (-1) = 0 + (-1) = -1. Which is h(-1). But also, since a <0, the minimum is at x=-1, which is h(-1) = -1. But let's compute h(x) at x = -1 and x =1. For a = -1, h(x) = (x - (-1))e^x + (-1) = (x +1)e^x -1. At x=-1: (0)e^{-1} -1 = -1. At x=1: (2)e^1 -1 ≈ 2e -1 ≈ 5.436. So minimum is indeed at x=-1, value -1.Similarly, take a =3 (a >2). Then minimum at x=1: h(1) = (1 -3)e +3 = (-2)e +3 ≈ -5.436 +3 ≈ -2.436. Let's check derivative: h'(x) = e^x(x -3 +1) = e^x(x -2). So on [-1,1], x -2 is always negative (since x ≤1 <2), so h'(x) <0, so function decreasing, so minimum at x=1. Correct.Thus, part (1) seems resolved.Problem (2): When a=3, if for all x1 ∈ [-1,1], there exists x2 ∈ [1,2] such that h(x1) ≥ x2² -2b x2 - a e + e +15/2. Find the range of b.First, let's substitute a=3 into h(x). So h(x) = (x -3)e^x +3.Given that for every x1 in [-1,1], there exists an x2 in [1,2] such that h(x1) ≥ x2² -2b x2 -3e +e +15/2. Simplify the right-hand side:-3e + e = -2e, so RHS is x2² -2b x2 -2e +15/2.Therefore, the inequality is h(x1) ≥ x2² -2b x2 -2e +15/2.So for each x1 ∈ [-1,1], we can find some x2 ∈ [1,2] so that this inequality holds. We need to find the range of b such that this condition is satisfied.First, let's think about what this means. For every x1 in [-1,1], the value h(x1) must be greater than or equal to the minimum value of the function x2² -2b x2 -2e +15/2 over x2 ∈ [1,2]. Wait, no. Because it's saying that there exists an x2 such that h(x1) ≥ ... So for each x1, there exists at least one x2 in [1,2] where the inequality holds. Therefore, for each x1, the value h(x1) must be greater than or equal to the minimum of x2² -2b x2 -2e +15/2 over x2 ∈ [1,2], but actually, since it's "there exists", it's sufficient that h(x1) is ≥ to the minimal value, but I need to think carefully.Wait, more precisely: For each x1 ∈ [-1,1], the value h(x1) must be at least as large as the minimal value of x2² -2b x2 -2e +15/2 over x2 ∈ [1,2], because if h(x1) is greater than or equal to the minimal value, then there exists an x2 (the one that attains the minimum) such that h(x1) ≥ that value. Conversely, if h(x1) is less than the minimal value of RHS over x2 ∈ [1,2], then there is no x2 that satisfies the inequality. Therefore, the condition is equivalent to:For all x1 ∈ [-1,1], h(x1) ≥ min_{x2 ∈ [1,2]} [x2² -2b x2 -2e +15/2].Therefore, the required condition is that the minimum of h(x1) over x1 ∈ [-1,1] is greater than or equal to the minimum of the quadratic function in x2 over x2 ∈ [1,2].Wait, let's verify:If for all x1, there exists x2 such that h(x1) ≥ f(x2), then it is necessary that h(x1) is ≥ the minimum of f(x2) over x2. Because if h(x1) is less than the minimum of f(x2), then even the smallest f(x2) is larger than h(x1), so no x2 would satisfy h(x1) ≥ f(x2). Conversely, if h(x1) ≥ min f(x2), then choosing x2 that gives the minimum f(x2), we have h(x1) ≥ f(x2). So indeed, the condition is equivalent to h(x1) ≥ min_{x2 ∈ [1,2]} f(x2) for all x1 ∈ [-1,1], which in turn is equivalent to min_{x1} h(x1) ≥ min_{x2} f(x2).Therefore, the problem reduces to:Find all b such that min_{x1 ∈ [-1,1]} h(x1) ≥ min_{x2 ∈ [1,2]} [x2² -2b x2 -2e +15/2].So first, compute min h(x1) when a=3. From part (1), when a=3 (>2), the minimum of h(x) on [-1,1] is at x=1, which is h(1) = (1 -3)e +3 = (-2)e +3 ≈ -5.436 +3 ≈ -2.436.So min h(x1) = -2e +3.Then, compute min_{x2 ∈ [1,2]} [x2² -2b x2 -2e +15/2].Let’s denote the quadratic function as f(x2) = x2² -2b x2 -2e +15/2.To find its minimum on [1,2], first, since it's a quadratic function in x2, we can find its vertex. The quadratic is f(x) = x² -2b x + C, where C = -2e +15/2.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). Here, a=1, so vertex at x = (2b)/(2*1) = b. Wait, wait: general form ax² + bx + c, vertex at x = -b/(2a). Wait, the function is f(x) = x² -2b x + C. So here, a=1, linear term coefficient is -2b, so vertex at x = -(-2b)/(2*1) = b.Therefore, the vertex is at x = b.Now, the interval [1,2]. So depending on where the vertex x = b is relative to [1,2], the minimum occurs either at the vertex (if it's inside the interval) or at the endpoints.So:Case 1: If the vertex is inside [1,2], i.e., if b ∈ [1,2], then the minimum is at x = b.Case 2: If b < 1, then the quadratic is decreasing on [1,2], so minimum at x=2.Case 3: If b > 2, then quadratic is increasing on [1,2], so minimum at x=1.Therefore, compute f(x2) accordingly:Case 1: b ∈ [1,2], minimum at x = b: f(b) = b² -2b*b -2e +15/2 = -b² -2e +15/2.Case 2: b <1, minimum at x=2: f(2) = 4 -4b -2e +15/2 = (4 + 7.5) -4b -2e = 11.5 -4b -2e.Case 3: b >2, minimum at x=1: f(1) = 1 -2b -2e +15/2 = 1 +7.5 -2b -2e = 8.5 -2b -2e.Therefore, the minimum of f(x2) is:- If b ∈ [1,2]: -b² -2e +15/2- If b <1: 11.5 -4b -2e- If b >2: 8.5 -2b -2eNow, according to the problem, we need:min h(x1) ≥ min f(x2)We have min h(x1) = -2e +3.So:-2e +3 ≥ [minimum of f(x2) based on b]Therefore, set:-2e +3 ≥ [case expression above]So let's handle each case.Case 1: b ∈ [1,2].Here, the minimum f(x2) is -b² -2e +15/2. So the inequality becomes:-2e +3 ≥ -b² -2e +15/2Simplify:Add 2e to both sides:3 ≥ -b² +15/2Subtract 15/2:3 -15/2 ≥ -b²Convert 3 to 6/2:6/2 -15/2 = -9/2 ≥ -b²Multiply both sides by -1 (inequality reverses):9/2 ≤ b²Thus,b² ≥ 9/2 ⇒ |b| ≥ 3/√2 ≈ 2.121But in this case, b ∈ [1,2], so |b| is between 1 and 2, but 3/√2 ≈2.121 >2, which is outside the interval [1,2]. Therefore, there is no solution in this case.Case 2: b <1.Minimum f(x2) =11.5 -4b -2e.So inequality:-2e +3 ≥11.5 -4b -2eSimplify: cancel -2e from both sides:3 ≥11.5 -4bSubtract 11.5:3 -11.5 ≥ -4b ⇒ -8.5 ≥ -4bMultiply by -1 (reverse inequality):8.5 ≤4b ⇒ b ≥8.5 /4 ⇒ b≥ 2.125But in this case, we assumed b <1. Therefore, 2.125 ≥ b and b <1, which is impossible. So no solution in this case.Case3: b >2.Minimum f(x2)=8.5 -2b -2e.Thus, inequality:-2e +3 ≥8.5 -2b -2eCancel -2e:3 ≥8.5 -2bSubtract8.5:3 -8.5 ≥-2b ⇒ -5.5 ≥ -2bMultiply by -1:5.5 ≤2b ⇒ b≥5.5/2=2.75But in this case, b>2. So combining with b>2, the solutions are b≥2.75.But we need to check if there are solutions in this case. So when b≥2.75, the inequality holds.However, but wait, in case3, when b>2, the minimum of f(x2) is at x=1: f(1)=8.5 -2b -2e. So we need:-2e +3 ≥8.5 -2b -2eWhich gives 3 ≥8.5 -2b ⇒ -5.5 ≥ -2b ⇒ b≥2.75.Therefore, in this case, if b≥2.75, then the inequality holds. Therefore, the range of b is b≥2.75, i.e., b∈[2.75, ∞).But wait, the original problem says "find the range of b", so we need to check if there are any other possibilities.But according to the three cases:- In case1 (b∈[1,2]): no solution.- In case2 (b<1): no solution.- In case3 (b>2): solutions when b≥2.75.But wait, in case3, when b>2, and b≥2.75, so b≥2.75.But we need to check if b can also be in [2,2.75) but that would be in case1 or case3? Wait, when b is in [2,2.75), but in our case3, we considered b>2, but perhaps the case division was b>2 as case3, but in reality, when b is in [2,2.75), it's still in case3 because b>2, but the inequality is only satisfied when b≥2.75. So in conclusion, the only solutions are b≥2.75.But wait, wait, let's verify with b=3, which is greater than 2.75. Let's compute f(x2)'s minimum.For b=3, which is in case3 (b>2), the minimum is at x=1: f(1)=8.5 -2*3 -2e=8.5 -6 -2e=2.5 -2e≈2.5 -5.436≈-2.936.Then, min h(x1)= -2e +3≈-5.436 +3≈-2.436.So -2.436 ≥ -2.936, which is true. So yes, works.For b=2.75, compute the RHS minimum:In case3, f(x2)=8.5 -2*2.75 -2e=8.5 -5.5 -2e=3 -2e≈3 -5.436≈-2.436, which is equal to the min h(x1). So equality holds at b=2.75.If we take b slightly less than 2.75, say b=2.7:f(x2)=8.5 -2*2.7 -2e=8.5 -5.4 -2e=3.1 -5.436≈-2.336. Then min h(x1)≈-2.436, so -2.436 ≥-2.336? No, because -2.436 is less than -2.336. So inequality does not hold. Therefore, only when b≥2.75 does the inequality hold.Therefore, the range of b is [2.75, ∞).But 2.75 is equal to 11/4, which is 2.75. However, in fractional terms, 2.75 is 11/4. Wait, 5.5 divided by 2 is 2.75, which is 11/4.Wait, but in the computation above, when we had:-2e +3 ≥8.5 -2b -2eWhich led to:3 ≥8.5 -2b ⇒ -5.5 ≥ -2b ⇒ 5.5 ≤2b ⇒ b≥5.5/2=2.75.So 5.5/2 is 11/4=2.75. So the exact value is 11/4.Therefore, the range of b is [11/4, ∞).But let's check if 11/4 is correct.Wait, 5.5 divided by 2 is indeed 2.75, which is 11/4. So yes, 11/4.But the problem says "find the range of b". So the answer is b≥11/4, which is [11/4, ∞).But let's check again the steps to be sure.We needed that for all x1 ∈ [-1,1], h(x1) ≥ x2² -2b x2 -2e +15/2 for some x2 ∈ [1,2].This reduces to min_{x1} h(x1) ≥ min_{x2} [x2² -2b x2 -2e +15/2].We found that min h(x1) when a=3 is h(1)= -2e +3.Then, min over x2 of the quadratic is:For b >2: min f(x2)=8.5 -2b -2e.Setting -2e +3 ≥8.5 -2b -2e ⇒ 3 ≥8.5 -2b ⇒ -5.5 ≥ -2b ⇒ 5.5 ≤2b ⇒ b≥5.5/2=2.75=11/4.Thus, the answer is b≥11/4, which is 11/4 ≤b<∞.Therefore, the range of b is [11/4, ∞).But the problem might expect the answer in fractions rather than decimals. So 11/4 is 2.75.Final Answer(1) The minimum value of ( h(x) ) is (boxed{ begin{cases} (-1 - a)e^{-1} + a & text{if } a < 0, a - e^{a - 1} & text{if } 0 leq a leq 2, (1 - a)e + a & text{if } a > 2.end{cases} })(2) The range of ( b ) is (boxed{[dfrac{11}{4}, +infty)}).</think>To find the minimum value of the function ( h(x) = (x - a)e^x + a ) on the interval ( x in [-1, 1] ):1. Critical Points Analysis: - The derivative ( h'(x) = e^x (x - a + 1) ). - Critical point at ( x = a - 1 ). - Depending on the value of ( a ): - If ( a in [0, 2] ), the minimum is at ( x = a - 1 ) with value ( a - e^{a - 1} ). - If ( a < 0 ), the minimum is at ( x = -1 ) with value ( (-1 - a)e^{-1} + a ). - If ( a > 2 ), the minimum is at ( x = 1 ) with value ( (1 - a)e + a ).2. When ( a = 3 ): - The minimum value of ( h(x) ) on ( [-1, 1] ) is at ( x = 1 ) with value ( -2e + 3 ). - For the inequality ( h(x_1) geq x_2^2 - 2bx_2 - 3e + e + frac{15}{2} ) to hold for all ( x_1 in [-1, 1] ) and some ( x_2 in [1, 2] ): - The quadratic function ( x_2^2 - 2bx_2 - 2e + frac{15}{2} ) is analyzed. - The minimum value of the quadratic function on ( [1, 2] ) is found to be ( 8.5 - 2b - 2e ) when ( b > 2 ). - Solving ( -2e + 3 geq 8.5 - 2b - 2e ) gives ( b geq frac{11}{4} ). Final Answer(1) The minimum value of ( h(x) ) is (boxed{ begin{cases} (-1 - a)e^{-1} + a & text{if } a < 0, a - e^{a - 1} & text{if } 0 leq a leq 2, (1 - a)e + a & text{if } a > 2.end{cases} })(2) The range of ( b ) is (boxed{[dfrac{11}{4}, +infty)}).]

