answer：Okay, so I've got this problem about two functions, f(x) = sin(πx) and g(x) = cos(πx). There are three parts to the problem, and I need to tackle each one step by step. Let me start with the first part.**Problem (1): Find the minimum value of |PQ| where P is a highest point on f(x) and Q is a lowest point on g(x).**Alright, so I need to find points P and Q on the graphs of f(x) and g(x) respectively, such that P is a maximum point and Q is a minimum point. Then, I have to find the minimum distance between such points P and Q.First, let's recall the general shapes of sine and cosine functions. Both are periodic with period 2π, but since the argument is πx, the period for both functions is 2. That is, the period T is given by T = 2π / |π| = 2. So, both functions repeat every 2 units along the x-axis.For f(x) = sin(πx), the maximum points occur where the sine function reaches 1. The sine function reaches its maximum at π/2, 5π/2, etc. So, solving πx = π/2 + 2πk, where k is an integer. Dividing both sides by π, we get x = 1/2 + 2k. So, the coordinates of the maximum points P are (1/2 + 2k, 1).Similarly, for g(x) = cos(πx), the minimum points occur where the cosine function reaches -1. The cosine function reaches its minimum at π, 3π, etc. So, solving πx = π + 2πk, which gives x = 1 + 2k. Therefore, the coordinates of the minimum points Q are (1 + 2k, -1).Now, I need to find the distance between a point P and a point Q. Let's denote P as (1/2 + 2k₁, 1) and Q as (1 + 2k₂, -1), where k₁ and k₂ are integers. The distance |PQ| can be calculated using the distance formula:|PQ| = sqrt[(x_Q - x_P)² + (y_Q - y_P)²]Plugging in the coordinates:|PQ| = sqrt[(1 + 2k₂ - (1/2 + 2k₁))² + (-1 - 1)²]= sqrt[(1 - 1/2 + 2k₂ - 2k₁)² + (-2)²]= sqrt[(1/2 + 2(k₂ - k₁))² + 4]Let me denote k = k₂ - k₁, which is also an integer. So, the expression becomes:|PQ| = sqrt[(1/2 + 2k)² + 4]Now, I need to find the minimum value of this expression. Since k is an integer, I can plug in different integer values for k and see which gives the smallest distance.Let's try k = 0:|PQ| = sqrt[(1/2 + 0)² + 4] = sqrt[(1/2)² + 4] = sqrt[1/4 + 4] = sqrt[17/4] = sqrt(17)/2 ≈ 2.0616Now, k = 1:|PQ| = sqrt[(1/2 + 2)² + 4] = sqrt[(5/2)² + 4] = sqrt[25/4 + 16/4] = sqrt[41/4] = sqrt(41)/2 ≈ 3.1902k = -1:|PQ| = sqrt[(1/2 - 2)² + 4] = sqrt[(-3/2)² + 4] = sqrt[9/4 + 16/4] = sqrt[25/4] = 5/2 = 2.5Hmm, so for k = -1, the distance is 2.5, which is larger than when k = 0. So, k = 0 gives a smaller distance.Wait, but maybe I should check more values? Let's try k = 2:|PQ| = sqrt[(1/2 + 4)² + 4] = sqrt[(9/2)² + 4] = sqrt[81/4 + 16/4] = sqrt[97/4] ≈ 4.924Similarly, k = -2:|PQ| = sqrt[(1/2 - 4)² + 4] = sqrt[(-7/2)² + 4] = sqrt[49/4 + 16/4] = sqrt[65/4] ≈ 4.031So, as k increases or decreases, the distance increases. Therefore, the minimum occurs at k = 0, giving |PQ| = sqrt(17)/2.Wait, but let me think again. When k = 0, the x-coordinate difference is 1 - 1/2 = 1/2. So, the horizontal distance is 1/2, and the vertical distance is 2 (from 1 to -1). So, the distance is sqrt((1/2)^2 + 2^2) = sqrt(1/4 + 4) = sqrt(17/4) = sqrt(17)/2, which is approximately 2.0616.Is there any closer point? For example, if k is not an integer? Wait, but k has to be an integer because P and Q are specific points on their respective functions. So, k must be integer, so k = 0 gives the closest possible points.Therefore, the minimum value of |PQ| is sqrt(17)/2.