🧠:[<think>Okay, let's tackle this problem step by step. First, part (1) asks to find the extreme values of the function f(x) when k=0. The function becomes f(x) = (x - 1)e^x + 2. To find the extreme values, I need to find the critical points by taking the derivative of f(x) and setting it equal to zero.So, let's compute f'(x). The function is (x - 1)e^x plus 2. The derivative of (x - 1)e^x can be found using the product rule. Let me recall: the derivative of u*v is u'v + uv'. Let u = (x - 1), so u' = 1, and v = e^x, so v' = e^x. Therefore, the derivative of (x - 1)e^x is 1*e^x + (x - 1)*e^x, which simplifies to e^x + (x - 1)e^x. Combining like terms, that's (1 + x - 1)e^x = x e^x. Then, the derivative of the constant term 2 is 0. So, f'(x) = x e^x.Now, set f'(x) = 0 to find critical points: x e^x = 0. Since e^x is always positive for any real x, the only solution is x = 0. Therefore, there's a critical point at x = 0.To determine if this critical point is a minimum or maximum, we can use the second derivative test. Let's compute f''(x). The first derivative is x e^x, so we take the derivative of that. Again, using the product rule: u = x, u' = 1; v = e^x, v' = e^x. So, f''(x) = 1*e^x + x*e^x = (1 + x)e^x.Evaluate f''(x) at x = 0: (1 + 0)e^0 = 1*1 = 1, which is positive. Therefore, the function has a local minimum at x = 0. There are no other critical points, so this is the only extreme value.Now, let's find the value of f at x = 0. f(0) = (0 - 1)e^0 + 2 = (-1)(1) + 2 = -1 + 2 = 1. So, the extreme value is 1, and since it's a local minimum, the function has a minimum value of 1 at x = 0 when k = 0.Moving on to part (2). The problem states that for any x in [0, +∞), f(x) ≥ 1 must hold. We need to find the range of k for which this inequality is always true.First, let's rewrite the function f(x) = (x - 1)e^x - kx^2 + 2. The requirement is that f(x) ≥ 1 for all x ≥ 0. So, subtract 1 from both sides to get (x - 1)e^x - kx^2 + 2 - 1 ≥ 0, which simplifies to (x - 1)e^x - kx^2 + 1 ≥ 0. So, we need to ensure that (x - 1)e^x - kx^2 + 1 ≥ 0 for all x ≥ 0.Alternatively, we can think of this as f(x) ≥ 1, so rearranged: (x - 1)e^x - kx^2 + 2 ≥ 1 ⇒ (x - 1)e^x - kx^2 + 1 ≥ 0. Let's denote this expression as g(x) = (x - 1)e^x - kx^2 + 1. We need g(x) ≥ 0 for all x ≥ 0.Our goal is to find all real numbers k such that g(x) ≥ 0 for all x ≥ 0.To approach this, we can analyze the function g(x) and find its minimum over [0, ∞) and then set that minimum to be greater than or equal to zero. To find the minimum, we can take the derivative of g(x), find its critical points, and evaluate the minima.First, let's compute the derivative g'(x):g(x) = (x - 1)e^x - kx^2 + 1g'(x) = derivative of (x - 1)e^x + derivative of (-kx^2) + derivative of 1.We already computed the derivative of (x - 1)e^x earlier, which is x e^x. Then, derivative of -k x^2 is -2k x. The derivative of 1 is 0. Therefore, g'(x) = x e^x - 2k x.Set this equal to zero to find critical points: x e^x - 2k x = 0 ⇒ x (e^x - 2k) = 0.So, critical points occur at x = 0 or e^x - 2k = 0 ⇒ e^x = 2k ⇒ x = ln(2k) (if 2k > 0, so k > 0).Therefore, critical points are at x = 0 (for all k) and x = ln(2k) (only when k > 0).Now, we need to analyze these critical points to find the minimum of g(x) over x ≥ 0.Case 1: k ≤ 0.If k ≤ 0, then the term -k x^2 becomes non-negative because k is non-positive, so -k x^2 is positive or zero. Therefore, the expression g(x) = (x -1)e^x -k x^2 +1 will have an added non-negative term. However, we still need to check if the entire expression is non-negative for all x ≥ 0.But let's think through. If k ≤ 0, then in the equation for critical points, e^x = 2k. Since k ≤ 0, 2k ≤ 0, but e^x is always positive. Therefore, e^x = 2k has no solution. Thus, the only critical point is at x = 0.Therefore, for k ≤ 0, the critical points are only at x = 0. Let's evaluate g(x) at x = 0:g(0) = (0 -1)e^0 -k*(0)^2 +1 = (-1)(1) -0 +1 = -1 +1 = 0.So, g(0) = 0. Then, since x = 0 is the only critical point, and for k ≤ 0, we need to check if the function is non-decreasing or non-increasing beyond x = 0. Let's compute the derivative at x > 0. For k ≤ 0, the derivative g'(x) = x e^x - 2k x. Since k ≤ 0, -2k x is non-negative (since k is non-positive, -2k is non-negative, multiplied by x ≥ 0). Therefore, g'(x) = x e^x + |something non-negative|. Hence, g'(x) > 0 for x > 0. Therefore, the function is increasing for x > 0. Since at x = 0, g(x) = 0, and it's increasing for x > 0, then g(x) ≥ 0 for all x ≥ 0. But wait, but in the problem statement, we need f(x) ≥ 1, which translates to g(x) ≥ 0. However, at x = 0, g(0) = 0, so the inequality holds as equality. So, for k ≤ 0, the function g(x) is 0 at x = 0 and increases afterwards. Therefore, it's non-negative for all x ≥ 0. So, k ≤ 0 is acceptable?Wait, but hold on. Let me verify this. Suppose k is negative, say k = -1. Then, the function is g(x) = (x -1)e^x - (-1)x^2 +1 = (x -1)e^x +x^2 +1. Let's see its behavior. At x =0: (0 -1)*1 +0 +1 = -1 +0 +1 =0. For x approaching infinity: (x -1)e^x grows exponentially, x^2 grows polynomially, so overall, g(x) tends to infinity. So, it's 0 at x=0 and increasing. So, yes, for k ≤0, g(x) ≥0 for all x ≥0. Hence, k ≤0 would be acceptable. But wait, the problem is part (2) says "If for any x ∈ [0,+∞), f(x) ≥1 always holds, find the range of values for k". So, according to this, k ≤0 would satisfy the condition. However, we need to check if this is entirely true.Wait, let's take k =0. Then, f(x) = (x-1)e^x +2. We already found that when k=0, the minimum is 1 at x=0, so f(x) =1 at x=0, and since it's increasing afterwards, so f(x) ≥1. Hence, k=0 is acceptable. For k negative, say k=-1, then f(x) = (x -1)e^x - (-1)x^2 +2 = (x -1)e^x +x^2 +2. At x=0: (-1)(1) +0 +2 =1, which meets the condition. Then, as x increases, since both (x -1)e^x (eventually) and x^2 grow, the function tends to infinity. So, perhaps k ≤0 is acceptable. But wait, but maybe for some negative k, even though the function tends to infinity, there could be a minimum somewhere below 1? Wait, but when k ≤0, we saw that the derivative g'(x) =x e^x -2k x. If k is negative, then -2k is positive, so g'(x) =x(e^x -2k). Since k is negative, -2k is positive, so e^x -2k is e^x + |something positive|. So, e^x -2k >0 for all x ≥0. Therefore, g'(x) =x*(something positive). Since x ≥0, then g'(x) ≥0. Therefore, the function g(x) is non-decreasing for x ≥0. Since at x=0, g(0)=0, and then it's non-decreasing, so g(x) ≥0 for all x ≥0. Therefore, indeed, for k ≤0, the condition holds.But wait, the problem says "find the range of values for k" such that f(x) ≥1 for all x in [0, ∞). So, k ≤0 is part of the solution. However, we also need to check k >0. Because for k >0, the critical points are at x=0 and x=ln(2k). So, in this case, the function could have a minimum at x=ln(2k), and we need to ensure that at that point, g(x) ≥0.So, let's proceed to analyze k >0.Case 2: k >0.In this case, the critical points are at x=0 and x=ln(2k). However, x=ln(2k) must be in the domain x ≥0, so ln(2k) ≥0 ⇒ 2k ≥1 ⇒k ≥1/2.Therefore, for 0 <k <1/2, ln(2k) is negative, which is not in the domain [0, ∞). Therefore, for 0 <k <1/2, the only critical point in the domain is x=0. Then, the derivative g'(x) =x(e^x -2k). For x >0, since 0 <k <1/2, 2k <1. So, e^x -2k. At x=0, e^0 -2k =1 -2k. Since k <1/2, 2k <1, so 1 -2k >0. Therefore, for x=0+, the derivative is x(e^x -2k). As x increases from 0, e^x increases from 1, so e^x -2k >1 -2k >0. Therefore, for x >0, g'(x) =x(e^x -2k) >0. Therefore, the function g(x) is increasing on [0, ∞). Then, since at x=0, g(0)=0, and increasing, so g(x) ≥0. Therefore, for 0 <k <1/2, the condition g(x) ≥0 holds?Wait, but when k is positive, the term -kx^2 is negative. So, even though the function is increasing, maybe at some points the negative term -kx^2 could make g(x) dip below zero. Wait, but according to the previous reasoning, if for 0 <k <1/2, the function g(x) is increasing, starting at 0, so it can't dip below zero. Hmm. Let me check with a concrete example. Let's take k=1/4 (which is 0.25, less than 1/2). Then, g(x) = (x -1)e^x - (1/4)x^2 +1.At x=0: (-1)e^0 -0 +1 = -1 +1=0.At x=1: (1 -1)e^1 - (1/4)(1) +1 =0 -0.25 +1=0.75.At x=2: (2 -1)e^2 - (1/4)(4) +1= e^2 -1 +1= e^2 ≈7.39.But wait, let's check x=0.5. Then, (0.5 -1)e^{0.5} - (1/4)(0.25) +1 = (-0.5)e^{0.5} -0.0625 +1 ≈-0.5*1.6487 -0.0625 +1 ≈-0.82435 -0.0625 +1≈0.113. So, positive.But maybe somewhere between x=0 and x approaching infinity, does it dip below zero? Since the derivative is always positive for x>0, then the function is increasing. Therefore, starting at 0 and increasing to infinity. So, g(x) ≥0. Therefore, for k ≤1/2 (i.e., k ≤0.5), the function g(x) is non-decreasing starting at 0, hence always non-negative. Therefore, the condition holds for k ≤1/2? Wait, but when k=1/2, let's check.Wait, when k=1/2, then 2k=1, so x=ln(2k)=ln(1)=0. So, critical points at x=0 and x=0 (duplicate). So, in this case, the critical point is at x=0. For k=1/2, the derivative is g'(x)=x(e^x -1). For x>0, e^x -1 >0, so g'(x) >0. Therefore, the function is increasing for x>0, so again, starting at g(0)=0, increasing. Hence, g(x)≥0.But wait, this seems conflicting with previous thoughts. Wait, when k=1/2, then the critical point x=ln(2k)=ln(1)=0. So, only one critical point at x=0. Then, for k>1/2, ln(2k) becomes positive. So, when k>1/2, there's another critical point at x=ln(2k). So, in this case, the function g(x) might have a minimum at x=ln(2k). Therefore, for k>1/2, we need to check the value of g at x=ln(2k) to ensure it's ≥0.Therefore, combining all this:For k ≤1/2, the function g(x) is non-decreasing on [0, ∞), starting at g(0)=0. Therefore, g(x)≥0 for all x≥0.For k >1/2, there is a critical point at x=ln(2k) >0. At this critical point, the function could have a minimum. If the minimum value is ≥0, then the condition holds. Otherwise, it doesn't.Thus, we need to analyze for k >1/2, find the value of g at x=ln(2k), set it to be ≥0, and solve for k.So, let's compute g(ln(2k)).First, compute x = ln(2k).So, g(ln(2k)) = (ln(2k) -1)e^{ln(2k)} -k*(ln(2k))^2 +1.Simplify term by term:e^{ln(2k)} = 2k.Therefore, first term: (ln(2k) -1)*2k.Second term: -k*(ln(2k))^2.Third term: +1.So, combining:g(ln(2k)) = 2k(ln(2k) -1) -k(ln(2k))^2 +1.Let's factor out k:= k[2(ln(2k) -1) - (ln(2k))^2] +1.Let me compute the expression inside the brackets:2(ln(2k) -1) - (ln(2k))^2 = 2 ln(2k) -2 - (ln(2k))^2.Therefore, g(ln(2k)) = k[ - (ln(2k))^2 +2 ln(2k) -2 ] +1.Let me denote t = ln(2k). Then, since k >1/2, 2k >1, so t = ln(2k) >0.Thus, g(ln(2k)) = k[ -t^2 + 2t -2 ] +1.But let's express everything in terms of t. Since t = ln(2k), we have 2k = e^t ⇒ k = e^t /2.Therefore, substituting back:g(ln(2k)) = (e^t /2)[ -t^2 +2t -2 ] +1.So, = (-e^t /2)(t^2 -2t +2) +1.So, this is the expression for g at the critical point.We need to ensure that this is ≥0.Therefore, (-e^t /2)(t^2 -2t +2) +1 ≥0 ⇒1 - (e^t /2)(t^2 -2t +2) ≥0 ⇒ (e^t /2)(t^2 -2t +2) ≤1.Multiply both sides by 2: e^t (t^2 -2t +2) ≤2.So, the inequality is e^t (t^2 -2t +2) ≤2.We need to find t >0 (since k>1/2 implies t=ln(2k)>0) such that e^t (t^2 -2t +2) ≤2.Let’s denote h(t) = e^t (t^2 -2t +2). We need h(t) ≤2.So, solve h(t) ≤2 for t >0.First, let's compute h(0). If t approaches 0+, then h(t) ≈ e^0 (0 -0 +2)=1*2=2. So, at t=0, h(t)=2. Then, as t increases from 0, what happens to h(t)?Let’s compute the derivative h’(t) to see if h(t) is increasing or decreasing.h(t) = e^t (t^2 -2t +2).h’(t) = e^t (t^2 -2t +2) + e^t (2t -2) = e^t [t^2 -2t +2 +2t -2] = e^t [t^2].So, h’(t) = t^2 e^t.Since t >0, h’(t) >0. Therefore, h(t) is increasing for t >0. Since h(0)=2 and h(t) is increasing, h(t) ≥2 for all t ≥0, with equality only at t=0.Therefore, the equation h(t)=2 has only the solution t=0. For t>0, h(t)>2. Therefore, the inequality h(t) ≤2 is only satisfied at t=0, which corresponds to k=1/2 (since t=ln(2k)=0 ⇒2k=1⇒k=1/2).Therefore, for k >1/2, the expression h(t) =e^t (t^2 -2t +2) >2, which means g(ln(2k)) =1 - (e^t /2)(t^2 -2t +2) =1 - h(t)/2 <1 -2/2=0. Hence, g(ln(2k)) <0 for k >1/2. Therefore, for k >1/2, there exists a point x=ln(2k) where g(x) <0, violating the condition g(x) ≥0. Therefore, k >1/2 is not allowed.For k =1/2, we have t=0, h(t)=2, so g(ln(2k))=1 -2/2=0. Therefore, the minimum at x=0 (which is also the only critical point) gives g(0)=0. For k=1/2, since the derivative is zero at x=0 and positive for x>0, the function is increasing, so g(x)≥0 for all x≥0.Putting it all together:- For k ≤1/2, g(x) ≥0 for all x ≥0.- For k >1/2, g(x) dips below zero at x=ln(2k), hence the condition fails.But wait, earlier, we considered k ≤0 and 0 <k ≤1/2. So, combining both cases, the range of k is (-∞,1/2].Wait, but earlier when analyzing k ≤0, we found that g(x) is non-decreasing from 0, hence g(x)≥0. Then, for 0 <k ≤1/2, even though k is positive, the function is still non-decreasing starting at 0, hence g(x)≥0. For k >1/2, the function has a minimum at x=ln(2k) where g(x) becomes negative, so it's invalid. Therefore, the valid range for k is k ≤1/2.But wait, let's confirm with k=1/2. At k=1/2, the critical point is x=ln(2*(1/2))=ln(1)=0. So, g(0)=0, and the derivative for x>0 is positive, so the function increases. Therefore, g(x)≥0 for all x≥0. So, k=1/2 is acceptable.Hence, the range is k ∈ (-∞, 1/2].But let's double-check with k=1/2 and x approaching infinity. For k=1/2, f(x) = (x-1)e^x - (1/2)x^2 +2. As x→∞, (x-1)e^x dominates, which goes to infinity, so f(x)→∞, which is ≥1. Therefore, yes, the function remains above 1.Another check: take k=1/2 and x=1. f(1)= (1-1)e^1 - (1/2)(1)^2 +2=0 -0.5 +2=1.5 ≥1. Okay. At x=0, f(0)= (-1)(1) -0 +2=1, which is exactly 1. So, works.Now, for k just above 1/2, say k=0.6. Then, the critical point is at x=ln(2*0.6)=ln(1.2)≈0.182. Let's compute g(ln(1.2)):g(0.182) = (0.182 -1)e^{0.182} -0.6*(0.182)^2 +1.Compute each term:First term: (-0.818)e^{0.182} ≈-0.818*1.2 ≈-0.9816.Second term: -0.6*(0.0331)≈-0.01986.Third term: +1.So, total≈-0.9816 -0.01986 +1≈-1.00146 +1≈-0.00146. Which is negative. Therefore, g(x) is slightly negative at x≈0.182, which violates the condition. Therefore, k=0.6 is invalid.Therefore, the maximum allowed k is 1/2.Thus, the answer for part (2) is k ≤1/2.Therefore, combining everything:(1) When k=0, the extreme value is a minimum of 1 at x=0.(2) The range of k is (-∞, 1/2].Final Answer(1) The extreme value of ( f(x) ) when ( k=0 ) is a minimum at (boxed{1}).(2) The range of values for ( k ) is (boxed{(-infty, dfrac{1}{2}]}).</think>(1) When ( k = 0 ), the function simplifies to ( f(x) = (x - 1)e^x + 2 ). To find the extreme values, we first compute the derivative:[f'(x) = frac{d}{dx}[(x - 1)e^x + 2] = x e^x]Setting ( f'(x) = 0 ) gives ( x e^x = 0 ). Since ( e^x ) is always positive, the only critical point is ( x = 0 ). To determine the nature of this critical point, we compute the second derivative:[f''(x) = frac{d}{dx}[x e^x] = (1 + x)e^x]Evaluating ( f''(0) ):[f''(0) = (1 + 0)e^0 = 1 > 0]Thus, ( x = 0 ) is a local minimum. Evaluating ( f(0) ):[f(0) = (0 - 1)e^0 + 2 = -1 + 2 = 1]Therefore, the extreme value of ( f(x) ) when ( k = 0 ) is a minimum at (boxed{1}).