🔑:Solution:(1) The derivative of h is h'(x)=(x-a+1)e^{x}. Setting h'(x)=0 gives x=a-1.When a-1 leqslant -1, i.e., a leqslant 0, on [-1,1], h'(x) geqslant 0, the function h(x)=(x-a)e^{x}+a is increasing, and the minimum value of h(x) is h(-1)=a- frac {1+a}{e}.When -1 < a-1 < 1, i.e., 0 < a < 2, on x in [-1,a-1], h'(x) leqslant 0, h(x) is a decreasing function, and on x in [a-1,1], h'(x) geqslant 0, h(x) is an increasing function. Therefore, the minimum value of h(x) is h(a-1)=-e^{a-1}+a.When a-1 geqslant 1, i.e., a geqslant 2, on [-1,1], h'(x) leqslant 0, h(x) is decreasing, and the minimum value of h(x) is h(1)=(1-a)e+a.In summary, when a leqslant 0, the minimum value of h(x) is a- frac {1+a}{e}; when a geqslant 2, the minimum value of h(x) is (1-a)e+a; when 0 < a < 2, the minimum value of h(x) is -e^{a-1}+a.(2) Let f(x)=x^{2}-2bx-ae+e+ frac {15}{2},According to the problem, "for forall x_{1} in [-1,1], exists x_{2} in [1,2], such that h(x_{1}) geqslant x_{2}^{2}-2bx_{2}-ae+e+ frac {15}{2} holds" is equivalent to "the minimum value of f(x) on [1,2] is not greater than the minimum value of h(x) on [-1,1]".That is, h(x)_{min} geqslant f(x)_{min}.From (1), when a=3, h(x)_{min}=h(1)=(1-a)e+a=-2e+3.When a=3, f(x)=x^{2}-2bx-2e+ frac {15}{2}=(x-b)^{2}-b^{2}-2e+ frac {15}{2}, for x in [1,2],(i) When b leqslant 1, f(x)_{min}=f(1)=-2b-2e+ frac {17}{2},From -2e+3 geqslant -2b-2e+ frac {17}{2}, we get b geqslant frac {11}{4}, which contradicts b leqslant 1, so we discard this case.(ii) When 1 < b < 2, f(x)_{min}=f(b)=-b^{2}-2e+ frac {15}{2},From -2e+3 geqslant -b^{2}-2e+ frac {15}{2}, we get b^{2} geqslant frac {9}{2}, which contradicts 1 < b < 2, so we discard this case.(iii) When b geqslant 2, f(x)_{min}=f(2)=-4b-2e+ frac {23}{2},From -2e+3 geqslant -4b-2e+ frac {23}{2}, we get b geqslant frac {17}{8}.Therefore, the range of b is boxed{[frac {17}{8},+infty)}.

🧠:[<think>Okay, let's tackle this problem step by step. It's about the power department's revenue after adjusting the electricity price. There are two parts: first, finding the revenue function in terms of the new price, and second, determining the minimum price that ensures a 20% increase in revenue. Let me start with part (1).First, the problem states that last year's electricity price was 0.80 yuan per kWh, and the consumption was 'a' kWh. This year, the price is being reduced to somewhere between 0.55 and 0.75 yuan/kWh, but users expect it to be 0.40 yuan/kWh. The key information here is about the additional electricity consumption after the price reduction. It says this additional consumption is inversely proportional to the difference between the actual price (x) and the user's expected price (0.40 yuan). The proportionality constant is k. Also, the cost of electricity is 0.30 yuan/kWh, which I assume is the cost per kWh for the power department.So, for part (1), we need to model the revenue y as a function of the actual price x. Revenue is generally calculated as (Price - Cost) multiplied by Quantity sold. Here, the price is x, and the cost is 0.30 yuan/kWh. The quantity sold is last year's consumption plus the additional consumption due to the price reduction.Last year's consumption was 'a' kWh. The additional consumption is given as inversely proportional to (x - 0.40). Since it's inversely proportional, the additional consumption should be k divided by (x - 0.40). So, total consumption this year would be a + k/(x - 0.40). Wait, but inverse proportionality: if additional consumption is inversely proportional to (x - 0.40), then if the difference (x - 0.40) increases, the additional consumption decreases, which makes sense. If the actual price is closer to the expected price (smaller difference), the additional consumption is higher. That seems right.But we need to check the units here. The proportionality constant k must have units that make sense. If additional consumption is in kWh, then k must be in kWh*(yuan/kWh) because the denominator (x - 0.40) is in yuan/kWh. So k would have units of kWh*(yuan/kWh) = yuan. Wait, but that's just the numerator. Wait, inverse proportionality would be additional consumption = k / (x - 0.40). So k must be in (kWh)*(yuan/kWh) = yuan. Hmm, maybe not. Let me think again.Actually, additional consumption (in kWh) is equal to k divided by (x - 0.40 yuan/kWh). So the units of k must be (kWh)*(yuan/kWh) to make the units work out. Because (k [units]) / (x - 0.40 [yuan/kWh]) would need to result in kWh. So k should have units of yuan. Hmm, but that might not be necessary for the problem. Maybe we don't need to worry about units here since they give k as a constant. The problem says the proportionality constant is k, so we can take it as given.Therefore, the total electricity consumption this year is last year's consumption 'a' plus the additional consumption k/(x - 0.40). So:Total consumption = a + k/(x - 0.40)But wait, the problem says "the additional electricity consumption after the price reduction is inversely proportional to the difference between the actual electricity price and the user's expected electricity price". The difference here is (x - 0.40), right? So if x is higher than 0.40, the difference is positive, and the additional consumption is k divided by that difference. But if x is less than 0.40, the difference would be negative, which would make the additional consumption negative? That doesn't make sense. So perhaps the problem assumes that the actual price x is still above the user's expected price of 0.40. Given that the plan is to reduce the price to between 0.55 and 0.75, which is still above 0.40, so x is between 0.55 and 0.75, so (x - 0.40) is positive. So we are safe.Therefore, the total consumption is a + k/(x - 0.40).Then, the revenue y would be (x - cost) multiplied by total consumption. The cost is 0.30 yuan/kWh. So:y = (x - 0.30) * [a + k/(x - 0.40)]So that's the function. Let me write that:y(x) = (x - 0.30)(a + k/(x - 0.40))I need to check if this makes sense. When the price x is higher, the per-unit profit (x - 0.30) is higher, but the consumption might decrease if x increases. Wait, but in this model, the consumption is a + k/(x - 0.40). So as x increases, (x - 0.40) increases, so k/(x - 0.40) decreases. Thus, total consumption decreases. That makes sense because if the price is higher, the additional consumption beyond last year's 'a' is less. Conversely, if x is lower (closer to 0.40), then the additional consumption is higher. That seems correct.Therefore, part (1) should be this function. Let me confirm once more. Yes, the problem states that the additional consumption is inversely proportional to the difference between actual price and expected price. So additional consumption = k / (x - 0.40). Total consumption is original a plus that. Then revenue is (price - cost) * total consumption. So yes, that's the function.So the answer to part (1) is:y = (x - 0.30)(a + k/(x - 0.40))Now moving on to part (2). We are told that k = 0.2a. We need to find the minimum electricity price x (within the planned range 0.55 to 0.75) such that the power department's revenue increases by at least 20% compared to last year.First, let's compute last year's revenue. Last year's price was 0.80 yuan/kWh, cost is 0.30 yuan/kWh, so profit per kWh is 0.80 - 0.30 = 0.50 yuan/kWh. The consumption was 'a' kWh, so revenue was 0.50 * a yuan.This year's revenue needs to be at least 20% higher, so:y >= 1.2 * 0.50a = 0.60a yuan.Given that k = 0.2a, substitute into the function:y = (x - 0.30)(a + 0.2a / (x - 0.40)) = (x - 0.30)(a + (0.2a)/(x - 0.40)) = a(x - 0.30)(1 + 0.2/(x - 0.40))Alternatively, factor out the a:y = a(x - 0.30)[1 + 0.2/(x - 0.40)]We need y >= 0.60a. Since a is positive, we can divide both sides by a:(x - 0.30)[1 + 0.2/(x - 0.40)] >= 0.60Let's simplify the left side. Let me compute 1 + 0.2/(x - 0.40):1 + 0.2/(x - 0.40) = (x - 0.40 + 0.2)/ (x - 0.40) ) = (x - 0.20)/(x - 0.40)Wait, let me check:1 is equal to (x - 0.40)/(x - 0.40), so:1 + 0.2/(x - 0.40) = [ (x - 0.40) + 0.2 ] / (x - 0.40 ) = (x - 0.40 + 0.2)/(x - 0.40) = (x - 0.20)/(x - 0.40)Therefore, the left side becomes:(x - 0.30) * (x - 0.20)/(x - 0.40)So the inequality is:(x - 0.30)(x - 0.20)/(x - 0.40) >= 0.60Now, we need to solve this inequality for x in [0.55, 0.75]. Let's first write the equation:[(x - 0.30)(x - 0.20)] / (x - 0.40) = 0.60Multiply both sides by (x - 0.40):(x - 0.30)(x - 0.20) = 0.60(x - 0.40)Expand the left side:x^2 - 0.20x - 0.30x + 0.06 = x^2 - 0.50x + 0.06Right side: 0.60x - 0.24Bring all terms to the left:x^2 - 0.50x + 0.06 - 0.60x + 0.24 = 0Combine like terms:x^2 - 0.50x - 0.60x + 0.06 + 0.24 = x^2 - 1.10x + 0.30 = 0So quadratic equation:x^2 - 1.10x + 0.30 = 0We can solve this quadratic equation. Let's compute the discriminant:D = (1.10)^2 - 4*1*0.30 = 1.21 - 1.20 = 0.01Square root of D is 0.10Thus, solutions are:x = [1.10 ± 0.10]/2So:x = (1.10 + 0.10)/2 = 1.20/2 = 0.60x = (1.10 - 0.10)/2 = 1.00/2 = 0.50So the solutions are x = 0.60 and x = 0.50. However, the planned price reduction is between 0.55 and 0.75. So x = 0.50 is below the lower limit of 0.55, so we can disregard that. The relevant solution is x = 0.60.But we need to check the inequality [(x - 0.30)(x - 0.20)]/(x - 0.40) >= 0.60. Let me analyze the sign of the left side.First, note that x is in [0.55, 0.75]. Therefore, x - 0.40 is positive (since x >= 0.55 > 0.40). So denominator is positive.Numerator: (x - 0.30)(x - 0.20). Both factors:x - 0.30: in [0.55 - 0.30, 0.75 - 0.30] = [0.25, 0.45], positive.x - 0.20: in [0.55 - 0.20, 0.75 - 0.20] = [0.35, 0.55], positive.Thus, numerator is positive, denominator positive, so the entire expression is positive. So the inequality is equivalent to [(x - 0.30)(x - 0.20)]/(x - 0.40) >= 0.60.We found that equality occurs at x = 0.60. To check where the left side is greater than or equal to 0.60, let's test intervals around x = 0.60.Take x = 0.55:Left side: (0.55 - 0.30)(0.55 - 0.20)/(0.55 - 0.40) = (0.25)(0.35)/0.15 ≈ 0.0875 / 0.15 ≈ 0.5833, which is less than 0.60.Take x = 0.60:Left side: (0.60 - 0.30)(0.60 - 0.20)/(0.60 - 0.40) = (0.30)(0.40)/0.20 = 0.12 / 0.20 = 0.60, which is equal.Take x = 0.65:Left side: (0.65 - 0.30)(0.65 - 0.20)/(0.65 - 0.40) = (0.35)(0.45)/0.25 ≈ 0.1575 / 0.25 ≈ 0.63, which is more than 0.60.Therefore, the expression increases as x increases beyond 0.60. Since at x = 0.60, the value is exactly 0.60, and for x > 0.60, it's higher. But at x = 0.55, it's lower. Therefore, the inequality [(x - 0.30)(x - 0.20)]/(x - 0.40) >= 0.60 holds for x >= 0.60.But we have to check if the revenue increases by at least 20% for x >= 0.60. Since the original revenue last year was 0.50a, this year's required is 0.60a. So, according to the equation, when x >= 0.60, the revenue meets or exceeds 0.60a.However, the problem asks for the minimum electricity price that can still ensure the revenue increases by at least 20%. The minimum x in the range [0.55, 0.75] where the revenue is at least 0.60a is x = 0.60.But let me confirm once more. If the price is set at 0.60 yuan/kWh, the revenue is exactly 0.60a, which is a 20% increase. If the price is set higher than 0.60, the revenue will be more than 0.60a, so the minimum price required is 0.60.But wait, the problem says "the plan is to reduce the electricity price to between 0.55 and 0.75". So they are lowering the price from last year's 0.80. Therefore, the price x is in [0.55, 0.75], which is lower than 0.80. So the minimum price in this range that still gives the required revenue is 0.60, which is within the planned range. Therefore, 0.60 is the answer.But let me double-check with another value. Suppose x = 0.60:Compute total consumption: a + 0.2a / (0.60 - 0.40) = a + 0.2a / 0.20 = a + a = 2a. Then revenue is (0.60 - 0.30)*2a = 0.30*2a = 0.60a, which is exactly 20% increase. Correct.If x is 0.60, that's the threshold. So the minimum price in the planned range is 0.60 yuan/kWh. Since the planned range is from 0.55 to 0.75, and 0.60 is within that, it's valid. Therefore, the answer is 0.60 yuan/kWh.Final Answer(1) boxed{y = (x - 0.30)left(a + dfrac{k}{x - 0.40}right)}(2) boxed{0.60}</think>To solve the problem, we need to determine the function for the power department's revenue and find the minimum electricity price that ensures a 20% increase