**Problem (2): Find the area of the triangle formed by three adjacent intersection points of the graphs of f(x) and g(x).**Alright, so I need to find where f(x) and g(x) intersect, then take three adjacent intersection points, form a triangle, and find its area.First, let's find the intersection points. We set f(x) = g(x):sin(πx) = cos(πx)Divide both sides by cos(πx) (assuming cos(πx) ≠ 0):tan(πx) = 1So, πx = π/4 + kπ, where k is integer.Therefore, x = 1/4 + k.So, the intersection points occur at x = 1/4 + k, where k is integer.Thus, the x-coordinates of the intersection points are ... -7/4, -3/4, 1/4, 5/4, 9/4, ...So, three adjacent intersection points would be, for example, x = -3/4, 1/4, 5/4.Wait, but let me confirm. If k = -1, x = 1/4 - 1 = -3/4. For k = 0, x = 1/4. For k = 1, x = 5/4.So, yes, three adjacent points are at x = -3/4, 1/4, 5/4.Now, I need to find the coordinates of these points.For x = -3/4:f(-3/4) = sin(π*(-3/4)) = sin(-3π/4) = -sin(3π/4) = -√2/2Similarly, g(-3/4) = cos(π*(-3/4)) = cos(-3π/4) = cos(3π/4) = -√2/2So, the point is (-3/4, -√2/2)For x = 1/4:f(1/4) = sin(π*(1/4)) = sin(π/4) = √2/2g(1/4) = cos(π*(1/4)) = cos(π/4) = √2/2So, the point is (1/4, √2/2)For x = 5/4:f(5/4) = sin(π*(5/4)) = sin(5π/4) = -√2/2g(5/4) = cos(π*(5/4)) = cos(5π/4) = -√2/2So, the point is (5/4, -√2/2)Therefore, the three points are:A: (-3/4, -√2/2)B: (1/4, √2/2)C: (5/4, -√2/2)Now, I need to find the area of triangle ABC.One way to find the area is to use the shoelace formula.First, let's list the coordinates:A: (-3/4, -√2/2)B: (1/4, √2/2)C: (5/4, -√2/2)Let me write them in order:A: (-3/4, -√2/2)B: (1/4, √2/2)C: (5/4, -√2/2)Back to A: (-3/4, -√2/2)Shoelace formula:Area = |(x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B))/2|Plugging in the values:x_A = -3/4, y_A = -√2/2x_B = 1/4, y_B = √2/2x_C = 5/4, y_C = -√2/2Compute each term:First term: x_A(y_B - y_C) = (-3/4)(√2/2 - (-√2/2)) = (-3/4)(√2/2 + √2/2) = (-3/4)(√2) = (-3√2)/4Second term: x_B(y_C - y_A) = (1/4)(-√2/2 - (-√2/2)) = (1/4)(-√2/2 + √2/2) = (1/4)(0) = 0Third term: x_C(y_A - y_B) = (5/4)(-√2/2 - √2/2) = (5/4)(-√2) = (-5√2)/4Now, sum these terms:(-3√2)/4 + 0 + (-5√2)/4 = (-8√2)/4 = -2√2Take absolute value and divide by 2:Area = | -2√2 | / 2 = 2√2 / 2 = √2So, the area is √2.Wait, let me double-check my calculations.First term: (-3/4)(√2/2 - (-√2/2)) = (-3/4)(√2) = -3√2/4Second term: (1/4)(-√2/2 - (-√2/2)) = (1/4)(0) = 0Third term: (5/4)(-√2/2 - √2/2) = (5/4)(-√2) = -5√2/4Sum: (-3√2/4) + 0 + (-5√2/4) = (-8√2)/4 = -2√2Absolute value: 2√2Divide by 