(2) To ensure ( f(x) geq 1 ) for all ( x in [0, +infty) ), we consider the function ( g(x) = (x - 1)e^x - kx^2 + 1 geq 0 ). We analyze the critical points of ( g(x) ) by computing its derivative:[g'(x) = x e^x - 2kx]Setting ( g'(x) = 0 ) gives ( x(e^x - 2k) = 0 ). The critical points are ( x = 0 ) and ( x = ln(2k) ) (if ( k > 0 )).For ( k leq 0 ), the term ( -kx^2 ) is non-negative, and ( g(x) ) is non-decreasing starting from ( g(0) = 0 ), ensuring ( g(x) geq 0 ).For ( 0 < k leq frac{1}{2} ), the critical point ( x = ln(2k) ) is non-positive or zero, so ( g(x) ) is non-decreasing starting from ( g(0) = 0 ), ensuring ( g(x) geq 0 ).For ( k > frac{1}{2} ), the critical point ( x = ln(2k) ) is positive. Evaluating ( g ) at this point:[g(ln(2k)) = k[ - (ln(2k))^2 + 2ln(2k) - 2 ] + 1]This expression is found to be negative, violating ( g(x) geq 0 ).Thus, the range of values for ( k ) ensuring ( f(x) geq 1 ) for all ( x geq 0 ) is (boxed{(-infty, dfrac{1}{2}]}).,<think>Okay, let's tackle this problem step by step. It's about finding extreme values and determining the range of a parameter to satisfy an inequality. Let's start with part (1).Problem (1): When ( k = 0 ), find the extreme values of ( f(x) ).First, let's rewrite the function when ( k = 0 ):[ f(x) = (x - 1)e^x + 2 ]Extreme values occur where the first derivative is zero or undefined. Since this function involves an exponential, which is differentiable everywhere, we just need to find where the derivative is zero.Let's compute the first derivative ( f'(x) ):[ f'(x) = frac{d}{dx} left[ (x - 1)e^x right] + frac{d}{dx}[2] ]The derivative of the constant term 2 is zero. For the first term, we'll use the product rule. If we let ( u = x - 1 ) and ( v = e^x ), then:[ u' = 1 ][ v' = e^x ]So, applying the product rule:[ frac{d}{dx}[uv] = u'v + uv' = 1 cdot e^x + (x - 1) cdot e^x = e^x + (x - 1)e^x ]Factor out ( e^x ):[ f'(x) = e^x (1 + x - 1) = e^x cdot x ]So, ( f'(x) = x e^x ). To find critical points, set this equal to zero:[ x e^x = 0 ]Since ( e^x ) is never zero for any real x, the only solution is ( x = 0 ).Now, we need to determine if this critical point is a minimum or maximum. Let's use the second derivative test.Compute the second derivative ( f''(x) ):First, ( f'(x) = x e^x ), so:[ f''(x) = frac{d}{dx}[x e^x] ]Again, use the product rule:Let ( u = x ), ( v = e^x ), so:[ u' = 1 ][ v' = e^x ][ f''(x) = u'v + uv' = e^x + x e^x = e^x (1 + x) ]Evaluate the second derivative at ( x = 0 ):[ f''(0) = e^0 (1 + 0) = 1 times 1 = 1 ]Since ( f''(0) > 0 ), the function has a local minimum at ( x = 0 ).Now, let's find the value of ( f(x) ) at this critical point:[ f(0) = (0 - 1)e^0 + 2 = (-1)(1) + 2 = -1 + 2 = 1 ]Therefore, when ( k = 0 ), the function ( f(x) ) has a local minimum at ( x = 0 ) with value 1. Are there any other critical points? We only found ( x = 0 ). Since the function is defined for all real numbers, but we only have one critical point. Let's check the behavior as ( x ) approaches infinity and negative infinity to ensure we haven't missed any global extrema.As ( x to +infty ):The term ( (x - 1)e^x ) dominates because exponential growth outpaces any polynomial. So, ( f(x) ) tends to ( +infty ).As ( x to -infty ):The term ( (x - 1)e^x ) approaches zero because ( e^x ) decays to zero faster than the linear term ( (x - 1) ) goes to negative infinity. So, ( f(x) ) approaches 2 (from the constant term). Wait, hold on. Let me check that again.Wait, actually, when ( x to -infty ), ( (x - 1) ) tends to ( -infty ), but ( e^x ) tends to 0. So, the product is ( -infty times 0 ), which is an indeterminate form. To resolve this, we can analyze the limit:[ lim_{x to -infty} (x - 1)e^x ]Let me rewrite this as:[ lim_{x to -infty} frac{x - 1}{e^{-x}} ]Because ( e^{-x} ) as ( x to -infty ) becomes ( e^{+infty} ), which is ( +infty ). So, we have ( frac{-infty}{infty} ), which is suitable for L’Hospital's Rule. Applying L’Hospital's:Differentiate numerator and denominator:Numerator derivative: 1Denominator derivative: ( -e^{-x} )So, the limit becomes:[ lim_{x to -infty} frac{1}{-e^{-x}} = lim_{x to -infty} -e^{x} = -0 = 0 ]Therefore, the term ( (x - 1)e^x ) approaches 0 as ( x to -infty ), so ( f(x) ) approaches ( 0 + 2 = 2 ).Therefore, as ( x to -infty ), ( f(x) to 2 ), and as ( x to +infty ), ( f(x) to +infty ). The only critical point is at ( x = 0 ), which is a local (and in this case, global) minimum. The function doesn't have a maximum because it goes to infinity as ( x ) increases. So, the extreme values are just the minimum at 1.Wait, but the problem says "extreme values", which usually refers to both minima and maxima. But since there is only a minimum here, and no maximum, we just report the minimum.So, part (1) answer: The function has a minimum value of 1 at ( x = 0 ).Problem (2): If for any ( x in [0, +infty) ), ( f(x) geqslant 1 ) always holds, find the range of values for ( k ).This part is more complex. We need to ensure that ( f(x) geq 1 ) for all ( x geq 0 ). So, the function ( f(x) = (x - 1)e^x - kx^2 + 2 ) must be at least 1 for all ( x geq 0 ).First, let's rearrange the inequality:[ (x - 1)e^x - kx^2 + 2 geq 1 ]Subtract 1 from both sides:[ (x - 1)e^x - kx^2 + 1 geq 0 ]Let me define a new function for simplicity:[ g(x) = (x - 1)e^x - kx^2 + 1 ]So, we need ( g(x) geq 0 ) for all ( x geq 0 ).Our task is to find all real numbers ( k ) such that ( g(x) geq 0 ) for all ( x geq 0 ).To find the range of ( k ), we can consider this as an optimization problem: find the maximum value of ( k ) such that the minimum of ( g(x) ) on ( [0, +infty) ) is non-negative. Because if we can find the minimal value of ( g(x) ) over ( x geq 0 ) and set it to be greater than or equal to zero, then the inequality will hold for all ( x geq 0 ).Therefore, our strategy is:1. Find the critical points of ( g(x) ) on ( [0, +infty) ).2. Ensure that at all critical points and at the boundary ( x = 0 ), ( g(x) geq 0 ).3. Determine the values of ( k ) that satisfy these conditions.Let me start by finding the derivative ( g'(x) ).First, compute ( g'(x) ):[ g(x) = (x - 1)e^x - kx^2 + 1 ]Differentiating term by term:- The derivative of ( (x - 1)e^x ): As before, we did this in part (1). It is ( x e^x ).- The derivative of ( -k x^2 ) is ( -2k x ).- The derivative of the constant 1 is 0.Therefore:[ g'(x) = x e^x - 2k x ]Factor out x:[ g'(x) = x (e^x - 2k) ]So, critical points occur where ( g'(x) = 0 ):Either ( x = 0 ) or ( e^x - 2k = 0 ).Thus, critical points are at:1. ( x = 0 )2. ( x = ln(2k) ) (if ( 2k > 0 ), so ( k > 0 ))Therefore, depending on the value of ( k ), we have different critical points.Case 1: ( k leq 0 )If ( k leq 0 ), then ( 2k leq 0 ), so ( e^x - 2k ) is always positive because ( e^x > 0 ) for all x, and subtracting a non-positive number (since ( 2k leq 0 )) will make ( e^x - 2k geq e^x > 0 ). Therefore, ( g'(x) = x (e^x - 2k) ). Since both ( x geq 0 ) and ( e^x - 2k > 0 ), ( g'(x) geq 0 ) for all ( x geq 0 ). Therefore, ( g(x) ) is non-decreasing on ( [0, +infty) ).Therefore, the minimal value of ( g(x) ) occurs at ( x = 0 ).Compute ( g(0) ):[ g(0) = (0 - 1)e^0 - k(0)^2 + 1 = (-1)(1) - 0 + 1 = -1 + 1 = 0 ]Therefore, when ( k leq 0 ), ( g(0) = 0 ), and since the function is non-decreasing, ( g(x) geq 0 ) for all ( x geq 0 ). So, for ( k leq 0 ), the inequality holds.Case 2: ( k > 0 )If ( k > 0 ), then there is a critical point at ( x = ln(2k) ). But we need to check if this critical point is in the domain ( [0, +infty) ).Since ( x = ln(2k) geq 0 iff 2k geq 1 iff k geq 1/2 ).Therefore, subcases:Subcase 2a: ( 0 < k < 1/2 )Then, ( 2k < 1 ), so ( ln(2k) < 0 ). But our domain is ( x geq 0 ), so the critical point ( x = ln(2k) ) is not in the domain. Therefore, the only critical point in the domain is at ( x = 0 ). However, since ( k > 0 ), let's check the derivative.For ( x > 0 ):Since ( k < 1/2 ), ( 2k < 1 ). Then, ( e^x ) starts at 1 when x=0 and increases. So, for x > 0, ( e^x > 1 > 2k ). Therefore, ( e^x - 2k > 0 ), so ( g'(x) = x (e^x - 2k) > 0 ). Therefore, ( g(x) ) is increasing on ( [0, +infty) ). Hence, the minimum occurs at x=0.Compute ( g(0) = 0 ) as before. Therefore, even for ( 0 < k < 1/2 ), the function is increasing, starting at 0 and increasing, so ( g(x) geq 0 ). But wait, the problem states that ( f(x) geq 1 ), which translates to ( g(x) geq 0 ). So, for ( k < 1/2 ), since the function starts at 0 and increases, the minimal value is 0, which satisfies the inequality ( g(x) geq 0 ). Therefore, perhaps even k up to 1/2 is allowed? Wait, but we need to check for all k. Wait, but when k increases, we might have a critical point entering the domain. Let's move to the next subcase.Subcase 2b: ( k = 1/2 )Then, ( 2k = 1 ), so ( x = ln(1) = 0 ). So, the critical point is at x=0. The derivative is:For x > 0, ( e^x - 1 ) since 2k=1. Since ( e^x > 1 ) for x > 0, ( g'(x) > 0 ). Therefore, the function is increasing for x > 0, with a critical point at x=0. Thus, the minimum is at x=0 with g(0)=0. So, again, the inequality holds.Subcase 2c: ( k > 1/2 )Here, 2k > 1, so ( x = ln(2k) > 0 ). Therefore, we have two critical points: x=0 and x=ln(2k). Now, we need to check the behavior of g(x) here.First, let's analyze the derivative:For x in [0, ln(2k)):Since x < ln(2k), so e^x < 2k. Therefore, e^x - 2k < 0. Hence, g'(x) = x (e^x - 2k) < 0, because x is positive (except at x=0) and (e^x - 2k) is negative. Therefore, on [0, ln(2k)), the function is decreasing.For x > ln(2k):e^x > 2k, so e^x - 2k > 0, and since x > 0, g'(x) > 0. Hence, the function is increasing on (ln(2k), +infty).Therefore, the critical point at x=ln(2k) is a local minimum. Therefore, the minimal value of g(x) on [0, +infty) is at x=ln(2k). Therefore, to ensure g(x) >= 0 everywhere, we need to ensure that this minimal value is >=0.Therefore, compute g(ln(2k)) and set it >=0.Compute g(ln(2k)):[ g(ln(2k)) = (ln(2k) - 1)e^{ln(2k)} - k (ln(2k))^2 + 1 ]Simplify:Since ( e^{ln(2k)} = 2k ):[ g(ln(2k)) = (ln(2k) - 1)(2k) - k (ln(2k))^2 + 1 ]Expand the first term:[ 2k ln(2k) - 2k - k (ln(2k))^2 + 1 ]Factor out k from the first three terms:But let's write all terms:= ( 2k ln(2k) - 2k - k (ln(2k))^2 + 1 )Let me rearrange:= ( -k (ln(2k))^2 + 2k ln(2k) - 2k + 1 )Factor out k from the first three terms:= ( k [ - (ln(2k))^2 + 2 ln(2k) - 2 ] + 1 )Hmm, perhaps we can write this as:Let’s let t = ln(2k). Then:= ( k [ -t^2 + 2t - 2 ] + 1 )But maybe this substitution complicates things. Alternatively, let's collect terms:Let me factor the expression:- The first three terms: -k (ln(2k))² + 2k ln(2k) - 2k = -k[ (ln(2k))² - 2 ln(2k) + 2 ]So:g(ln(2k)) = -k [ (ln(2k))² - 2 ln(2k) + 2 ] + 1We need this to be >= 0:So:- k [ (ln(2k))² - 2 ln(2k) + 2 ] + 1 >= 0Rearranged:1 >= k [ (ln(2k))² - 2 ln(2k) + 2 ]Or,k [ (ln(2k))² - 2 ln(2k) + 2 ] <= 1This seems complicated. Let's denote u = ln(2k). Since k > 1/2, 2k > 1, so u = ln(2k) > 0.Then, the inequality becomes:k [ u² - 2u + 2 ] <= 1But u = ln(2k), so 2k = e^u, so k = e^u / 2.Substitute into the inequality:(e^u / 2) [ u² - 2u + 2 ] <= 1Multiply both sides by 2:e^u [ u² - 2u + 2 ] <= 2So,e^u (u² - 2u + 2) <= 2We need to solve this inequality for u > 0 (since k > 1/2 implies u > 0).Let’s define the function h(u) = e^u (u² - 2u + 2). We need h(u) <= 2.So, find all u > 0 such that h(u) <= 2.Let’s analyze h(u):Compute h(0):u = 0: h(0) = e^0 (0 - 0 + 2) = 2Compute derivative h’(u):h(u) = e^u (u² - 2u + 2)Differentiate using product rule:h’(u) = e^u (u² - 2u + 2) + e^u (2u - 2)Factor out e^u:h’(u) = e^u [ u² - 2u + 2 + 2u - 2 ] = e^u [ u² ]So, h’(u) = u² e^uSince u > 0, h’(u) > 0. Therefore, h(u) is strictly increasing for u > 0.At u = 0, h(u) = 2, and as u increases, h(u) increases beyond 2. Therefore, the equation h(u) = 2 has only the solution u = 0. Hence, for u > 0, h(u) > 2.But our inequality is h(u) <= 2. Therefore, the only solution is u = 0, but u > 0 (since k > 1/2 implies u > 0). Therefore, there are no solutions for u > 0. This suggests that when k > 1/2, the minimal value at x = ln(2k) gives g(ln(2k)) < 0, which violates the inequality. However, this contradicts the previous conclusion. Wait, this is critical.Wait, when k > 1/2, the minimal value is at x = ln(2k). However, our analysis shows that for u > 0, h(u) = e^u (u² - 2u + 2) > 2, which would mean that:h(u) = e^u (u² - 2u + 2) > 2 => k [ (ln(2k))² - 2 ln(2k) + 2 ] > 1Therefore, substituting back:g(ln(2k)) = -k [ ... ] +1 = - [ something > 1 ] +1 < 0Thus, when k > 1/2, the minimal value g(ln(2k)) is negative, which violates the inequality. Therefore, for k > 1/2, the inequality does not hold.But when k = 1/2, the critical point is at x = 0, which we saw earlier gives g(0) = 0. However, when k = 1/2, 2k = 1, so ln(2k) = 0. Then, the critical point is at x=0. However, when k = 1/2, what is the behavior?Wait, let's check k = 1/2:g(x) = (x - 1)e^x - (1/2)x² + 1At x=0:g(0) = (-1)(1) - 0 + 1 = 0Derivative:g'(x) = x e^x - 2*(1/2)x = x e^x - x = x (e^x - 1)For x > 0, e^x -1 > 0, so g'(x) > 0. Hence, the function is increasing for x > 0. Therefore, the minimal value is at x=0 with g(0)=0, and the function increases afterwards. Thus, for k=1/2, the inequality holds as equality at x=0 and is greater elsewhere.But for k > 1/2, as shown, the minimal value is negative. Hence, the maximum allowed k is 1/2.But wait, when k=1/2, the minimal value is zero. So, perhaps the allowable range of k is k <= 1/2.But earlier, when k < 1/2, even though there's no critical point in x >=0, the function is increasing, starting at g(0)=0, so it's okay. Wait, but in the problem statement, the original inequality is f(x) >= 1, which translates to g(x) >=0. However, when k <=1/2, the minimal value of g(x) is zero (achieved at x=0), so g(x) >=0 for all x >=0. However, when k >1/2, the minimal value is negative, hence violating the inequality.Wait, but the analysis in the case of k <=0 also gives g(0)=0, but in that case, the function is non-decreasing. Wait, but if k <=0, the derivative is non-negative for x >=0, hence the function is increasing. Therefore, starting at g(0)=0 and increasing to infinity. So, for k <=0, it's okay. For 0 < k <=1/2, the function is still increasing (even though when k <1/2, the critical point is at x=ln(2k) <0, outside the domain). Therefore, the minimal value is at x=0, g(0)=0. For k =1/2, the critical point is at x=0, and the function is increasing for x >0, so again minimal at x=0. For k >1/2, the critical point is at x=ln(2k) >0, and the function first decreases to that point and then increases. However, the minimal value at x=ln(2k) is negative, so the inequality fails.Therefore, combining all these cases, the allowable values for k are k <=1/2.But wait, but when k=1/2, the minimal value is zero. So, does k=1/2 work? Yes, because g(x) >=0 for all x >=0. For k >1/2, it fails. For k <=1/2, it works.Therefore, the range of k is (-infty, 1/2].Wait, but let's verify with an example. Take k=1/2:g(x) = (x -1)e^x - (1/2)x² +1At x=0: 0, as before.At x approaching infinity: (x -1)e^x dominates, which tends to infinity. So, the function goes to infinity. At x=1:g(1) = (1 -1)e^1 - (1/2)(1) +1 = 0 - 0.5 +1 = 0.5 >=0.At x=2:g(2) = (2 -1)e^2 - (1/2)(4) +1 = e^2 -2 +1 = e^2 -1 ≈7.389 -1=6.389 >=0.So, seems okay. For k=1/2, the function is okay.For k=0.6 (>1/2):g(x) = (x -1)e^x -0.6x² +1Find its minimum. The critical point is at x=ln(1.2) ≈0.182.Compute g(ln(1.2)):First, ln(1.2) ≈0.1823Compute:g(0.1823) = (0.1823 -1)e^{0.1823} -0.6*(0.1823)^2 +1Compute term by term:(0.1823 -1) = -0.8177e^{0.1823} ≈1.200So, first term: -0.8177*1.200 ≈-0.9812Second term: -0.6*(0.0332) ≈-0.0199Third term: +1Total ≈ -0.9812 -0.0199 +1 ≈-1.0011 +1≈-0.0011 <0So, g(ln(1.2))≈-0.0011 <0, which violates the inequality. Thus, even slightly above 0.5, the function dips below zero.Therefore, confirming that k must be <=1/2.Hence, the answer to part (2) is k ≤1/2.But let's check k=0.5 again with x approaching ln(2*0.5)=ln(1)=0. So, minimal at x=0, which is zero.But what if k is slightly less than 0.5?Take k=0.4:Critical point is at x=ln(0.8)≈-0.223, outside the domain. Hence, the function is increasing on x >=0, starting at g(0)=0. Therefore, even for k=0.4, the function is increasing from 0 upwards. So, g(x) >=0.Therefore, the allowable k is k <=1/2.Therefore, combining all cases, the range of k is (-infty, 1/2].Final Answer(1) The extreme value is a minimum of boxed{1} at ( x = 0 ).(2) The range of values for ( k ) is boxed{(-infty, dfrac{1}{2}]}).</think>Given the function ( f(x) = (x - 1)e^x - kx^2 + 2 ):1. When ( k = 0 ), find the extreme values of ( f(x) ): The function simplifies to ( f(x) = (x - 1)e^x + 2 ). - Compute the first derivative: ( f'(x) = x e^x ). - Set the derivative to zero: ( x e^x = 0 ). The only critical point is ( x = 0 ). - Compute the second derivative: ( f''(x) = e^x (1 + x) ). - Evaluate the second derivative at ( x = 0 ): ( f''(0) = 1 > 0 ), indicating a local minimum. - Evaluate ( f(x) ) at ( x = 0 ): ( f(0) = 1 ). Therefore, the function has a local minimum at ( x = 0 ) with value 1. As ( x to +infty ), ( f(x) to +infty ), and as ( x to -infty ), ( f(x) to 2 ). The only extreme value is the minimum at 1. [ boxed{1} ]2. If for any ( x in [0, +infty) ), ( f(x) geq 1 ) always holds, find the range of values for ( k ): Rewrite the inequality as ( (x - 1)e^x - kx^2 + 1 geq 0 ). - Define ( g(x) = (x - 1)e^x - kx^2 + 1 ). - Compute the derivative: ( g'(x) = x(e^x - 2k) ). - Critical points occur at ( x = 0 ) and ( x = ln(2k) ) (if ( k > 0 )). Analyze different cases for ( k ): - For ( k leq 0 ): - ( g(x) ) is non-decreasing on ( [0, +infty) ). - Minimum value at ( x = 0 ): ( g(0) = 0 ). - Inequality holds as ( g(x) geq 0 ). - For ( 0 < k leq frac{1}{2} ): - Critical point ( x = ln(2k) ) is negative or zero, not in the domain. - ( g(x) ) is increasing on ( [0, +infty) ). - Minimum value at ( x = 0 ): ( g(0) = 0 ). - Inequality holds as ( g(x) geq 0 ). - For ( k > frac{1}{2} ): - Critical point ( x = ln(2k) ) is positive. - Minimum value at ( x = ln(2k) ) is negative, violating the inequality. Therefore, the range of ( k ) is ( (-infty, frac{1}{2}] ). [ boxed{(-infty, dfrac{1}{2}]} ]]

🔑:Solution: (1) When k=0, f(x)=(x-1)e^{x}+2, f'(x)=xe^{x}, Let f'(x)=xe^{x}=0, we get x=0. therefore When x=0, the function f(x) attains a minimum value, f(0)=1; there is no maximum value. (2) For any xin[0,+infty), f(x)geqslant 1 always holds, Leftrightarrow (x-1)e^{x}-kx^{2}+1geqslant 0, Let g(x)=(x-1)e^{x}-kx^{2}+1, xin[0,+infty), And g(0)=0, Therefore, the function g(x) is monotonically increasing on xin[0,+infty), therefore g'(x)=xe^{x}-2kx=x(e^{x}-2k)geqslant 0, therefore kleqslant dfrac {e^{x}}{2}, we can obtain kleqslant dfrac {1}{2}. therefore The range of values for k is boxed{(-infty, dfrac {1}{2}]}.

