answer：Okay, so I have this function f(x) = sin(2x)cosφ + cos(2x)sinφ, and it's given that |φ| < π/2. The graph of this function is shifted to the right by π/3 units and is symmetric about the origin. I need to find the minimum value of f(x) on the interval [0, π/2]. The options are A: -√3/2, B: √3/2, C: -1/2, D: 1/2.First, I remember that sin(A + B) = sinA cosB + cosA sinB. So, looking at f(x), it seems like it can be rewritten using this identity. Let me try that:f(x) = sin(2x)cosφ + cos(2x)sinφ = sin(2x + φ).So, f(x) simplifies to sin(2x + φ). That makes things easier.Now, the graph of this function is shifted to the right by π/3 units. Shifting a function to the right by π/3 means replacing x with x - π/3. So, the shifted function becomes:f_shifted(x) = sin(2(x - π/3) + φ) = sin(2x - 2π/3 + φ).It's also given that this shifted graph is symmetric about the origin. Symmetric about the origin means it's an odd function. An odd function satisfies the condition f(-x) = -f(x). So, let's apply this condition to f_shifted(x):f_shifted(-x) = sin(-2x - 2π/3 + φ) = -sin(2x + 2π/3 - φ).For the function to be odd, f_shifted(-x) should equal -f_shifted(x). Let's write that:-sin(2x + 2π/3 - φ) = -sin(2x - 2π/3 + φ).Simplify both sides by multiplying both sides by -1:sin(2x + 2π/3 - φ) = sin(2x - 2π/3 + φ).Now, when do two sine functions equal each other? The general solution is that their arguments are either equal modulo 2π or supplementary angles modulo 2π. So:Case 1: 2x + 2π/3 - φ = 2x - 2π/3 + φ + 2πk, where k is an integer.Simplify this:2x + 2π/3 - φ = 2x - 2π/3 + φ + 2πk.Subtract 2x from both sides:2π/3 - φ = -2π/3 + φ + 2πk.Bring like terms together:2π/3 + 2π/3 = φ + φ + 2πk.4π/3 = 2φ + 2πk.Divide both sides by 2:2π/3 = φ + πk.So, φ = 2π/3 - πk.But we have |φ| < π/2, so let's find integer k such that φ is within this range.If k = 0: φ = 2π/3 ≈ 120 degrees, which is greater than π/2 (≈90 degrees). Not acceptable.If k = 1: φ = 2π/3 - π = -π/3 ≈ -60 degrees, which is within |φ| < π/2. So φ = -π/3.If k = 2: φ = 2π/3 - 2π = -4π/3, which is less than -π/2. Not acceptable.So, the only possible value is φ = -π/3.Case 2: 2x + 2π/3 - φ = π - (2x - 2π/3 + φ) + 2πk.Let me check this case too, just to be thorough.So:2x + 2π/3 - φ = π - 2x + 2π/3 - φ + 2πk.Simplify:2x + 2π/3 - φ = π - 2x + 2π/3 - φ + 2πk.Bring like terms together:2x + 2x = π + 2πk.4x = π + 2πk.x = π/4 + (π/2)k.But this must hold for all x, which is not possible unless the coefficients of x are equal on both sides. But on the left side, the coefficient is 2, and on the right side, it's -2. So, unless 2 = -2, which is not true, this case doesn't provide a solution. So, we can disregard this case.Therefore, the only valid solution is φ = -π/3.So, now, the function f(x) is sin(2x + φ) = sin(2x - π/3).Now, we need to find the minimum value of f(x) = sin(2x - π/3) on the interval [0, π/2].First, let's find the range of 2x - π/3 when x is in [0, π/2].When x = 0: 2(0) - π/3 = -π/3.When x = π/2: 2(π/2) - π/3 = π - π/3 = 2π/3.So, the argument of the sine function, 2x - π/3, ranges from -π/3 to 2π/3.We need to find the minimum value of sinθ where θ ∈ [-π/3, 2π/3].The sine function reaches its minimum at θ = -π/2, but in our interval, the lowest θ is -π/3, which is greater than -π/2. So, the minimum will occur either at θ = -π/3 or at the critical points within the interval.Let's find the critical points by taking the derivative of f(x):f'(x) = 2cos(2x - π/3).Set f'(x) = 0:2cos(2x - π/3) = 0 ⇒ cos(2x - π/3) = 0.Solutions are:2x - π/3 = π/2 + πk, where k is integer.So,2x = π/2 + π/3 + πk = (3π/6 + 2π/6) + πk = 5π/6 + πk.Thus,x = 5π/12 + (π/2)k.Now, find x in [0, π/2]:For k = 0: x = 5π/12 ≈ 1.308 radians, which is