2: √2Yes, that seems correct.Alternatively, I could have noticed that the three points form an isosceles triangle with a base between A and C, and the height from B.The distance between A and C is the base.Compute distance AC:x_A = -3/4, y_A = -√2/2x_C = 5/4, y_C = -√2/2Distance AC = sqrt[(5/4 - (-3/4))² + (-√2/2 - (-√2/2))²] = sqrt[(8/4)² + 0] = sqrt[2²] = 2So, the base is 2 units.The height is the vertical distance from point B to the base AC.Since AC is a horizontal line at y = -√2/2, and point B is at (1/4, √2/2), the vertical distance is |√2/2 - (-√2/2)| = |√2| = √2Therefore, area = (base * height)/2 = (2 * √2)/2 = √2Same result. So, the area is indeed √2.**Problem (3): Find the analytical expression of the function h(x) whose graph is symmetric to the graph of f(x) about the line x = 1/4, and find the range of h(x) for x ∈ [-2/3, 1/3].**Alright, so h(x) is the reflection of f(x) over the line x = 1/4. So, for every point (a, b) on f(x), there is a corresponding point (c, b) on h(x) such that the line x = 1/4 is the perpendicular bisector of the segment joining (a, b) and (c, b).Wait, no, reflection over a vertical line x = a affects the x-coordinate, leaving the y-coordinate unchanged. So, if we reflect a point (x, y) over x = 1/4, the reflected point will be (2*(1/4) - x, y) = (1/2 - x, y).Therefore, for any x, h(x) = f(1/2 - x).Given that f(x) = sin(πx), then h(x) = sin(π*(1/2 - x)) = sin(π/2 - πx)Using the identity sin(π/2 - θ) = cosθ, so h(x) = cos(πx)Wait, is that correct? Let me verify.Yes, because reflecting over x = 1/4 would replace x with 1/2 - x, so f(1/2 - x) = sin(π*(1/2 - x)) = sin(π/2 - πx) = cos(πx). So, h(x) = cos(πx).Wait, but let me think again. If h(x) is the reflection of f(x) over x = 1/4, then for each x, h(x) = f(2*(1/4) - x) = f(1/2 - x). So, yes, h(x) = sin(π*(1/2 - x)) = sin(π/2 - πx) = cos(πx). So, h(x) = cos(πx).Wait, but that seems too straightforward. Let me confirm with an example. Suppose x = 0. Then, h(0) = cos(0) = 1. The reflection of (0, sin(0)) = (0, 0) over x = 1/4 would be (1/2, 0). But h(1/2) = cos(π*(1/2)) = 0, which matches. Wait, but h(0) = 1, which is the reflection of (1/2, 1) over x = 1/4. Wait, no, (1/2, 1) is a point on f(x). Reflecting it over x = 1/4 would give (0, 1), which is on h(x). So, yes, h(0) = 1, which is correct.Similarly, take x = 1/4. Then, h(1/4) = cos(π*(1/4)) = √2/2. The reflection of (1/4, sin(π*(1/4))) = (1/4, √2/2) over x = 1/4 is the same point, so h(1/4) = √2/2, which is correct.So, yes, h(x) = cos(πx).Now, I need to find the range of h(x) for x ∈ [-2/3, 1/3].Since h(x) = cos(πx), let's analyze cos(πx) over x ∈ [-2/3, 1/3].First, let's find the range of πx when x ∈ [-2/3, 1/3].