🧠:[<think>Okay, let's tackle this problem step by step. The question is asking for the smallest positive integer n that is a multiple of 75 and has exactly 75 positive integral divisors. Then, we need to find n divided by 75, which is essentially finding the smallest such n and then dividing it by 75. First, let me recall some number theory concepts. The number of divisors of a positive integer is determined by its prime factorization. If a number n has a prime factorization of the form ( n = p_1^{e_1} times p_2^{e_2} times dots times p_k^{e_k} ), then the number of divisors of n is ( (e_1 + 1)(e_2 + 1) dots (e_k + 1) ). So, our goal is to find the smallest multiple of 75 such that when we factor it into primes, the product of one more than each of the exponents in its prime factorization equals 75.Now, 75 itself factors into primes as ( 75 = 3 times 5^2 ). Therefore, if a number is a multiple of 75, its prime factorization must include at least the primes 3 and 5 with exponents at least 1 and 2 respectively. So, the number n must have a prime factorization that includes 3^1 and 5^2, and possibly other primes as well. But since we want the smallest such n, we need to minimize the overall product, which suggests that we should use the smallest possible primes and balance their exponents appropriately.The number of divisors is 75, so we need to find exponents ( e_1, e_2, dots, e_k ) such that ( (e_1 + 1)(e_2 + 1) dots (e_k + 1) = 75 ). Let's factor 75 into its prime components to understand the possible combinations. The prime factorization of 75 is ( 3 times 5 times 5 ). So, 75 can be written as 75, 25×3, 15×5, 5×5×3. Since the number of divisors is a product of exponents plus one, we need to partition the exponents such that when each is increased by one, their product is 75.Let me list all the possible factorizations of 75 into integers greater than 1 (since each ( e_i + 1 ) must be at least 2, because exponents ( e_i ) are at least 1 for primes present in the factorization):1. 752. 25 × 33. 15 × 54. 5 × 5 × 3These correspond to different possible exponent combinations:1. A single prime with exponent 74 (since 74 + 1 = 75). But n must be a multiple of 75, which requires at least two primes (3 and 5), so this is invalid.2. Two primes with exponents 24 and 2 (since 24 + 1 = 25 and 2 + 1 = 3). So exponents 24 and 2.3. Two primes with exponents 14 and 4 (14 + 1 = 15, 4 + 1 = 5). Exponents 14 and 4.4. Three primes with exponents 4, 4, and 2 (4 + 1 = 5, 4 + 1 = 5, 2 + 1 = 3). Exponents 4, 4, 2.Now, since n must be a multiple of 75, which is 3 × 5², we need to have at least 3¹ and 5² in the prime factors. Therefore, in the exponent combinations above, the primes 3 and 5 must be assigned exponents such that 3 has at least 1 and 5 has at least 2. The remaining exponents can be assigned to other primes, but to minimize n, we should assign the larger exponents to the smaller primes. That is, after 3 and 5, the next smallest primes are 2, 7, 11, etc. So, larger exponents should be assigned to smaller primes to keep the overall product smaller.Let me analyze each possible factorization:1. Case 1: Single prime with exponent 74. Not possible, as discussed.2. Case 2: Two primes with exponents 24 and 2. Since we need at least two primes (3 and 5), but here we have only two primes. So, one of them is 3 and the other is 5. Let's see:But 5 needs to have exponent at least 2, so if one prime is 5 with exponent 24, and the other is 3 with exponent 2? Wait, but that would mean the exponents are 24 and 2. So, which prime gets which exponent?If we assign the higher exponent to the smaller prime, that would give a smaller n. For example, if prime 2 had exponent 24 and prime 3 had exponent 2, but wait, but n has to be a multiple of 75, which requires factors 3 and 5. So in this case, with two primes, we need one of them to be 3 and the other to be 5. But if we do that, then the exponents would be 24 and 2, but which one is which?Wait, let's clarify. The number n must have primes 3 and 5. If we have two primes total, then those are the only primes. So, one is 3 and the other is 5. Then, their exponents are 24 and 2. So, which one is which? Assigning the higher exponent to the smaller prime (3) would make n smaller. Because 3^24 * 5^2 is smaller than 3^2 * 5^24. Let's verify:3^24 * 5^2 vs 3^2 * 5^24. Since 3 < 5, raising 3 to a higher power will result in a smaller number. So yes, 3^24 * 5^2 is smaller. But hold on, is 3^24 * 5^2 a multiple of 75? Let's check. 75 is 3 * 5^2. So, 3^24 * 5^2 is 3^24 * 5^2, which certainly includes 3^1 * 5^2. So, yes, it's a multiple. So in this case, n would be 3^24 * 5^2. But let's hold on—the problem is that 75 is 3 * 5^2, so n must have at least those exponents. However, 3^24 * 5^2 is okay, but 5^2 is already there, but 3^24 is way more than needed. But is that the minimal?Wait, but if we have two primes, 3 and 5, with exponents 24 and 2, which would lead to a number of divisors (24 + 1)(2 + 1) = 25 * 3 = 75. So that works. But is there a smaller n possible with more primes?3. Case 3: Two primes with exponents 14 and 4. Similarly, if we have two primes, 3 and 5, with exponents 14 and 4. Assigning higher exponent to the smaller prime: 3^14 * 5^4. Let's compute the number of divisors: (14 + 1)(4 + 1) = 15 * 5 = 75. That's correct. Now, comparing 3^24 * 5^2 vs 3^14 * 5^4. Which is smaller?Compute the ratio: (3^14 * 5^4) / (3^24 * 5^2) = 3^(14-24) * 5^(4-2) = 3^-10 * 5^2 = (5^2)/(3^10). Since 3^10 is 59049, and 5^2 is 25, so the ratio is 25 / 59049 ≈ 0.000423, which is less than 1. Therefore, 3^14 * 5^4 is smaller. So Case 3 gives a smaller n than Case 2.4. Case 4: Three primes with exponents 4, 4, and 2. So, three primes. Since n must include 3 and 5, the third prime would be the next smallest prime, which is 2. Assigning exponents to minimize n: assign the largest exponents to the smallest primes. The exponents are 4, 4, 2. So, assign 4 to 2, 4 to 3, and 2 to 5. Then, the number would be 2^4 * 3^4 * 5^2. Let's check the number of divisors: (4 + 1)(4 + 1)(2 + 1) = 5 * 5 * 3 = 75. Correct. Now, let's compare this with Case 3's 3^14 * 5^4.Compute the values:Case 3: 3^14 * 5^4. Let's compute approximate values:3^14: 3^10 = 59049, 3^14 = 59049 * 81 = 47829695^4 = 625So, 4782969 * 625 ≈ let's compute 4782969 * 600 = 2,869,781,400 and 4782969 * 25 = 119,574,225. Total ≈ 2,869,781,400 + 119,574,225 = 2,989,355,625.Case 4: 2^4 * 3^4 * 5^2. Compute:2^4 = 16; 3^4 = 81; 5^2 = 25. Multiply all together: 16 * 81 = 1296; 1296 * 25 = 32,400.Wow, 32,400 is way smaller than approximately 3 billion. So, Case 4 gives a much smaller n. Therefore, the three-prime case is better.Therefore, among the possible factorizations, the three-prime case gives the smallest n. However, we need to verify if there's a way to arrange the exponents in Case 4 even better.Wait, in Case 4, we have exponents 4, 4, 2. If we assign the exponents to the primes as 2^4, 3^4, 5^2. Alternatively, could we assign 2^4, 3^2, 5^4? Wait, but 5 needs to have exponent at least 2. So if we assign 5^4, that's okay. But 3 must have at least exponent 1, so even if we assigned 3^2, that's still acceptable. Wait, but in this case, the number of divisors would still be (4 + 1)(2 + 1)(4 + 1) = 5 * 3 * 5 = 75. Wait, no, the exponents must correspond to the factorization of 75 as 5 * 5 * 3. Wait, but in Case 4, we had exponents 4,4,2, which gives (4+1)(4+1)(2+1) = 5 * 5 * 3 = 75. If instead, we assigned exponents 4,2,4, that would be (4+1)(2+1)(4+1) = same thing. So regardless of the order, the product is the same. However, the actual value of n would be different depending on how we assign the exponents to the primes.So, to minimize n, we need to assign the largest exponents to the smallest primes. Because the smaller primes raised to higher exponents contribute less to the overall size of the number compared to larger primes. For example, 2^4 is 16, whereas 5^4 is 625, so it's better to have the higher exponent on the smaller prime.Therefore, in Case 4, the optimal assignment is to assign the largest exponents to the smallest primes. The exponents we have are 4, 4, 2. The two largest exponents are 4 and 4, so assign those to the two smallest primes (2 and 3), and the smallest exponent (2) to the next prime (5). Therefore, 2^4 * 3^4 * 5^2. That gives n = 16 * 81 * 25 = 32,400. Then, n / 75 = 32,400 / 75. Let me compute that. 75 times 400 is 30,000. 32,400 - 30,000 = 2,400. 75 * 32 = 2,400. So total is 400 + 32 = 432. Therefore, n / 75 = 432. But before concluding, let's make sure there isn't a smaller n. Let's check if we can have more primes. For example, could we have four primes? The number of divisors is 75, which factors into 3 * 5 * 5. To have four primes, the exponents plus one would have to multiply to 75, but 75 as a product of four integers greater than 1 would require 3 * 5 * 5 * 1, but 1 is not allowed since each factor is e_i + 1, which must be at least 2. Therefore, four primes are not possible. So the maximum number of primes is three. Therefore, the three-prime case is the one with the most primes, which allows for smaller exponents assigned to larger primes, thus minimizing the overall product. So 32,400 seems to be the minimal n.But let me verify once again. Are there other exponent combinations? For example, in the three-prime case, are there other exponent configurations?Wait, 75 factors as 5 * 5 * 3. So each e_i + 1 must be 5, 5, 3. Therefore, the exponents are 4, 4, 2. There's no other way to factor 75 into three factors greater than 1. 75 = 5 * 5 * 3. So that's the only way. Therefore, exponents must be 4, 4, 2. So the three-prime case is fixed. Therefore, the minimal n is 2^4 * 3^4 * 5^2, which is 16 * 81 * 25. Let me compute that again step by step:16 * 81 = 1,2961,296 * 25 = 32,40032,400 divided by 75 is 432. But wait, let me confirm if 32,400 is indeed a multiple of 75. 75 is 25 * 3. 32,400 divided by 25 is 1,296, which is an integer. Then 1,296 divided by 3 is 432, which is also an integer. So yes, 32,400 is divisible by 75. Additionally, does 32,400 have exactly 75 divisors? Let's verify. The prime factorization is 2^4 * 3^4 * 5^2. Therefore, the number of divisors is (4 + 1)(4 + 1)(2 + 1) = 5 * 5 * 3 = 75. Correct. Therefore, this seems to check out. However, let's make sure we considered all possibilities. For example, could there be a case with three primes where one of the exponents is higher, but another is lower, leading to a smaller number? Let's see.Suppose we tried exponents other than 4,4,2. For example, is there a way to have exponents such that the product (e1+1)(e2+1)(e3+1) = 75, but with different exponents?But since 75 factors as 3 * 5 * 5, the exponents plus one must be 3,5,5. Therefore, exponents must be 2,4,4. There's no other factorization. Therefore, all three-prime cases must have exponents 2,4,4. Therefore, we can't get different exponents.Alternatively, if we considered using more primes beyond three, but as we saw earlier, 75 can't be expressed as a product of four numbers greater than 1. Because 75 is 3 * 5 * 5, so you can't split it into four factors greater than 1. Therefore, four primes are impossible. So the three-prime case is the best.Alternatively, let's check if maybe there's a two-prime case with a different exponent combination. Wait, in the two-prime case, the possible exponent combinations are:- 24 and 2 (from 25 * 3)- 14 and 4 (from 15 * 5)We saw that 3^14 * 5^4 is smaller than 3^24 * 5^2. Let's compute 3^14 * 5^4 and compare it to 32,400.3^14 is 4782969 as computed before. 5^4 is 625. So 4782969 * 625 is approximately 2,989,355,625, which is way larger than 32,400. Therefore, the three-prime case gives a much smaller n. So indeed, 32,400 is the minimal n.Wait, but hold on. What if in the two-prime case, we use different primes? For example, if we have primes 2 and 3, but then 5 is missing. But n has to be a multiple of 75, which requires 5^2. Therefore, in two-prime case, the primes must include 3 and 5. Therefore, we can't use primes 2 and 3, because then 5 would not be included. So, the two primes must be 3 and 5. Therefore, in the two-prime case, we can't have primes 2,3,5. So, in two-prime case, it's 3 and 5, but with exponents 24 and 2 or 14 and 4.But as we saw, those numbers are way larger than the three-prime case. Therefore, the three-prime case is the way to go.Thus, the minimal n is 2^4 * 3^4 * 5^2 = 32,400. Therefore, n / 75 = 32,400 / 75 = 432.But let me just check once more for any possible oversight. For example, is there a way to have three primes, but using a different third prime? Like 2, 3, and 7? Then assign exponents to them. Wait, but using a larger prime like 7 would make the number larger. So, even if we assigned the smallest exponent to 7, but since 7 is larger than 5, it's better to have exponents on smaller primes. For example, if we had 2^4 * 3^4 * 7^2, that would be 16 * 81 * 49 = 16 * 81 = 1,296; 1,296 * 49 = 63,504, which is larger than 32,400. Therefore, definitely worse. So, no, using a larger prime would only increase the value of n.Therefore, the minimal n must use the smallest possible primes, which are 2, 3, and 5, with exponents assigned as 4,4,2. So, 2^4 * 3^4 * 5^2 = 32,400. Dividing by 75 gives 432. So the answer should be 432.But wait, just to make sure, let's see if there is another way to assign exponents with three primes. Suppose we have exponents 4, 2, 4. Assigning 2^4, 3^2, 5^4. But 5^4 is 625, which is much larger. So 2^4 * 3^2 * 5^4 = 16 * 9 * 625 = 144 * 625 = 90,000. Which is bigger than 32,400. So definitely worse. Similarly, if we assign 2^2, 3^4, 5^4, that's 4 * 81 * 625 = 202,500, which is even worse. So the minimal is indeed when we assign the higher exponents to the smaller primes.Therefore, I'm confident that the answer is 432.Final Answerboxed{432}</think>To find the smallest positive integer ( n ) that is a multiple of 75 and has exactly 75 positive integral divisors, we start by noting that ( 75 = 3 times 5^2 ). Therefore, ( n ) must include at least the primes 3 and 5 with exponents at least 1 and 2, respectively. The number of divisors of ( n ) is given by the product of one more than each exponent in its prime factorization.We need the product of these terms to equal 75. The prime factorization of 75 is ( 3 times 5 times 5 ), leading to possible factorizations of the form ( (e_1 + 1)(e_2 + 1)(e_3 + 1) ). The valid factorizations for 75 are ( 3 times 5 times 5 ), corresponding to exponents 2, 4, and 4.To minimize ( n ), we assign the largest exponents to the smallest primes. Thus, the prime factorization of ( n ) should be ( 2^4 times 3^4 times 5^2 ). Calculating this, we get:[2^4 times 3^4 times 5^2 = 16 times 81 times 25 = 32,400]We verify that ( 32,400 ) is a multiple of 75 and has exactly 75 divisors. Dividing ( 32,400 ) by 75 gives:[frac{32,400}{75} = 432]Thus, the final answer is (boxed{432}).,<think>Alright, so I need to find the smallest positive integer n that is a multiple of 75 and has exactly 75 positive integral divisors. Then, I have to find n divided by 75, which is essentially finding the smallest such n and then dividing out the 75. Let's start by understanding what the problem is asking for.First, n must be a multiple of 75. That means 75 divides n, or in other words, n is a multiple of 75. The prime factorization of 75 is 3 * 5^2. So, n must have at least the primes 3 and 5 in its prime factorization with exponents at least 1 and 2 respectively.Second, n must have exactly 75 positive integral divisors. The number of divisors of a number is determined by its prime factorization. If a number n has the prime factorization n = p1^a1 * p2^a2 * ... * pk^ak, then the number of divisors d(n) is (a1 + 1)*(a2 + 1)*...*(ak + 1). So, we need to arrange the exponents in the prime factors of n such that when each exponent is increased by 1 and multiplied together, the product is 75.Given that, let's first factorize 75 to understand the possible combinations for the exponents. The prime factors of 75 are 3 * 5^2. So, 75 can be written as 75 = 3 * 5 * 5, or 75 = 75 * 1, but since we're dealing with exponents +1, we need to see how 75 factors into integers greater than 1. Let's list all the possible factorizations:1. 75 = 752. 75 = 25 * 33. 75 = 15 * 54. 75 = 5 * 5 * 3These are the possible ways to write 75 as a product of integers greater than 1. Each of these factors corresponds to (exponent + 1) in the prime factorization of n. Therefore, the exponents themselves would be one less than each factor. For example, if we take the factorization 75 = 75, that would correspond to a single prime raised to the 74th power (since 74 + 1 = 75). However, since n must be a multiple of 75, which is 3 * 5^2, we need at least two primes: 3 and 5. So, using a single prime factor is impossible here because n would have to be a power of a single prime, but 75 requires multiple primes.Similarly, other factorizations correspond to different numbers of primes. Let's consider each factorization:1. 75 = 75: As discussed, one prime with exponent 74. Not possible here.2. 75 = 25 * 3: This would correspond to two primes, one with exponent 24 (25 -1) and another with exponent 2 (3 -1). So, possible exponents (24, 2)3. 75 = 15 * 5: Two primes, exponents 14 and 44. 75 = 5 * 5 * 3: Three primes, each with exponents 4, 4, 2 (since 5-1=4, 5-1=4, 3-1=2)These are the possible exponent combinations. Now, since we need n to be as small as possible, we need to assign the largest exponents to the smallest primes. Because smaller primes raised to higher powers contribute less to the overall size of the number than larger primes. For example, 2^4 is 16, while 5^4 is 625. So, assigning higher exponents to smaller primes will keep n smaller.So, we need to consider each possible exponent combination and assign the exponents to primes in descending order (largest exponents to the smallest primes) to minimize the product.Let's analyze each possible factorization:First, the two-prime factorizations:Case 1: 25 * 3: exponents (24, 2). Assigning the larger exponent (24) to the smaller prime (2) and the smaller exponent (2) to the next prime (3). Wait, but n needs to be a multiple of 75, which is 3 * 5^2. So, in the two-prime case, if we have primes 2 and 3, but we need to have 5^2. Therefore, the two-prime case might not work unless 5 is one of the primes. Wait, this is a key point.Hold on. If n is a multiple of 75, which is 3 * 5^2, then n must have at least the primes 3 and 5. So, if we are considering two primes in the factorization, they must be 3 and 5. But 75 is 3 * 5^2, so if n is a multiple of 75, but if we have two primes, 3 and 5, then the exponents would have to be at least 1 for 3 and at least 2 for 5. However, in the case of the two-prime factorization for the number of divisors, we have exponents (24, 2) or (14,4). Wait, but in both cases, if we need to have 3 and 5 as the primes, then the exponents for 3 and 5 need to be at least 1 and 2, respectively.So, for example, in the first two-prime case, 25 * 3, if we have exponents 24 and 2, then one prime has exponent 24, and the other has exponent 2. To satisfy the multiple of 75, we need to have at least 3^1 and 5^2. So, if we assign exponent 24 to 3, then 3^24 * 5^2. But 5^2 is required. Alternatively, if we assign exponent 24 to 5 and 2 to 3, then we have 5^24 * 3^2. However, in this case, 3^2 is more than the minimum required 3^1, and 5^24 is way more than 5^2. But both are acceptable, as long as they meet the minimum. However, since we need the minimal n, we need to see which arrangement is smaller.Similarly, in the second two-prime case, 15 *5: exponents 14 and 4. Assigning the exponents to 3 and 5. So either 3^14 *5^4 or 3^4 *5^14. Again, which is smaller?Then, the three-prime case: 5*5*3: exponents 4,4,2. Assigning these exponents to primes. Since we need primes 3 and 5, but also possibly other primes. However, since n must include 3 and 5, the three primes could be 2,3,5. So, assign the largest exponents to the smallest primes. So, exponents 4,4,2. So, assign 4 to 2, 4 to 3, and 2 to 5? Wait, but 5 is already required to have exponent at least 2, which would be satisfied here. But if we assign exponents as 4,4,2 to primes 2,3,5 respectively, then n would be 2^4 *3^4 *5^2. Alternatively, maybe arrange the exponents in descending order. The exponents sorted descendingly would be 4,4,2. So, the two highest exponents (4 and 4) assigned to the two smallest primes (2 and 3), and the lowest exponent (2) assigned to the next prime (5). That would give 2^4 *3^4 *5^2. Let's compute the value of that.Alternatively, if we have three primes, we need to include 2,3,5? Wait, n is a multiple of 75, which is 3*5^2. It doesn't need to be a multiple of 2. But since we want the minimal n, perhaps introducing the prime 2 could allow for a smaller n. For example, 2^a *3^b *5^c, with b >=1, c >=2. So, including 2 might allow smaller n.But let's check each possible case.First, the two-prime cases:Case 1: Exponents (24,2). Assigning these to primes 3 and 5. Let's compute both possibilities:Option 1: 3^24 *5^2. Then, n = 3^24 *5^2. That's a very large number.Option 2: 3^2 *5^24. Even larger. So both are very large. Hence, two-prime case here would result in a very large n, which is probably not the minimal.Case 2: Exponents (14,4). Similarly:Option 1: 3^14 *5^4. Again, a large number.Option 2: 3^4 *5^14. Even larger. So, again, these are big numbers.Now, the three-prime case: exponents (4,4,2). Assigning to primes 2,3,5. So, 2^4 *3^4 *5^2. Let's compute that:2^4 = 16, 3^4 = 81, 5^2 = 25. So, n = 16 *81 *25.16 *25 = 400, 400 *81 = 32,400.Alternatively, maybe another arrangement. If we assign exponents differently. For example, if we assign the highest exponent (4) to prime 2, next highest (4) to prime 3, and lowest (2) to prime 5. Which is what we did. Alternatively, if we assign 4 to 2, 4 to 5, and 2 to 3. Then, n = 2^4 *3^2 *5^4. Then, 2^4=16, 3^2=9, 5^4=625. Then, 16 *9=144, 144*625=90,000. Which is larger. So the first arrangement is better.Alternatively, assign 4 to 3, 4 to 5, 2 to 2. Then n= 3^4 *5^4 *2^2. 3^4=81, 5^4=625, 2^2=4. 