less than π/2 ≈ 1.5708. So, x = 5π/12 is within the interval.For k = 1: x = 5π/12 + π/2 = 5π/12 + 6π/12 = 11π/12 ≈ 2.879, which is greater than π/2. So, not in the interval.For k = -1: x = 5π/12 - π/2 = 5π/12 - 6π/12 = -π/12, which is negative. Not in the interval.So, the only critical point in [0, π/2] is x = 5π/12.Now, evaluate f(x) at the endpoints and at the critical point:1. x = 0:f(0) = sin(2*0 - π/3) = sin(-π/3) = -√3/2 ≈ -0.866.2. x = π/2:f(π/2) = sin(2*(π/2) - π/3) = sin(π - π/3) = sin(2π/3) = √3/2 ≈ 0.866.3. x = 5π/12:f(5π/12) = sin(2*(5π/12) - π/3) = sin(5π/6 - π/3) = sin(5π/6 - 2π/6) = sin(3π/6) = sin(π/2) = 1.So, the function reaches a maximum of 1 at x = 5π/12 and has values of -√3/2 and √3/2 at the endpoints.Therefore, the minimum value on [0, π/2] is -√3/2.Wait, but let me double-check. The sine function is increasing from -π/3 to π/2, right? Because the derivative is positive there. So, from x = 0 to x = 5π/12, the function increases from -√3/2 to 1, and then from x = 5π/12 to x = π/2, it decreases back to √3/2. So, the minimum is indeed at x = 0, which is -√3/2.So, the answer should be A: -√3/2.**Final Answer**boxed{A}

answer：Okay, so I have this problem about propositions related to π and some rational numbers a, b, c, d. The proposition p is: If aπ + b = cπ + d, then a = c and b = d. First, I need to understand what this proposition is saying. It seems like it's saying that if two expressions involving π and some constants are equal, then the coefficients of π and the constants themselves must be equal. That makes sense because π is an irrational number, and if you have a rational number times π plus another rational number equaling another such expression, the only way that can happen is if the coefficients of π and the constants are the same. So, I think this proposition is true.Now, part (1) asks for the negation of proposition p and to judge its truthfulness. The negation of an implication "If P, then Q" is "P and not Q." So, in this case, the negation would be: "aπ + b = cπ + d and (a ≠ c or b ≠ d)." So, the negation is saying that there exist rational numbers a, b, c, d such that aπ + b equals cπ + d, but either a is not equal to c or b is not equal to d. Is this possible? Well, since π is irrational, the only way aπ + b can equal cπ + d is if a = c and b = d. Otherwise, you would have a non-zero multiple of π equaling a rational number, which is impossible because π is irrational. So, the negation would be false because such a case cannot happen. Therefore, the negation is false.Moving on to part (2), I need to write the converse, inverse, and contrapositive of proposition p and judge their truthfulness.Starting with the converse. The converse of "If P, then Q" is "If Q, then P." So, in this case, the converse would be: "If a = c and b = d, then aπ + b = cπ + d." That seems straightforward. If a equals c and b equals d, then substituting them into the expressions would naturally make the expressions equal. So, the converse is true.Next, the inverse. The inverse of "If P, then Q" is "If not P, then not Q." So, the inverse here would be: "If aπ + b ≠ cπ + d, then a ≠ c or b ≠ d." Is this true? Well, if the expressions are not equal, then it must be that either the coefficients of π are different or the constants are different. Since π is irrational, you can't have the expressions unequal without at least one of the coefficients or constants being different. So, the inverse is also true.Finally, the contrapositive. The contrapositive of "If P, then Q" is "If not Q, then not P." So, the contrapositive here is: "If a ≠ c or b ≠ d, then aπ + b ≠ cπ + d." This is essentially the same as the inverse, but phrased differently. Since