πx ∈ [π*(-2/3), π*(1/3)] = [-2π/3, π/3]So, we need to find the range of cos(θ) where θ ∈ [-2π/3, π/3].The cosine function is even, so cos(-θ) = cosθ. Therefore, cos(θ) over [-2π/3, π/3] is the same as over [0, 2π/3] and [π/3, 0], but since it's even, it's symmetric around θ=0.Wait, actually, let's plot cos(θ) from θ = -2π/3 to θ = π/3.At θ = -2π/3, cos(-2π/3) = cos(2π/3) = -1/2At θ = 0, cos(0) = 1At θ = π/3, cos(π/3) = 1/2So, the maximum value of cos(θ) in this interval is 1 (at θ=0), and the minimum value is -1/2 (at θ=-2π/3).Therefore, the range of h(x) = cos(πx) over x ∈ [-2/3, 1/3] is [-1/2, 1].Wait, let me confirm:When x = -2/3, πx = -2π/3, cos(-2π/3) = -1/2When x = 1/3, πx = π/3, cos(π/3) = 1/2But wait, at x = 0, πx = 0, cos(0) = 1, which is the maximum.So, the function h(x) = cos(πx) reaches its maximum at x=0, which is 1, and its minimum at x=-2/3, which is -1/2.Therefore, the range is [-1/2, 1].Wait, but let me check if there's any point in between where the function might go lower than -1/2 or higher than 1. Since cos(πx) is continuous and differentiable, and we've checked the endpoints and the critical point at x=0, which gives the maximum. The minimum occurs at x=-2/3, which is -1/2.So, yes, the range is [-1/2, 1].Wait, but let me think again. The interval is x ∈ [-2/3, 1/3], so πx ∈ [-2π/3, π/3]. The cosine function on this interval starts at -1/2, increases to 1 at θ=0, then decreases to 1/2 at θ=π/3. So, the minimum is indeed -1/2, and the maximum is 1.Therefore, the range of h(x) over x ∈ [-2/3, 1/3] is [-1/2, 1].**Summary of Answers:**1. The minimum value of |PQ| is sqrt(17)/2.2. The area of the triangle is sqrt(2).3. The function h(x) is cos(πx), and its range over x ∈ [-2/3, 1/3] is [-1/2, 1].**Final Answer**(1) The minimum value of (|PQ|) is (boxed{dfrac{sqrt{17}}{2}}).(2) The area of the triangle is (boxed{sqrt{2}}).(3) The analytical expression of (h(x)) is (cos(pi x)), and the range of (h(x)) for (x in left[-dfrac{2}{3}, dfrac{1}{3}right]) is (boxed{left[-dfrac{1}{2}, 1right]}).

answer：First, I need to solve the equation (8^y - 8^{y-1} = 56) to find the value of (y).I can rewrite (8^{y-1}) as (frac{8^y}{8}), which simplifies the equation to:[8^y - frac{8^y}{8} = 56]Combining the terms on the left side gives:[frac{7}{8} cdot 8^y = 56]Solving for (8^y), I find:[8^y = 64]Since (8^2 = 64), it follows that (y = 2).Next, I need to calculate ((3y)^y). Substituting (y = 2) into the expression:[(3 cdot 2)^2 = 6^2 = 36]Therefore, the value of ((3y)^y) is 36.

answer：First, I need to determine the initial number of days the food was meant to last for 150 men. Let's denote this number of days as ( x ).The total amount of food available can be represented as ( 150 times x ).After 10 days, 25 men left the fort, leaving 125 men. The remaining food lasted for 42 days for these 125 men.The amount of food consumed by the 150 men in the first 10 days is ( 150 times 10 ).The remaining food after 10 days is ( 150 times x - 150 times 10 ).This remaining food is equal to the amount of food consumed by 125 men over 42 days, which is ( 125 times 42 ).Setting up the equation:[150x - 1500 = 5250]Solving for ( x ):[150x = 6750][x = frac{6750}{150}][x = 45]Therefore, the initial food provision was meant to last for 45 days.