in revenue compared to last year. Part (1)The revenue ( y ) is calculated as the product of the profit per kWh (price minus cost) and the total electricity consumption. The total electricity consumption this year is last year's consumption ( a ) plus the additional consumption due to the price reduction, which is inversely proportional to the difference between the actual price ( x ) and the user's expected price (0.40 yuan/kWh). The proportionality constant is ( k ).1. Profit per kWh: ( x - 0.30 )2. Total consumption: ( a + frac{k}{x - 0.40} )Thus, the revenue function ( y ) is:[y = (x - 0.30) left( a + frac{k}{x - 0.40} right)] Part (2)Given ( k = 0.2a ), we need to find the minimum price ( x ) such that the revenue increases by at least 20% compared to last year. Last year's revenue was ( 0.50a ) yuan (profit per kWh 0.50 yuan, consumption ( a ) kWh). This year's required revenue is ( 0.60a ) yuan.Substituting ( k = 0.2a ) into the revenue function:[y = (x - 0.30) left( a + frac{0.2a}{x - 0.40} right)]Simplifying:[y = a(x - 0.30) left( 1 + frac{0.2}{x - 0.40} right)]Dividing both sides by ( a ):[(x - 0.30) left( 1 + frac{0.2}{x - 0.40} right) geq 0.60]Simplifying further:[(x - 0.30) left( frac{x - 0.20}{x - 0.40} right) geq 0.60]Solving the quadratic equation derived from the inequality:[x^2 - 1.10x + 0.30 = 0]The solutions are ( x = 0.60 ) and ( x = 0.50 ). Since ( x ) must be between 0.55 and 0.75, the valid solution is ( x = 0.60 ). Final Answers(1) boxed{y = (x - 0.30)left(a + dfrac{k}{x - 0.40}right)}(2) boxed{0.60},<think>Okay, let me try to tackle this problem step by step. First, I need to understand what's being asked here. There are two parts: part (1) is to find the function y(x) which represents the power department's revenue in terms of the actual electricity price x after the reduction. Part (2) is to find the minimum electricity price x that ensures the revenue increases by at least 20% compared to last year, given that k = 0.2a.Starting with part (1). Let me parse the problem again. Last year's price was 0.80 yuan per kWh, and the consumption was a kWh. This year, the price is being reduced to between 0.55 and 0.75 yuan/kWh. The user's expected price is 0.40 yuan/kWh. The additional electricity consumption after the price reduction is inversely proportional to the difference between the actual price x and the user's expected price (0.40). The proportionality constant is k. Also, the cost of electricity is 0.30 yuan/kWh.So the power department's revenue y would be the actual price x minus the cost (0.30) multiplied by the total electricity consumption this year. The total electricity consumption this year is last year's consumption a plus the additional consumption due to the price reduction. The additional consumption is inversely proportional to (x - 0.40), so it's k / (x - 0.40). Since it's inversely proportional, if the difference (x - 0.40) is larger, the additional consumption is smaller, and vice versa.Therefore, the total consumption this year should be a + k / (x - 0.40). Then, the revenue y would be (x - 0.30) multiplied by this total consumption. So, y = (x - 0.30)(a + k / (x - 0.40)).Wait, let me check if that's correct. The revenue is (price - cost) * quantity. Yes, because the cost is 0.30 per kWh, so the profit per kWh is x - 0.30. Then, total profit is (x - 0.30) multiplied by total kWh sold. The total kWh sold is last year's a plus the additional consumption. The additional consumption is given as inversely proportional to (x - 0.40), which is the difference between the actual price and the user's expected price. So yes, the additional is k / (x - 0.40). Therefore, the function should be y = (x - 0.30)(a + k / (x - 0.40)).But let me make sure about the inverse proportionality. The problem states: "the additional electricity consumption after the price reduction is inversely proportional to the difference between the actual electricity price and the user's expected electricity price (the proportionality constant is k)". So, additional consumption ΔE = k / (x - 0.40). So total consumption is a + ΔE = a + k / (x - 0.40). Therefore, revenue y is (x - 0.30)(a + k / (x - 0.40)). That seems right.So for part (1), the answer is y = (x - 0.30)(a + k / (x - 0.40)).Now part (2): Let k = 0.2a, find the minimum electricity price x that can still ensure the power department's revenue increases by at least 20% compared to last year.First, let's compute last year's revenue. Last year's price was 0.80 yuan/kWh, cost is 0.30 yuan/kWh, so profit per kWh was 0.80 - 0.30 = 0.50 yuan. The consumption was a kWh, so revenue was 0.50 * a yuan.This year, the revenue needs to be at least 20% more than last year's. So 0.50 * a * 1.20 = 0.60 * a yuan.Given that k = 0.2a, substitute into the function y(x) from part (1):y = (x - 0.30)(a + 0.2a / (x - 0.40)).Simplify this expression:First, factor out a in the second term:y = (x - 0.30)(a + (0.2a)/(x - 0.40)) = a(x - 0.30)(1 + 0.2 / (x - 0.40)).So, y = a(x - 0.30)[1 + 0.2 / (x - 0.40)].We need to set this greater than or equal to 0.60a.Divide both sides by a (assuming a > 0, which makes sense as it's electricity consumption):(x - 0.30)[1 + 0.2 / (x - 0.40)] ≥ 0.60.Now, let's compute [1 + 0.2 / (x - 0.40)]. Let's write that as 1 + 0.2/(x - 0.40).Multiply (x - 0.30) by that:(x - 0.30)[1 + 0.2/(x - 0.40)] = (x - 0.30) + 0.2(x - 0.30)/(x - 0.40).So the inequality becomes:(x - 0.30) + 0.2(x - 0.30)/(x - 0.40) ≥ 0.60.Let me combine these terms. Let me denote x - 0.40 as t to simplify the expression. Let t = x - 0.40. Then x = t + 0.40.So, substituting:(x - 0.30) = (t + 0.40 - 0.30) = t + 0.10.Similarly, 0.2(x - 0.30)/(x - 0.40) becomes 0.2(t + 0.10)/t.So the inequality becomes:(t + 0.10) + 0.2(t + 0.10)/t ≥ 0.60.But t = x - 0.40, and since the price after reduction is between 0.55 and 0.75, x ∈ [0.55, 0.75]. Therefore, t = x - 0.40 ∈ [0.15, 0.35].So t is positive, which is good because we don't have division by zero or negative denominators here.So the inequality is:t + 0.10 + (0.2(t + 0.10))/t ≥ 0.60.Let me compute each term:First, t + 0.10 is straightforward.Second term: 0.2(t + 0.10)/t = 0.2*(t + 0.10)/t.So, let's write the entire left side:t + 0.10 + 0.2*(t + 0.10)/t.Let me combine these terms. Let's find a common denominator for the terms involving t.Multiply all terms by t:t*(t + 0.10) + 0.2*(t + 0.10) ≥ 0.60*t.But actually, maybe it's better to combine the terms first. Let's write:Left side: t + 0.10 + 0.2(t + 0.10)/t.Let me factor 0.2(t + 0.10)/t as 0.2*(1 + 0.10/t).But perhaps another approach: Let me compute the entire expression.Left side = t + 0.10 + 0.2*(t + 0.10)/t.Let me write 0.2*(t + 0.10) as 0.2t + 0.02.Then, divide by t: (0.2t + 0.02)/t = 0.2 + 0.02/t.So the left side becomes:t + 0.10 + 0.2 + 0.02/t = t + 0.30 + 0.02/t.Therefore, the inequality is:t + 0.30 + 0.02/t ≥ 0.60.Subtract 0.60:t + 0.30 + 0.02/t - 0.60 ≥ 0.Simplify:t - 0.30 + 0.02/t ≥ 0.So, t - 0.30 + 0.02/t ≥ 0.Multiply both sides by t (since t > 0):t^2 - 0.30t + 0.02 ≥ 0.So, the quadratic inequality is t^2 - 0.30t + 0.02 ≥ 0.Let me solve the quadratic equation t^2 - 0.30t + 0.02 = 0.Use the quadratic formula:t = [0.30 ± sqrt(0.30^2 - 4*1*0.02)] / 2.Compute discriminant:0.09 - 0.08 = 0.01.So sqrt(0.01) = 0.10.Therefore, roots are [0.30 ± 0.10]/2.First root: (0.30 + 0.10)/2 = 0.40/2 = 0.20.Second root: (0.30 - 0.10)/2 = 0.20/2 = 0.10.So the quadratic is t^2 -0.30t +0.02, which factors as (t - 0.10)(t - 0.20). Wait, let's check:(t - 0.10)(t - 0.20) = t^2 -0.30t +0.02. Yes, correct.So the quadratic is positive outside the roots, since the coefficient of t^2 is positive. So the inequality t^2 -0.30t +0.02 ≥ 0 holds when t ≤ 0.10 or t ≥ 0.20.But t = x - 0.40, and x ∈ [0.55, 0.75], so t ∈ [0.15, 0.35]. Therefore, in the interval t ∈ [0.15, 0.35], we need to check where t^2 -0.30t +0.02 ≥ 0.Since the roots are at t = 0.10 and t = 0.20, between t=0.10 and t=0.20, the quadratic is negative. But our t is in [0.15,0.35]. So for t ∈ [0.15,0.20), quadratic is negative; for t ∈ [0.20,0.35], quadratic is positive.Therefore, the inequality t^2 -0.30t +0.02 ≥ 0 holds when t ≥0.20 or t ≤0.10, but in our case, t is between 0.15 and 0.35. Therefore, the inequality holds when t ≥0.20. So t ≥0.20 implies x -0.40 ≥0.20 → x ≥0.60.But x is in [0.55,0.75], so the solution is x ∈ [0.60,0.75]. Therefore, the minimum x that satisfies the inequality is 0.60 yuan/kWh.Wait, but let me check this. Let me verify.Original inequality after substitution: t^2 -0.30t +0.02 ≥0. So in the interval t ∈ [0.15,0.35], the quadratic is positive when t ≥0.20. Therefore, x ≥0.60.Therefore, the minimum electricity price x is 0.60 yuan/kWh.But wait, let's check with x=0.60. Let's compute y(0.60) and see if it's at least 0.60a.Given k=0.2a.Compute y = (0.60 -0.30)(a +0.2a/(0.60 -0.40)) = 0.30*(a +0.2a/0.20) =0.30*(a + a) =0.30*2a=0.60a. Exactly 0.60a, which is the required 20% increase.If x is less than 0.60, say x=0.59, then t=0.19, which is less than 0.20. Then quadratic would be negative, so the inequality is not satisfied. Therefore, the minimal x is 0.60.Therefore, the answer is 0.60 yuan/kWh.But let me check if all steps are correct.Starting from the original function y = (x -0.30)(a +0.2a/(x -0.40)). We set this ≥ 1.2*(0.80 -0.30)*a =1.2*0.50a=0.60a.So, (x -0.30)(1 +0.2/(x -0.40)) ≥0.60.After substitution t =x -0.40, leading to t ∈[0.15,0.35], and after simplifying, we obtained t ≥0.20, so x ≥0.60.Therefore, the minimum x is 0.60. Therefore, the answer is 0.60 yuan per kWh.Yes, this seems correct.Final Answer(1) The function is boxed{y = (x - 0.30)left(a + dfrac{k}{x - 0.40}right)};(2) The minimum electricity price is boxed{0.60} yuan/(kW・h).</think>To solve the given problem, we need to determine the function relationship between the power department's revenue ( y ) and the actual electricity price ( x ), and then find the minimum electricity price that ensures a 20% increase in revenue compared to last year. Part (1)The power department's revenue ( y ) is calculated as the profit per kWh (actual price ( x ) minus cost 0.30 yuan/kWh) multiplied by the total electricity consumption. The total electricity consumption this year is last year's consumption ( a ) plus the additional consumption due to the price reduction, which is inversely proportional to the difference between the actual price ( x ) and the user's expected price 0.40 yuan/kWh (with proportionality constant ( k )).The function relationship is:[y = (x - 0.30)left(a + frac{k}{x - 0.40}right)] Part (2)Given ( k = 0.2a ), we need to find the minimum electricity price ( x ) such that the revenue increases by at least 20% compared to last year. Last year's revenue was ( 0.50a ) yuan, so this year's revenue needs to be at least ( 0.60a ) yuan.Substituting ( k = 0.2a ) into the revenue function:[y = (x - 0.30)left(a + frac{0.2a}{x - 0.40}right)]Simplifying, we get:[y = a(x - 0.30)left(1 + frac{0.2}{x - 0.40}right)]Dividing both sides by ( a ):[(x - 0.30)left(1 + frac{0.2}{x - 0.40}right) geq 0.60]Let ( t = x - 0.40 ), then ( x = t + 0.40 ) and the inequality becomes:[(t + 0.10) + frac{0.2(t + 0.10)}{t} geq 0.60]Simplifying further:[t - 0.30 + frac{0.02}{t} geq 0]Multiplying by ( t ) (since ( t > 0 )):[t^2 - 0.30t + 0.02 geq 0]Solving the quadratic inequality, we find the roots ( t = 0.10 ) and ( t = 0.20 ). The inequality holds for ( t leq 0.10 ) or ( t geq 0.20 ). Given ( t in [0.15, 0.35] ), the valid solution is ( t geq 0.20 ), which translates to ( x geq 0.60 ).Thus, the minimum electricity price is:[boxed{0.60}] Final Answers(1) The function is boxed{y = (x - 0.30)left(a + dfrac{k}{x - 0.40}right)};(2) The minimum electricity price is boxed{0.60} yuan/(kW・h).]

🔑:(1) Let the adjusted electricity price be x yuan/(kW·h), and the revenue of the power department be y=left(frac{k}{x-0.40}+aright)(x-0.30)(0.55 leqslant x leqslant 0.75).(2) According to the problem, we have: left(frac{0.2 a}{x-0.40}+aright)(x-0.30) geqslant a cdot(0.80-0.30)(1+20 %), where 0.55 leqslant x leqslant 0.75, solving this gives 0.60 leqslant x leqslant 0.75.Therefore, when the electricity price is set to at least 0.60 yuan/(kW·h), it can still ensure that the revenue of the power department increases by at least 20 % compared to the previous year.

