answer：Okay, so I have this function f(x) = -x² + a x - ln(x), where a is a real number. There are two parts to this problem. Let me tackle them one by one.Starting with part (1): I need to find the necessary and sufficient condition for f(x) to have both a maximum and a minimum value. Hmm, okay. So, for a function to have both a maximum and a minimum, it must have critical points where the derivative is zero, and these critical points must be such that one is a maximum and the other is a minimum. That means the function must have at least two critical points, right?So, first, I should find the derivative of f(x). Let me compute that:f'(x) = d/dx (-x² + a x - ln(x)) = -2x + a - 1/x.Alright, so f'(x) = -2x + a - 1/x. To find critical points, I set this equal to zero:-2x + a - 1/x = 0.Let me rearrange this equation to make it easier to handle. Multiply both sides by x to eliminate the denominator:-2x² + a x - 1 = 0.So, that's a quadratic equation in terms of x: -2x² + a x - 1 = 0. Alternatively, I can write it as 2x² - a x + 1 = 0 by multiplying both sides by -1.Now, for this quadratic equation to have two distinct real roots, the discriminant must be positive. The discriminant D of a quadratic equation ax² + bx + c = 0 is D = b² - 4ac.In this case, the quadratic is 2x² - a x + 1 = 0, so a (the coefficient) is 2, b is -a, and c is 1. Therefore, the discriminant D is:D = (-a)² - 4 * 2 * 1 = a² - 8.For two distinct real roots, we need D > 0, so:a² - 8 > 0 => a² > 8 => |a| > 2√2.But since we are dealing with x > 0 (because of the ln(x) term in the original function), we need to ensure that the roots are positive. The quadratic equation is 2x² - a x + 1 = 0. For both roots to be positive, certain conditions must be met. Specifically, the sum of the roots is a/2, and the product is 1/2.Since the product is positive, both roots must be either positive or negative. But since x > 0, we only consider positive roots. The sum of the roots is a/2, which must be positive. Therefore, a must be positive. So, combining this with the discriminant condition, we have a > 2√2.Wait, hold on. The discriminant condition gives |a| > 2√2, but since a must be positive for the sum of the roots to be positive, it's sufficient to say a > 2√2.So, the necessary and sufficient condition is that a > 2√2.Moving on to part (2): When the function f(x) is monotonic on the interval [1, 2], find the range of values for a.Monotonic means that the function is either entirely non-increasing or non-decreasing on that interval. So, for f(x) to be monotonic on [1, 2], its derivative f'(x) must not change sign on that interval. That is, f'(x) must be either always non-negative or always non-positive on [1, 2].From part (1), we have f'(x) = -2x + a - 1/x. So, let's write that again:f'(x) = -2x + a - 1/x.We need f'(x) ≥ 0 for all x in [1, 2] or f'(x) ≤ 0 for all x in [1, 2].Let me analyze f'(x). Let's consider f'(x) as a function on [1, 2]. To ensure it doesn't change sign, we need to find the maximum and minimum of f'(x) on [1, 2] and ensure that either the minimum is non-negative or the maximum is non-positive.Alternatively, since f'(x) is continuous on [1, 2], if it doesn't change sign, then either f'(x) ≥ 0 for all x in [1, 2] or f'(x) ≤ 0 for all x in [1, 2].So, let's find the extrema of f'(x) on [1, 2]. To do that, we can take the derivative of f'(x), which is f''(x):f''(x) = d/dx (-2x + a - 1/x) = -2 + 1/x².Set f''(x) = 0 to find critical points:-2 + 1/x² = 0 => 1/x² = 2 => x² = 1/2 => x = 1/√2 ≈ 0.707.But 1/√2 is approximately 0.707, which is less than 1. So, on the interval [1, 