answer：First, calculate the initial expenditure on sugar when the price is Rs. 2 per kg. If the person consumes ( X ) kg, the initial expenditure is ( 2X ) rupees.Next, when the price increases to Rs. 5 per kg, the new expenditure should remain the same to avoid an increase. Let ( Y ) be the new quantity consumed. The new expenditure is ( 5Y ) rupees.Set the new expenditure equal to the initial expenditure:[5Y = 2X]Solve for ( Y ):[Y = frac{2}{5}X]Determine the percentage reduction in consumption by comparing the change in quantity to the initial quantity:[text{Percentage Reduction} = left( frac{X - Y}{X} right) times 100 = left( frac{X - frac{2}{5}X}{X} right) times 100 = left( frac{3}{5} right) times 100 = 60%]

answer：Okay, so I have this problem here: the function is y equals a x cubed plus b x squared. When x is 1, the function has a maximum value of 3. I need to find the values of a and b, and then find the minimum value of the function y.Alright, let's start with part (1). I remember that to find maxima or minima, we need to use calculus, specifically the derivative. So, first, I should find the derivative of y with respect to x.The function is y = a x³ + b x². The derivative of this with respect to x is y' = 3a x² + 2b x. Okay, that seems right.Now, the problem says that when x = 1, the function has a maximum value of 3. So, two things here: at x = 1, the function value is 3, and the derivative at x = 1 should be zero because it's a maximum point.Let me write that down:1. y(1) = 32. y'(1) = 0So, substituting x = 1 into the function:y(1) = a(1)³ + b(1)² = a + b = 3.That's one equation: a + b = 3.Now, the derivative at x = 1:y'(1) = 3a(1)² + 2b(1) = 3a + 2b = 0.So, another equation: 3a + 2b = 0.Now, I have a system of two equations:1. a + b = 32. 3a + 2b = 0I need to solve for a and b. Let me use substitution or elimination. Maybe elimination is easier here.If I multiply the first equation by 2, I get:2a + 2b = 6.Now, subtract the second equation from this:(2a + 2b) - (3a + 2b) = 6 - 0Simplify:2a + 2b - 3a - 2b = 6Which simplifies to:- a = 6So, a = -6.Now, substitute a = -6 into the first equation:-6 + b = 3So, b = 3 + 6 = 9.Alright, so a is -6 and b is 9. Let me just double-check that.Substitute a and b back into the original function:y = -6x³ + 9x².At x = 1:y = -6(1) + 9(1) = -6 + 9 = 3. That's correct.Now, check the derivative:y' = 3*(-6)x² + 2*9x = -18x² + 18x.At x = 1:y' = -18(1) + 18(1) = -18 + 18 = 0. Perfect, that's a critical point.So, part (1) is done: a = -6, b = 9.Now, part (2): Find the minimum value of the function y.Hmm, so the function is y = -6x³ + 9x². I need to find its minimum value. Since it's a cubic function, it can have a local maximum and a local minimum. We already know that at x = 1, it's a local maximum. So, there must be another critical point which is a local minimum.To find the minimum, I need to find all critical points by setting the derivative equal to zero and then determine which one is the minimum.We already have the derivative:y' = -18x² + 18x.Set y' = 0:-18x² + 18x = 0.Factor out 18x:18x(-x + 1) = 0.So, 18x = 0 => x = 0, or -x + 1 = 0 => x = 1.So, critical points at x = 0 and x = 1.We already know that at x = 1, it's a local maximum. So, the other critical point at x = 0 must be a local minimum or a saddle point.Wait, but let me confirm whether x = 0 is a minimum or not.To do that, I can use the second derivative test.First, find the second derivative:y'' = derivative of y' = derivative of (-18x² + 18x) = -36x + 18.Now, evaluate y'' at x = 0:y''(0) = -36(0) + 18 = 18, which is positive. So, the function is concave up at x = 0, meaning it's a local minimum.Similarly, evaluate y'' at x = 1:y''(1) = -36(1) + 18 = -18, which is negative. So, concave down, meaning it's a