🧠:[<think>Okay, let's see. I need to figure out if the function f(x) = |x + 2| is increasing or decreasing on the interval (-∞, -4). Hmm, absolute value functions usually have a V-shape, right? So the graph of |x + 2| is a V with the vertex at x = -2. That shifts the standard absolute value graph two units to the left. Now, the interval we're looking at is from negative infinity up to -4. So that's way to the left of the vertex at x = -2. Since the vertex is at x = -2, and the interval is (-∞, -4), which is to the left of -4. Wait, hold on. -4 is further left on the number line than -2. So the interval is all numbers less than -4. So we're looking at the part of the graph that's to the left of x = -4. But the vertex is at x = -2. So between -∞ and -4, how does the function behave? Let me think. The absolute value function, when you're to the left of the vertex, which is at x = -2, the function is decreasing. But wait, hold on. If the vertex is at x = -2, then for x < -2, the function f(x) = |x + 2| simplifies to -(x + 2) because the expression inside the absolute value is negative. So f(x) = -x - 2 when x < -2. The slope here is -1, which is negative, so the function is decreasing on (-∞, -2). But the question is about the interval (-∞, -4). Since -4 is less than -2, the interval (-∞, -4) is entirely within the region where x < -2. So in that interval, the function f(x) is linear with a slope of -1. Therefore, since the slope is negative, the function is decreasing on that interval. Wait, but let me verify. Let's take two points in (-∞, -4), say x1 = -5 and x2 = -6. Then f(x1) = |-5 + 2| = |-3| = 3. f(x2) = |-6 + 2| = |-4| = 4. So as x decreases from -5 to -6, f(x) increases from 3 to 4. Wait, that seems like when x decreases, f(x) increases. But that would mean the function is increasing as x decreases, which would correspond to being decreasing as x increases. Wait, maybe I need to clarify. Monotonicity is about how the function behaves as x increases. So if as x increases, f(x) increases, then it's increasing. If as x increases, f(x) decreases, then it's decreasing. So let's check with x increasing. Take x1 = -6 and x2 = -5. x1 < x2. Then f(x1) = 4 and f(x2) = 3. So as x increases from -6 to -5, f(x) decreases from 4 to 3. Therefore, on the interval (-∞, -4), as x increases, f(x) decreases. Therefore, the function is monotonically decreasing on that interval. But wait, another example: take x = -10 and x = -9. f(-10) = |-10 + 2| = |-8| = 8. f(-9) = |-9 + 2| = |-7| = 7. So again, as x increases from -10 to -9, f(x) decreases from 8 to 7. So yeah, it's decreasing. Therefore, even though the interval is further left, the function is decreasing there because as x increases (moves to the right), the value of f(x) decreases. Therefore, the monotonicity on (-∞, -4) is decreasing. Alternatively, since f(x) = |x + 2|, let's find its derivative where it's differentiable. The derivative of |x + 2| is (x + 2)/|x + 2)| when x ≠ -2. For x < -2, (x + 2) is negative, so the derivative is -1. So the slope is -1 on (-∞, -2), which includes (-∞, -4). So since the derivative is negative there, the function is decreasing. So both by checking specific points and by computing the derivative, the function is decreasing on (-∞, -4). Therefore, the answer should be "decreasing".Final AnswerThe monotonicity of the function ( f(x) = |x+2| ) on the interval ( (-infty, -4) ) is monotonically boxed{decreasing}.</think>To determine the monotonicity of the function ( f(x) = |x + 2| ) on the interval ( (-infty, -4) ):1. The function ( f(x) = |x + 2| ) is an absolute value function with a vertex at ( x = -2 ).2. For ( x < -2 ), the expression inside the absolute value ( x + 2 ) is negative, so the function simplifies to ( f(x) = -(x + 2) = -x - 2 ). This is a linear function with a slope of -1.3. The interval ( (-infty, -4) ) is entirely to the left of -4, which is further left than the vertex at ( x = -2 ). Therefore, in this interval, the function is linear with a slope of -1.4. A negative slope (-1) indicates that as ( x ) increases, ( f(x) ) decreases.5. Checking specific points: For ( x_1 = -6 ) and ( x_2 = -5 ), we find ( f(-6) = 4 ) and ( f(-5) = 3 ). As ( x ) increases from -6 to -5, ( f(x) ) decreases from 4 to 3.6. The derivative of ( f(x) ) for ( x < -2 ) is -1, confirming the function is decreasing in this interval.Thus, the monotonicity of the function ( f(x) = |x + 2| ) on the interval ( (-infty, -4) ) is monotonically boxed{decreasing}.,<think>Okay, let's see. The problem is asking about the monotonicity of the function f(x) = |x + 2| on the interval (-∞, -4). Monotonicity means whether the function is increasing or decreasing over that interval. Hmm, absolute value functions can be a bit tricky because they have a V-shape, so their behavior changes depending on which side of the vertex you're on. Let me break this down step by step.First, I need to recall what the graph of f(x) = |x + 2| looks like. The general form of an absolute value function is f(x) = |x - h| + k, where (h, k) is the vertex. In this case, it's |x + 2|, which can be written as |x - (-2)|, so the vertex is at (-2, 0). That means the V-shape has its point at (-2, 0). To the left of x = -2, the function is decreasing, and to the right of x = -2, it's increasing. Wait, but the interval given here is (-∞, -4). So -4 is further to the left on the number line compared to -2. So the interval in question is entirely to the left of the vertex at -2.Now, if the interval (-∞, -4) is to the left of the vertex, I need to figure out if the function is increasing or decreasing there. Let me think. For absolute value functions, left of the vertex (for a standard V-shape opening upwards), the function is decreasing. Because as x increases towards the vertex, the value of the function decreases until it hits the vertex, then starts increasing again. Wait, no. Let me verify.Take an example. Let's consider x values less than -2. Let's pick x = -3. Then f(-3) = |-3 + 2| = |-1| = 1. If I take x = -4, f(-4) = |-4 + 2| = |-2| = 2. So as x decreases from -3 to -4, the function's value increases from 1 to 2. Conversely, if x increases from -4 to -3, moving towards -2, the function's value decreases from 2 to 1. So on the interval (-∞, -2), as x increases, the function decreases. Therefore, the function is decreasing on (-∞, -2). But the interval here is (-∞, -4), which is a subset of (-∞, -2). So the same behavior should hold, right?Therefore, on the interval (-∞, -4), since we're still to the left of the vertex at -2, the function is decreasing. So as x increases towards -4, the function decreases, and as x decreases towards -∞, the function increases. Wait, but the question is about monotonicity on (-∞, -4). So when they ask if it's monotonically increasing or decreasing, they want to know the behavior as x moves from left to right across that interval. So as x increases from -∞ to -4, what does the function do?Wait, let's check with specific numbers. Let me take x1 = -5 and x2 = -4.5, which are both in (-∞, -4). x1 is less than x2, right? -5 < -4.5. Then f(x1) = |-5 + 2| = |-3| = 3. f(x2) = |-4.5 + 2| = |-2.5| = 2.5. So as x increases from -5 to -4.5, f(x) decreases from 3 to 2.5. Therefore, when x increases, f(x) decreases. Hence, the function is decreasing on that interval.Alternatively, if we move from left to right on the interval (-∞, -4), x is increasing, but the function's output is decreasing. So the function is monotonically decreasing there.Let me verify another pair. Let x1 = -10 and x2 = -9. Then f(x1) = |-10 + 2| = 8, f(x2) = |-9 + 2| = 7. So again, as x increases from -10 to -9, f(x) decreases from 8 to 7. So same result.Therefore, even though the entire left side of the vertex is decreasing, the interval in question is a part of that left side. So the function is decreasing on (-∞, -4). Therefore, the answer should be decreasing.But just to be thorough, let's think about the derivative. The function f(x) = |x + 2|. The derivative of |x| is x/|x|, which is -1 when x < 0 and 1 when x > 0. So for f(x) = |x + 2|, the derivative is (x + 2)/|x + 2|. When x + 2 < 0, which is when x < -2, the derivative is -1. When x + 2 > 0, the derivative is 1. So on the interval (-∞, -2), the derivative is -1, which is negative, meaning the function is decreasing. Since (-∞, -4) is a subset of (-∞, -2), the derivative is also -1 there, hence the function is decreasing.Therefore, both by testing specific points and by computing the derivative, the function is monotonically decreasing on (-∞, -4).Final AnswerThe function is monotonically boxed{decreasing} on the interval ((-infty, -4)).</think>To determine the monotonicity of the function ( f(x) = |x+2| ) on the interval ((-infty, -4)), we first identify the vertex of the absolute value function. The vertex is at ((-2, 0)). For values of (x) less than (-2), the function ( f(x) = |x+2| ) is decreasing because the absolute value function forms a V-shape that opens upwards, decreasing to the left of the vertex.The interval ((-infty, -4)) is entirely to the left of the vertex at (x = -2). To confirm the behavior, we can test specific points within the interval:- For ( x = -5 ), ( f(-5) = |-5 + 2| = 3 )- For ( x = -4.5 ), ( f(-4.5) = |-4.5 + 2| = 2.5 )As (x) increases from (-5) to (-4.5), the function value decreases from 3 to 2.5, indicating a decreasing trend.Additionally, the derivative of ( f(x) = |x+2| ) is (-1) for ( x < -2 ), confirming that the function is decreasing on this interval. Since ((-infty, -4)) is a subset of ((-infty, -2)), the derivative remains (-1), indicating a monotonically decreasing function.Thus, the function ( f(x) = |x+2| ) is monotonically decreasing on the interval ((-infty, -4)).[boxed{decreasing}]]

🔑:When x in (-infty, -2), f(x) = -x - 2 is a decreasing function. When x in (2, +infty), f(x) = x + 2 is an increasing function. Since (-infty, -4) subseteq (-infty, -2), thus, the function f(x) = |x+2| is monotonically decreasing on (-infty, -4). Therefore, the answer is: boxed{text{decreasing}} Analysis: By using the zero point to divide the function f(x) = |x+2| into segments and analyzing the monotonicity of the function, and then analyzing the inclusion relationship between the given interval and the monotonic interval of the function, the answer can be obtained.

🧠:[<think>Okay, let's see. The problem says that the radius of circle ⊙O is 3, and point P is inside the circle. Then we have to determine which of the options A, B, C, or D is a possible length for OP. The options are 2, 3, 4, or all of the above.First, I need to recall the definition of a circle. A circle is the set of all points in a plane that are at a fixed distance (the radius) from a given point (the center). So, if the radius is 3, then any point on the circle is exactly 3 units away from the center O. Now, point P is inside the circle. That means the distance from O to P must be less than the radius. Wait, but if the radius is 3, then OP has to be less than 3, right? Because if it's equal to 3, then P would be on the circle, not inside. And if it's more than 3, then P would be outside. So, inside means strictly less than the radius.Looking at the options: A is 2, which is less than 3, so that's possible. B is 3, which would be exactly on the circle. But the problem states that P is inside the circle, so B might not be correct. C is 4, which is greater than the radius, so that's outside, which definitely can't be. So then, only A is possible? But let me check the options again. The options are A: 2, B: 3, C: 4, D: All of the above. So if the answer is only A, then D is not correct. But maybe there's something I'm missing here.Wait, the problem says "could be". So if point P is inside the circle, OP can be any value from 0 up to but not including 3. Wait, hold on. If the circle has radius 3, then the set of all points inside the circle are those points where OP < 3. Therefore, OP can be 0 (if P is the center) up to, but not reaching, 3. So 3 itself is excluded. So then OP cannot be 3, because that's on the circle. Similarly, 4 is outside. So only OP=2 is possible. Therefore, the answer should be A.But let me check again. Maybe there's a different interpretation. Sometimes in math problems, the term "inside" might include the boundary? But usually, in geometry, "inside" a circle refers to the interior, not including the boundary. The boundary is the circle itself. So, if the problem had said "on or inside the circle", then 3 would be allowed. But since it specifically says "inside", 3 is excluded.So then, OP can be 2, which is less than 3, so possible. 3 is on the circle, so not inside. 4 is outside. Therefore, only option A is correct. Therefore, the answer should be A.But wait, the options given are A: 2, B:3, C:4, D: All of the above. So, if only A is correct, then the answer is A. But maybe the problem is tricking me. Let me think again.Is there a case where point P is inside the circle but OP is 3? For example, maybe the circle has radius 3, but if P is on the circumference, then OP is 3. But if P is inside, then OP must be less than 3. So 3 is not allowed. So answer A.Alternatively, perhaps in some contexts, "inside" can be inclusive? Hmm. Let me check standard definitions. In topology, the interior of a circle (a disk) does not include the boundary. So, if we are talking strictly inside, then OP must be less than the radius. But if the problem is using a different definition, maybe? But in standard geometry, inside a circle usually means the interior, not the boundary. So, unless specified otherwise, 3 is not included.Therefore, the answer should be A: 2.But wait, in some math problems, especially in exams, sometimes they might consider the boundary as part of the "inside" if not specified. Is that possible? Let me think. For example, if you have a region defined as "inside and on the circle", then 3 is included. But if it's just "inside", then it's only the interior. So, given that the problem says "point P is inside circle ⊙O", then OP must be less than 3. Therefore, OP cannot be 3 or 4. Therefore, the correct answer is A:2.But the options are presented as A:2, B:3, C:4, D: All of the above. So why is B:3 an option? Maybe the problem is in translation or there's a misinterpretation. Alternatively, maybe there's a different way to interpret "inside". Wait, the problem says "the radius of circle ⊙O is 3". So the circle has radius 3. Then, the set of points inside the circle is the open disk of radius 3 centered at O. Therefore, any point P inside must satisfy OP < 3. Therefore, 3 is excluded, and 4 is excluded. So only 2 is possible, answer A.Therefore, the answer is A.But let me check once again. Suppose the problem had said "the radius is 3, and point P is inside or on the circle". Then, OP could be up to 3, inclusive. But since it's only "inside", OP must be less than 3. So 3 is not possible. So answer A.Therefore, the answer is boxed{A}.Final Answerboxed{A}</think>The radius of circle (odot O) is 3. Point (P) is inside the circle, which means the distance from (O) to (P) (denoted as (OP)) must be less than the radius of the circle. - The radius of the circle is 3, so any point inside the circle must be at a distance less than 3 from the center (O).- Option A (2) is valid because 2 is less than 3.- Option B (3) is not valid because a distance of 3 would place point (P) on the circle, not inside it.- Option C (4) is not valid because 4 is greater than the radius, placing point (P) outside the circle.Therefore, the correct answer is (boxed{A}).,<think>Okay, let me try to figure out this problem. So, we have a circle called ⊙O with a radius of 3. There's a point P inside this circle, and we need to determine which of the options A, B, C, or D could be the length of OP. The options are 2, 3, 4, or all of the above.First, let me recall some basic facts about circles. The radius of a circle is the distance from the center to any point on the circle. So, if the radius is 3, then any point on the circle is exactly 3 units away from the center O. Now, the problem says point P is inside the circle. Hmm, so if P is inside the circle, the distance from O to P must be less than the radius, right? Because if it's equal to the radius, then P would be on the circle, not inside. And if it's more than the radius, then P would be outside the circle.So, the key here is understanding the definition of the interior of a circle. The interior consists of all points whose distance from the center is less than the radius. Therefore, OP must be less than 3. Let me check the options again. Option A is 2, which is less than 3. Option B is 3, which is equal to the radius. Option C is 4, which is greater than 3. And Option D is all of the above.But wait, the problem says "point P is inside the circle." So, if OP is 3, then P would be on the circumference, not inside. Similarly, if OP is 4, P would be outside. Therefore, only OP = 2 would satisfy the condition that P is inside the circle. That would make the answer A.But let me make sure I didn't misinterpret the problem. Is there any other consideration here? For instance, sometimes in math problems, depending on the context, "inside" might be considered inclusive of the boundary. But in standard geometry terminology, the interior of a circle does not include the boundary (the circumference). The boundary points are considered part of the circle itself, not the interior. So, the interior points are strictly those with distance from the center less than the radius.Therefore, OP must be strictly less than 3. So, 2 is acceptable, 3 is not, 4 is not. Hence, the correct answer is A. But let me check the options again. The options are A: 2, B: 3, C: 4, D: All of the above.So, since only A is possible, the answer should be A. However, maybe the problem uses a different definition? Let me check again. If the problem states that P is inside the circle, then yes, OP < 3. If P were allowed to be on the circle, then B would be acceptable. But the problem specifies "inside," so B is excluded. Similarly, C is definitely excluded because 4 is greater than 3. Therefore, only A is possible.But wait, I just realized that in some contexts, especially in certain textbooks or regions, "inside" might include the boundary. But in standard mathematical terms, the interior of a circle is the set of points with distance from the center strictly less than the radius. The term "inside" usually refers to the interior, not including the boundary. The boundary is a separate set of points.Therefore, if the problem had said "inside or on the circle," then B would be acceptable. But since it specifically says "inside," B is not. So, the answer must be A.Therefore, the options are:A: 2 – PossibleB: 3 – Not possible (since that's on the circle)C: 4 – Not possible (outside the circle)D: All of the above – Not possibleHence, the correct answer is A.But wait, hold on, the options are given as A, B, C, D. Let me check once more. Maybe there's a trick here. The problem says "the length of OP could be ( )". So, they are asking for possible values. So, if OP can be 2, 3, or 4, but since P is inside, only 2 is possible. Therefore, only A is the answer.Alternatively, maybe the problem is in Chinese, since the original mentions "⊙O". But the translation seems accurate. The radius is 3, point P is inside. Therefore, OP must be less than 3. So, 2 is possible, 3 is on the circle, 4 is outside. Therefore, only A. So the answer is A.But the options given are A:2, B:3, C:4, D: All of the above. So, since only A is possible, the answer is A. Therefore, the final answer is boxed{A}.Final Answerboxed{A}</think>The radius of circle (odot O) is 3. Point (P) is inside the circle, so the length of (OP) must be less than the radius. - If (OP = 2), since 2 is less than 3, point (P) is inside the circle.- If (OP = 3), point (P) would be on the circumference of the circle, not inside.- If (OP = 4), point (P) would be outside the circle.Therefore, the only possible length for (OP) that places point (P) inside the circle is 2. (boxed{A})]

🔑:Given that the radius of circle odot O is 3, and point P is inside circle odot O, we analyze the possible lengths of OP.- For point P to be inside circle odot O, the distance OP must be less than the radius of the circle. Thus, we have OP 3. This indicates that point P is outside the circle, which again does not satisfy the condition.- Option D: Suggests that P could be inside or on the circle. However, based on our analysis, only being inside satisfies the condition, not on or outside.Therefore, the only option that satisfies the condition that point P is inside circle odot O with a radius of 3 is option A.Hence, the correct answer is boxed{A}.