2], f''(x) is always negative because:At x = 1, f''(1) = -2 + 1 = -1 < 0.At x = 2, f''(2) = -2 + 1/4 = -1.75 < 0.Therefore, f'(x) is concave down on [1, 2], meaning its maximum occurs at one of the endpoints.Wait, but f'(x) is concave down, so its maximum is at the endpoints, and it has a minimum somewhere inside the interval if there's a critical point. However, since the critical point of f'(x) is at x = 1/√2, which is outside [1, 2], f'(x) is strictly decreasing on [1, 2] because f''(x) < 0 there.Therefore, f'(x) is decreasing on [1, 2]. So, the maximum of f'(x) on [1, 2] is at x = 1, and the minimum is at x = 2.Thus, to ensure f'(x) does not change sign on [1, 2], we need either:1. f'(1) ≥ 0 and f'(2) ≥ 0, meaning f'(x) is non-decreasing throughout, or2. f'(1) ≤ 0 and f'(2) ≤ 0, meaning f'(x) is non-increasing throughout.But wait, since f'(x) is decreasing, if f'(1) ≥ 0 and f'(2) ≥ 0, then f'(x) is always non-negative on [1, 2]. Similarly, if f'(1) ≤ 0 and f'(2) ≤ 0, then f'(x) is always non-positive on [1, 2].So, let's compute f'(1) and f'(2):f'(1) = -2(1) + a - 1/1 = -2 + a - 1 = a - 3.f'(2) = -2(2) + a - 1/2 = -4 + a - 0.5 = a - 4.5.So, for f'(x) to be non-negative on [1, 2], we need:a - 3 ≥ 0 and a - 4.5 ≥ 0.Which simplifies to:a ≥ 3 and a ≥ 4.5.So, the stricter condition is a ≥ 4.5.Similarly, for f'(x) to be non-positive on [1, 2], we need:a - 3 ≤ 0 and a - 4.5 ≤ 0.Which simplifies to:a ≤ 3 and a ≤ 4.5.The stricter condition here is a ≤ 3.Therefore, for f(x) to be monotonic on [1, 2], a must satisfy either a ≤ 3 or a ≥ 4.5.Wait, let me verify this. If a = 4.5, then f'(2) = 0, and since f'(x) is decreasing, f'(x) would be non-negative on [1, 2]. Similarly, if a = 3, f'(1) = 0, and f'(x) would be non-positive on [1, 2]. So, the endpoints are included.Hence, the range of a is a ≤ 3 or a ≥ 4.5.I think that makes sense. Let me just recap:1. For f(x) to have both a maximum and a minimum, the derivative must have two real positive roots, which requires a > 2√2.2. For f(x) to be monotonic on [1, 2], the derivative must not change sign, which happens when a ≤ 3 or a ≥ 4.5.Yeah, that seems consistent.**Final Answer**(1) The necessary and sufficient condition is boxed{a > 2sqrt{2}}.(2) The range of values for ( a ) is boxed{(-infty, 3] cup [4.5, infty)}.

answer：Alright, so I have this problem about a regular square pyramid. The base edge length is given as (4sqrt{2}), and the volume is 32. I need to find the length of its side edge. Hmm, okay, let me think about how to approach this.First, I remember that a regular square pyramid has a square base, and all the side edges are equal in length. The volume of a pyramid is given by the formula:[V = frac{1}{3} times text{Base Area} times text{Height}]So, if I can find the height of the pyramid, I can then use that to find the side edge length. Let me start by calculating the base area.The base is a square with edge length (4sqrt{2}). So, the area of the base is:[text{Base Area} = (4sqrt{2})^2 = 16 times 2 = 32]Okay, so the base area is 32. Now, plugging this into the volume formula:[32 = frac{1}{3} times 32 times text{Height}]Wait, hold on. That doesn't seem right. If the volume is 32, then:[32 = frac{1}{3} times 32 times h]So, solving for (h):[32 = frac{32}{3}h h = 32 times frac{3}{32} = 3]Got it, so the height of the pyramid is 3 units.Now, to find the side edge length, which is the distance from the apex of the pyramid to one of the base vertices. Let me visualize the pyramid: it's a square base, so the apex is directly above the center of the square.Let me denote the apex as point (P), and the base