81*625=50,625, 50,625*4=202,500. Which is even larger.Therefore, the minimal three-prime case is 2^4 *3^4 *5^2=32,400.Now, compare this to the two-prime cases. The two-prime cases gave n=3^14 *5^4, which is 3^14 is 4,782,969, and 5^4=625. Multiplying those gives a gigantic number. Similarly for the other options. So, three-prime case is much better.But wait, could there be a case with more than three primes? The number 75 factors into 3*5*5, so three factors. However, if we consider that 75 can also be written as, say, 5*3*5, but that's the same as before. Alternatively, is there a way to have more factors? For example, 75= 3*5*5, which is three factors. If we want to write 75 as a product of more factors, we can break down the exponents further. For example, 75=3*5*5=3*5*5*1*1..., but since each factor must be at least 2, we can't have 1s. So, the maximum number of factors (exponents +1) is 3. Therefore, the number of primes in the factorization can be at most 3. Wait, no. Wait, the number of exponents is equal to the number of prime factors. So, if we have three factors in the divisor count formula, then we have three primes. If we have two factors, two primes, and so on. Therefore, since 75 factors into three numbers (5*5*3), then the maximum number of primes is 3. Therefore, the three-prime case is the maximum here. So, we can't have more primes than three. Therefore, the possible cases are two primes or three primes, but as we saw, the three-prime case gives a smaller n.Wait, but let me verify. Suppose we tried to break 75 into more factors. For example, if we write 75 as 3*5*5, that's three factors. If we try to factorize 75 into four factors, that would require writing 75 as a product of four numbers each at least 2, but 2*2*2*2=16, which is less than 75. However, 75 divided by 2*2*2 is 75/8 ≈9.375, which is not integer. So, 2*2*3*something? Let's see: 2*3*5*5=150, which is more than 75. So, actually, 75 can't be expressed as a product of four integers each at least 2. Because the minimal product with four factors each at least 2 is 16, but 75 isn't a multiple of 16. Alternatively, 2*2*3* something. 2*2*3=12, 75/12=6.25, not integer. 2*3*5=30, 75/30=2.5, not integer. So, actually, 75 cannot be expressed as a product of four integers each greater than or equal to 2. Therefore, the maximum number of prime factors for n is 3. So, we can't have four primes. Therefore, the only possible cases are two primes or three primes.But in the two-prime cases, as we saw, the numbers are way too big. The three-prime case gives n=32,400. Wait, but let's check if 32,400 is a multiple of 75. 32,400 divided by 75: 75*432=32,400. So, yes, 432*75=32,400. So, n=32,400. Then, n/75=432. Therefore, the answer would be 432. But before concluding, let me make sure there isn't a smaller n.Wait, but perhaps there is another combination of exponents that I missed? Let me check the factorization of 75 again.75=3*5*5. So, in terms of exponents +1, that would be exponents 2,4,4 (since 3-1=2, 5-1=4, 5-1=4). Wait, but earlier, I considered the exponents as 4,4,2. But since the order doesn't matter in terms of multiplication, but when assigning to primes, the order matters for the size of n.Wait, so the exponents (from the divisor formula) are (2,4,4). Since the formula is (e1+1)(e2+1)(e3+1)=75, so e1,e2,e3 are 2,4,4. So, sorted in ascending order: 2,4,4. Assigning these exponents to the primes. To minimize n, we assign the largest exponents to the smallest primes. So, the exponents 4,4,2 assigned to primes 2,3,5. Therefore, 2^4 *3^4 *5^2. That gives 16*81*25=32,400.Alternatively, if we assigned the exponents 4,2,4 to primes 2,3,5, we get 2^4 *3^2 *5^4=16*9*625=90,000. Which is larger.Similarly, 2^2 *3^4 *5^4=4*81*625=202,500. Even worse.Hence, the minimal is 32,400.But let's check if there's a two-prime case that I might have miscalculated. For example, if n is a multiple of 75, and has two prime factors, which must be 3 and 5. Then, the number of divisors is (a+1)(b+1)=75, where a is the exponent of 3 and b is the exponent of 5. Given that n is a multiple of 75, a >=1, b >=2.So, solve (a+1)(b+1)=75 with a >=1, b >=2.Possible factor pairs of 75:1 and 75: a+1=1, a=0 (invalid since a>=1)3 and 25: a+1=3, a=2; b+1=25, b=24. So exponents (2,24). So n=3^2 *5^24. Which is 9 *5^24, which is huge.5 and 15: a+1=5, a=4; b+1=15, b=14. So n=3^4 *5^14. 81*5^14. Still enormous.25 and 3: a+1=25, a=24; b+1=3, b=2. So n=3^24 *5^2. 3^24 is a very large number.15 and 5: a+1=15, a=14; b+1=5, b=4. n=3^14 *5^4. Also huge.So, all two-prime cases result in very large n. Therefore, the minimal n must be in the three-prime case.Therefore, n=2^4 *3^4 *5^2=32,400. Then, n/75=32,400 /75=432. Therefore, the answer is 432.But let me check once more to ensure there are no mistakes in reasoning.First, verify that 32,400 is indeed a multiple of 75. 75*432=32,400. Correct.Second, check the number of divisors. The prime factorization of 32,400 is 2^4 *3^4 *5^2. The number of divisors is (4+1)*(4+1)*(2+1)=5*5*3=75. Correct.Third, check if there is a smaller number that meets the criteria. Suppose someone tries to use another prime, like 7. For example, 2^4 *3^2 *5^4 *7^something. Wait, but adding another prime would require the exponents to multiply to 75/(something). Wait, actually, if we add another prime, the number of divisors would be multiplied by (1 + exponent of that prime). But since 75 is 3*5*5, which is three factors, adding another prime would require another factor, but 75 can't be divided by another factor. Wait, for example, if we have four primes, then the number of divisors would be (e1 +1)(e2 +1)(e3 +1)(e4 +1). But 75 factors into 3*5*5, which is three factors, so unless we have a 1 in the factorization, which isn't allowed (since exponents are at least 1, so e_i +1 >=2), we can't have four primes. Therefore, the number of primes must be three at maximum.Alternatively, if we tried to include another prime with exponent 0, but exponents must be at least 1 for primes present in the factorization. Wait, but if a prime is not in the factorization, its exponent is 0. But in that case, the number of divisors would be calculated as product over primes present. So, if we have three primes, we have three exponents. If we have four primes, each with exponent at least 1, then the number of divisors would be (e1 +1)(e2 +1)(e3 +1)(e4 +1). But since 75 can't be expressed as a product of four numbers greater than 1, we can't have four primes. Therefore, the minimal number of primes is three. Therefore, we can't get away with fewer than three primes, and we can't have more. Therefore, the three-prime case is necessary.Therefore, n=2^4 *3^4 *5^2=32,400 is indeed the minimal n. Therefore, n/75=432. So the answer is 432.But just to be thorough, let's check if there's a different exponent arrangement with three primes that could result in a smaller n. For example, perhaps instead of exponents 4,4,2, we can have different exponents that multiply to 75. Wait, 75=3*5*5, which are the (e1 +1)(e2 +1)(e3 +1). So, e1 +1=3, e2 +1=5, e3 +1=5. So, e1=2, e2=4, e3=4. So, exponents 2,4,4. So, sorted in descending order, 4,4,2. Assign to primes 2,3,5. So, 2^4 *3^4 *5^2. Which is what we did.Alternatively, if we assigned 4,2,4, we get 2^4 *3^2 *5^4=16*9*625=90,000, which is bigger. So, no.Alternatively, if we use different exponents. For example, is there another way to factor 75 as three numbers? 75=1*3*25, but 1 is not allowed because exponents +1 must be at least 2. So, that's invalid. 75=3*5*5, which we have. So, no other factorizations.Therefore, the exponents must be 2,4,4. Assigning them to the smallest primes in descending order. Therefore, 2^4 *3^4 *5^2 is the minimal.Hence, the answer is 432.Final Answerboxed{432}</think>To find the smallest positive integer ( n ) that is a multiple of 75 and has exactly 75 positive integral divisors, we start by noting that 75 factors as ( 3 times 5^2 ). Therefore, ( n ) must include the primes 3 and 5 in its prime factorization with exponents at least 1 and 2, respectively.The number of divisors of ( n ) is given by the product of one more than each exponent in its prime factorization. We need this product to be 75. The factorization of 75 is ( 3 times 5 times 5 ), which corresponds to exponents (2, 4, 4) in the prime factorization of ( n ) (since each factor is one more than the exponent).To minimize ( n ), we assign the largest exponents to the smallest primes. Thus, we use the primes 2, 3, and 5 with exponents 4, 4, and 2 respectively. This gives the prime factorization ( 2^4 times 3^4 times 5^2 ).Calculating this value:[2^4 = 16, quad 3^4 = 81, quad 5^2 = 25][n = 16 times 81 times 25 = 32,400]We verify that ( 32,400 ) is a multiple of 75:[frac{32,400}{75} = 432]The number of divisors of ( 32,400 ) is:[(4+1)(4+1)(2+1) = 5 times 5 times 3 = 75]Thus, the smallest ( n ) is 32,400, and dividing by 75 gives the final answer:[boxed{432}],<think>Okay, let's see. The problem is asking for the smallest positive integer n that is a multiple of 75 and has exactly 75 positive integral divisors. Then, we need to find n divided by 75. Hmm, so first, n has to be a multiple of 75, which means 75 must divide n. Also, n must have exactly 75 divisors. We need to find the smallest such n and then compute n/75. Alright, let's start by recalling that the number of divisors of a number is related to its prime factorization. Specifically, if a number n has a prime factorization of n = p₁^a₁ * p₂^a₂ * ... * p_k^a_k, then the number of divisors d(n) is (a₁ + 1)(a₂ + 1)...(a_k + 1). So, our task is to find the smallest n divisible by 75 such that this product of exponents plus one equals 75.First, let's factorize 75. 75 is 3 * 5 * 5, so in prime factors, that's 3^1 * 5^2. Therefore, any multiple of 75 must have at least the primes 3 and 5 in its factorization with exponents at least 1 and 2 respectively. So, in n's prime factorization, the exponent of 3 must be at least 1, and the exponent of 5 must be at least 2.Now, the number of divisors is 75. So, we need (a₁ + 1)(a₂ + 1)...(a_k + 1) = 75. Let's factorize 75 to see possible exponents. 75 factors into 3 * 5 * 5, or 15 * 5, or 25 * 3, or 75 * 1. However, since 75 is the product of exponents plus one, we need to represent 75 as a product of integers greater than 1, each of which is one more than the exponent in the prime factorization. So, possible factorizations of 75 into integers greater than 1 (since each (a_i + 1) must be at least 2, given that exponents a_i are at least 1 for primes in n's factorization). Let's list all possible ways to write 75 as a product of integers greater than 1:1) 75 = 752) 75 = 25 * 33) 75 = 15 * 54) 75 = 5 * 5 * 3These are the possible factorizations. Each of these corresponds to different possible exponent configurations. Our goal is to arrange these factors such that when each is subtracted by 1, they give the exponents in the prime factorization. Also, since we need the number n to be as small as possible, we should assign higher exponents to smaller primes. That's because smaller primes raised to higher powers contribute less to the overall size of the number than larger primes. For example, 2^3 is 8, whereas 3^3 is 27, which is larger. So, to minimize n, we want to allocate the higher exponents to the smaller primes.First, let's consider the factorization 75 = 75. Then, (a₁ + 1) = 75, so a₁ = 74. This would correspond to a prime raised to the 74th power. But since n must be a multiple of 75 (which is 3*5^2), this prime would have to be 3 or 5. If we take 3^74, then it's not a multiple of 5^2, so that's invalid. Similarly, 5^74 is not a multiple of 3. So, this single prime case doesn't work. So, factorization 75 is invalid because n needs to have at least two primes (3 and 5). So we can disregard this case.Next, factorization 75 = 25 * 3. Then, (a₁ + 1) = 25 and (a₂ + 1) = 3. Therefore, exponents a₁ = 24 and a₂ = 2. Assigning these exponents to primes. To minimize n, we should assign the larger exponent (24) to the smaller prime. So, 2^24 * 3^2. Wait, but n must include 3 and 5. So, if we use primes 2 and 3 here, then n would be 2^24 * 3^2, but this is not a multiple of 5^2. So, invalid. Alternatively, if we use primes 3 and 5: 3^24 * 5^2. But this includes 3 and 5, which is necessary. But we need to check if this is the minimal. Wait, but is this the only way? Alternatively, maybe we need more primes? Because if we have 25 * 3, that would correspond to two primes. But n must have at least two primes (3 and 5). However, 3^24 * 5^2 would have the required factors. But maybe there's a way to include more primes but with smaller exponents to get a smaller n?Wait, let's hold that thought. Let's check the next factorization: 75 = 15 * 5. Then, (a₁ +1) =15, (a₂ +1)=5, so exponents a₁=14, a₂=4. Assigning these to primes. Again, assign higher exponents to smaller primes. So, 2^14 * 3^4. But this lacks 5^2. If we use 3 and 5, then 3^14 * 5^4. But this would be a larger number. Alternatively, if we need to include 3 and 5, perhaps the exponents would be split as 2^14 * 3^4 * 5^2. Wait, but that's three primes. However, the number of divisors would be (14 +1)(4 +1)(2 +1) =15*5*3=225, which is more than 75. So that's not correct. Hmm, maybe the 15*5 factorization must correspond to two primes. If we have two primes, 2 and 3, then 2^14 *3^4, but that's not multiple of 5. If we have two primes 3 and 5, then 3^14 *5^4, which is multiple of 3 and 5^4. But is 5^4 okay? Since we need at least 5^2. So yes, but perhaps that's a higher power than necessary. However, the problem is that this approach would not be minimal. Let's check the actual value.Alternatively, the third factorization: 75 = 5*5*3. So, (a₁+1)(a₂+1)(a₃+1)=5*5*3. Therefore, exponents a₁=4, a₂=4, a₃=2. Assigning these exponents to primes. Since we need to include 3 and 5, the third prime would be 2. To minimize n, assign the largest exponents to the smallest primes. So, 2^4 * 3^4 * 5^2. Let's check the number of divisors: (4+1)(4+1)(2+1)=5*5*3=75. Perfect. So that works. Now, the number n in this case is 2^4 *3^4 *5^2. Let's compute that:2^4 =16, 3^4=81, 5^2=25. Multiply them together: 16*81=1296, 1296*25=32400. So, 32,400. Wait, but we need to check if this is indeed a multiple of 75. 32,400 divided by 75: 32,400 /75= 432. So, yes, it's a multiple. Now, is this the minimal n? Let's check the other cases.Earlier, we had the factorization 25*3, which would lead to exponents 24 and 2. Assigning to primes 3 and 5, 3^24 *5^2. Let's compute that. 3^24 is a huge number, way larger than 32,400, so that's not minimal. Similarly, the 15*5 factorization gives exponents 14 and 4, which would be 3^14 *5^4. 3^14 is 4,782,969 and 5^4 is 625. Multiplying those gives 4,782,969 *625, which is about 3 billion, which is way larger. So, 32,400 is better.The other factorization, 75=75, which gives a single prime with exponent 74, but as discussed, that doesn't work. So, the only viable factorization that gives exactly 75 divisors is 5*5*3, which gives exponents 4,4,2. Assigning those to primes 2,3,5 gives 2^4 *3^4 *5^2=32,400. But wait, let's make sure there isn't another arrangement of exponents with more primes but smaller exponents. For example, if we take the factorization 75=3*5*5. If we have four primes, would that be possible? For example, 3*5*5*1, but the factors have to be greater than 1. So, 3*5*5 is the maximum number of factors here. So, it's three primes. So, if we have three primes, the exponents are 4,4,2. Assigning to primes 2,3,5 gives the minimal number. If we instead assigned the 2 exponent to a larger prime, say 2^4 *3^2 *5^4, that would be 16*9*625=16*5625=90,000, which is larger than 32,400. So, that's worse. Similarly, if we assigned the exponents as 2^2 *3^4 *5^4, that's 4*81*625=4*50,625=202,500, which is even worse. So, 32,400 seems better. Alternatively, could we have more primes? For example, 75=3*5*5, which is three factors. If we tried to split into more factors, but since 75 factors as 3*5*5, which are primes, we can't split further. So, we need three primes. So, n must have three distinct prime factors. But 75 is 3*5*5, so (4,4,2). Wait, but the problem states that n must be a multiple of 75. 75 is 3*5^2. So, the prime factors of n must include 3 and 5. The third prime factor is 2. So, 2^4 *3^4 *5^2. Is there a way to have a prime factor other than 2? For example, if we use 7 instead of 2? Then, 7^4 *3^4 *5^2. But 7^4 is much larger than 2^4, so that would make n larger. So, 2 is the smallest prime available, so using 2 is optimal.Therefore, 2^4 *3^4 *5^2= 16*81*25= 32,400. So, n=32,400. Then, n/75=32,400 /75. Let's compute that. 75*400=30,000. 32,400 -30,000=2,400. 75*32=2,400. So, total is 400+32=432. So, n/75=432. Therefore, the answer is 432.Wait, but before finalizing, let's check once again if there is a smaller n. Suppose we try another exponent combination. For example, perhaps 2^2 *3^4 *5^4 *7^? But that would require more primes, which would increase the number. Wait, but 75=3*5*5. If we have four primes, then we need to have (a1 +1)(a2 +1)(a3 +1)(a4 +1)=75. However, 75 factors into 3*5*5, so we can't split it into four factors each greater than 1. Because 3*5*5 is already three factors. If we tried to write 75 as 3*5*5*1, but 1 is not allowed, as each factor must be at least 2. So, the maximum number of primes is three. Therefore, we can't have four primes. Therefore, the minimal n is indeed 2^4 *3^4 *5^2. Wait, another thought. Let's verify the divisor count. For n=2^4 *3^4 *5^2, the number of divisors is (4+1)*(4+1)*(2+1)=5*5*3=75. Correct. And n is 16*81*25=32,400. And 32,400 divided by 75 is 432. Is there a possibility that by permuting exponents among primes, even with the same number of primes, we could get a smaller number? Let's check. For example, if we have exponents 4,2,4 instead of 4,4,2. Assigning to primes 2,3,5 as 2^4 *3^2 *5^4. Then, n=16*9*625=144*625=90,000, which is larger. Similarly, 2^2 *3^4 *5^4=4*81*625=4*50,625=202,500. So, definitely larger. So, the original assignment is better. Another check: if we use primes 2,3,5 with exponents 4,4,2, n=16*81*25=32,400. If we consider exponents 2,4,4, but assign 2 to a different prime. Wait, no, 2 is the smallest prime, so assigning the largest exponent to the smallest prime gives the minimal n. Therefore, I think 32,400 is indeed the minimal n. Then, n/75=432. Therefore, the answer is 432. Just to make sure, let's see if there's another factorization of 75 that we might have missed. 75=5*15. Wait, we considered that as 15*5. If we take exponents 14 and 4, but as discussed, that would be a two-prime case. But n needs to have three primes (2,3,5). Wait, no, in the two-prime case, if we have exponents 14 and 4, but since n must include 3 and 5, the two primes would have to be 3 and 5. So, 3^14 *5^4. Let's compute that. 3^14 is 4782969, 5^4 is 625. Multiplying those gives 4782969 *625. Let me compute that:First, 4,782,969 * 600 = 2,869,781,400Then, 4,782,969 *25 = 119,574,225Adding those together: 2,869,781,400 + 119,574,225 = 2,989,355,625That's about 3 billion, which is way bigger than 32,400. So, no, that's not minimal.Alternatively, if we use 3 and 5 as primes, but with exponents 4 and 14. So, 3^4 *5^14. Then, 81 * (5^14). 5^14 is 6103515625. 81 * 6,103,515,625 = 494,384,765,625. Even larger. So, definitely worse.Therefore, the minimal n is indeed 32,400, so n/75 is 432. Therefore, the answer is 432.Final Answerboxed{432}</think>To find the smallest positive integer ( n ) that is a multiple of 75 and has exactly 75 positive integral divisors, we start by noting that 75 factors as ( 3 times 5^2 ). Therefore, ( n ) must include the primes 3 and 5 with exponents at least 1 and 2, respectively. The number of divisors of ( n ) is given by the product of one more than each exponent in its prime factorization. We need this product to be 75. The factorization of 75 is ( 3 times 5 times 5 ). This corresponds to exponents in the prime factorization such that when each is increased by 1, their product is 75. Considering the factorization ( 75 = 3 times 5 times 5 ), we get exponents 4, 4, and 2 (since ( (4+1)(4+1)(2+1) = 5 times 5 times 3 = 75 )). To minimize ( n ), we assign the largest exponents to the smallest primes. Thus, the prime factorization of ( n ) is ( 2^4 times 3^4 times 5^2 ).Calculating ( n ):[2^4 = 16, quad 3^4 = 81, quad 5^2 = 25][n = 16 times 81 times 25 = 32400]Finally, we find ( frac{n}{75} ):[frac{32400}{75} = 432]Thus, the answer is (boxed{432}).,<think>Okay, let's tackle this problem step by step. The problem says:"Let ( n ) be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find ( frac{n}{75} )."So, I need to find the smallest number ( n ) such that:1. ( n ) is a multiple of 75.2. ( n ) has exactly 75 positive divisors.3. Then compute ( frac{n}{75} ).First, let's recall some number theory concepts. The number of divisors of a positive integer is determined by its prime factorization. If ( n ) has a prime factorization of ( n = p_1^{e_1} p_2^{e_2} dotsm p_k^{e_k} ), then the number of divisors ( d(n) ) is given by:[d(n) = (e_1 + 1)(e_2 + 1) dotsm (e_k + 1)]So, the problem reduces to finding the smallest ( n ) divisible by 75 (which is ( 3 times 5^2 )) such that the product of one more than each of the exponents in its prime factorization is 75.First, let's factorize 75 (the number of divisors) to understand the possible exponents. The prime factorization of 75 is ( 75 = 3 times 5 times 5 ). So, the product ( (e_1 + 1)(e_2 + 1)dotsm ) must equal 75. Therefore, the exponents ( e_i ) in the prime factorization of ( n ) must be such that when each is increased by one and multiplied together, the result is 75.We need to consider the possible ways to express 75 as a product of integers greater than 1 (since each ( e_i + 1 ) must be at least 2, as exponents are at least 1). The order of the factors matters because different arrangements correspond to different prime bases, which will affect the size of ( n ). To minimize ( n ), we need to assign the largest exponents to the smallest primes. This is due to the fact that smaller primes raised to higher powers contribute less to the overall size of the number compared to larger primes.Let's list the possible factorizations of 75 into integers greater than 1:1. 75 = 752. 75 = 25 × 33. 75 = 15 × 54. 