🧠:[<think>Okay, so I have this problem here: solve the equation x² + 2·]x[ = 6, where ]x[ is the semi-integer part of x. First, I need to make sure I understand what a semi-integer is. The problem says a number x is semi-integer if 2x is an integer. So, if 2x is an integer, then x is a semi-integer. That means x can be written as k/2 where k is an integer. So semi-integers are numbers like ..., -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, etc. Got it.Then, the semi-integer part of x, denoted ]x[, is the greatest semi-integer not exceeding x. So it's like the floor function but instead of rounding down to the nearest integer, we round down to the nearest semi-integer. For example, if x is 1.2, then the semi-integer part would be 1, because 1 is a semi-integer (since 2*1=2 is integer) and the next semi-integer up would be 1.5, which is greater than 1.2. Similarly, if x is 2.7, then ]x[ would be 2.5, since 2.5 is the greatest semi-integer less than or equal to 2.7.So, to solve x² + 2·]x[ = 6, I need to consider the value of ]x[ based on where x is in the real number line. Since semi-integers are spaced every 0.5 units, the semi-integer part function ]x[ will be constant on intervals between semi-integers. For example, between 1 and 1.5, ]x[ is 1. Between 1.5 and 2, ]x[ is 1.5, and so on.So, maybe I should break the problem into cases based on the interval in which x lies. Each interval will be between two consecutive semi-integers, say k/2 and (k+1)/2 for some integer k. In each interval, ]x[ is equal to k/2, so I can substitute that into the equation and solve for x, then check if the solution is within the interval.Alternatively, since semi-integers are every 0.5, maybe I can let ]x[ = n/2 where n is an integer, and then x must satisfy n/2 ≤ x < (n+1)/2. Then, substituting ]x[ = n/2 into the equation gives x² + 2*(n/2) = x² + n = 6, so x² = 6 - n. Then, since x is in [n/2, (n+1)/2), x must satisfy sqrt(6 - n) being in that interval, but also considering negative roots?Wait, but x² is always non-negative, so 6 - n must be non-negative. Therefore, n ≤ 6. Also, since x is in [n/2, (n+1)/2), x could be negative or positive. Hmm, but if x is negative, then ]x[ is the greatest semi-integer less than or equal to x. For example, if x is -0.3, then ]x[ would be -0.5, because -0.5 is the greatest semi-integer not exceeding -0.3. Similarly, if x is -1.2, ]x[ would be -1.5. So, in the negative direction, each semi-integer interval is from (k-1)/2 to k/2 where k is a negative integer. Maybe I need to adjust how I define n. Let me think.Alternatively, maybe instead of using n as an integer, let me consider that ]x[ = m where m is a semi-integer, so m = k/2 for some integer k. Then, m ≤ x < m + 0.5. Then, substituting m into the equation gives x² + 2m = 6. So, x² = 6 - 2m. Therefore, x = sqrt(6 - 2m) or x = -sqrt(6 - 2m). But x must lie in [m, m + 0.5). Also, since x² is non-negative, 6 - 2m ≥ 0 ⇒ m ≤ 3.But m is a semi-integer, so m can be ..., 1.5, 2, 2.5, 3, 3.5, ... but wait, if m is a semi-integer, m can be written as k/2 where k is an integer. Then, m ≤ 3 implies k/2 ≤ 3 ⇒ k ≤ 6. So k can be any integer less than or equal to 6. But we also need to consider negative values of m.Wait, for example, if m is negative, then 6 - 2m could be larger. For instance, if m = -1.5, then 6 - 2*(-1.5) = 6 + 3 = 9, so x² = 9, so x = 3 or x = -3. But then we have to check if x is in [m, m + 0.5). If m = -1.5, then the interval is [-1.5, -1.0). So x = 3 is not in that interval, and x = -3 is in [-1.5, -1.0)? No, -3 is less than -1.5, so that's not in the interval. So maybe m cannot be too negative? Let me check.So, the idea is that for each semi-integer m, we can solve x² + 2m = 6, so x = ±sqrt(6 - 2m). Then, we check if x is in [m, m + 0.5). But for x to be in that interval, depending on the sign of m.Wait, but x could be positive or negative, but the interval [m, m + 0.5) is in the positive side if m is positive, and in the negative side if m is negative. For example, if m is negative, say m = -1.5, then the interval is [-1.5, -1). So x would be negative in that case, so x = -sqrt(6 - 2m). Let's see.Wait, let's take m = -1.5. Then x² = 6 - 2*(-1.5) = 6 + 3 = 9, so x = 3 or x = -3. The interval is [-1.5, -1). So x = -3 is not in that interval, so m = -1.5 is not possible.Similarly, if m = -1.0, then x² = 6 - 2*(-1) = 6 + 2 = 8, so x = ±2√2 ≈ ±2.828. The interval for m = -1.0 is [-1.0, -0.5). So x = -2.828 is less than -1.0, so not in the interval. So that's not valid.Similarly, m = -0.5: x² = 6 - 2*(-0.5) = 6 + 1 = 7, x ≈ ±2.645. The interval is [-0.5, 0). x = -2.645 is less than -0.5, so not in the interval.m = 0: x² = 6 - 0 = 6, x ≈ ±2.449. The interval [0, 0.5). But x is ±2.449, which are not in [0, 0.5). So no solution here.m = 0.5: x² = 6 - 2*0.5 = 6 - 1 = 5, x ≈ ±2.236. The interval [0.5, 1.0). But x = ±2.236 are not in [0.5, 1.0). So no.m = 1.0: x² = 6 - 2*1 = 4, x = ±2. The interval [1.0, 1.5). x = 2 is not in [1.0, 1.5); x = -2 is not in [1.0, 1.5). So no solution.m = 1.5: x² = 6 - 2*1.5 = 6 - 3 = 3, x ≈ ±1.732. The interval [1.5, 2.0). x = 1.732 is approximately 1.732, which is in [1.5, 2.0). So this is a candidate. Let me check. If x ≈ 1.732, then ]x[ is the greatest semi-integer not exceeding x, which is 1.5. So substituting back into the equation: x² + 2*(1.5) = 3 + 3 = 6. That works. So x = sqrt(3) ≈ 1.732 is a solution. Similarly, x = -sqrt(3) ≈ -1.732. Let's check if that works. ]x[ for x = -1.732 would be the greatest semi-integer not exceeding -1.732. The semi-integers less than or equal to -1.732 are ..., -2.0, -2.5, ... but wait, the semi-integers are spaced by 0.5. So -2.0 is a semi-integer (since 2*(-2.0) = -4 is integer), and -1.732 is between -2.0 and -1.5. The greatest semi-integer not exceeding -1.732 is -2.0. So ]x[ = -2.0. Then, substituting into the equation: (-sqrt(3))² + 2*(-2.0) = 3 - 4 = -1, which is not equal to 6. So x = -sqrt(3) is not a solution. So only the positive sqrt(3) is a solution here.Continuing with m = 2.0: x² = 6 - 2*2 = 6 - 4 = 2, so x ≈ ±1.414. The interval [2.0, 2.5). But x ≈ 1.414 is less than 2.0, so not in the interval. So no solution here.m = 2.5: x² = 6 - 2*2.5 = 6 - 5 = 1, so x = ±1. The interval [2.5, 3.0). But x = 1 is not in [2.5, 3.0). So no.m = 3.0: x² = 6 - 2*3 = 0, so x = 0. The interval [3.0, 3.5). But x = 0 is not in [3.0, 3.5). So no solution.m = 3.5: x² = 6 - 2*3.5 = 6 - 7 = -1. That's impossible. So m cannot be greater than 3.0.Wait, but m is a semi-integer, so m can go up to 3.0. So beyond that, 6 - 2m becomes negative, which is invalid. So m must be ≤ 3.But we saw that for m = 1.5, we have a solution. Let's check m = 1.5 again. x = sqrt(3) ≈ 1.732, which is in [1.5, 2.0). So that's valid.Is there any other m where x could be in [m, m + 0.5)?Wait, what about m = 2.0. x² = 2, x ≈ ±1.414, which is less than m = 2.0, so not in [2.0, 2.5). So no.Similarly, m = 2.5: x = ±1, which is less than 2.5. Not valid.Wait, maybe negative m? Let's check m = -2.0: x² = 6 - 2*(-2.0) = 6 + 4 = 10, so x ≈ ±3.162. The interval for m = -2.0 is [-2.0, -1.5). So x = -3.162 is less than -2.0, so not in the interval. So no.m = -1.5: x² = 6 - 2*(-1.5) = 9, x = ±3. Interval [-1.5, -1.0). x = -3 is not in that interval.m = -1.0: x² = 8, x ≈ ±2.828. Interval [-1.0, -0.5). x = -2.828 is less than -1.0, so not in the interval.m = -0.5: x² = 7, x ≈ ±2.645. Interval [-0.5, 0). x = -2.645 is less than -0.5, so not in the interval.m = 0.0: x² = 6, x ≈ ±2.449. Interval [0.0, 0.5). Negative x is not in that interval, positive x is 2.449 not in [0.0, 0.5). So no.m = 0.5: x² = 5, x ≈ ±2.236. Interval [0.5, 1.0). x is not in that interval.m = 1.0: x² = 4, x = ±2. Interval [1.0, 1.5). x = 2 is not in that interval. x = -2 not relevant.m = 1.5: x = sqrt(3) ≈ 1.732 in [1.5, 2.0). So that's valid. Then, m = 2.0 as before, no.Wait, but what about m = 1.0? Let me check again. If m = 1.0, then x² = 4, x = ±2. The interval [1.0, 1.5). So x = 2 is not in [1.0, 1.5). So no solution here.Wait, but could there be another m where x is in the interval? Let's see. Suppose m = 1.5, x = sqrt(3) ≈ 1.732. Then ]x[ = 1.5, correct. What if x is in [1.5, 2.0), which is m = 1.5. So that's the only possible m here.But wait, could there be another semi-integer part? For example, if x is exactly 1.5, then ]x[ is 1.5. If x is 2.0, ]x[ is 2.0. So for x in [1.5, 2.0), ]x[ is 1.5. So sqrt(3) ≈ 1.732 is in that interval, so m = 1.5 gives x = sqrt(3) ≈ 1.732, which is valid.Are there any other possible solutions?Wait, let me check for m = 2.5. x² = 6 - 2*2.5 = 1, so x = ±1. But the interval [2.5, 3.0). x = 1 is not in there. Similarly, m = 3.0: x = 0, not in [3.0, 3.5). So no.Alternatively, maybe if x is in a different interval where ]x[ is a different semi-integer. Let's check for x in [1.0, 1.5). Then ]x[ = 1.0. Then the equation becomes x² + 2*1 = 6 ⇒ x² = 4 ⇒ x = ±2. But in this interval [1.0, 1.5), x = 2 is outside. So no solution here.Similarly, x in [2.0, 2.5): ]x[ = 2.0, equation becomes x² + 4 = 6 ⇒ x² = 2 ⇒ x ≈ ±1.414, but x must be in [2.0, 2.5), so no.x in [2.5, 3.0): ]x[ = 2.5, equation becomes x² + 5 = 6 ⇒ x² = 1 ⇒ x = ±1. Not in the interval.x in [3.0, 3.5): ]x[ = 3.0, x² + 6 = 6 ⇒ x = 0. Not in the interval.How about negative x? Let's say x is in [-2.0, -1.5): ]x[ = -2.0. Then the equation becomes x² + 2*(-2.0) = x² - 4 = 6 ⇒ x² = 10 ⇒ x ≈ ±3.162. So x = -3.162. Is that in [-2.0, -1.5)? No, because -3.162 < -2.0. So not valid.x in [-1.5, -1.0): ]x[ = -1.5. Equation: x² + 2*(-1.5) = x² - 3 = 6 ⇒ x² = 9 ⇒ x = ±3. So x = -3. Not in [-1.5, -1.0). No.x in [-1.0, -0.5): ]x[ = -1.0. Equation: x² - 2 = 6 ⇒ x² = 8 ⇒ x ≈ ±2.828. x = -2.828 is less than -1.0, not in [-1.0, -0.5). No.x in [-0.5, 0.0): ]x[ = -0.5. Equation: x² + 2*(-0.5) = x² -1 = 6 ⇒ x² = 7 ⇒ x ≈ ±2.645. x = -2.645 is less than -0.5, so not in [-0.5, 0.0). No.x in [0.0, 0.5): ]x[ = 0.0. Equation: x² = 6 ⇒ x ≈ ±2.449. Not in [0.0, 0.5). No.x in [0.5, 1.0): ]x[ = 0.5. Equation: x² + 1 = 6 ⇒ x² = 5 ⇒ x ≈ ±2.236. Not in the interval.x in [1.5, 2.0): as before, x = sqrt(3) ≈1.732. Valid.x in [2.0, 2.5): ]x[ = 2.0. Equation: x² +4 =6 ⇒ x²=2. x ≈1.414. Not in [2.0, 2.5).Wait, so the only solution is x = sqrt(3)? But let me check another interval. For example, maybe x is exactly a semi-integer. For example, x = 1.5. Then ]x[ = 1.5. Then the equation becomes (1.5)^2 + 2*(1.5) = 2.25 + 3 = 5.25 ≠6. So not a solution.x = 2.0: ]x[ = 2.0. Equation: 4 + 4 =8 ≠6.x = 1.0: ]x[ =1.0. 1 + 2 =3 ≠6.x = sqrt(3): as checked before, works.Wait, are there any other intervals where x could be? For example, maybe x is in [sqrt(3), 2.0). But ]x[ is still 1.5. So even if x is like 1.8, ]x[ is still 1.5, so the equation would be x² +3 =6 ⇒x²=3⇒x=sqrt(3)≈1.732. But if x is 1.8, which is greater than sqrt(3), then x² is 3.24, so 3.24 +3=6.24≠6. So that doesn't work. So only x=sqrt(3) in [1.5,2.0) satisfies x²=3.Wait, but hold on, when m =1.5, x is sqrt(6 - 2*1.5) = sqrt(3). So that's the only solution in that interval.Wait, but maybe there's another semi-integer part where m is a semi-integer such that x is in [m, m +0.5), and x² =6 -2m. So, in other words, for x to be in [m, m+0.5), we need m ≤x <m+0.5. But x²=6 -2m, so m must be ≤ sqrt(6 -2m) <m +0.5 or m ≤ -sqrt(6 -2m) <m +0.5. But since x can be positive or negative. Let me think again.Alternatively, for each semi-integer m, check if there exists x in [m, m+0.5) such that x²=6 -2m. So if x is positive, then x = sqrt(6 -2m). So we need sqrt(6 -2m) ∈ [m, m +0.5). Similarly, if x is negative, then x = -sqrt(6 -2m). Then we need -sqrt(6 -2m) ∈ [m, m +0.5). Let's check for positive x first.Positive x: sqrt(6 -2m) ∈ [m, m +0.5)So sqrt(6 -2m) ≥m and sqrt(6 -2m) <m +0.5Similarly, for negative x: -sqrt(6 -2m) ∈ [m, m +0.5)But if m is negative, the interval [m, m +0.5) is on the negative side, so x must be negative. Let's handle positive and negative x separately.First, positive x:sqrt(6 -2m) ∈ [m, m +0.5)But m must be a semi-integer, and since x is positive, m must be non-negative? Because if m is negative, then the interval [m, m +0.5) would include negative numbers, but x is positive here. Wait, no. If we are considering positive x, then m could be negative, but x is positive. But the interval [m, m +0.5) for m negative would include some negative numbers and maybe some positive numbers if m is close to zero. Wait, but sqrt(6 -2m) is always positive. So for sqrt(6 -2m) to be in [m, m +0.5), m must satisfy m ≤ sqrt(6 -2m) <m +0.5. But since sqrt(6 -2m) is positive, m must be less than sqrt(6 -2m). If m is negative, this is automatically true because sqrt(6 -2m) is positive and m is negative. Let me check.Case 1: m is negative.So m = -k/2 where k is a positive integer. Then, sqrt(6 -2m) = sqrt(6 +k). So x = sqrt(6 +k) must be in [m, m +0.5). But m is negative, and x is positive, so [m, m +0.5) is on the negative side. So x being positive cannot be in that interval. Therefore, for positive x, m must be non-negative.So, m ≥0.So, m is a semi-integer (i.e., m =0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0) and we need sqrt(6 -2m) ∈ [m, m +0.5).Let's check for each m:m =0: sqrt(6) ≈2.449. Check if 2.449 ∈ [0,0.5). No.m =0.5: sqrt(5)≈2.236. Check if 2.236 ∈ [0.5,1.0). No.m =1.0: sqrt(4)=2.0. Check if 2.0 ∈ [1.0,1.5). No.m =1.5: sqrt(3)≈1.732. Check if 1.732 ∈ [1.5,2.0). Yes. So this is valid.m =2.0: sqrt(2)≈1.414. Check if 1.414 ∈ [2.0,2.5). No.m =2.5: sqrt(1)=1.0. Check if 1.0 ∈ [2.5,3.0). No.m =3.0: sqrt(0)=0. Check if 0 ∈ [3.0,3.5). No.So only m=1.5 gives a valid positive x.For negative x: x = -sqrt(6 -2m). Then, x must be in [m, m +0.5). Since x is negative, m must be ≤ x <m +0.5. Since x is negative, m must also be negative or zero.But m is a semi-integer. Let's take m as negative semi-integers: m = -0.5, -1.0, -1.5, -2.0, etc.So for each m ≤0 (considering m as negative semi-integers), check if -sqrt(6 -2m) ∈ [m, m +0.5).Let's check:m =0: x = -sqrt(6)≈-2.449. Check if -2.449 ∈ [0,0.5). No.m =-0.5: x = -sqrt(6 -2*(-0.5))= -sqrt(7)≈-2.645. Check if -2.645 ∈ [-0.5,0.0). No.m =-1.0: x = -sqrt(6 -2*(-1.0))= -sqrt(8)≈-2.828. Check if -2.828 ∈ [-1.0,-0.5). No.m =-1.5: x = -sqrt(6 -2*(-1.5))= -sqrt(9)= -3.0. Check if -3.0 ∈ [-1.5,-1.0). No.m =-2.0: x = -sqrt(6 -2*(-2.0))= -sqrt(10)≈-3.162. Check if -3.162 ∈ [-2.0,-1.5). No.m =-2.5: x = -sqrt(6 -2*(-2.5))= -sqrt(11)≈-3.316. Check if ∈ [-2.5,-2.0). No.And so on. As m becomes more negative, 6 -2m becomes larger, so sqrt(6 -2m) becomes larger, so x = -sqrt(6 -2m) becomes more negative, moving further left from m. Therefore, none of these will satisfy x ∈ [m, m +0.5) because x is more negative than m.Thus, there are no solutions for negative x.Therefore, the only solution is x = sqrt(3).But let me check this solution again. If x = sqrt(3) ≈1.732, then ]x[ is the greatest semi-integer not exceeding x. The semi-integers less than or equal to 1.732 are 1.5, 1.0, 0.5, etc. So the greatest one is 1.5. Therefore, ]x[=1.5. Then, x² + 2*1.5 =3 +3=6, which matches. So that's correct.Is there any other possible solution? Let me check m=1.5 again. x=sqrt(3) is in [1.5,2.0). So yes, and that's the only solution. So the equation has only one solution: x=sqrt(3).Wait, but the problem says "solve the equation", so maybe there's another solution. Let me check m=1.5 again. Is there another x in [1.5,2.0) such that x² +2*1.5 =6? So x²=3. So x=sqrt(3)≈1.732 and x=-sqrt(3)≈-1.732. But -sqrt(3) is not in [1.5,2.0). So only positive sqrt(3) is valid. So only one solution.Alternatively, maybe if m= something else, but I don't think so. For example, let's suppose m=2.0. Then x²=6-4=2, x≈1.414. But x must be in [2.0,2.5). Not valid. Similarly, m=2.5, x=1, not in [2.5,3.0). So no. So only solution is x=sqrt(3).Wait, but let me check if there's another semi-integer m where x could be in [m, m+0.5). For example, m=1.0, x=2.0, but x=2.0 is not in [1.0,1.5). So no.Alternatively, m=1.5, x=sqrt(3)≈1.732. That's correct.Wait, let me check for x=1.732. ]x[=1.5, so 1.732² +2*1.5≈3+3=6. Correct.Is there a possibility that x is a semi-integer? For example, x=1.5. Then ]x[=1.5, so x² +2*1.5=2.25+3=5.25≠6. Not a solution. x=2.0: ]x[=2.0, equation is 4 +4=8≠6. x=0.5: ]x[=0.5, equation is0.25 +1=1.25≠6. So no.Thus, the only solution is x=sqrt(3). Therefore, the answer is sqrt(3), which is approximately 1.732, but exact value is sqrt(3).Final AnswerThe solution to the equation is boxed{sqrt{3}}.</think>To solve the equation (x^2 + 2 cdot ]x[ = 6), where (]x[) denotes the semi-integer part of (x), we need to consider the definition of semi-integers and the semi-integer part function. A semi-integer is a number (x) such that (2x) is an integer, and the semi-integer part (]x[) is the greatest semi-integer not exceeding (x).1. Understanding Semi-Integers: Semi-integers are numbers of the form (frac{k}{2}) where (k) is an integer. The semi-integer part function (]x[) rounds (x) down to the nearest semi-integer.2. Interval Analysis: For each semi-integer (m), we consider the interval ([m, m + 0.5)). Within each interval, (]x[ = m), so we substitute (m) into the equation and solve for (x).3. Substituting and Solving: The equation becomes (x^2 + 2m = 6), leading to (x^2 = 6 - 2m). We need to check if the solutions (x = pmsqrt{6 - 2m}) lie within the interval ([m, m + 0.5)).4. Checking Valid Solutions: - For positive (x), we check if (sqrt{6 - 2m}) lies in ([m, m + 0.5)). - For negative (x), we check if (-sqrt{6 - 2m}) lies in ([m, m + 0.5)).5. Finding the Valid Interval: - When (m = 1.5), the equation becomes (x^2 = 6 - 2 cdot 1.5 = 3), leading to (x = sqrt{3}). The interval ([1.5, 2.0)) includes (sqrt{3} approx 1.732). - Negative solutions and other intervals do not yield valid results as they fall outside their respective intervals.6. Verification: Substituting (x = sqrt{3}) and (]x[ = 1.5) into the equation confirms ( (sqrt{3})^2 + 2 cdot 1.5 = 3 + 3 = 6 ).Thus, the solution to the equation is (boxed{sqrt{3}}).,<think>Alright, so I have this equation to solve: (x^2 + 2 cdot ]x[ = 6). The problem defines the semi-integer part of a number (x) as the greatest semi-integer not exceeding (x), denoted by (]x[). A semi-integer is a number (x) such that (2x) is an integer. First, let me make sure I understand what a semi-integer is. If (2x) is an integer, then (x) must be a multiple of (0.5). For example, numbers like 0, 0.5, 1, 1.5, 2, etc., are semi-integers because doubling them gives integers. Conversely, numbers like 0.3 or 1.2 are not semi-integers because doubling them gives 0.6 and 2.4, which aren't integers. So semi-integers are all numbers of the form (k/2) where (k) is an integer.Now, the semi-integer part (]x[) is the greatest semi-integer not exceeding (x). So this is similar to the floor function, but instead of rounding down to the nearest integer, we're rounding down to the nearest half-integer. For example, if (x = 1.3), the semi-integers less than or equal to 1.3 are 1.0, 0.5, 0.0, etc. So the greatest one is 1.0, hence (]1.3[ = 1.0). If (x = 1.5), then (]x[ = 1.5) because 1.5 is a semi-integer. If (x = 1.6), then (]x[ = 1.5). So, basically, (]x[) is the largest number (n) such that (n leq x) and (2n) is an integer.So, to solve the equation (x^2 + 2 cdot ]x[ = 6), we need to consider the possible values of (]x[) based on where (x) lies. Since (]x[) depends on (x), we can break down the problem into intervals where (]x[) is constant and solve the equation within each interval.Let me start by considering how (]x[) behaves. Suppose (x) is in the interval ([n, n + 0.5)), where (n) is a semi-integer. Then (]x[ = n). For example, if (x) is in ([1.0, 1.5)), then (]x[ = 1.0). If (x) is exactly (1.5), then (]x[ = 1.5), and the next interval would start at (1.5).Therefore, the semi-integer part function (]x[) steps by 0.5 each time. So, the intervals where (]x[) is constant are of the form ([k/2, (k+1)/2)) for integers (k). Therefore, for each integer (k), if (x in [k/2, (k+1)/2)), then (]x[ = k/2). So, for example, if (x) is in [1.0, 1.5), then (]x[ = 1.0); if (x) is in [1.5, 2.0), then (]x[ = 1.5); and so on.Given that, we can consider each interval where (]x[ = n) (where (n) is a semi-integer) and solve the equation (x^2 + 2n = 6) for (x) in that interval. Then, check if the solution (x) actually lies within the interval.So, step by step:1. For each semi-integer (n = k/2) (where (k) is an integer), consider the interval ([n, n + 0.5)).2. In this interval, the equation becomes (x^2 + 2n = 6).3. Solve for (x): (x^2 = 6 - 2n).4. Find the solutions (x = sqrt{6 - 2n}) and (x = -sqrt{6 - 2n}).5. Check if these solutions lie within the interval ([n, n + 0.5)).6. If they do, then those are valid solutions; otherwise, discard them.But first, we need to figure out the possible values of (n) such that (6 - 2n geq 0), because (x^2) can't be negative. So, (6 - 2n geq 0 implies n leq 3). Since (n) is a semi-integer, the maximum possible (n) is 3.0. Then, (n) can be 3.0, 2.5, 2.0, 1.5, 1.0, 0.5, 0.0, -0.5, etc. But we need to consider all (n) such that (n leq 3.0), but also such that the interval ([n, n + 0.5)) is valid. However, since (n) is a semi-integer, the possible (n) are all numbers of the form (k/2) where (k) is an integer, and (k/2 leq 3.0), so (k leq 6). Therefore, (k) can be from, say, some lower bound up to 6. But we might need to consider negative (n) as well if (x) is negative.But before that, let's consider the possible intervals for (x). Since (x) can be both positive and negative, we need to check both sides. However, since (x^2) is always non-negative, let's see how the equation behaves. The equation is (x^2 + 2n = 6). So, if (x) is negative, (n) would be a semi-integer less than or equal to (x). But if (x) is negative, then (n) would also be negative or zero.Let me start by considering positive (x) and then negative (x).Starting with positive (x):First, consider (n = 3.0). The interval is [3.0, 3.5). Then the equation becomes (x^2 + 2*3 = x^2 + 6 = 6), so (x^2 = 0), which implies (x = 0). But (x = 0) is not in [3.0, 3.5), so no solution here.Next, (n = 2.5): interval [2.5, 3.0). Equation: (x^2 + 2*2.5 = x^2 + 5 = 6), so (x^2 = 1). Solutions: (x = 1) and (x = -1). Check if these are in [2.5, 3.0). But (1) and (-1) are not in that interval. So no solution here.Next, (n = 2.0): interval [2.0, 2.5). Equation: (x^2 + 4 = 6), so (x^2 = 2). Solutions (x = sqrt{2} approx 1.414) and (x = -sqrt{2} approx -1.414). Check if these are in [2.0, 2.5). But 1.414 is less than 2.0, so no solution here.Next, (n = 1.5): interval [1.5, 2.0). Equation: (x^2 + 3 = 6), so (x^2 = 3). Solutions (x = sqrt{3} approx 1.732) and (x = -sqrt{3} approx -1.732). Check if (x) is in [1.5, 2.0). (sqrt{3} approx 1.732) is in [1.5, 2.0). So, that's a valid solution. The negative solution, (-sqrt{3}), is approximately -1.732. Let's check where that would fall. For negative (x), the intervals would be different. Let's hold onto that thought for a moment. So for positive (x), in the interval [1.5, 2.0), we have a solution (x = sqrt{3}). Let's verify if this is correct.Wait, hold on. If (x) is in [1.5, 2.0), then (]x[ = 1.5). So substituting into the equation: (x^2 + 2*1.5 = x^2 + 3 = 6), so (x^2 = 3), so (x = sqrt{3}) or (x = -sqrt{3}). But since we are considering the interval [1.5, 2.0), which is for positive (x), (x = sqrt{3} approx 1.732) is indeed in [1.5, 2.0). So that's a valid solution. However, (x = -sqrt{3}) is in the negative side, so we have to check that separately when considering negative intervals.So moving on. Next interval for positive (x): (n = 1.0), interval [1.0, 1.5). Equation: (x^2 + 2*1 = x^2 + 2 = 6), so (x^2 = 4), solutions (x = 2) and (x = -2). Check if (x) is in [1.0, 1.5). (2) is not in that interval, so no solution here.Next, (n = 0.5): interval [0.5, 1.0). Equation: (x^2 + 2*0.5 = x^2 + 1 = 6), so (x^2 = 5), solutions (x = sqrt{5} approx 2.236) and (x = -sqrt{5} approx -2.236). Check if they're in [0.5, 1.0). (sqrt{5}) is approximately 2.236, which is not in [0.5, 1.0). So no solution here.Next, (n = 0.0): interval [0.0, 0.5). Equation: (x^2 + 0 = 6), so (x^2 = 6), solutions (x = sqrt{6} approx 2.449) and (x = -sqrt{6} approx -2.449). Check if in [0.0, 0.5). Positive solution is (sqrt{6}), which is not in [0, 0.5). So no solution here.So for positive (x), we found one solution in the interval [1.5, 2.0): (x = sqrt{3}). Now, let's check negative (x). For negative (x), the semi-integer part (]x[) is the greatest semi-integer not exceeding (x). Since (x) is negative, the semi-integers less than or equal to (x) are going to be negative or zero. For example, if (x = -1.3), the semi-integers less than or equal to -1.3 are -1.5, -2.0, -2.5, etc., but the greatest one is -1.5. Wait, no. Wait, hold on. If (x = -1.3), the semi-integers are numbers like -1.0, -1.5, -2.0, etc. But semi-integers are spaced by 0.5. So, the semi-integers less than or equal to -1.3 would be -1.5, -2.0, etc., but the greatest one is -1.5. Because -1.5 is less than -1.3, and the next semi-integer is -2.0, which is even smaller. So, (]x[ = -1.5) if (x in [-1.5, -1.0)). Wait, similar