vertices as (A), (B), (C), and (D). The center of the base is point (O). So, the height (PO) is 3 units.Now, the side edge is (PA), which is the distance from (P) to (A). To find this, I can consider the right triangle (PAO), where (PO) is one leg (height), (AO) is the other leg (distance from center to vertex), and (PA) is the hypotenuse (side edge).So, I need to find (AO). Since the base is a square with edge length (4sqrt{2}), the diagonal of the base is:[text{Diagonal} = 4sqrt{2} times sqrt{2} = 4 times 2 = 8]Wait, that seems off. Let me recall, the diagonal of a square with side length (s) is (ssqrt{2}). So, if (s = 4sqrt{2}), then the diagonal is:[text{Diagonal} = 4sqrt{2} times sqrt{2} = 4 times 2 = 8]Yes, that's correct. So, the diagonal is 8 units. Therefore, the distance from the center (O) to a vertex (A) is half of the diagonal:[AO = frac{8}{2} = 4]Okay, so (AO = 4). Now, in triangle (PAO), we have:[PA = sqrt{AO^2 + PO^2} = sqrt{4^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5]So, the length of the side edge (PA) is 5 units.Wait, let me double-check my steps to make sure I didn't make a mistake. First, base edge is (4sqrt{2}), so base area is indeed (32). Volume is (32), so height comes out to 3. Then, the diagonal of the base is (8), so (AO) is 4. Then, using Pythagoras, (PA = 5). That seems consistent.I think that's correct. So, the side edge length is 5.**Final Answer**The length of the side edge is boxed{5}.

answer：Okay, so I have this problem here: If 30% of 40% of a number is 36, what is 40% of 30% of the same number? Hmm, let me try to figure this out step by step.First, I need to understand what the problem is asking. It mentions percentages of a number, so I think I should let that number be something, maybe x. Yeah, let's say the number is x.The problem says 30% of 40% of x is 36. So, in math terms, that would be 0.3 times 0.4 times x equals 36. Let me write that down:0.3 * 0.4 * x = 36Now, I can simplify 0.3 times 0.4. Let me calculate that. 0.3 multiplied by 0.4 is... 0.12. So, the equation becomes:0.12 * x = 36Okay, so to find x, I need to divide both sides of the equation by 0.12. Let me do that:x = 36 / 0.12Hmm, what's 36 divided by 0.12? Let me think. 0.12 times 300 is 36 because 0.12 * 300 = 36. So, x must be 300.Alright, so the number is 300. Now, the question is asking for 40% of 30% of the same number. So, that would be 0.4 times 0.3 times x. Let me write that:0.4 * 0.3 * xAgain, I can simplify 0.4 times 0.3 first. 0.4 multiplied by 0.3 is... 0.12. So, the equation becomes:0.12 * xBut wait, x is 300, right? So, plugging that in:0.12 * 300What's 0.12 times 300? Well, 0.1 times 300 is 30, and 0.02 times 300 is 6, so adding those together, 30 + 6 = 36.So, 40% of 30% of the number is also 36. That's interesting because switching the order of the percentages didn't change the result. I guess that's because multiplication is commutative, so 0.3 * 0.4 is the same as 0.4 * 0.3.Let me just double-check my steps to make sure I didn't make any mistakes. I set up the equation correctly, solved for x, and then used that x to find the second percentage. Everything seems to add up. Yeah, I think I did it right.

answer：First, let's denote the cost price of the item as ( x ) yuan.The merchant sells the item for 48 yuan, which includes a 20% profit. This means the selling price is 120% of the cost price.We can set up the equation:[x + 20% cdot x = 48]Simplifying the equation:[1x + 0.20x = 48][1.20x = 48]To find ( x ), divide both sides by 1.20:[x = frac{48}{1.20}][x = 40]Therefore, the cost price of the item is 40 yuan.