75 = 5 × 5 × 3These correspond to different configurations of exponents:1. One exponent: 74 (since 74 + 1 = 75)2. Two exponents: 24 and 2 (since 24 + 1 = 25 and 2 + 1 = 3)3. Two exponents: 14 and 4 (since 14 + 1 = 15 and 4 + 1 = 5)4. Three exponents: 4, 4, and 2 (since 4 + 1 = 5, 4 + 1 = 5, and 2 + 1 = 3)Now, since ( n ) must be a multiple of 75, which is ( 3 times 5^2 ), the prime factors of ( n ) must include at least 3 and 5 with exponents at least 1 and 2 respectively. Also, ( n ) can have other prime factors as well, but we need to minimize ( n ).We need to consider the possible exponent configurations and map them to primes, ensuring that the exponents for primes 3 and 5 meet the minimum requirements. Let's analyze each case.Case 1: 75 = 75 (single exponent)This would mean ( n = p^{74} ). However, since ( n ) must be divisible by 3 and 5^2, this would require that ( p = 3 ) or ( p = 5 ), but even if we take the smallest prime, 3, ( n = 3^{74} ), which is way larger than necessary. Alternatively, if we take 5, ( 5^{74} ), which is even larger. This is clearly not the minimal ( n ), so we can discard this case.Case 2: 75 = 25 × 3 (two exponents)This corresponds to exponents 24 and 2. Therefore, ( n ) would be ( p^{24} q^{2} ), where ( p ) and ( q ) are primes. To satisfy divisibility by 75, ( n ) must include 3 and 5 with exponents at least 1 and 2. So, we need to assign the exponents 24 and 2 to the primes 3 and 5. However, we need to check which assignment gives a smaller number.If we assign the larger exponent (24) to the smaller prime (3) and the smaller exponent (2) to 5, then ( n = 3^{24} times 5^2 ). Alternatively, if we swap them, ( n = 5^{24} times 3^2 ). The first option is ( 3^{24} times 5^2 ), which is enormous, but the second is even larger because 5 is a bigger prime. However, maybe we can introduce another prime here? Wait, no—since the number of exponents is two, which corresponds to two primes. But since n must include 3 and 5, we can't have another prime here. So, if we use two primes, they have to be 3 and 5. Therefore, we must assign the exponents 24 and 2 to 3 and 5. However, the problem is that 5 needs to have at least exponent 2, so we have to assign exponent 24 to 3 and 2 to 5. But 3^24 * 5^2 is a huge number. Maybe there's a better way with more primes?Wait, but in this case, since the number of exponents is two, we can only have two primes. But 75 is 25 × 3, so two factors. So, n is a product of two primes, each raised to exponents 24 and 2. But since we need to cover 3 and 5, we have to use both primes. So this would result in n = 3^24 * 5^2 or 3^2 *5^24. The first is smaller. But even that is massive. Let's see if other cases can give a smaller n.Case 3: 75 = 15 × 5 (two exponents)This corresponds to exponents 14 and 4. So, ( n = p^{14} q^4 ). Again, we need to use primes 3 and 5. Assigning exponents:Option 1: 3^14 * 5^4Option 2: 5^14 * 3^4Which is smaller? Let's compute both.Compute 3^14 vs 5^4 and vice versa. However, 3^14 is 4,782,969 and 5^4 is 625. So 3^14 * 5^4 ≈ 4,782,969 * 625 ≈ 2,989,355,625On the other hand, 5^14 is a huge number (5^14 = 6103515625) and 3^4 = 81. So 6103515625 * 81 is way larger. So the first option is smaller. So n would be 3^14 * 5^4. But still, that's over 2 billion. Maybe we can do better with the next case.Case 4: 75 = 5 × 5 × 3 (three exponents)This corresponds to exponents 4, 4, 2. So, ( n = p^4 q^4 r^2 ). Here, we can have three primes. Since n must be divisible by 3 and 5^2, two of the primes must be 3 and 5, and the third can be another prime (preferably the smallest available, which is 2). Assigning the exponents:We need to assign exponents to primes 2, 3, and 5. To minimize n, we should assign the largest exponents to the smallest primes. The exponents we have are 4, 4, 2. So, the largest exponent is 4, so assign that to the smallest primes. Assign 4 to 2, 4 to 3, and 2 to 5. Therefore, n = 2^4 * 3^4 * 5^2. Let's compute that:2^4 = 163^4 = 815^2 = 25Multiply them together: 16 * 81 = 1296; 1296 * 25 = 32,400Alternatively, if we assigned the exponents differently, say 4 to 2, 2 to 3, and 4 to 5: n = 2^4 * 3^2 * 5^4. Then:2^4 = 163^2 = 95^4 = 62516 * 9 = 144; 144 * 625 = 90,000, which is larger.Similarly, assigning 2 to 2, 4 to 3, 4 to 5: 2^2 * 3^4 * 5^4 = 4 * 81 * 625 = 4 * 50,625 = 202,500. So the first assignment is the smallest.Thus, n = 2^4 * 3^4 * 5^2 = 32,400. However, wait, but we need to check if this n is actually a multiple of 75. 75 is 3 * 5^2. Let's check:32,400 divided by 75: 32,400 / 75 = 432. So yes, it is a multiple. But does it have exactly 75 divisors?Let's verify the number of divisors. The prime factorization is 2^4 * 3^4 * 5^2. So, the number of divisors is (4+1)(4+1)(2+1) = 5*5*3 = 75. Yes, that's correct. So, this n is 32,400. Then, n / 75 = 432. So is this the answer? Wait, but let's check the other cases to make sure there isn't a smaller n.In Case 3, we had n as 3^14 * 5^4 ≈ 2.989 billion, which is way larger than 32,400. In Case 2, even larger. Case 1 is even worse. So Case 4 gives the smallest n. However, wait—maybe there's a different arrangement of exponents with more primes? Let's think.Wait, 75 factors as 5 × 5 × 3, so that gives three exponents. But what if we considered more factors? For example, 75 can also be written as 5 × 5 × 3 × 1, but the 1 would correspond to an exponent of 0, which is allowed only if we have a prime raised to the 0 power (which is 1). But including 1s in the factorization would not contribute to the product, so we can ignore them because adding more primes with exponent 0 doesn't change the number. Therefore, the minimal number of primes needed is three. So, if we have three primes, with exponents 4, 4, 2, then n is 2^4 * 3^4 * 5^2. If we try adding a fourth prime, that would require that 75 is factored into four integers, which would require introducing a 1, but since 75 = 5×5×3×1, but the 1 would correspond to an exponent of 0, which would mean adding a prime raised to the 0 power (i.e., not including it). So, adding a fourth prime would require an exponent of 0, which doesn't contribute. Therefore, the minimal number of primes is three. So, Case 4 is the minimal.But just to make sure, let's check another possibility. Suppose instead of 4,4,2, maybe another combination? Wait, 75 = 5×5×3, which are the factors for the number of divisors. So each factor corresponds to (e_i +1). Therefore, the exponents would need to be 4,4,2. There's no other way to factor 75 into three numbers greater than 1. Because 75 divided by 3 is 25, and 25 factors into 5×5. Therefore, 75 = 3×5×5. So that's the only way to factor 75 into three numbers. Therefore, the exponents are 2,4,4 (since 3-1=2, 5-1=4, 5-1=4). Wait, but in our case, we have (e1 +1)(e2 +1)(e3 +1) = 75. So, the exponents are 4,4,2 (since 4+1=5, 4+1=5, 2+1=3). So, 5×5×3=75. Therefore, the exponents are 4,4,2. Assigning these exponents to the primes in ascending order (to minimize the product), we get 2^4 * 3^4 *5^2. Which is 32,400. So that seems to be the minimal.But let's check if there's a different exponent arrangement. For example, if we have exponents 4,2,4 instead. But since the primes are assigned from smallest to largest, swapping exponents between primes would only increase the number. For instance, if we assign 4 to 2, 2 to 3, and 4 to 5, which would be 2^4 *3^2 *5^4. As calculated before, that's 90,000, which is larger. So, keeping the higher exponents on the smaller primes is better.Another thought: What if we use more primes? For example, if we have four primes. Then, the number of divisors would be (e1+1)(e2+1)(e3+1)(e4+1). But 75 factors into four numbers as 5×5×3×1. However, the 1 corresponds to (e4 +1)=1 => e4=0. So, that would mean having a prime raised to the 0 power, which is 1, so effectively not including that prime. Therefore, we can't actually get a fourth prime because the exponent would be 0. So, the minimal number of primes is three. Hence, Case 4 is indeed the way to go.Therefore, n=2^4 *3^4 *5^2=16*81*25=32,400. Then, n/75=32,400/75=432. So the answer should be 432. But let me double-check to ensure there's no mistake in the calculation.Calculating 2^4 *3^4 *5^2:2^4=163^4=815^2=2516*81=1,2961,296*25=32,40032,400 divided by 75: 32,400 ÷ 75.75*400=30,00032,400 -30,000=2,40075*32=2,400So total is 400+32=432. Correct.Therefore, the answer is 432. However, let's confirm once more that there isn't a smaller number that meets the criteria. Suppose someone tried to use another prime, say 7, but with exponents. For example, if we have primes 2, 3, 5, 7, but that would require exponents such that the product of (e_i +1) is 75. Let's see:If we have four primes, then the factorization of 75 would need to be into four factors. Since 75=3×5×5×1, but again, the 1 corresponds to an exponent of 0. So, we can have exponents like 2,4,4,0. Assigning 0 to 7, which would mean not including it. So, back to three primes. Therefore, introducing more primes doesn't help, as the exponent for the new prime would have to be 0. Hence, the minimal is three primes, as we had.Another check: Suppose we have exponents in a different order. For instance, if we take the exponents as 2,4,4, but assign them to primes 2,5,3. So, 2^2 *3^4 *5^4. Then, this is 4 * 81 * 625 = 4*50,625=202,500. Which is larger than 32,400. So, that's worse.Alternatively, 2^4 *5^4 *3^2. That's 16 * 625 *9=16*5,625=90,000. Still larger.Therefore, the minimal n is indeed 32,400. Hence, the answer is 432.Final Answerboxed{432}</think>To solve the problem, we need to find the smallest positive integer ( n ) that is a multiple of 75 and has exactly 75 positive integral divisors. We then need to compute ( frac{n}{75} ).1. Prime Factorization and Divisors: - The number of divisors of ( n ) is given by the product of one more than each exponent in its prime factorization. For ( n ) to have 75 divisors, the product of these terms must be 75. - The prime factorization of 75 (the number of divisors) is ( 75 = 3 times 5 times 5 ). This means the exponents in the prime factorization of ( n ) must satisfy ( (e_1 + 1)(e_2 + 1)(e_3 + 1) = 75 ).2. Possible Exponent Configurations: - The factorization of 75 into integers greater than 1 gives possible configurations: ( 75 = 3 times 5 times 5 ). This corresponds to exponents 2, 4, and 4 (since ( 3 - 1 = 2 ), ( 5 - 1 = 4 ), and ( 5 - 1 = 4 )).3. Assigning Exponents to Primes: - To minimize ( n ), we assign the largest exponents to the smallest primes. The prime factorization of ( n ) must include at least 3 and ( 5^2 ) (since 75 is ( 3 times 5^2 )). - Assigning the exponents 4, 4, and 2 to the primes 2, 3, and 5 respectively gives the smallest ( n ). Thus, ( n = 2^4 times 3^4 times 5^2 ).4. Calculating ( n ): - Compute ( 2^4 = 16 ), ( 3^4 = 81 ), and ( 5^2 = 25 ). - Multiply these values: ( 16 times 81 = 1296 ) and ( 1296 times 25 = 32400 ).5. Verification: - Check that ( 32400 ) is a multiple of 75: ( 32400 div 75 = 432 ). - Check the number of divisors: ( (4+1)(4+1)(2+1) = 5 times 5 times 3 = 75 ).Thus, the smallest ( n ) is 32400, and ( frac{n}{75} = 432 ).[boxed{432}]]

🔑:The [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) of 75 = 3^15^2 = (2+1)(4+1)(4+1). For n to have exactly 75 integral divisors, we need to have n = p_1^{e_1-1}p_2^{e_2-1}cdots such that e_1e_2 cdots = 75. Since 75|n, two of the [prime](https://artofproblemsolving.com/wiki/index.php/Prime) [factors](https://artofproblemsolving.com/wiki/index.php/Factor) must be 3 and 5. To minimize n, we can introduce a third prime factor, 2. Also to minimize n, we want 5, the greatest of all the factors, to be raised to the least power. Therefore, n = 2^43^45^2 and frac{n}{75} = frac{2^43^45^2}{3 cdot 5^2} = 16 cdot 27 = boxed{432}.

🧠:[<think>Alright, let's tackle this problem step by step. The function f(x) is defined piecewise, with two parts: one for the interval [a, 0) and another for [0, 4]. The given range of the entire function is [-8, 1], and we need to find the range of the real number a. Hmm, okay. Let's break this down.First, let me understand the function better. For the interval from a to 0 (not including 0), the function is - (1/2)^x. Then, from 0 to 4 (including both ends), it's a quadratic function: -x² + 2x. The range of the entire function is given as [-8, 1], which means that the lowest value the function reaches is -8, and the highest is 1. We need to figure out what values of a would make the combined range of both pieces [-8, 1].Let me start by analyzing each piece of the function separately. Maybe if I can find the range of each part, I can see how they contribute to the overall range and then determine what a needs to be for the total range to be [-8, 1].Starting with the quadratic part: f(x) = -x² + 2x on [0, 4]. Let's find its range. Since this is a quadratic function, it's a parabola. The coefficient of x² is -1, so it opens downward, meaning the vertex is the maximum point. The vertex of a quadratic ax² + bx + c is at x = -b/(2a). Here, a = -1 and b = 2, so the x-coordinate of the vertex is -2/(2*(-1)) = -2/(-2) = 1. Plugging x = 1 into the function: f(1) = -(1)^2 + 2*1 = -1 + 2 = 1. So the maximum value on this interval is 1, which matches the given upper bound of the range. Good.Now, let's check the endpoints of the quadratic's interval. At x = 0: f(0) = -0 + 0 = 0. At x = 4: f(4) = -(16) + 8 = -8. So on the interval [0, 4], the quadratic function ranges from -8 to 1. That's exactly the given range of the entire function. That suggests that the quadratic part alone already covers the entire range [-8, 1]. But then why is there another piece of the function? Maybe the exponential part also contributes to the range, but in this case, since the quadratic already reaches -8 and 1, perhaps the exponential part's range is within [-8, 1], but we need to ensure that it doesn't extend beyond that. Or maybe, depending on the value of a, the exponential part could affect the lower or upper bounds. Wait, let me check.The other part of the function is - (1/2)^x defined on [a, 0). Let's analyze its range. Let me first consider the expression (1/2)^x. Since the base is between 0 and 1, (1/2)^x is an exponential decay function when x increases, but since here x is negative (since the interval is [a, 0), and a is a real number presumably less than 0), then x can be negative. Let me recall that (1/2)^x is the same as 2^(-x). So when x is negative, say x = -b where b > 0, then (1/2)^(-b) = 2^(b), which is an exponential growth. Therefore, as x approaches 0 from the left, (1/2)^x approaches 1, and as x becomes more negative (i.e., moving left from 0), (1/2)^x grows exponentially. However, the function is - (1/2)^x, so it will be negative, approaching -1 as x approaches 0 from the left, and approaching negative infinity as x approaches negative infinity. But since the domain of this piece is [a, 0), the value of a will affect the lower bound of this piece's range.Wait a second, so the exponential part, - (1/2)^x, when x is in [a, 0). Let's consider x approaching 0 from the left: x approaches 0, so (1/2)^x approaches 1, so the function approaches -1. At x = a, the left endpoint, the value is - (1/2)^a. Since (1/2)^a is equal to 2^(-a). If a is negative, then -a is positive, so 2^(-a) is 2^(positive), which is greater than 1. Therefore, - (1/2)^a is -2^(-a), which is less than -1. Therefore, the range of the exponential part is (- (1/2)^a, -1], since as x increases from a to 0, (1/2)^x decreases from (1/2)^a to 1, so - (1/2)^x increases from - (1/2)^a to -1.Wait, hold on. Let me clarify. Let's say a is some number less than 0. Then, as x increases from a to 0, the exponent x increases. Since the base is 1/2, which is less than 1, as x increases, (1/2)^x decreases. Therefore, - (1/2)^x increases. So the function - (1/2)^x is increasing on the interval [a, 0). Therefore, its minimum value is at x = a: - (1/2)^a, and its maximum value is approaching -1 as x approaches 0 from the left.So the range of the exponential part is [- (1/2)^a, -1), but since x approaches 0 but doesn't reach it, the upper limit is approaching -1. But since the quadratic part includes x=0, f(0) = 0. Wait, but in the quadratic part, at x=0, f(0) is 0, which is higher than -1. Therefore, the upper bound of the exponential part is -1, but the quadratic part at x=0 is 0. So when combined, the upper bound of the entire function is still 1, as given.But the lower bound of the entire function is -8, which comes from the quadratic part at x=4. However, the exponential part could potentially have lower values than -8 if a is too small. For example, if a is very negative, then - (1/2)^a = -2^(-a). If a is very negative, say a approaches negative infinity, then -a approaches positive infinity, so 2^(-a) approaches infinity, making -2^(-a) approach negative infinity. But the overall function's range is given as [-8, 1], which includes -8 as the lowest value. Therefore, we need to ensure that the exponential part doesn't go below -8. In other words, the minimum of the exponential part must be greater than or equal to -8. Otherwise, the overall function's range would be less than -8, which contradicts the given range.Therefore, we need to set the lower bound of the exponential part, which is - (1/2)^a, to be greater than or equal to -8. Wait, but since the exponential part is decreasing (since the function is increasing, but the negative of a decreasing function is increasing), the minimum value of the exponential part on [a, 0) is at x = a, which is - (1/2)^a. Therefore, to prevent the exponential part from going below -8, we need:- (1/2)^a ≥ -8Multiplying both sides by -1 (which reverses the inequality):(1/2)^a ≤ 8Since (1/2)^a = 2^(-a), so:2^(-a) ≤ 8But 8 is 2^3, so:2^(-a) ≤ 2^3Since the base 2 is greater than 1, the inequality implies:- a ≤ 3Multiply both sides by -1 (reversing the inequality again):a ≥ -3Therefore, a must be greater than or equal to -3. However, we also need to remember that a must be less than 0, since the exponential part is defined on [a, 0). So a is in [-3, 0).Wait, but let me check this again. If a is -3, then the lower bound of the exponential part is - (1/2)^(-3) = - (2^3) = -8. If a is greater than -3, say a = -2, then the lower bound is - (1/2)^(-2) = - (2^2) = -4, which is higher than -8. Therefore, as a increases from -3 to 0, the lower bound of the exponential part increases from -8 to -1. Since the quadratic part already reaches -8 at x=4, the minimum of the entire function is -8. But if a is less than -3, then the exponential part would have a lower bound less than -8, which would contradict the given range. Therefore, a must be at least -3. So the range of a is [-3, 0).But let me confirm this with an example. Let's take a = -4. Then, the exponential part on [-4, 0) would have a lower bound at x = -4: - (1/2)^(-4) = - (16) = -16. That's lower than -8, which would mean the function's range would be [-16, 1], which is not the given [-8, 1]. Therefore, a cannot be less than -3. If a = -3, the lower bound is -8, which matches the quadratic's minimum. Therefore, a must be greater than or equal to -3. Since a is the left endpoint of the domain, which is less than 0, the range of a is [-3, 0).But let's also check the upper end. If a approaches 0 from the left, say a = -0.1, then the exponential part is defined on [-0.1, 0). The lower bound would be - (1/2)^(-0.1) = -2^{0.1} ≈ -1.071, which is greater than -8. Therefore, the exponential part's range is approximately [-1.071, -1), and the quadratic part covers from -8 to 1. So the overall range is still [-8, 1]. Therefore, as long as a is ≥ -3, the exponential part doesn't introduce any values below -8, and the quadratic part takes care of the rest. Hence, a can be as close to 0 as desired, but must be at least -3.Therefore, the range of a is [-3, 0). But let me check once more.Wait, the problem states that the range of the function is [-8,1]. So we need to ensure that all values from -8 to 1 are covered. The quadratic part alone gives us from -8 to 1. The exponential part gives us from - (1/2)^a down to approaching -1. But if a is greater than -3, then the exponential part's minimum is greater than -8, which is already covered by the quadratic part. So as long as the exponential part doesn't go below -8, the overall range is still [-8,1]. Therefore, a must be ≥ -3.Additionally, since the exponential part is defined for a ≤ x < 0, a must be ≤ 0. But since a is a real number, the upper limit for a is 0. But since the interval is a ≤ x < 0, if a were 0, the interval would collapse to a single point, which is x=0, but x=0 is already included in the quadratic part. So likely a must be less than 0, but the problem statement says a ≤ x < 0, so a can be equal to 0? Wait, no. Because if a is 0, then the first case would be for 0 ≤ x < 0, which is an empty set. So actually, a must be less than 0. The original function is defined as:f(x) = - (1/2)^x , a ≤ x < 0,andf(x) = -x² + 2x, 0 ≤ x ≤ 4.So if a were 0, then the first case would be for 0 ≤ x < 0, which is empty, so the function would just be the quadratic part. But the problem states that the range is [-8,1], which is exactly the quadratic part's range. However, the problem includes both cases, so likely a is intended to be less than 0, so that the exponential part is non-empty. Therefore, a must be in [-3, 0).But the problem says "the range of the real number a". So they want all real numbers a such that the function's range is [-8,1]. Therefore, a can be any real number from -3 up to, but not including, 0. Hence, the range of a is [-3, 0).Let me double-check. If a is exactly -3, then the exponential part's minimum is -8, which matches the quadratic's minimum. So the overall function's range is [-8,1], as required. If a is greater than -3, the exponential part's minimum is higher than -8, but the quadratic part still reaches -8, so the overall range is still [-8,1]. If a is less than -3, the exponential part would have values less than -8, which would extend the range below -8, which is not allowed. Therefore, a must be ≥ -3 and < 0. So the range of a is [-3, 0).Therefore, the answer is a ∈ [-3, 0), so the range of a is from -3 to 0, not including 0. Therefore, in boxed notation, it's boxed{[-3, 0)}.Final AnswerThe range of the real number ( a ) is boxed{[-3, 0)}.</think>Given the piecewise function ( f(x) ):[f(x) = begin{cases} -left( frac{1}{2} right)^x & text{if } a leq x < 0, -x^2 + 2x & text{if } 0 leq x leq 4 end{cases}]we need to determine the range of the real number ( a ) such that the range of the function is ([-8, 1]).1. Quadratic Part Analysis: - The quadratic function ( -x^2 + 2x ) on the interval ([0, 4]) has a maximum value at ( x = 1 ) which is ( f(1) = 1 ). - The minimum value occurs at ( x = 4 ), which is ( f(4) = -8 ). - Therefore, the range of the quadratic part is ([-8, 1]).2. Exponential Part Analysis: - The exponential function ( -left( frac{1}{2} right)^x ) on the interval ([a, 0)) can be rewritten as ( -2^{-x} ). - As ( x ) approaches 0 from the left, ( -2^{-x} ) approaches (-1). - The minimum value of the exponential part occurs at ( x = a ), which is ( -2^{-a} ).3. Range Consideration: - The given range of the entire function is ([-8, 1]). The quadratic part already covers this range. - To ensure the exponential part does not extend the range below (-8), the minimum value of the exponential part must be at least (-8). - This gives the inequality: ( -2^{-a} geq -8 ). - Solving this inequality: ( 2^{-a} leq 8 ) which simplifies to ( -a leq 3 ) or ( a geq -3 ).4. Domain Consideration: - The interval for the exponential part is ([a, 0)), so ( a ) must be less than 0.Combining these results, the range of ( a ) is ([-3, 0)).