to the positive side, but shifted.Wait, let me formalize this. For negative (x), the semi-integer part (]x[) is the greatest semi-integer less than or equal to (x). So, for example:- If (x in [-0.5, 0)), then (]x[ = -0.5)? Wait, no. Wait, if (x) is in [-0.5, 0), the semi-integers less than or equal to (x) are -0.5, -1.0, -1.5, etc. But the greatest semi-integer less than or equal to (x) would be -0.5 if (x geq -0.5). Wait, no. Wait, if (x in [-0.5, 0)), then the semi-integers less than or equal to (x) would be -0.5, -1.0, etc., but the greatest one is -0.5. So yes, (]x[ = -0.5) for (x in [-0.5, 0)). Similarly, for (x in [-1.0, -0.5)), (]x[ = -1.0). For (x in [-1.5, -1.0)), (]x[ = -1.5), and so on.So, in general, for negative (x), the semi-integer part (]x[) is the next lower semi-integer. So, intervals for negative (x) would be:- ([-0.5, 0)): (]x[ = -0.5)- ([-1.0, -0.5)): (]x[ = -1.0)- ([-1.5, -1.0)): (]x[ = -1.5)- Etc.So, similar to the positive side, but each interval is of length 0.5, starting at each semi-integer going downward. So, for each semi-integer (n = -k/2) where (k) is a positive integer, the interval is ([n, n + 0.5)), but since (n) is negative, (n + 0.5) is a less negative number. For example, for (n = -1.5), the interval is ([-1.5, -1.0)).Therefore, to solve for negative (x), we can iterate through these intervals, set (n) accordingly, solve (x^2 + 2n = 6), and check if the solutions lie within the interval.Let's proceed step by step for negative (x):First interval: (n = -0.5), interval ([-0.5, 0)). Equation: (x^2 + 2*(-0.5) = x^2 - 1 = 6), so (x^2 = 7), solutions (x = sqrt{7} approx 2.645) and (x = -sqrt{7} approx -2.645). Check if in ([-0.5, 0)). The negative solution is approximately -2.645, which is not in ([-0.5, 0)). So no solution here.Next, (n = -1.0), interval ([-1.0, -0.5)). Equation: (x^2 + 2*(-1.0) = x^2 - 2 = 6), so (x^2 = 8), solutions (x = sqrt{8} approx 2.828) and (x = -sqrt{8} approx -2.828). Check if in ([-1.0, -0.5)). The negative solution is approximately -2.828, which is less than -1.0, so not in the interval. No solution here.Next, (n = -1.5), interval ([-1.5, -1.0)). Equation: (x^2 + 2*(-1.5) = x^2 - 3 = 6), so (x^2 = 9), solutions (x = 3) and (x = -3). Check if in ([-1.5, -1.0)). The negative solution is -3, which is less than -1.5, so not in the interval. No solution here.Next, (n = -2.0), interval ([-2.0, -1.5)). Equation: (x^2 + 2*(-2.0) = x^2 - 4 = 6), so (x^2 = 10), solutions (x = sqrt{10} approx 3.162) and (x = -sqrt{10} approx -3.162). Check if in ([-2.0, -1.5)). -3.162 is less than -2.0, so no solution here.Continuing this pattern, as (n) becomes more negative, (x^2 = 6 - 2n) becomes larger, leading to more negative solutions which are outside the interval. For example, (n = -2.5), interval ([-2.5, -2.0)). Equation: (x^2 + 2*(-2.5) = x^2 -5 =6), so (x^2=11), solutions (pm sqrt{11} approx pm3.316). The negative solution is -3.316, which is less than -2.5, so not in the interval.Therefore, for negative (x), even though we can continue checking, it's clear that as (n) decreases, the required (x) becomes more negative, but the equation (x^2 = 6 - 2n) with (n) negative leads to (x^2 = 6 + |2n|), so (x) becomes more negative as (n) decreases, but the interval we are considering is moving left (more negative) as well. However, since the left endpoint of each interval is (n), and (x) must be (geq n), but (x = -sqrt{6 - 2n}), which equals (-sqrt{6 + |2n|}). So as (n) becomes more negative (i.e., ( |n| ) increases), (x = -sqrt{6 + |2n|}) becomes more negative. Let's check whether (x) can be in the interval ([n, n + 0.5)) for some (n).Suppose (n = -k/2) where (k) is a positive integer. Then, (x = -sqrt{6 - 2*(-k/2)} = -sqrt{6 + k}). The interval is ([-k/2, -k/2 + 0.5)). So we need:(-k/2 leq -sqrt{6 + k} < -k/2 + 0.5)But let's substitute (n = -k/2):(n leq x < n + 0.5)But (x = -sqrt{6 - 2n}). Since (n) is negative, let's substitute (n = -m) where (m > 0). Then:(x = -sqrt{6 - 2*(-m)} = -sqrt{6 + 2m})And the interval becomes:(-m leq -sqrt{6 + 2m} < -m + 0.5)Multiply all parts by -1, which reverses inequalities:(m geq sqrt{6 + 2m} > m - 0.5)But this seems complicated. Let's check for possible integer (m) such that (k = 2m), but maybe it's better to test specific values.Let me try (n = -3.0) (so (m = 3.0)), interval ([-3.0, -2.5)). Equation: (x^2 + 2*(-3) = x^2 -6 =6), so (x^2 =12), solutions (x = sqrt{12} approx 3.464) and (x = -sqrt{12} approx -3.464). Check if (x in [-3.0, -2.5)). -3.464 is less than -3.0, so not in the interval.Next, (n = -3.5), interval ([-3.5, -3.0)). Equation: (x^2 + 2*(-3.5) = x^2 -7 =6), so (x^2=13), solutions (x approx pm3.605). Check (x = -3.605), which is less than -3.5, not in the interval.Continuing, (n = -4.0), interval ([-4.0, -3.5)). Equation: (x^2 -8 =6), so (x^2=14), solutions (pmsqrt{14} approx pm3.741). Negative solution is -3.741, which is less than -4.0? No, -3.741 is greater than -4.0, so it is in [-4.0, -3.5). Wait, hold on. If (x = -3.741), then (x approx -3.741). The interval is ([-4.0, -3.5)). So, -4.0 ≤ x < -3.5. -3.741 is approximately -3.741, which is greater than -3.75? Wait, -3.741 is approximately -3.741, which is greater than -3.75 (since -3.741 is -3 and 0.741, whereas -3.75 is -3 and 0.75). So -3.741 is actually greater than -3.75, which is in the interval [-4.0, -3.5)? Wait, no. Wait, the interval [-4.0, -3.5) includes numbers from -4.0 up to but not including -3.5. So, -3.741 is between -4.0 and -3.5. Because -4.0 is less than -3.741 (since it's more negative), but wait: -3.741 is greater than -4.0. So, -4.0 ≤ -3.741 < -3.5 is false because -3.741 is approximately -3.741, which is less than -3.5? Wait, no. Wait, -3.5 is -3.5, and -3.741 is less than -3.5? No, -3.741 is more negative than -3.5, so it's less than -3.5. Wait, but in terms of magnitude: 3.741 is greater than 3.5, but as negatives, -3.741 is less than -3.5. Therefore, -3.741 is in [-4.0, -3.5). Because -4.0 ≤ -3.741 < -3.5. So yes, -3.741 is in that interval.Therefore, when (n = -4.0), solving (x = -sqrt{6 - 2n} = -sqrt{6 - 2*(-4)} = -sqrt{6 +8} = -sqrt{14} approx -3.741). Check if (x in [-4.0, -3.5)). Since -4.0 ≤ -3.741 < -3.5, yes, it is. Therefore, (x = -sqrt{14}) is a valid solution in this interval.Wait, this is interesting. So, even though we went down several intervals, we found a solution here. Let me verify this.For (n = -4.0), which is a semi-integer (since 2*(-4.0) = -8, which is an integer). The interval is ([-4.0, -3.5)). The equation becomes (x^2 + 2*(-4.0) = x^2 -8 =6), so (x^2 =14), solutions (x = sqrt{14}) and (x = -sqrt{14}). Since we are considering negative (x), (x = -sqrt{14} approx -3.741). Check if this is in ([-4.0, -3.5)). -3.741 is between -4.0 and -3.5, so yes. Therefore, (x = -sqrt{14}) is a valid solution.Similarly, let's check the next interval (n = -4.5), interval ([-4.5, -4.0)). Equation: (x^2 + 2*(-4.5) = x^2 -9 =6), so (x^2 =15), solutions (x approx pm3.872). Negative solution (x approx -3.872). Check if in ([-4.5, -4.0)). -3.872 is greater than -4.0, so not in the interval. No solution here.Continuing to (n = -5.0), interval ([-5.0, -4.5)). Equation: (x^2 -10 =6), so (x^2=16), solutions (x = pm4). Check (x = -4), which is in ([-5.0, -4.5))? No, -4 is greater than -4.5, so not in the interval. No solution here.Next, (n = -5.5), interval ([-5.5, -5.0)). Equation: (x^2 -11 =6), so (x^2=17), solutions (x approx pm4.123). Negative solution (x approx -4.123). Check if in ([-5.5, -5.0)). -4.123 is greater than -5.0, so not in the interval.Similarly, as (n) decreases further, the solutions (x = -sqrt{6 - 2n}) become more negative but the intervals are moving left even faster, so the solutions will not lie in the intervals. For example, (n = -6.0), interval ([-6.0, -5.5)). Equation: (x^2 -12 =6), so (x^2=18), solutions (x approx pm4.242). Negative solution is -4.242, which is not in ([-6.0, -5.5)).Therefore, the only negative solution we found is (x = -sqrt{14}) in the interval ([-4.0, -3.5)) with (n = -4.0). Let me check if there are any other solutions.Wait, going back, when (n = -1.5), we had (x^2 =9), so (x = pm3), but (x = -3) was not in the interval ([-1.5, -1.0)). However, when (n = -4.0), (x = -sqrt{14} approx -3.741) is in ([-4.0, -3.5)). So, this is a valid solution.Let me check if there's a solution in another interval. For example, let me check (n = -2.5), interval ([-2.5, -2.0)). Equation: (x^2 + 2*(-2.5) = x^2 -5 =6), so (x^2=11), solutions (x = pmsqrt{11} approx pm3.316). The negative solution is -3.316. Check if in ([-2.5, -2.0)). -3.316 is less than -2.5, so no.Similarly, (n = -3.0), interval ([-3.0, -2.5)). Equation: (x^2 -6 =6), so (x^2=12), solutions (x = pmsqrt{12} approx pm3.464). Negative solution is -3.464. Check if in ([-3.0, -2.5)). -3.464 is less than -3.0, so no.So, seems like the only negative solution is (x = -sqrt{14}). Wait, but let me check for (n = -4.0):Equation: (x^2 + 2*(-4.0) = x^2 -8 =6 implies x^2=14 implies x = pmsqrt{14}). Then (x = -sqrt{14} approx -3.741). Check if in ([-4.0, -3.5)). Yes, since -4.0 ≤ -3.741 < -3.5. So that works.So, so far, we have two solutions: (x = sqrt{3}) and (x = -sqrt{14}). Wait, but let's check if there are any other solutions. Wait, going back to the positive side: when (n = 1.5), the solution was (x = sqrt{3} approx1.732), which is in [1.5, 2.0). Correct. Is there another positive solution? Let's check (n = 3.0), but the solution was 0, which wasn't in [3.0, 3.5). So no. Similarly, for other (n), the solutions fell outside the intervals.What about (n = 0.0)? The solution was (x = sqrt{6} approx 2.449), which is not in [0.0, 0.5). So no.Therefore, the only positive solution is (x = sqrt{3}) and the only negative solution is (x = -sqrt{14}).Wait, but let me verify these solutions in the original equation.First, check (x = sqrt{3}). Compute (]x[). Since (x approx1.732), which is in [1.5, 2.0), so (]x[ =1.5). Then, the equation becomes:((sqrt{3})^2 + 2*1.5 = 3 + 3 =6). Which matches. So correct.Now check (x = -sqrt{14} approx -3.741). Compute (]x[). Since (x approx -3.741), which is in [-4.0, -3.5), so (]x[ =-4.0). Then, the equation becomes:((-sqrt{14})^2 + 2*(-4.0) =14 -8 =6). Which is correct. So that works.Are there any other solutions? Let me check another interval in the negative side. For example, (n = -0.5), interval [-0.5, 0). Equation: (x^2 -1 =6), so (x^2=7), solutions (x approx pm2.645). Negative solution is -2.645, which is not in [-0.5, 0). No.(n = -1.0), interval [-1.0, -0.5). Equation: (x^2 -2 =6), (x^2=8), (x approx pm2.828). Negative solution is -2.828, not in [-1.0, -0.5). No.(n = -1.5), interval [-1.5, -1.0). Equation: (x^2 -3 =6), (x^2=9), (x = pm3). Negative solution -3, not in [-1.5, -1.0). No.(n = -2.0), interval [-2.0, -1.5). Equation: (x^2 -4 =6), (x^2=10), (x approx pm3.162). Negative solution -3.162, which is in [-3.162, -3.0)? Wait, interval is [-2.0, -1.5). -3.162 is less than -2.0, so no.Similarly, (n = -2.5), interval [-2.5, -2.0). Equation: (x^2 -5 =6), (x^2=11), (x approx pm3.316). Negative solution -3.316, not in [-2.5, -2.0). So no.Wait, so only (x = -sqrt{14}) is valid for negative (x). Therefore, the solutions are (x = sqrt{3}) and (x = -sqrt{14}).Wait, but wait a second. Let me check for (n = -3.5), interval [-3.5, -3.0). Equation: (x^2 +2*(-3.5)=x^2 -7=6), so (x^2=13), (x = pmsqrt{13} approx pm3.605). The negative solution is -3.605. Check if in [-3.5, -3.0). -3.605 is less than -3.5? No, because -3.605 is more negative than -3.5, so it is in [-3.5, -3.0)? Wait, no. -3.5 ≤ x < -3.0. -3.605 is less than -3.5, so not in the interval. So, no.Similarly, for (n = -4.5), interval [-4.5, -4.0). Equation: (x^2 -9 =6), (x^2=15), solutions (x approx pm3.872). Negative solution -3.872. Check if in [-4.5, -4.0). -3.872 is greater than -4.0, so not in the interval. So no.Thus, seems like the only solutions are (x = sqrt{3}) and (x = -sqrt{14}).But wait, let's check (n = 3.0) again. Interval [3.0, 3.5). Equation: (x^2 +6 =6), so (x=0). Not in the interval. Correct. What about (n = 0.5), interval [0.5, 1.0). Equation gives (x = sqrt{5} approx2.236). Not in [0.5,1.0). So no.Wait, another thought: could there be a solution where (x) is a semi-integer? For example, if (x = 1.5), then (]x[ =1.5). Plug into the equation: (1.5^2 + 2*1.5 =2.25 +3 =5.25 ≠6). So that's not a solution. What about (x = 2.0). Then (]x[ =2.0). Equation: (4 +4=8≠6). Not a solution.Similarly, (x = -3.5). Then (]x[ =-3.5). Equation: ((-3.5)^2 +2*(-3.5)=12.25 -7=5.25≠6). Not a solution.Therefore, all solutions are non-semi-integer.Thus, the solutions are (x = sqrt{3}) and (x = -sqrt{14}). Let me verify once more.For (x = sqrt{3}):- (x approx1.732), which is in [1.5,2.0), so (]x[=1.5).