answer：Alright, so I have this expression to simplify: ((a - frac{2a - 1}{a}) div frac{a^2 - 1}{a}). Hmm, okay, let's break this down step by step. First, I need to simplify the numerator and the denominator separately before dealing with the division.Starting with the numerator: (a - frac{2a - 1}{a}). I think I can combine these terms by finding a common denominator. Since the first term is just (a), I can write it as (frac{a}{1}) to have the same denominator as the second term. So, (frac{a}{1} - frac{2a - 1}{a}). To combine these, I'll need to multiply the first fraction by (frac{a}{a}) to get (frac{a^2}{a}). Now, the expression becomes (frac{a^2}{a} - frac{2a - 1}{a}). Subtracting these gives (frac{a^2 - (2a - 1)}{a}). Simplifying the numerator inside the fraction: (a^2 - 2a + 1). So, the numerator simplifies to (frac{a^2 - 2a + 1}{a}).Now, looking at the denominator of the original expression: (frac{a^2 - 1}{a}). I recognize that (a^2 - 1) is a difference of squares, which factors into ((a + 1)(a - 1)). So, the denominator becomes (frac{(a + 1)(a - 1)}{a}).Now, the expression is (frac{frac{a^2 - 2a + 1}{a}}{frac{(a + 1)(a - 1)}{a}). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes (frac{a^2 - 2a + 1}{a} times frac{a}{(a + 1)(a - 1)}). The (a) in the numerator and denominator cancels out, leaving (frac{a^2 - 2a + 1}{(a + 1)(a - 1)}).Looking at the numerator (a^2 - 2a + 1), I notice that this is a perfect square trinomial, which factors into ((a - 1)^2). So, the expression now is (frac{(a - 1)^2}{(a + 1)(a - 1)}). One of the ((a - 1)) terms cancels out from the numerator and denominator, simplifying to (frac{a - 1}{a + 1}).Okay, so the simplified expression is (frac{a - 1}{a + 1}). Now, I need to choose a suitable number within the range (-3 < a < 3) to substitute in and evaluate. I should pick a number that doesn't make any denominators zero in the original expression. So, (a) can't be 0, 1, or -1 because those would make some denominators zero.Let me choose (a = 2) because it's within the range and doesn't cause any issues. Plugging (a = 2) into the simplified expression: (frac{2 - 1}{2 + 1} = frac{1}{3}).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, combining the terms in the numerator correctly, factoring, canceling out, and substituting. It seems consistent. So, I think the final answer is (frac{1}{3}).

answer：Okay, so I have this inequality to prove: for all x between 0 and π/2, tan(tan(x)) + sin(x) is greater than sin(sin(x)) + tan(x). Hmm, that's interesting. I need to show that tan(tan(x)) + sin(x) > sin(sin(x)) + tan(x) for all x in (0, π/2).First, maybe I should try plugging in some values to see if it holds. Let me try x = π/4, which is in the middle of the interval. So tan(π/4) is 1, so tan(tan(π/4)) is tan(1). Let me calculate tan(1) in radians. I think it's approximately 1.5574. Then sin(π/4) is √2/2, which is about 0.7071. So sin(sin(π/4)) is sin(0.7071). Let me compute that. Sin(0.7071) is approximately 0.6496. So putting it all together: tan(tan(π/4)) + sin(π/4) ≈ 1.5574 + 0.7071 ≈ 2.2645. On the other side, sin(sin(π/4)) + tan(π/4) ≈ 0.6496 + 1 ≈ 1.6496. So 2.2645 > 1.6496, which is true. Okay, so it holds for x = π/4.Maybe try another value, like x = π/6. So tan(π/6) is 1/√3 ≈ 0.5774. Then tan(tan(π/6)) is tan(0.5774). Let me compute that. Tan(0.5774) is approximately 0.6421. Sin(π/6) is 0.5, so sin(sin(π/6)) is sin(0.5) ≈ 0.4794. So tan(tan(π/6)) + sin(π/6) ≈ 0.6421 + 0.5 ≈ 1.1421. On the other side, sin(sin(π/6)) + tan(π/6) ≈ 0.4794 + 0.5774 ≈ 1.0568. So 1.1421 > 