[boxed{[-3, 0)}],<think>Okay, so I need to figure out the range of the real number 'a' for this piecewise function f(x). The function is defined in two parts: one from 'a' to 0, and the other from 0 to 4. The overall range of the function is given as [-8, 1], and I need to find what values 'a' can take for this to be true. Let me start by understanding each part of the function separately.First, let's look at the first part of the function: f(x) = - (1/2)^x for a ≤ x < 0. Let me analyze its behavior. The base of the exponent is 1/2, which is between 0 and 1, so as x increases towards 0, (1/2)^x decreases. Wait, actually, since the base is less than 1, (1/2)^x would increase as x increases because the exponent is becoming less negative. For example, if x is a large negative number, like -3, then (1/2)^(-3) is 8, and as x approaches 0, (1/2)^x approaches 1. So, (1/2)^x is a decreasing function when the exponent is positive, but since here x is negative (since a ≤ x < 0), it's actually increasing as x approaches 0. But the function f(x) is the negative of that, so f(x) = - (1/2)^x would be decreasing as x approaches 0.Wait, let me confirm that. Let's take two points in the interval a ≤ x < 0. Suppose x1 and x2 are in this interval, and x1 < x2 (since x is approaching 0 from the left). Then, (1/2)^x1 and (1/2)^x2. Since x1 < x2 (closer to 0), (1/2)^x1 = (1/2)^{x1} which, since x1 is more negative, is larger. For example, x1 = -2, x2 = -1: (1/2)^(-2) = 4, and (1/2)^(-1) = 2. So, as x increases from a to 0, (1/2)^x decreases. Therefore, (1/2)^x is decreasing over the interval a ≤ x < 0. Then, f(x) = - (1/2)^x is increasing over that interval because we are multiplying by -1, which reverses the order. So, when x is more negative (closer to a), (1/2)^x is larger, so - (1/2)^x is more negative, and as x approaches 0, (1/2)^x approaches 1, so f(x) approaches -1. Therefore, the first part of the function, f(x) = - (1/2)^x, on the interval [a, 0), is increasing from - (1/2)^a up to -1.Wait, hold on. If x is approaching 0 from the left, then (1/2)^x approaches 1, so f(x) approaches -1. When x is at 'a', the left endpoint, the value is - (1/2)^a. Since a ≤ x < 0, 'a' is some number less than or equal to 0. But depending on the value of 'a', the function's minimum and maximum on that interval could vary.Now, the second part of the function is f(x) = -x^2 + 2x for 0 ≤ x ≤ 4. Let me analyze that. Let's rewrite this as f(x) = -x^2 + 2x. That's a quadratic function opening downward because the coefficient of x^2 is negative. The vertex of this parabola will be at x = -b/(2a) where the quadratic is in the form ax^2 + bx + c. Here, a = -1, b = 2, so the vertex is at x = -2/(2*(-1)) = -2 / (-2) = 1. So the vertex is at x = 1. Plugging that into the function, f(1) = -(1)^2 + 2*(1) = -1 + 2 = 1. So the maximum value of this part is 1 at x = 1. Then, the function decreases on either side of x = 1. At x = 0, f(0) = -0 + 0 = 0. At x = 4, f(4) = -(16) + 8 = -8. So on the interval [0,4], the function ranges from 0 down to -8, peaking at 1 in the middle. Therefore, the range for the second part is [-8, 1], which matches the given overall range. So the entire function's range is the combination of the ranges from both parts.But wait, the problem states that the overall range of the function is [-8, 1]. The second part already achieves -8 at x=4 and 1 at x=1, so the first part's range must not exceed these bounds. However, the first part might have values that could potentially be lower than -8 or higher than 1, which would affect the overall range. But since the given total range is [-8,1], that means the first part must not go below -8 or above 1. Let me check the first part.The first part is f(x) = - (1/2)^x on [a, 0). Since x is in [a, 0), so x is negative or zero. Let's let t = -x, so t is positive when x is negative. Then (1/2)^x = (1/2)^{-t} = 2^t. Therefore, f(x) = -2^t, where t = -x, and since x is in [a, 0), t is in (0, -a]. Therefore, as x ranges from a (which is ≤ 0) up to 0, t ranges from -a down to 0. Wait, no. If x is approaching 0 from the left, then t = -x is approaching 0 from the right. If x is a, then t = -a. So as x increases from a to 0, t decreases from -a to 0. Therefore, 2^t decreases from 2^{-a} to 2^0 = 1. So, f(x) = -2^t, which would go from -2^{-a} to -1. Therefore, the first part of the function, f(x) on [a, 0), has a range from -2^{-a} to -1. Since 2^{-a} is equal to (1/2)^{a}, but since a is ≤ 0, -a is ≥ 0. So 2^{-a} is ≥ 1. Therefore, -2^{-a} is ≤ -1. Wait, that seems conflicting with my earlier analysis. Wait, hold on.Wait, x is in [a, 0). So let's take a specific example: suppose a = -2. Then, x ranges from -2 to 0. So t = -x ranges from 2 to 0, so 2^t goes from 4 to 1. Therefore, f(x) = -2^t goes from -4 to -1. Therefore, the range is [-4, -1]. Similarly, if a = -3, x ranges from -3 to 0, t ranges from 3 to 0, 2^t from 8 to 1, so f(x) ranges from -8 to -1. If a is closer to 0, say a = -1, then t ranges from 1 to 0, so 2^t ranges from 2 to 1, so f(x) ranges from -2 to -1. If a is 0, then the interval [a,0) is just the point 0, but since it's up to 0, not including 0, maybe it's undefined? Wait, but in the problem statement, the first case is a ≤ x < 0, so if a = 0, then the interval is 0 ≤ x <0, which is empty. So probably a has to be less than 0. But let's check.So the first part is defined for a ≤ x <0. So for a to have a valid interval, a must be less than 0. If a were equal to 0, then the interval would be 0 ≤ x <0, which is empty. So the first part is only non-empty when a <0. So we can assume a <0. Therefore, -a is positive, and t ranges from -a down to 0. Therefore, 2^t decreases from 2^{-a} (which is 2^{positive number}, so greater than 1) down to 1. Therefore, f(x) = -2^t ranges from -2^{-a} (which is a more negative number) up to -1. Therefore, the range of the first part is [-2^{-a}, -1].Now, the second part of the function has range [-8,1]. The first part's range is [-2^{-a}, -1]. The overall range of the function is the union of these two ranges. Since the problem states that the overall range is [-8,1], which is exactly the range of the second part, that means the first part's range must not extend beyond [-8,1]. However, the first part's lower bound is -2^{-a}, which must be greater than or equal to -8 (since the overall range's lower bound is -8). Similarly, the upper bound of the first part is -1, which is within the overall upper bound of 1, so that's fine. Therefore, we need to ensure that -2^{-a} ≥ -8. Let's solve that inequality.So, -2^{-a} ≥ -8. Multiply both sides by -1 (remembering to reverse the inequality sign):2^{-a} ≤ 8Since 2^{-a} is equal to (2)^{-a} = 2^{-a}, and 8 is 2^3. Therefore:2^{-a} ≤ 2^3Since the base 2 is greater than 1, the inequality holds when the exponent on the left is ≤ the exponent on the right:- a ≤ 3Multiply both sides by -1 (again, reversing the inequality):a ≥ -3So, a must be greater than or equal to -3. But we also know that a must be less than 0 for the first interval to be non-empty. Therefore, the range of a is -3 ≤ a < 0.Wait, but let me confirm this. Let me check with a = -3. Then, the first part's range is [-2^{-(-3)}, -1] = [-2^{3}, -1] = [-8, -1]. Then, the overall range would be the union of [-8, -1] and [-8,1], which is [-8,1], which matches the given range. If a is greater than -3, say a = -2, then the first part's range is [-2^{-(-2)}, -1] = [-2^{2}, -1] = [-4, -1]. Then, the overall range is the union of [-4, -1] and [-8,1], which is [-8,1], still matching. If a is between -3 and 0, say a = -1, then the first part's range is [-2^{1}, -1] = [-2, -1], and the union is [-8,1], which is okay. However, if a is less than -3, say a = -4, then the first part's range would be [-2^{4}, -1] = [-16, -1], which would make the overall range [-16,1], but the given range is [-8,1], so that's not allowed. Therefore, a must be ≥ -3.Therefore, combining this with the necessity that a < 0, the range of a is [-3, 0). However, let's check if when a = -3, the function's range is exactly [-8,1]. As we saw earlier, when a = -3, the first part's range is [-8, -1], and the second part's range is [-8,1], so the union is indeed [-8,1]. Similarly, for any a between -3 and 0, the first part's lower bound is between -8 and -1, which is within the overall range. Therefore, the range of a is [-3, 0).But wait, let me check again. Suppose a is greater than -3, say a = -2. Then the first part's range is [-4, -1], which is within the overall range. The second part's maximum is 1, and minimum is -8, so the union is still [-8,1]. Therefore, any a ≥ -3 and <0 is acceptable. Therefore, the range of a is [-3,0). So the answer is a ∈ [-3,0). So in boxed notation, that would be boxed{[-3, 0)}.But let me just confirm once more. Suppose a is exactly -3. Then the first part of the function when x = a = -3 is f(-3) = - (1/2)^{-3} = - (2^3) = -8. Then, as x approaches 0 from the left, f(x) approaches -1. So the first part's range is [-8, -1]. The second part's range is [-8,1], so combining both, the total range is [-8,1], which is exactly what's given. If a is less than -3, say a = -4, then the first part's range would be from - (1/2)^{-4} = -16 up to -1, which introduces -16 into the range, but the problem states the range is [-8,1], so that's invalid. Therefore, a cannot be less than -3. Similarly, if a is between -3 and 0, the first part's lower bound is between -8 and -2, which is within the overall range. Therefore, the allowable values of a are from -3 up to but not including 0. Hence, the range of a is [-3,0).Yes, that makes sense. So the answer should be boxed{[-3, 0)}.Final AnswerThe range of the real number ( a ) is boxed{[-3, 0)}.</think>Given the piecewise function ( f(x) = begin{cases} -left( frac{1}{2} right)^x & text{if } a leq x < 0, -x^2 + 2x & text{if } 0 leq x leq 4 end{cases} ) with the range ([-8, 1]), we need to determine the range of the real number ( a ).1. Analyzing the first part of the function ( f(x) = -left( frac{1}{2} right)^x ) for ( a leq x < 0 ): - Let ( t = -x ). Then ( t ) ranges from ( -a ) to 0 as ( x ) ranges from ( a ) to 0. - The function can be rewritten as ( f(x) = -2^t ), which ranges from ( -2^{-a} ) to ( -1 ). - The range of this part is ( [-2^{-a}, -1] ).2. Analyzing the second part of the function ( f(x) = -x^2 + 2x ) for ( 0 leq x leq 4 ): - This is a quadratic function with a maximum value of 1 at ( x = 1 ) and a minimum value of -8 at ( x = 4 ). - The range of this part is ( [-8, 1] ).3. Combining the ranges: - The overall range of the function is the union of the ranges of both parts. - The given range is ([-8, 1]), so the first part's range must not extend below -8. - For the first part's lower bound ( -2^{-a} ) to be within ([-8, 1]), we need ( -2^{-a} geq -8 ). - Solving ( -2^{-a} geq -8 ) gives ( 2^{-a} leq 8 ), which simplifies to ( -a leq 3 ) or ( a geq -3 ).4. Conclusion: - Since ( a ) must be less than 0 (to have a non-empty interval for the first part), the range of ( a ) is ([-3, 0)).Thus, the range of the real number ( a ) is (boxed{[-3, 0)}).]

🔑:This problem involves finding the range of a parameter using the range of a function, and it covers piecewise functions, exponential functions, and quadratic functions. It is a basic-level problem. From the properties of a quadratic function, when 0leqslant xleqslant 4, the range of the function is exactly [-8,1]. We only need y={left( frac{1}{2}right)}^{x}, aleqslant x < 0, to have a range that is a subset of [-8,1], from which we can find the inequality for a by utilizing the monotonicity of the exponential function.Solution Enhancement:1. Consider the quadratic part of the function, f(x)=-x^2+2x=-(x-1)^2+1 when 0leq x leq 4.2. The graph of this function is a downward-facing parabola with a symmetry axis at x=1.3. The function is increasing on [0,1] and decreasing on [1,4].4. The maximum value of the function occurs at x=1, which is 1, and the minimum value at x=4, which is -8.5. Therefore, the range of f(x) is [-8,1].6. Now consider the exponential part of the function, y=-{left( frac{1}{2}right)}^{x} when aleq x < 0.7. This function is strictly increasing, so its range should be a subset of [-8,1], i.e., y in left[-{left( frac{1}{2}right)}^{a}, -1 right).8. To satisfy this condition, we need -{left( frac{1}{2}right)}^{a} geq -8.9. Solving this inequality, we get -3 leq a < 0.Thus, the range of a is boxed{[-3,0)}.

🧠:[<think>Okay, let's tackle this problem step by step. First, let's make sure I understand the question correctly. We have the function f(x) = -x² - 2x. Then we're given three values: a = ln 2, b = log base 1/3 of 2, and c = 3^(1/2), which is the square root of 3. The question is asking us to compare f(a), f(b), and f(c) and determine which order they fall into, with the options given as A, B, C, D.So the key here is to figure out the numerical values of a, b, and c, plug them into the function f(x), compute f(a), f(b), and f(c), and then compare their magnitudes. Then, based on that, choose the correct option among A to D.Let me start by recalling the function f(x) = -x² - 2x. That's a quadratic function, and since the coefficient of x² is negative (-1), the parabola opens downward. Its vertex is the maximum point. But maybe we don't need the vertex here; perhaps just evaluating the function at the given points will suffice. Let's see.First, let's compute the values of a, b, and c.Starting with a = ln 2. The natural logarithm of 2 is approximately 0.693. I remember that ln 2 is about 0.693 because it's a common value. Let me verify that mentally: yes, e^0.7 is approximately 2.013, so ln 2 is just under 0.7, so 0.693 is correct.Next, b = log_{1/3} 2. Hmm, logarithm with base 1/3 of 2. To compute this, I can use the change of base formula: log_b a = (ln a)/(ln b). So log_{1/3} 2 = (ln 2)/(ln (1/3)). Let's compute that.First, ln (1/3) is ln 1 - ln 3 = 0 - ln 3 = -ln 3 ≈ -1.0986. So log_{1/3} 2 = (ln 2)/(-ln 3) ≈ (0.693)/(-1.0986) ≈ -0.6309. So b is approximately -0.6309.Then, c = 3^(1/2), which is the square root of 3. That's approximately 1.732. I know that sqrt(3) is about 1.732 because 1.732 squared is approximately 3.0 (since 1.732 * 1.732 ≈ 3.0). So that's correct.So, summarizing:a ≈ 0.693b ≈ -0.631c ≈ 1.732Now, we need to compute f(a), f(b), and f(c). Let's compute each one step by step.Starting with f(a):f(a) = -a² - 2aSubstituting a ≈ 0.693:First compute a²: (0.693)^2 ≈ 0.693 * 0.693. Let's compute that:0.693 * 0.693:First, 0.7 * 0.7 = 0.49But since it's 0.693, which is 0.7 - 0.007, so perhaps (0.7 - 0.007)^2 = 0.49 - 2*0.7*0.007 + 0.007² ≈ 0.49 - 0.0098 + 0.000049 ≈ 0.49 - 0.0098 = 0.4802 + 0.000049 ≈ 0.480249. So approximately 0.4802.Alternatively, 0.693 * 0.693: Let's compute 0.693 * 0.6 = 0.4158; 0.693 * 0.09 = 0.06237; 0.693 * 0.003 = 0.002079. Adding those up: 0.4158 + 0.06237 = 0.47817 + 0.002079 ≈ 0.480249. So yeah, approximately 0.4802.So a² ≈ 0.4802. Then -a² ≈ -0.4802.Then 2a = 2 * 0.693 = 1.386. So -2a = -1.386.Therefore, f(a) = -a² -2a ≈ -0.4802 -1.386 ≈ -1.8662.So f(a) ≈ -1.8662.Next, f(b):b ≈ -0.631f(b) = -b² - 2bFirst compute b²: (-0.631)^2 = 0.631². Let's compute that:0.6 * 0.6 = 0.360.6 * 0.031 = 0.01860.031 * 0.6 = 0.01860.031 * 0.031 ≈ 0.000961So (0.6 + 0.031)^2 = 0.6² + 2*0.6*0.031 + 0.031² ≈ 0.36 + 0.0372 + 0.000961 ≈ 0.36 + 0.0372 = 0.3972 + 0.000961 ≈ 0.398161.So b² ≈ 0.398161. Then -b² ≈ -0.398161.Then -2b = -2*(-0.631) = 1.262.So f(b) = -b² -2b ≈ -0.398161 +1.262 ≈ 0.863839.So f(b) ≈ 0.8638.Then f(c):c ≈ 1.732f(c) = -c² -2cFirst compute c²: (1.732)^2. Since sqrt(3) is approximately 1.732, (sqrt(3))² = 3. So c² = 3 exactly. Therefore, -c² = -3.Then 2c = 2*1.732 ≈ 3.464. So -2c ≈ -3.464.Therefore, f(c) = -3 -3.464 ≈ -6.464.So f(c) ≈ -6.464.Now, let's list the computed values:f(a) ≈ -1.8662f(b) ≈ 0.8638f(c) ≈ -6.464So now we need to compare f(a), f(b), and f(c). The problem is asking which of the options A-D is correct, where each option orders these three from largest to smallest.Looking at the computed values:f(b) ≈ 0.8638 is the largest.Then f(a) ≈ -1.8662 is next.Then f(c) ≈ -6.464 is the smallest.So ordering from largest to smallest: f(b) > f(a) > f(c). That corresponds to option A: f(b) > f(a) > f(c).But wait, let me verify these calculations again to be sure. Sometimes when dealing with negatives and squares, it's easy to make a mistake.First, f(b):b ≈ -0.631f(b) = -(-0.631)^2 -2*(-0.631) = -0.631² +1.262Yes, 0.631 squared is approximately 0.398, so -0.398 +1.262 ≈ 0.864. That's correct.f(a) = -(0.693)^2 -2*(0.693) ≈ -0.480 -1.386 ≈ -1.866. Correct.f(c) = -(sqrt(3))^2 -2*sqrt(3) = -3 - 3.464 ≈ -6.464. Correct.So yes, the order is f(b) > f(a) > f(c), which is option A.Wait, but the answer options are:A: f(b) > f(a) > f(c)B: f(c) > f(a) > f(b)C: f(a) > f(b) > f(c)D: f(b) > f(c) > f(a)So based on our calculations, A is the correct answer. However, let me check again because sometimes with negative values, especially when dealing with quadratic functions, the order might not be as straightforward.Wait, but f(b) is positive, f(a) is negative, and f(c) is more negative. So the order is indeed f(b) (positive) first, then f(a) (less negative), then f(c) (most negative). So yes, A is correct.But let me double-check the computation for a, b, c values.First, a = ln 2 ≈ 0.693. Correct.b = log_{1/3} 2. Let me verify that calculation again. log base 1/3 of 2 is equal to (ln 2)/(ln (1/3)) = (ln 2)/(-ln 3) ≈ 0.6931 / (-1.0986) ≈ -0.6309. Yes, that's correct. So b ≈ -0.6309.c = sqrt(3) ≈ 1.732. Correct.So plugging into f(x) = -x² -2x:For b ≈ -0.6309:f(b) = -(-0.6309)^2 -2*(-0.6309) = -0.398 +1.2618 ≈ 0.8638. Correct.For a ≈ 0.693:f(a) = -0.693² -2*0.693 ≈ -0.480 -1.386 ≈ -1.866. Correct.For c ≈ 1.732:f(c) = -3 -3.464 ≈ -6.464. Correct.Therefore, the order is f(b) ≈ 0.8638 > f(a) ≈ -1.866 > f(c) ≈ -6.464. So the correct answer is A.Wait, but let me check the options again. The options are:A: f(b) > f(a) > f(c)B: f(c) > f(a) > f(b)C: f(a) > f(b) > f(c)D: f(b) > f(c) > f(a)So A is exactly what we have. Therefore, the answer must be A.But wait, hold on. Let me check if the function f(x) is indeed -x² -2x. The function is given as f(x)=-x²-2x. So yes, correct. So when we plug in a positive number, say c = sqrt(3) ≈1.732, then f(c) = - (1.732)^2 -2*(1.732) = -3 -3.464 ≈ -6.464. Correct. For a positive number x, as x increases, f(x) becomes more negative because of the -x² term. Similarly, for a negative number, like b ≈ -0.631, plugging into f(x) gives us - (negative number squared) which is negative, but then subtract 2 times a negative number, which becomes adding a positive. So in this case, since b is negative, -2b is positive. Depending on the magnitude of b, this can make f(b) positive or negative.In our case, b ≈ -0.631. So the term -2b is ≈1.262, and -b² is ≈-0.398. The sum is positive, so f(b) is positive, which is indeed higher than f(a) and f(c).But just to be thorough, let me consider the function's behavior. Since f(x) is a downward-opening parabola, its maximum is at the vertex. The vertex occurs at x = -b/(2a) for a quadratic ax² + bx + c. In our case, f(x) = -x² -2x, so a = -1, b = -2. Wait, actually, in standard form ax² + bx + c, so here, the coefficient of x² is -1, and the coefficient of x is -2. Therefore, the vertex is at x = -b/(2a) = -(-2)/(2*(-1)) = 2/(-2) = -1. So the vertex is at x = -1.Therefore, the function f(x) has its maximum at x = -1. So as we move left or right from x = -1, the function decreases. So for x < -1, the function is decreasing, and for x > -1, the function is also decreasing (since it's a downward parabola). Wait, actually, since it's a maximum at x = -1, the function is increasing for x < -1 and decreasing for x > -1. Wait, no. Wait, for a downward opening parabola, the function increases to the vertex and then decreases after the vertex. So left of the vertex (x < -1), the function is increasing, and right of the vertex (x > -1), the function is decreasing.Given that, let's see where our points a, b, c are in relation to the vertex at x = -1.a = ln 2 ≈ 0.693. So this is to the right of the vertex (x = -1). Since the function is decreasing for x > -1, as x increases beyond -1, f(x) decreases. So the larger x is to the right of -1, the smaller f(x) becomes. So since a is ≈0.693, which is greater than -1, and c ≈1.732 is further to the right, so f(c) should be less than f(a), which aligns with our previous calculation where f(a) ≈-1.866 and f(c)≈-6.464.Now, for b ≈ -0.631. Since this is greater than -1 (because -0.631 is to the right of -1 on the number line), so also in the decreasing part of the parabola (for x > -1). Wait, but wait, x = -0.631 is greater than -1, so it's still in the decreasing part. Therefore, as x increases from -1 onwards, f(x) decreases. So the order between a and b: since b is at x ≈ -0.631, and a is at x ≈0.693, which is further to the right. Therefore, since the function is decreasing for x > -1, the further right you go, the smaller the function value. So f(b) should be greater than f(a), because x = -0.631 is closer to the vertex at x = -1 than x =0.693. Therefore, since the function is decreasing past the vertex, closer x is to the vertex, the higher the value. Therefore, f(b) > f(a) > f(c). Which is exactly what we found numerically. So this confirms our earlier calculation.Additionally, since b is between -1 and 0, and a and c are positive, with c further out, so the order f(b) > f(a) > f(c) makes sense.Therefore, the correct answer is A.Final Answerboxed{A}</think>Given the function ( f(x) = -x^2 - 2x ), we need to compare the values ( f(a) ), ( f(b) ), and ( f(c) ) where ( a = ln 2 ), ( b = log_{frac{1}{3}} 2 ), and ( c = 3^{frac{1}{2}} ).First, we compute the values of ( a ), ( b ), and ( c ):- ( a = ln 2 approx 0.693 )- ( b = log_{frac{1}{3}} 2 = frac{ln 2}{ln frac{1}{3}} = frac{ln 2}{-ln 3} approx -0.631 )- ( c = sqrt{3} approx 1.732 )Next, we evaluate the function ( f(x) ) at these points:- For ( a approx 0.693 ): [ f(a) = -(0.693)^2 - 2(0.693) approx -0.480 - 1.386 = -1.866 ]- For ( b approx -0.631 ): [ f(b) = -(-0.631)^2 - 2(-0.631) approx -0.398 + 1.262 = 0.864 ]- For ( c approx 1.732 ): [ f(c) = -(1.732)^2 - 2(1.732) approx -3 - 3.464 = -6.464 ]Comparing the values ( f(a) approx -1.866 ), ( f(b) approx 0.864 ), and ( f(c) approx -6.464 ), we find the order to be ( f(b) > f(a) > f(c) ).Thus, the correct answer is (boxed{A}).,<think>Okay, let's try to solve this problem step by step. So, we have the function f(x) = -x² - 2x. Then we're given three values: a = ln 2, b = log base 1/3 of 2, and c = 3^(1/2), which is the square root of 3. We need to compare f(a), f(b), and f(c) and see which option (A, B, C, D) is correct.First, let's recall what the function f(x) looks like. It's a quadratic function, and since the coefficient of x² is negative (-1), it opens downward, meaning it has a maximum point. The vertex form of a quadratic function is f(x) = a(x - h)² + k, where (h, k) is the vertex. Let me try to find the vertex here.The standard form is f(x) = -x² - 2x. To find the vertex, we can use the formula for the x-coordinate of the vertex, which is -b/(2a). In this case, the coefficient of x² is -1 (a = -1), and the coefficient of x is -2 (b = -2). Wait, hold on, maybe I need to write it in the form ax² + bx + c. Let's see:f(x) = -x² - 2x. So, a = -1, b = -2. Then the x-coordinate of the vertex is -b/(2a) = -(-2)/(2*(-1)) = 2 / (-2) = -1. So the vertex is at x = -1. Let's find f(-1): f(-1) = -(-1)^2 - 2*(-1) = -1 + 2 = 1. So the maximum point is at (-1, 1). Since the parabola opens downward, the function is increasing to the left of x = -1 and decreasing to the right of x = -1.Therefore, for values of x less than -1, the function is increasing, and for values greater than -1, the function is decreasing. So, as x increases beyond -1, f(x) decreases. Similarly, as x decreases below -1, f(x) increases. Hmm, so depending on where a, b, and c are in relation to -1, their f(x) values will be ordered differently.So, first, we need to figure out the values of a, b, c and see where they are relative to -1.Given:a = ln 2. Let's approximate that. Since ln 1 = 0, ln e = 1, and e ≈ 2.718. So ln 2 is approximately 0.693. So a ≈ 0.693.b = log_{1/3} 2. To evaluate this, remember that log_b a = ln a / ln b. So log_{1/3} 2 = ln 2 / ln(1/3). ln(1/3) is ln 1 - ln 3 = 0 - ln 3 = -ln 3. Therefore, log_{1/3} 2 = ln 2 / (-ln 3) = - (ln 2)/(ln 3). Let's compute that numerically. ln 2 ≈ 0.693, ln 3 ≈ 1.098. So this is approximately -0.693 / 1.098 ≈ -0.631. So b ≈ -0.631.c = 3^(1/2) = sqrt(3) ≈ 1.732.So, summarizing:a ≈ 0.693b ≈ -0.631c ≈ 1.732Now, let's see where these are relative to the vertex at x = -1.a ≈ 0.693 is greater than -1.b ≈ -0.631 is also greater than -1 (since -0.631 is to the right of -1 on the number line).c ≈ 1.732 is also greater than -1.So all three values are to the right of the vertex at x = -1. Since the function is decreasing for x > -1, the larger the x value, the smaller the f(x). Therefore, the order of f(a), f(b), f(c) will be the reverse order of a, b, c.But wait, let's check if they are all on the same side of the vertex. Since all are greater than -1, yes. So since the function is decreasing there, the largest x will give the smallest f(x), and the smallest x (closest to -1) will give the largest f(x).So let's order a, b, c:First, let's list their approximate numerical values:b ≈ -0.631a ≈ 0.693c ≈ 1.732So from smallest to largest: b ≈ -0.631, a ≈ 0.693, c ≈ 1.732.But since the function is decreasing for x > -1, the order of f(x) will be f(b) > f(a) > f(c). Because the smallest x (b) gives the largest f(x), the next (a) gives a middle f(x), and the largest x (c) gives the smallest f(x). So that would be option A: f(b) > f(a) > f(c). But let's verify this by actually computing f(a), f(b), and f(c) to make sure.Wait, maybe the approximation is misleading. Let's compute f(a), f(b), f(c) more precisely.First, compute a = ln 2 ≈ 0.6931.f(a) = - (ln 2)^2 - 2*(ln 2). Let's compute this.(ln 2)^2 ≈ (0.6931)^2 ≈ 0.4804.2*(ln 2) ≈ 1.3862.So f(a) ≈ -0.4804 - 1.3862 ≈ -1.8666.Next, compute b = log_{1/3} 2 ≈ -0.6309.f(b) = - (b)^2 - 2*(b) = - (-0.6309)^2 - 2*(-0.6309).First, (-0.6309)^2 ≈ 0.398.So -0.398 + 1.2618 ≈ 0.8638.Wait, that seems positive. Wait, but f(x) is -x² -2x, so for x = b:f(b) = - (b)^2 - 2*b. Since b is negative (-0.6309), so:- ( (-0.6309)^2 ) - 2*(-0.6309) = -0.398 + 1.2618 ≈ 0.8638. Hmm, that's a positive value.But according to the vertex, the maximum value of f(x) is 1 at x = -1. So 0.8638 is less than 1, which is possible.But wait, let's check the calculation again. Let's take more decimal places for accuracy.Compute b = log_{1/3} 2. Let's compute it exactly first. Since log_{1/3} 2 = ln2 / ln(1/3) = ln2 / (-ln3) ≈ -0.6931 / 1.0986 ≈ -0.6309. So that's accurate.So f(b) = - (b)^2 - 2b = - ( (-0.6309)^2 ) - 2*(-0.6309) = -0.398 + 1.2618 ≈ 0.8638.Now, f(a) ≈ -1.8666 and f(c) we haven't computed yet. Let's compute c = sqrt(3) ≈ 1.732.f(c) = - (sqrt(3))^2 - 2*sqrt(3) = -3 - 2*1.732 ≈ -3 - 3.464 ≈ -6.464.So putting it all together:f(b) ≈ 0.8638f(a) ≈ -1.8666f(c) ≈ -6.464Therefore, ordering from largest to smallest: f(b) > f(a) > f(c). So option A is correct.But let me cross-verify. Since f(x) is a quadratic function opening downward, with vertex at x=-1, and the values of b, a, c are all to the right of -1 (since -0.63, 0.69, 1.73 are all greater than -1), then as x increases beyond -1, f(x) decreases. Therefore, the order of f(x) should be in reverse order of the x-values. Since b ≈ -0.63 is the smallest x among the three, then f(b) should be the largest. Then a ≈ 0.69 is next, so f(a) is next, and c ≈ 1.73 is the largest x, so f(c) is the smallest. Therefore, f(b) > f(a) > f(c), which is option A.But wait, wait, the answer in the options is A: f(b) > f(a) > f(c). But according to our calculations, f(b) is positive, f(a) is negative, and f(c) is more negative. So yes, 0.8638 > -1.8666 > -6.464 is correct. So A is the correct answer.Wait, but let's check once again, maybe the signs.Wait, when I computed f(b):x = b ≈ -0.6309f(b) = -x² -2x = - (x² + 2x). Let's compute x² + 2x:x² = (-0.6309)^2 ≈ 0.3982x ≈ 2*(-0.6309) ≈ -1.2618So x² + 2x ≈ 0.398 -1.2618 ≈ -0.8638Therefore, f(b) = - (x² + 2x) ≈ - (-0.8638) ≈ 0.8638. That's correct.Similarly, for f(a):x = a ≈ 0.6931f(a) = -x² -2x = - (x² + 2x)x² ≈ 0.48042x ≈ 1.3862x² + 2x ≈ 0.4804 + 1.3862 ≈ 1.8666Therefore, f(a) = -1.8666. Correct.For f(c):x = sqrt(3) ≈ 1.732x² = (sqrt(3))² = 32x ≈ 3.464x² + 2x ≈ 3 + 3.464 ≈ 6.464Therefore, f(c) = -6.464. Correct.So yes, f(b) ≈ 0.8638, f(a) ≈ -1.8666, f(c) ≈ -6.464. So ordering is f(b) > f(a) > f(c), which is option A. Therefore, the answer should be A.Wait, but let me check the options again:A: f(b) > f(a) > f(c)B: f(c) > f(a) > f(b)C: f(a) > f(b) > f(c)D: f(b) > f(c) > f(a)So according to our calculation, A is correct.But wait, the problem says "it must be true that ( )". So in the options, A is the correct one.Wait, but just to make sure, maybe there's a mistake in my calculations. Let's check once more.Alternatively, maybe the problem is expecting us to compute without numerical approximation? Let's see.Alternatively, maybe we can compare a, b, c without approximating their decimal values. Let's try that.Given:a = ln 2b = log_{1/3} 2 = - log_3 2 (change of base formula)c = sqrt(3) = 3^{1/2}We need to compare a, b, c in terms of their numerical values.First, compare a and b:a = ln 2 ≈ 0.693log_3 2 is log base 3 of 2, which is approximately 0.6309. Therefore, b = -0.6309.So a is positive, b is negative. Therefore, a > 0 > b. But wait, b is negative, but how does it compare to -1?Wait, since log_3 2 is less than 1 because 3^1 = 3 > 2. Therefore, log_3 2 ≈ 0.6309, so b = -0.6309, which is greater than -1.Therefore, the order in terms of magnitude:From smallest to largest:b ≈ -0.6309, a ≈ 0.693, c ≈ 1.732.But since the function f(x) is decreasing for x > -1, the larger x is, the smaller f(x). So the order of f(x) is reverse order of x. So f(b) > f(a) > f(c).Alternatively, since a, b, c are all greater than -1, and f is decreasing on that interval, then f(b) > f(a) > f(c) because b < a < c.Therefore, the answer is A.Alternatively, if we can compare them without approximation:First, note that a = ln 2. Since e ≈ 2.718, so ln 2 is less than 1, because ln e = 1. So a ≈ 0.693 < 1.c = sqrt(3) ≈ 1.732, so c is greater than 1.b is log_{1/3} 2. Since (1/3)^1 = 1/3 < 2, and (1/3)^0 = 1 < 2, but (1/3)^{-1} = 3 > 2. Therefore, log_{1/3} 2 is between -1 and 0. So b is between -1 and 0.Therefore, ordering: b (-1 < b < 0) < a (0 < a < 1) < c (c > 1). Therefore, the order from smallest to largest is b < a < c. Therefore, since f is decreasing for x > -1, f(b) > f(a) > f(c). So A is correct.Therefore, the answer is A.Final Answerboxed{A}</think>Given the function ( f(x) = -x^2 - 2x ), we need to compare ( f(a) ), ( f(b) ), and ( f(c) ) where ( a = ln 2 ), ( b = log_{frac{1}{3}} 2 ), and ( c = 3^{frac{1}{2}} ).First, we note that the function ( f(x) ) is a quadratic function opening downwards with its vertex at ( x = -1 ). For ( x > -1 ), the function is decreasing.Next, we approximate the values of ( a ), ( b ), and ( c ):- ( a = ln 2 approx 0.693 )- ( b = log_{frac{1}{3}} 2 = -log_3 2 approx -0.631 )- ( c = sqrt{3} approx 1.732 )Since all values ( a ), ( b ), and ( c ) are greater than -1, the function ( f(x) ) is decreasing in this interval. Therefore, the order of ( f(a) ), ( f(b) ), and ( f(c) ) will be the reverse of the order of ( a ), ( b ), and ( c ).The values ( a approx 0.693 ), ( b approx -0.631 ), and ( c approx 1.732 ) are ordered as ( b < a < c ). Since ( f(x) ) is decreasing, the corresponding function values will be ordered as ( f(b) > f(a) > f(c) ).To verify, we compute:- ( f(b) approx -(-0.631)^2 - 2(-0.631) approx 0.8638 )- ( f(a) approx -(0.693)^2 - 2(0.693) approx -1.8666 )- ( f(c) approx -(1.732)^2 - 2(1.732) approx -6.464 )Thus, the order is ( f(b) > f(a) > f(c) ), which corresponds to option A.[boxed{A}]]

🔑:The graph of the function f(x)=-x^{2}-2x is a parabola opening downwards, symmetric about the line x=-1. Therefore, the function f(x) is decreasing on [-1,+infty). Since a=ln 2 in (0,1), b=log_{frac{1}{3}}2 in (-1,0), c=3^{frac{1}{2}} in (1,2), it follows that f(b) > f(a) > f(c). Thus, the correct choice is: boxed{text{A}} Analyzing the graph and properties of the function f(x)=-x^{2}-2x allows us to determine the relationship between the values of the three expressions. This question tests knowledge of the graph and properties of quadratic functions, and the properties of logarithmic operations, with a medium level of difficulty.

🧠:[<think>Okay, so I need to find the set M of all values x such that both "p and q" and "not q" are false propositions. Let me break this down step by step. First, let's recall what each proposition means.Proposition p is "x² - x ≥ 6", and proposition q is "x is an integer", so x ∈ ℤ. The problem states that both "p and q" and "not q" must be false. So, we need to find all x where these two statements are false.Let me first consider what "p and q" being false means. The conjunction "p and q" is false when at least one of p or q is false. So, if either p is false, or q is false, or both are false, then "p and q" is false. But the problem says that "p and q" is false and also "not q" is false. So both these statements must be false.Wait, no. Wait, the problem says that both "p and q" and "not q" are false propositions. So, two conditions:1. "p and q" is false.2. "not q" is false.So, both of these have to be false. So first, let's handle each condition.First, "not q" is false. The proposition "not q" is equivalent to "x is not an integer". If "not q" is false, then q must be true. Because the negation of q is false, so q is true. Wait, but this is a contradiction. Wait, hold on: If "not q" is false, then q is true. Because "not q" is false means that it's not the case that x is not an integer, so x is an integer. So, q must be true.But wait, the problem says both "p and q" and "not q" are false. So, if "not q" is false, then q is true. But then, since "p and q" is false, and q is true, then p must be false. Because if q is true, then "p and q" is equivalent to p. So, "p and q" is false and q is true implies that p is false.Therefore, combining these two conditions:1. q is true (since "not q" is false).2. p is false (since "p and q" is false and q is true).Therefore, x must be an integer (q is true) and x² - x < 6 (p is false). So, the set M should be all integers x such that x² - x < 6.So, now my task is to solve the inequality x² - x < 6 for integer x.Let me solve the inequality step by step.First, x² - x < 6.Subtract 6 from both sides:x² - x - 6 < 0.Now, factor the quadratic expression:x² - x - 6 = (x - 3)(x + 2).So, (x - 3)(x + 2) < 0.To find where this product is negative, we can analyze the critical points. The roots are x = 3 and x = -2.The quadratic opens upwards (since the coefficient of x² is positive), so the graph is a parabola opening upwards, crossing the x-axis at -2 and 3.The product is negative between the roots. So, the solution to (x - 3)(x + 2) < 0 is x ∈ (-2, 3).Therefore, the inequality x² - x < 6 is true for x between -2 and 3. However, since we are looking for integer solutions (because q is true, so x ∈ ℤ), we need to list all integers between -2 and 3.But wait, x must be strictly between -2 and 3. So, the integers in the open interval (-2, 3) are -1, 0, 1, 2.Therefore, the integers x such that x² - x < 6 are x = -1, 0, 1, 2.Hence, the set M is {-1, 0, 1, 2}.But let me verify this. Let's check each integer in that range.For x = -1:x² - x = (-1)² - (-1) = 1 + 1 = 2 < 6: True.x = 0:0² - 0 = 0 - 0 = 0 < 6: True.x = 1:1² - 1 = 1 - 1 = 0 < 6: True.x = 2:2² - 2 = 4 - 2 = 2 < 6: True.What about x = -2?x² - x = (-2)² - (-2) = 4 + 2 = 6. So, 6 is not less than 6, so it's not included. Similarly, x = 3:3² - 3 = 9 - 3 = 6, again not less than 6, so excluded. So yes, the integers between -2 and 3, not including -2 and 3, are -1, 0, 1, 2. That seems right.So, according to the problem, the set M is these integers. Let me just confirm again the logic.We needed both "p and q" and "not q" to be false. "not q" being false means q is true, so x is integer. Then, "p and q" being false, with q being true, implies p is false, so x² - x < 6. Therefore, x must be integer and x² - x < 6. Which gives us {-1, 0, 1, 2}.Wait, but let me check if there's another interpretation. Suppose the problem said that both "p and q" is false and "not q" is false. So, maybe "both" is not connecting two separate propositions, but the set M must consist of x such that each of these two propositions is false. Which is what we did. So, that would require "p and q" is false and "not q" is false. So, both "p and q" and "not q" are false. So, that's correct. So, "not q" is false implies q is true. Then "p and q" is false, but since q is true, p must be false. So, x is integer and p is false, which is x is integer and x² - x < 6. Hence, M = {-1, 0, 1, 2}.Alternatively, let me think if there's another way. Suppose someone thought that "both 'p and q' and 'not q'" being false means that the statement "(p and q) and (not q)" is false. But that would be different. But the problem states: "both 'p and q' and 'not q' are false propositions". So, each of them is false. So, both propositions have to be false. So, "p and q" is false, and "not q" is false. So, that's how we interpreted it. So, that's correct.Alternatively, if it was phrased as "the proposition 'p and q and not q' is false", that would be a different case, but that's not the case here. The problem says both "p and q" and "not q" are false. So, each one is false.Therefore, the conclusion is that M = {-1, 0, 1, 2}.But let me check once more with an example. Take x = -1. Then, q is true (since -1 is integer). p is x² - x = 1 +1 = 2 < 6, so p is false. Then "p and q" is false (since p is false). "not q" is false (since q is true). So, both are false. So, x = -1 is in M. Similarly, x = 0,1,2.If x = -2: q is true (integer). p: x² -x = 4 + 2 = 6 ≥6, so p is true. Then "p and q" is true, which would not satisfy the first condition. So, x = -2 is excluded.x = 3: similar to above. p is true, so "p and q" is true, excluded.If x = 0.5: then q is false (not integer). "not q" is true, but we need "not q" to be false. So, x must make "not q" false, which requires x to be integer. So, non-integer x are excluded because "not q" would be true, but we need "not q" to be false. So, M only includes integers. So, non-integers are out.Therefore, all steps seem correct. So, the answer is M = {-1, 0, 1, 2}.Final AnswerThe set ( M ) is boxed{{-1, 0, 1, 2}}.</think>Given the propositions ( p ): ( x^2 - x geq 6 ) and ( q ): ( x in mathbb{Z} ), we need to determine the set ( M ) of all values ( x ) such that both "p and q" and "not q" are false propositions.1. Analyzing "not q" is false: - "not q" being false means ( q ) is true. Therefore, ( x ) must be an integer.2. Analyzing "p and q" is false: - Since ( q ) is true, "p and q" being false implies ( p ) must be false. Therefore, ( x^2 - x < 6 ).3. Solving the inequality ( x^2 - x < 6 ): - Rewrite the inequality: ( x^2 - x - 6 < 0 ). - Factor the quadratic: ( (x - 3)(x + 2) < 0 ). - The solution to this inequality is ( x in (-2, 3) ).4. Finding integer solutions within the interval: - The integers in the interval ( (-2, 3) ) are ( -1, 0, 1, 2 ).5. Verification: - Check each integer to ensure ( x^2 - x < 6 ): - For ( x = -1 ): ( (-1)^2 - (-1) = 2 < 6 ) - For ( x = 0 ): ( 0^2 - 0 = 0 < 6 ) - For ( x = 1 ): ( 1^2 - 1 = 0 < 6 ) - For ( x = 2 ): ( 2^2 - 2 = 2 < 6 )Thus, the set ( M ) of all values ( x ) such that both "p and q" and "not q" are false propositions is (boxed{{-1, 0, 1, 2}}).,<think>Okay, so I need to figure out the set M where both "p and q" and "not q" are false. Let me start by breaking down what each part means.First, proposition p is "x² - x ≥ 6", and q is "x is an integer". The problem says that both "p ∧ q" and "¬q" are false. So I need to find all x such that these two propositions are false.Let me recall some logic. If both "p ∧ q" and "¬q" are false, then:1. "p ∧ q" is false means that it's not the case that both p and q are true. So either p is false, or q is false, or both.2. "¬q" is false means that q is true. Because the negation of q being false is equivalent to q being true.Wait, but the second point says that "¬q" is false. So that means q is true. But if q must be true, then how does "p ∧ q" being false interact with that?Let me write this out:Given that "p ∧ q" is false and "¬q" is false. Since "¬q" is false, q is true. So, if q is true, then "p ∧ q" simplifies to "p". Therefore, "p ∧ q" being false when q is true implies that p must be false. Therefore, the conditions are:1. q is true (since ¬q is false)2. p is false (since p ∧ q is false and q is true)Therefore, M should be the set of all x where q is true (so x is an integer) and p is false (so x² - x < 6). Therefore, M = {x ∈ ℤ | x² - x < 6}So now I need to solve the inequality x² - x < 6 for integers x.Let me solve x² - x - 6 < 0. First, find the roots of the quadratic equation x² - x - 6 = 0. Using the quadratic formula:x = [1 ± √(1 + 24)] / 2 = [1 ± 5]/2So, x = (1 + 5)/2 = 6/2 = 3, and x = (1 -5)/2 = -4/2 = -2.Therefore, the quadratic x² - x -6 factors as (x - 3)(x + 2). So the inequality (x - 3)(x + 2) < 0 holds when x is between -2 and 3. Because the parabola opens upwards, the quadratic is negative between its roots.So, the solution to the inequality is -2 < x < 3. But since x must be an integer (from q being true), x can be -1, 0, 1, 2.Therefore, the set M is {-1, 0, 1, 2}.Wait, let me verify that. If x is an integer and between -2 and 3, not including -2 and 3. So integers from -1 up to 2. Yes, that's correct.But let me double-check. Let me test each integer in that interval:For x = -1: (-1)^2 - (-1) = 1 + 1 = 2 < 6. Correct.x = 0: 0 - 0 = 0 < 6. Correct.x = 1: 1 - 1 = 0 < 6. Correct.x = 2: 4 - 2 = 2 < 6. Correct.What about x = -2? Let's check: (-2)^2 - (-2) = 4 + 2 = 6, which is not less than 6, so -2 is excluded.x = 3: 9 - 3 = 6, same thing, so 3 is excluded.So yes, the integers between -2 and 3, not including the endpoints, are -1, 0, 1, 2. Therefore, M is {-1, 0, 1, 2}.Wait a second, but let me make sure that the logic steps are correct. The problem states that both "p and q" and "not q" are false. So:"p ∧ q" is false and "¬q" is false.But "¬q" is false implies q is true. Therefore, if q is true, then "p ∧ q" simplifies to p. Therefore, "p ∧ q" being false implies p is false. Therefore, when q is true, p must be false.Therefore, all x where q is true (x is integer) and p is false (x² - x < 6). That's exactly the set we found. So that seems correct.Alternatively, another way to approach is to consider the truth values:We need both "p ∧ q" and "¬q" to be false.So, for "¬q" to be false, q must be true. So x must be in integers. Then, given that q is true, "p ∧ q" is equivalent to p. So "p ∧ q" being false implies p is false. Therefore, x must be integers where p is false, i.e., x² - x < 6. So same result.Alternatively, if we construct a truth table for p and q:But since q can be true or false, but "¬q" is false only when q is true. So the only rows of the truth table we need to consider are where q is true, and in those rows, "p ∧ q" is false, which would require p to be false.Hence, the same conclusion. So the set M is integers where x² - x < 6, which is {-1, 0, 1, 2}.I think that's solid. Let me check if there's any integer outside that range where x² - x < 6.Take x = -3: (-3)^2 - (-3) = 9 + 3 = 12 ≥6, so no.x = 4: 16 -4 = 12 ≥6, so no.x = -1: 1 +1 = 2 <6, which is in the set.So yes, the set is correct.Therefore, the answer is M = {-1, 0, 1, 2}, which in boxed form is boxed{{-1, 0, 1, 2}}.Final AnswerThe set ( M ) is boxed{{-1, 0, 1, 2}}.</think>Given propositions ( p ): ( x^2 - x geq 6 ) and ( q ): ( x in mathbb{Z} ), we need to determine the set ( M ) of all values of ( x ) such that both "p and q" and "not q" are false propositions.1. Analyzing "not q" being false: - "¬q" is false implies ( q ) is true. Therefore, ( x ) must be an integer.2. Analyzing "p and q" being false: - Since ( q ) is true, "p and q" simplifies to ( p ). Therefore, ( p ) must be false. - ( p ) being false means ( x^2 - x < 6 ).3. Solving the inequality ( x^2 - x < 6 ): - Rewrite the inequality: ( x^2 - x - 6 < 0 ). - Factor the quadratic: ( (x - 3)(x + 2) < 0 ). - The roots of the quadratic equation are ( x = 3 ) and ( x = -2 ). - The solution to the inequality is ( -2 < x < 3 ).4. Considering ( x ) as integers: - The integers within the interval ( -2 < x < 3 ) are ( -1, 0, 1, 2 ).Thus, the set ( M ) of all values of ( x ) such that both "p and q" and "not q" are false is (boxed{{-1, 0, 1, 2}}).]