local maximum. That's consistent with what we knew.So, the function has a local minimum at x = 0.Therefore, the minimum value of y is y(0).Compute y(0):y(0) = -6(0)³ + 9(0)² = 0 + 0 = 0.So, the minimum value is 0.But wait, let me think again. Is this the global minimum? Since it's a cubic function, as x approaches positive infinity, y approaches negative infinity because the leading term is -6x³. Similarly, as x approaches negative infinity, y approaches positive infinity. So, the function doesn't have a global minimum or maximum, but it does have a local minimum at x = 0.But the question says "Find the minimum value of the function y." So, maybe it's referring to the local minimum. So, the minimum value is 0.Alternatively, if it's asking for the global minimum, it might not exist because as x approaches infinity, y goes to negative infinity. But in the context of the problem, since we found a local minimum at x = 0, and the function does have a maximum at x = 1, perhaps the question is referring to the local minimum.Alternatively, maybe the function is considered over a certain interval, but the problem doesn't specify. It just says "the function y." So, perhaps it's expecting the local minimum.Therefore, the minimum value is 0.Wait, but just to make sure, let me analyze the behavior of the function.Given y = -6x³ + 9x², as x increases beyond 1, since the leading term is negative, the function will decrease to negative infinity. So, the function doesn't have a global minimum because it goes to negative infinity. However, it does have a local minimum at x = 0.So, depending on the interpretation, if the question is asking for the local minimum, it's 0. If it's asking for the global minimum, it doesn't exist. But since the function does have a local minimum, and the question is phrased as "the minimum value," it's likely referring to the local minimum.So, the minimum value is 0.Alternatively, maybe the function is being considered over the domain where it's defined, which is all real numbers, but as we saw, it goes to negative infinity. So, perhaps the question is expecting the local minimum.Alternatively, maybe I made a mistake in thinking that x = 0 is a local minimum. Let me double-check.We found that y''(0) = 18, which is positive, so it's a local minimum. So, yes, x = 0 is a local minimum, and the value is 0.Therefore, the minimum value is 0.So, summarizing:(1) a = -6, b = 9.(2) The minimum value of y is 0.I think that's it.

answer：First, I recognize that a three-digit number ( M ) with a ones digit of 5 can be expressed as ( M = 100a + 10b + 5 ), where ( a ) and ( b ) are digits. To determine if ( M ) is divisible by 4, I need to examine the last two digits of ( M ), which form the number ( 10b + 5 ).Next, I set up the congruence for divisibility by 4:[10b + 5 equiv 0 pmod{4}]Simplifying this, I find:[10b equiv -5 pmod{4} quad Rightarrow quad 10b equiv 3 pmod{4}]Since ( 10 equiv 2 pmod{4} ), the equation becomes:[2b equiv 3 pmod{4}]I then test each possible value of ( b ) from 0 to 9 to find which satisfy the congruence:- ( b = 0 ): ( 2 times 0 = 0 notequiv 3 pmod{4} )- ( b = 1 ): ( 2 times 1 = 2 notequiv 3 pmod{4} )- ( b = 2 ): ( 2 times 2 = 4 equiv 0 pmod{4} )- ( b = 3 ): ( 2 times 3 = 6 equiv 2 pmod{4} )- ( b = 4 ): ( 2 times 4 = 8 equiv 0 pmod{4} )- ( b = 5 ): ( 2 times 5 = 10 equiv 2 pmod{4} )- ( b = 6 ): ( 2 times 6 = 12 equiv 0 pmod{4} )- ( b = 7 ): ( 2 times 7 = 14 equiv 2 pmod{4} )- ( b = 8 ): ( 2 times 8 = 16 equiv 0 pmod{4} )- ( b = 9 ): ( 2 times 9 = 18 equiv 2 pmod{4} )Only ( b = 4 ) and ( b = 8 ) satisfy the condition ( 2b equiv 3 pmod{4} ). Therefore, there are 2 favorable outcomes out of 10 possible values for ( b ). The probability that ( M ) is divisible by 4 is:[frac{2}{10} = frac{1}{5}]