1.0568, which is also true.Hmm, seems like it's holding for these specific values. Maybe I should try to see if I can generalize this.Let me define a function f(x) = tan(tan(x)) + sin(x) - sin(sin(x)) - tan(x). So I need to show that f(x) > 0 for all x in (0, π/2). If I can show that f(x) is always positive in this interval, that would prove the inequality.To analyze f(x), maybe I can look at its derivative. If I can show that f'(x) is positive, then f(x) is increasing, and since f(0) is 0 (well, approaching 0 as x approaches 0), then f(x) would be positive for all x in (0, π/2).So let's compute f'(x). The derivative of tan(tan(x)) is sec²(tan(x)) * sec²(x), right? Because derivative of tan(u) is sec²(u) * u', and here u = tan(x), so u' = sec²(x). So derivative of tan(tan(x)) is sec²(tan(x)) * sec²(x).Derivative of sin(x) is cos(x). Derivative of sin(sin(x)) is cos(sin(x)) * cos(x), by the chain rule. Derivative of tan(x) is sec²(x).So putting it all together:f'(x) = sec²(tan(x)) * sec²(x) + cos(x) - cos(sin(x)) * cos(x) - sec²(x).Let me write that more neatly:f'(x) = sec²(tan(x)) * sec²(x) + cos(x) - cos(x) * cos(sin(x)) - sec²(x).Hmm, maybe I can factor some terms here. Let's see, the first term is sec²(tan(x)) * sec²(x), and the last term is -sec²(x). So factor out sec²(x):f'(x) = sec²(x) [sec²(tan(x)) - 1] + cos(x) [1 - cos(sin(x))].Now, sec²(tan(x)) - 1 is equal to tan²(tan(x)), since sec²(u) - 1 = tan²(u). So:f'(x) = sec²(x) * tan²(tan(x)) + cos(x) [1 - cos(sin(x))].Now, both terms here are positive. Because tan²(tan(x)) is always positive for x in (0, π/2), and sec²(x) is positive. Similarly, 1 - cos(sin(x)) is positive because cos(sin(x)) is less than 1 for x in (0, π/2) (since sin(x) is between 0 and 1, and cos(z) < 1 for z in (0, π/2)). Also, cos(x) is positive in this interval.Therefore, both terms in f'(x) are positive, which means f'(x) > 0 for all x in (0, π/2). So f(x) is strictly increasing on this interval.Now, let's check the limit as x approaches 0 from the right. What is f(0)? Well, tan(tan(0)) is tan(0) = 0, sin(0) is 0, sin(sin(0)) is sin(0) = 0, and tan(0) is 0. So f(0) = 0 + 0 - 0 - 0 = 0. But since f(x) is increasing and f(0) = 0, then for all x > 0 in (0, π/2), f(x) must be greater than 0.Therefore, the inequality holds for all x in (0, π/2).Wait, let me double-check my derivative calculation. I had f'(x) = derivative of tan(tan(x)) + derivative of sin(x) - derivative of sin(sin(x)) - derivative of tan(x). So:- d/dx tan(tan(x)) = sec²(tan(x)) * sec²(x)- d/dx sin(x) = cos(x)- d/dx sin(sin(x)) = cos(sin(x)) * cos(x)- d/dx tan(x) = sec²(x)So f'(x) = sec²(tan(x)) * sec²(x) + cos(x) - cos(sin(x)) * cos(x) - sec²(x). That seems correct.Then I factored out sec²(x) from the first and last terms:sec²(x) [sec²(tan(x)) - 1] + cos(x) [1 - cos(sin(x))]Which simplifies to sec²(x) * tan²(tan(x)) + cos(x) [1 - cos(sin(x))]. That looks right.Since both terms are positive, f'(x) is positive, so f(x) is increasing. And since f(0) = 0, f(x) > 0 for x > 0 in (0, π/2). That makes sense.I think this approach is solid. I started by testing specific values, then moved on to defining the function and analyzing its derivative. The key was recognizing that the derivative is always positive, which implies the function is increasing. Since it starts at zero and increases, it must be positive throughout the interval.I don't see any mistakes in my reasoning. The derivative steps seem correct, and the conclusions follow logically. So I feel confident that this proves the inequality holds for all x in (0, π/2).