🔑:Starting with proposition p: x^2 - x geq 6, we can rewrite it in an equivalent form as p: x geq 3 or x leq -2. Since "not q" is a false proposition, the proposition q: x in mathbb{Z} must be true. Given that "p and q" is a false proposition, we can deduce that the proposition p: x geq 3 or x leq -2 must be false as well. Therefore, we are looking for integers x such that -2 < x < 3. Thus, the set of integers that satisfy this condition is {-1, 0, 1, 2}.Hence, the set M consists of: M = boxed{{-1, 0, 1, 2}}.

🧠:[<think>Okay, so I need to figure out which is greater between √2 + √7 and √3 + √6. Hmm, let's see. Both expressions are sums of two square roots, and they look pretty similar, but the numbers inside the radicals are different. I need a way to compare them without just calculating their decimal approximations because I don't have a calculator here. Maybe there's a clever algebraic method to compare them?First, let me write down both expressions clearly:Expression A: √2 + √7Expression B: √3 + √6I need to determine if A > B, A < B, or A = B. Since both expressions are positive, squaring them might help because squaring is a monotonic function for non-negative numbers. That means if A² > B², then A > B, and vice versa. Let's try that.Compute A²: (√2 + √7)² = (√2)² + 2*√2*√7 + (√7)² = 2 + 2√14 + 7 = 9 + 2√14Compute B²: (√3 + √6)² = (√3)² + 2*√3*√6 + (√6)² = 3 + 2√18 + 6 = 9 + 2√18So now we have:A² = 9 + 2√14B² = 9 + 2√18Therefore, to compare A² and B², we can just compare 2√14 and 2√18, since the rest of the terms are the same (both have 9). Dividing both sides by 2, we get:Compare √14 and √18But √14 and √18 are both radicals. Since 14 < 18, √14 < √18. Therefore, 2√14 < 2√18, so A² < B². Which implies that A < B.Wait, so that means √2 + √7 is less than √3 + √6? Let me verify this because sometimes squaring can be tricky if the original numbers are close. But in this case, since both numbers are positive and we squared them, the conclusion should hold. Let me check with approximate decimal values just to be sure.Calculating approximate values:√2 ≈ 1.4142√7 ≈ 2.6458So √2 + √7 ≈ 1.4142 + 2.6458 ≈ 4.06√3 ≈ 1.7320√6 ≈ 2.4495So √3 + √6 ≈ 1.7320 + 2.4495 ≈ 4.1815Comparing 4.06 and 4.1815, indeed 4.06 < 4.1815. So the second expression is larger. Therefore, the greater value is √3 + √6.But wait, just to make sure there isn't a mistake in my algebra. Let's go back through the steps.First, expanding the squares:(√2 + √7)² = 2 + 7 + 2*√(2*7) = 9 + 2√14. Correct.(√3 + √6)² = 3 + 6 + 2*√(3*6) = 9 + 2√18. Correct.Since √14 ≈ 3.7417 and √18 ≈ 4.2426, then 2√14 ≈ 7.4834 and 2√18 ≈ 8.4852. Adding 9 to both gives approximately 16.4834 vs. 17.4852, so A² ≈ 16.48 and B² ≈ 17.49. Therefore, indeed, A² < B². So the conclusion holds. Therefore, √3 + √6 is greater.But let me consider another approach in case there's a different method. Maybe comparing term by term? But how?We have:√2 + √7 vs. √3 + √6If we subtract √2 from both sides, we get:√7 vs. √3 + √6 - √2Not sure if that helps. Alternatively, subtract √3 from both sides:√2 + √7 - √3 vs. √6Still not obvious. Maybe compare differences.Alternatively, let's denote x = √2 + √7 and y = √3 + √6. Then, as above, x² - y² = (9 + 2√14) - (9 + 2√18) = 2(√14 - √18) = 2(√14 - √(14 + 4)) which is negative, so x² - y² < 0, so x < y.Alternatively, perhaps using the property that if a + b = c + d, then compare a² + b² vs. c² + d² or something else. But in this case, the sums are not equal. Wait, but in our case, when we squared them, the linear terms became part of the expression.Alternatively, maybe approximate each square root and see.√2 ≈ 1.414√3 ≈ 1.732√6 ≈ 2.449√7 ≈ 2.645So expression A: 1.414 + 2.645 ≈ 4.059Expression B: 1.732 + 2.449 ≈ 4.181So 4.059 vs. 4.181, so B is larger. Therefore, confirms the previous result.Alternatively, think of the differences between the radicals. For example, between √7 and √6, √7 is larger by about 0.2. Similarly, √3 is larger than √2 by about 0.318. So, in expression A, the larger term is √7, and in expression B, the larger term is √6, but √3 is also larger than √2. So maybe the combination of a slightly smaller large term and a bigger small term in B leads to a larger total?But that's more of a heuristic. The exact method via squaring seems solid.Another way: Let's see, suppose we want to compare √2 + √7 vs. √3 + √6.Let me try moving terms around:√7 - √6 vs. √3 - √2If we can compare these two differences, maybe that would help. Let's compute both sides.Left side: √7 - √6. Right side: √3 - √2.Compute left side: √7 - √6 ≈ 2.6458 - 2.4495 ≈ 0.1963Right side: √3 - √2 ≈ 1.7320 - 1.4142 ≈ 0.3178So, √7 - √6 ≈ 0.1963 and √3 - √2 ≈ 0.3178. Therefore, the right side is larger. So, √3 - √2 > √7 - √6Therefore, adding √6 + √2 to both sides:√3 + √6 > √7 + √2Which is the result we want. Therefore, this approach also shows that √3 + √6 is greater.Alternatively, maybe rationalizing the differences:For example, √7 - √6 can be rationalized by multiplying numerator and denominator (if considering as a fraction) by √7 + √6:√7 - √6 = (√7 - √6)(√7 + √6)/(√7 + √6) = (7 - 6)/(√7 + √6) = 1/(√7 + √6) ≈ 1/(2.6458 + 2.4495) ≈ 1/5.0953 ≈ 0.1963Similarly, √3 - √2 = (√3 - √2)(√3 + √2)/(√3 + √2) = (3 - 2)/(√3 + √2) = 1/(√3 + √2) ≈ 1/(1.732 + 1.414) ≈ 1/3.146 ≈ 0.318So again, showing that √3 - √2 > √7 - √6, leading to the conclusion that √3 + √6 > √2 + √7.Another angle: Maybe using the concavity of the square root function? Since square root is concave, the difference in outputs relates to the difference in inputs. But I think that might complicate things here. The previous methods seem sufficient.Alternatively, considering the function f(x) = √x + √(9 - x). Let's see, if we set 2 + 7 = 9, and 3 + 6 = 9. So maybe comparing f(2) vs. f(3). Then, we can analyze how f(x) behaves. Let's check.Define f(x) = √x + √(9 - x) for x in [0,9]Then, f'(x) = (1/(2√x)) - (1/(2√(9 - x)))Setting derivative to zero for maxima/minima:(1/(2√x)) = (1/(2√(9 - x))) => √x = √(9 - x) => x = 9 - x => 2x =9 => x=4.5Therefore, the function f(x) has a maximum at x=4.5, and it's symmetric around 4.5. Since x=2 and x=3 are equidistant from 4.5 by 2.5 and 1.5? Wait, no.Wait, 4.5 -2 = 2.5 and 4.5 -3 =1.5, so they are not equidistant. Wait, maybe not. Wait, the function is defined from 0 to9. Let me check the values.Wait, f(2)=√2 + √7 ≈1.414 +2.645≈4.059f(3)=√3 +√6≈1.732 +2.449≈4.181f(4.5)=√4.5 +√4.5≈2.121 +2.121≈4.242So, as we can see, the function f(x) reaches its maximum at x=4.5, and as we move away from 4.5 towards 0 or 9, the value decreases. Since 3 is closer to 4.5 than 2 is (distance from 3 to 4.5 is 1.5, while from 2 to4.5 is2.5), then f(3) should be greater than f(2). Therefore, f(3) > f(2), which means √3 + √6 > √2 + √7.That's another way to see it. Since 3 is closer to the midpoint 4.5 than 2 is, the sum of square roots is larger.This makes sense because the function f(x) is symmetric around 4.5 and reaches its maximum there. Points closer to the center have higher values. So, since 3 is closer to 4.5 than 2 is, f(3) > f(2). Therefore, this method also confirms the result.So, multiple methods lead to the same conclusion: √3 + √6 is greater than √2 + √7.Just to ensure there's no mistake, let me cross-verify once more with another technique.Suppose we let a = √2 + √7 and b = √3 + √6.Suppose we assume that a < b, which we need to verify.If a < b, then a - b < 0.Compute a - b = (√2 + √7) - (√3 + √6) = (√2 - √3) + (√7 - √6)We can approximate each difference:√2 - √3 ≈1.414 -1.732≈-0.318√7 - √6≈2.645 -2.449≈0.196Adding them: -0.318 +0.196≈-0.122, which is negative. Therefore, a - b ≈-0.122 <0, so a <b.Therefore, again, the conclusion holds.Alternatively, perhaps use inequalities between the radicals. For example, compare √7 - √6 and √3 - √2. As done before, which shows that √3 - √2 is bigger, which when rearranged gives the result.Alternatively, consider the function g(x) = √(x +5) - √x. Let's see, does this function increase or decrease as x increases?Compute derivative: g'(x) = (1/(2√(x +5))) - (1/(2√x))Which is positive or negative?If x increases, √(x +5) increases slower than √x, since the derivative of √(x +5) is less than derivative of √x. Wait, let's compute:g'(x) = [1/(2√(x +5))] - [1/(2√x)]So, if [1/(2√(x +5))] < [1/(2√x)], which is true because √(x +5) > √x, so 1/√(x +5) <1/√x. Therefore, g'(x) is negative, meaning g(x) is decreasing. Therefore, as x increases, g(x) decreases. So, since 2 <3, then g(2) >g(3). But wait, let's check:g(2) =√7 -√2Wait, no, g(x) was defined as √(x +5) -√x. So, if x=2, then √7 -√2. If x=1, then √6 -√1. Wait, but we were comparing √7 -√6 and √3 -√2. Hmm, maybe not directly applicable.Alternatively, perhaps another function. Let me think.But perhaps this is overcomplicating. The previous methods have sufficiently shown that √3 +√6 is greater. All right, so I think I'm confident in the result. The answer should be √3 +√6.Final AnswerThe greater value is boxed{sqrt{3} + sqrt{6}}.</think>To determine which is greater between (sqrt{2} + sqrt{7}) and (sqrt{3} + sqrt{6}), we can compare their squares:1. Compute the square of (sqrt{2} + sqrt{7}): [ (sqrt{2} + sqrt{7})^2 = (sqrt{2})^2 + 2 cdot sqrt{2} cdot sqrt{7} + (sqrt{7})^2 = 2 + 2sqrt{14} + 7 = 9 + 2sqrt{14} ]2. Compute the square of (sqrt{3} + sqrt{6}): [ (sqrt{3} + sqrt{6})^2 = (sqrt{3})^2 + 2 cdot sqrt{3} cdot sqrt{6} + (sqrt{6})^2 = 3 + 2sqrt{18} + 6 = 9 + 2sqrt{18} ]3. Compare (9 + 2sqrt{14}) and (9 + 2sqrt{18}): - Since (sqrt{14} < sqrt{18}), it follows that (2sqrt{14} < 2sqrt{18}). - Therefore, (9 + 2sqrt{14} < 9 + 2sqrt{18}).This implies that ((sqrt{2} + sqrt{7})^2 < (sqrt{3} + sqrt{6})^2), so (sqrt{2} + sqrt{7} < sqrt{3} + sqrt{6}).Additionally, approximating the decimal values:- (sqrt{2} + sqrt{7} approx 1.4142 + 2.6458 approx 4.06)- (sqrt{3} + sqrt{6} approx 1.7320 + 2.4495 approx 4.1815)Thus, (sqrt{3} + sqrt{6}) is indeed greater.The greater value is (boxed{sqrt{3} + sqrt{6}}).,<think>Okay, so I need to figure out which is bigger: √2 + √7 or √3 + √6. Hmm, both are sums of square roots, so comparing them directly might be tricky. Let me think of some strategies. Maybe squaring both expressions to eliminate the square roots? That might work because squaring is a monotonic function for non-negative numbers, which these are. Let me write that down.Let me denote the first expression as A = √2 + √7 and the second as B = √3 + √6. I want to compare A and B. If I square both, I can compare A² and B² instead. Calculating A²: (√2 + √7)² = (√2)² + 2*√2*√7 + (√7)² = 2 + 2√14 + 7 = 9 + 2√14.Similarly, B²: (√3 + √6)² = (√3)² + 2*√3*√6 + (√6)² = 3 + 2√18 + 6 = 9 + 2√18.So now I need to compare 9 + 2√14 and 9 + 2√18. Since both have 9 added to twice a square root, the one with the larger square root term will be larger. So which is larger, √14 or √18?Well, √14 is approximately 3.7417 and √18 is approximately 4.2426. So √18 is larger. Therefore, 2√18 is larger than 2√14, so B² is larger than A². Since both A and B are positive, their squares preserve the inequality, so B must be larger than A. Therefore, √3 + √6 is greater than √2 + √7.Wait, but let me double-check my steps. Squaring can sometimes be tricky if the numbers are close, but in this case, since √18 is clearly larger than √14, and the rest of the terms are the same (9 and 2 times), the conclusion should hold. Let me verify with approximate values.Calculating A: √2 ≈ 1.4142, √7 ≈ 2.6458, so A ≈ 1.4142 + 2.6458 ≈ 4.06.Calculating B: √3 ≈ 1.732, √6 ≈ 2.4495, so B ≈ 1.732 + 2.4495 ≈ 4.1815.Comparing 4.06 and 4.1815, indeed B is larger. So that confirms it. Therefore, the greater value is √3 + √6.But just to ensure there's no miscalculation, let me check the squaring again. A² = 9 + 2√14 ≈ 9 + 2*3.7417 ≈ 9 + 7.4834 ≈ 16.4834. B² = 9 + 2√18 ≈ 9 + 2*4.2426 ≈ 9 + 8.4852 ≈ 17.4852. Yes, 17.4852 is greater than 16.4834, so B² > A², hence B > A.Alternatively, another approach might be to compare the differences. Let me see. Let's compute A - B = (√2 + √7) - (√3 + √6). If this is positive, then A is larger; otherwise, B. But dealing with the difference of square roots might not be straightforward. But maybe we can manipulate it.Alternatively, perhaps consider the difference squared? Hmm, but that might get complicated. Let me try.Compute (A - B)² = (√2 + √7 - √3 - √6)². Expanding this would give terms like (√2 - √3)² + (√7 - √6)² + 2*(√2 - √3)*(√7 - √6). Not sure if that helps. Let's compute it step by step.First, expand (√2 + √7 - √3 - √6)^2:= (√2 - √3)^2 + (√7 - √6)^2 + 2*(√2 - √3)*(√7 - √6)Calculating each term:(√2 - √3)^2 = 2 + 3 - 2√6 = 5 - 2√6 ≈ 5 - 4.899 ≈ 0.101.(√7 - √6)^2 = 7 + 6 - 2√42 = 13 - 2√42 ≈ 13 - 12.961 ≈ 0.039.Now, the cross term: 2*(√2 - √3)*(√7 - √6). Let's compute the product inside first.(√2 - √3)*(√7 - √6) = √2√7 - √2√6 - √3√7 + √3√6.Simplify each term:√14 - √12 - √21 + √18.Which is √14 - 2√3 - √21 + 3√2.Hmm, this seems messy. Let me approximate each term:√14 ≈ 3.7417, √12 ≈ 3.4641, √21 ≈ 4.5837, √18 ≈ 4.2426.So substituting:≈ 3.7417 - 3.4641 - 4.5837 + 4.2426 ≈ (3.7417 + 4.2426) - (3.4641 + 4.5837) ≈ (7.9843) - (8.0478) ≈ -0.0635.Multiply by 2: ≈ -0.127.Therefore, adding all the terms together:First two squared terms: 0.101 + 0.039 ≈ 0.14.Then cross term: -0.127.Total (A - B)^2 ≈ 0.14 - 0.127 ≈ 0.013.So (A - B)^2 ≈ 0.013, which is positive. That means A - B is either approximately √0.013 ≈ 0.114 or -√0.013 ≈ -0.114. Since our approximated A was 4.06 and B was 4.1815, A - B ≈ -0.1215, so squaring gives approximately 0.0147, which matches our calculation here. So since (A - B)^2 is positive, and A - B is negative (as B > A), so the square of a negative number is positive. Therefore, this method also shows that B > A, but it's a bit more involved.So both methods confirm that √3 + √6 is larger. Thus, the answer should be √3 + √6.Final AnswerThe greater value is boxed{sqrt{3} + sqrt{6}}.</think>To determine which is greater between (sqrt{2} + sqrt{7}) and (sqrt{3} + sqrt{6}), we can square both expressions and compare the results.Let (A = sqrt{2} + sqrt{7}) and (B = sqrt{3} + sqrt{6}).First, we calculate (A^2):[A^2 = (sqrt{2} + sqrt{7})^2 = (sqrt{2})^2 + 2 cdot sqrt{2} cdot sqrt{7} + (sqrt{7})^2 = 2 + 2sqrt{14} + 7 = 9 + 2sqrt{14}]Next, we calculate (B^2):[B^2 = (sqrt{3} + sqrt{6})^2 = (sqrt{3})^2 + 2 cdot sqrt{3} cdot sqrt{6} + (sqrt{6})^2 = 3 + 2sqrt{18} + 6 = 9 + 2sqrt{18}]Comparing (A^2) and (B^2), we need to determine which is larger between (2sqrt{14}) and (2sqrt{18}). Since (sqrt{18} approx 4.2426) is greater than (sqrt{14} approx 3.7417), we have (2sqrt{18} > 2sqrt{14}). Therefore, (B^2 > A^2).Since both (A) and (B) are positive, it follows that (B > A).Thus, the greater value is (boxed{sqrt{3} + sqrt{6}}).]

🔑:Solution: ( sqrt {3}+ sqrt {6})^{2}=9+2 sqrt {18},( sqrt {2}+ sqrt {7})^{2}=9+2 sqrt {14},therefore( sqrt {3}+ sqrt {6})^{2} > ( sqrt {2}+ sqrt {7})^{2},therefore sqrt {3}+ sqrt {6} > sqrt {2}+ sqrt {7},Hence, the answer is: boxed{sqrt {3}+ sqrt {6}}.Square each number and compare them to find the greater value.This problem tests the comparison of inequalities, with squaring being a common method, making it a basic question.