- (x^2 +2*1.5 =3 +3=6). Correct.For (x = -sqrt{14}):- (x approx-3.741), which is in [-4.0,-3.5), so (]x[=-4.0).- (x^2 +2*(-4.0)=14 -8=6). Correct.Therefore, both solutions satisfy the equation.But wait, let me check if there are any other intervals that might have been missed. For example, (n = -4.0) gave us a solution. Let's check (n = -4.5): equation gives (x = -sqrt{6 -2*(-4.5)} = -sqrt{6 +9} = -sqrt{15} approx -3.872). The interval is [-4.5, -4.0). So, is -3.872 in [-4.5, -4.0)? No, because -3.872 is greater than -4.0. So, not in the interval.Similarly, if I check (n = -6.0): equation gives (x = -sqrt{6 -2*(-6)} = -sqrt{6 +12} = -sqrt{18} approx -4.242). Interval is [-6.0, -5.5). -4.242 is not in that interval. So no.Therefore, only two solutions: (sqrt{3}) and (-sqrt{14}).Wait, but wait another moment. Let me check if the equation could have more solutions.Suppose (x) is a semi-integer. Then, (]x[ =x). So, the equation becomes (x^2 +2x =6). Solve: (x^2 +2x -6 =0). Solutions: (x = [-2 pm sqrt{4 +24}]/2 = [-2 pm sqrt{28}]/2 = [-2 pm 2sqrt{7}]/2 = -1 pm sqrt{7}). (sqrt{7} approx2.645). So, solutions: (x approx -1 +2.645 =1.645) and (x approx -1 -2.645 =-3.645). Are these semi-integers?1.645 is approximately 1.645, which is not a semi-integer. Semi-integers are 1.5, 2.0, etc. Similarly, -3.645 is not a semi-integer. Therefore, even if we considered the case where (x) is a semi-integer, the solutions are not semi-integers, so this case doesn't contribute any solutions.Therefore, the only solutions are (x = sqrt{3}) and (x = -sqrt{14}).But let me check once again for possible intervals. Maybe there's a semi-integer (n) such that the equation (x^2 =6 -2n) gives a solution (x) in the interval ([n, n +0.5)) that I missed.For positive (x):- (n = 1.5): solution (x = sqrt{3} approx1.732), in [1.5,2.0): yes.For negative (x):- (n = -4.0): solution (x = -sqrt{14} approx-3.741), in [-4.0, -3.5): yes.All other (n) gave solutions outside their respective intervals. Therefore, these are the only solutions.Thus, the final answers are (x = sqrt{3}) and (x = -sqrt{14}). But let me check if the problem expects multiple solutions. The problem says "Solve the equation", so we need to present all real solutions. Therefore, the solutions are (x = sqrt{3}) and (x = -sqrt{14}).Final AnswerThe solutions are (boxed{sqrt{3}}) and (boxed{-sqrt{14}}).</think>To solve the equation (x^2 + 2 cdot ]x[ = 6), we need to consider the semi-integer part (]x[), which is the greatest semi-integer not exceeding (x). A semi-integer is a number (x) such that (2x) is an integer, meaning (x) is a multiple of (0.5).1. Understanding Semi-integer Part: - The semi-integer part (]x[) of (x) is the largest semi-integer less than or equal to (x). - For example, if (x) is in the interval ([1.5, 2.0)), then (]x[ = 1.5).2. Interval Analysis: - For each semi-integer (n = k/2) (where (k) is an integer), consider the interval ([n, n + 0.5)). - Solve the equation (x^2 + 2n = 6) within each interval and check if the solution lies within the interval.3. Positive (x) Analysis: - For (n = 1.5) (interval ([1.5, 2.0))): - Equation: (x^2 + 2 cdot 1.5 = 6) - Solution: (x^2 = 3 implies x = sqrt{3} approx 1.732) - Check: (sqrt{3} approx 1.732) is in ([1.5, 2.0)), so valid.4. Negative (x) Analysis: - For (n = -4.0) (interval ([-4.0, -3.5))): - Equation: (x^2 + 2 cdot (-4.0) = 6) - Solution: (x^2 = 14 implies x = -sqrt{14} approx -3.741) - Check: (-sqrt{14} approx -3.741) is in ([-4.0, -3.5)), so valid.5. Verification: - For (x = sqrt{3}): (]x[ = 1.5), equation holds as ((sqrt{3})^2 + 2 cdot 1.5 = 3 + 3 = 6). - For (x = -sqrt{14}): (]x[ = -4.0), equation holds as ((- sqrt{14})^2 + 2 cdot (-4.0) = 14 - 8 = 6).Thus, the solutions to the equation are (boxed{sqrt{3}}) and (boxed{-sqrt{14}}).]

🔑:Solution: Consider two cases.1) The number x- is half-integer, then ] xleft[=xright. and the original equation will take the form x^{2}+2 x-6=0. The roots of this equation are x_{1,2}=-1 pm sqrt{7}, but then the numbers 2 x_{1,2} are not integers, so there are no solutions.2) There is an equality x=frac{n}{2}+r, where n in mathbb{Z} and 00 and can only have a solution if the inequalityfrac{n}{2}0end{array}right.end{gathered}The first inequality of the system is satisfied when -2-2 sqrt{7}-3+4 sqrt{2} or n4(6-n)>(n+1)^{2} left{begin{array}{l}n^{2}+4 n-24>0 n^{2}+6 n-23-2+2 sqrt{7} or n<-2-2 sqrt{7}. Since -9<-3-4 sqrt{2}<-8, -8<-2-2 sqrt{7}<-7, and -3+4 sqrt{2}<-2+2 sqrt{7}, then n=-8- is the only integer value satisfying the system. In this case, x=frac{n}{2}+r=-sqrt{6-n}=-sqrt{14}.Answer: sqrt{3},-sqrt{14}

🧠:[<think>Okay, let's see. So, the problem is about translating a point inside a triangle. The translation moves any point P(a, b) to P1(a+2, b-6). After this translation, point A becomes A1(4, -3). We need to find the original coordinates of point A from the given options.Hmm, translation in geometry usually means sliding the point without rotating or resizing. So, if every point is translated by adding 2 to the x-coordinate and subtracting 6 from the y-coordinate, then to find the original point, we should reverse this operation, right? That is, if after translation the point is (a+2, b-6), then the original point must be (a1 - 2, b1 + 6), where (a1, b1) are the coordinates after translation.Wait, let me make sure. Suppose the original point is A(x, y). After translating it by (2, -6), meaning moving 2 units to the right and 6 units down, it becomes A1(x + 2, y - 6). According to the problem, A1 is (4, -3). So, we can set up equations:x + 2 = 4y - 6 = -3Solving for x and y:From the first equation: x = 4 - 2 = 2From the second equation: y = -3 + 6 = 3So, the original point A is (2, 3). Let me check the options:Option D is (2,3). So, that should be the answer.Wait a second, let me verify again. If we take point (2,3) and apply the translation: x becomes 2 + 2 = 4, y becomes 3 - 6 = -3. Which matches A1(4, -3). Yep, that seems right.Let me just make sure none of the other options work. Let's check option A: (6, -1). If we translate that, we get (6+2, -1-6) = (8, -7). Not (4, -3). So not A.Option B: (2, -6). Translating gives (2+2, -6-6) = (4, -12). Not -3. Not B.Option C: (-9,6). Translating: (-9+2,6-6) = (-7,0). Not matching. So, only D works.Therefore, the correct answer is D: (2,3).Final Answerboxed{D}</think>Given that any point ( P(a, b) ) inside (triangle ABC) is translated to point ( P_1(a+2, b-6) ). After this translation, point ( A ) corresponds to point ( A_1(4, -3) ). We need to find the coordinates of point ( A ).To reverse the translation, we subtract 2 from the x-coordinate and add 6 to the y-coordinate of the translated point ( A_1 ).1. Let the coordinates of point ( A ) be ( (x, y) ).2. After translation, the coordinates become ( (x+2, y-6) ).3. Given that the translated coordinates are ( (4, -3) ), we set up the equations: [ x + 2 = 4 quad text{and} quad y - 6 = -3 ]4. Solving these equations: [ x = 4 - 2 = 2 quad text{and} quad y = -3 + 6 = 3 ]Thus, the coordinates of point ( A ) are ( (2, 3) ), which corresponds to option D.[boxed{D}],<think>Okay, so I have this problem here about translating a point inside a triangle. Let me try to understand what it's asking. Alright, the problem says that for any point P(a, b) inside triangle ABC, after a translation, it corresponds to point P₁(a+2, b-6). So, translation means moving the point without rotating or resizing, right? So, translating a point (a, b) by adding 2 to the x-coordinate and subtracting 6 from the y-coordinate gives the new point (a+2, b-6). That makes sense. So, the translation vector here is (2, -6). Because in translations, you move each point by a certain vector. So, if the original point is (a, b), after translation, it's (a + h, b + k) where (h, k) is the translation vector. In this case, h is 2 and k is -6. Got it.Now, the problem states that point A corresponds to point A₁(4, -3) after this translation. So, we need to find the original coordinates of point A before the translation. Let's denote the original coordinates of point A as (x, y). After applying the translation, which is adding 2 to the x-coordinate and subtracting 6 from the y-coordinate, we get (x + 2, y - 6). According to the problem, this translated point is (4, -3). So, we can set up equations here.Let me write that down:x + 2 = 4y - 6 = -3So, solving for x and y. Starting with the x-coordinate:x + 2 = 4Subtract 2 from both sides:x = 4 - 2x = 2Okay, so the x-coordinate of point A is 2.Now for the y-coordinate:y - 6 = -3Add 6 to both sides:y = -3 + 6y = 3So, the y-coordinate of point A is 3.Therefore, the original coordinates of point A are (2, 3). Let me check the answer choices:A: (6, -1)B: (2, -6)C: (-9, 6)D: (2, 3)So, D is (2, 3), which matches what I found. Let me double-check my work to make sure I didn't make any mistakes. If point A is (2, 3), then translating it by adding 2 to the x-coordinate gives 2 + 2 = 4, and subtracting 6 from the y-coordinate gives 3 - 6 = -3. So, the translated point is indeed (4, -3), which matches the given A₁. That seems right. Just to be thorough, let's check the other options to see if any of them could also result in (4, -3) after the translation. Option A: (6, -1). Translating this would give (6 + 2, -1 -6) = (8, -7), which is not (4, -3). So that's not correct.Option B: (2, -6). Translating gives (2 + 2, -6 -6) = (4, -12). Not (4, -3). Nope.Option C: (-9, 6). Translating gives (-9 + 2, 6 -6) = (-7, 0). Not matching. So, none of the other options work. Therefore, D is the correct answer. Hmm, I think that's all. The key here was understanding that translation moves each point by a fixed vector, and since we know the image point after translation, we can reverse the translation to find the original coordinates. Instead of adding 2 and subtracting 6, we subtract 2 and add 6 to the image coordinates. Let me verify that again.If the translation is (a+2, b-6), then to reverse it, you take the image coordinates and subtract 2 from the x-coordinate and add 6 to the y-coordinate. So, if A₁ is (4, -3), then original A is (4 - 2, -3 + 6) = (2, 3). Yep, exactly what I did earlier. So that's correct. Therefore, the answer is D.Final Answerboxed{D}</think>Given that for any point ( P(a, b) ) inside (triangle ABC), after a translation, it corresponds to point ( P_1(a+2, b-6) ). This translation vector is ((2, -6)). Point ( A ) corresponds to point ( A_1(4, -3) ) after this translation. To find the original coordinates of point ( A ), we reverse the translation by subtracting 2 from the x-coordinate and adding 6 to the y-coordinate of ( A_1 ).Setting up the equations for the original coordinates ((x, y)) of point ( A ):[x + 2 = 4 quad text{and} quad y - 6 = -3]Solving for ( x ):[x + 2 = 4 implies x = 4 - 2 = 2]Solving for ( y ):[y - 6 = -3 implies y = -3 + 6 = 3]Thus, the original coordinates of point ( A ) are ((2, 3)). The correct answer is (boxed{D}).]

🔑:To solve this problem, we need to understand how the translation affects the coordinates of a point. Given that any point P(a,b) is translated to P_{1}(a+2,b-6), we can deduce the transformation rule applied to point A to get to its translated point A_{1}(4,-3).1. The translation moves point A 2 units to the right. This means we add 2 to the x-coordinate of A to get the x-coordinate of A_{1}.2. The translation moves point A 6 units down. This means we subtract 6 from the y-coordinate of A to get the y-coordinate of A_{1}.Given A_{1}(4,-3), we can reverse the translation to find the coordinates of A:- For the x-coordinate: 4 = a + 2 Rightarrow a = 4 - 2- For the y-coordinate: -3 = b - 6 Rightarrow b = -3 + 6Therefore, the coordinates of point A are (a,b) = (4-2, -3+6) = (2,3).Hence, the correct answer is boxed{D}.Explanation:This step-by-step solution demonstrates how to apply the translation rules in reverse to find the original coordinates of a point before it was translated. By understanding that translating a point to the right increases its x-coordinate and translating it down decreases its y-coordinate, we can easily reverse these operations to find the original point's coordinates.Review:This problem tests the concept of translation in the Cartesian plane, a fundamental aspect of coordinate geometry. It emphasizes the importance of understanding how to manipulate and reverse transformations to find original positions of geometric figures or points.