if either a ≠ c or b ≠ d, then the expressions involving π would not be equal because π is irrational. Therefore, the contrapositive is true.Wait, let me double-check. The contrapositive is logically equivalent to the original proposition, so since the original proposition is true, the contrapositive must also be true. Similarly, the inverse is logically equivalent to the converse, so since the converse is true, the inverse is also true. The negation, as we saw, is false because it's impossible for aπ + b to equal cπ + d without a = c and b = d.So, summarizing:1. The negation is false.2. The converse, inverse, and contrapositive are all true.I think that's it. I don't see any mistakes in my reasoning, but maybe I should think about specific examples to confirm.For the negation: Suppose a = 1, b = 2, c = 1, d = 2. Then aπ + b = cπ + d, and a = c, b = d. If I try to find a case where aπ + b = cπ + d but a ≠ c or b ≠ d, I can't because π is irrational. So, no such example exists, meaning the negation is false.For the converse: If a = c and b = d, then obviously aπ + b = cπ + d. Simple substitution.For the inverse: If aπ + b ≠ cπ + d, then either a ≠ c or b ≠ d. Suppose a = c but b ≠ d. Then aπ + b would not equal cπ + d because b ≠ d. Similarly, if a ≠ c, then even if b = d, aπ + b would not equal cπ + d because the coefficients of π are different. So, the inverse holds.For the contrapositive: If a ≠ c or b ≠ d, then aπ + b ≠ cπ + d. This is the same as the inverse, so it's true.Yeah, I think I'm confident with these conclusions.

answer：To simplify the expression (1 - (2 - (3 - (4 - (5 - y))))), I'll start by addressing the innermost parentheses.First, I'll simplify (4 - (5 - y)):[4 - (5 - y) = 4 - 5 + y = y - 1]Next, I'll substitute this result into the next layer:[3 - (y - 1) = 3 - y + 1 = 4 - y]Then, I'll substitute this into the following layer:[2 - (4 - y) = 2 - 4 + y = y - 2]Finally, I'll substitute this into the outermost expression:[1 - (y - 2) = 1 - y + 2 = 3 - y]Therefore, the simplified form of the expression is (3 - y).

answer：Okay, so I have this problem here with two propositions, p and q. Proposition p is 2^x < 2^y, and proposition q is log base 2 of x < log base 2 of y. The question is asking how proposition p relates to proposition q in terms of being a sufficient or necessary condition. The options are:A: A sufficient but not necessary condition B: A necessary but not sufficient condition C: A sufficient and necessary condition D: Neither sufficient nor necessary conditionAlright, let me try to break this down step by step. I need to figure out whether p is a sufficient condition for q, a necessary condition, both, or neither. First, let's recall what sufficient and necessary conditions mean. A sufficient condition means that if p is true, then q must be true. But it doesn't necessarily mean that q can only be true if p is true. On the other hand, a necessary condition means that for q to be true, p must be true, but p being true doesn't guarantee q is true. If p is both sufficient and necessary, then p and q imply each other. If p is neither, then p doesn't guarantee q, and q doesn't require p.So, starting with proposition p: 2^x < 2^y. Since the base 2 is greater than 1, the exponential function is increasing. That means if 2^x < 2^y, then x must be less than y. So, p simplifies to x < y.Now, proposition q is log base 2 of x < log base 2 of y. The logarithm base 2 is also an increasing function, but it's only defined for positive numbers. So, for log base 2 of x and log base 2 of y to be defined, x and y must be greater than 0. Therefore, if log base 2 of x < log base 2 of y, then x < y, but with the added condition that x and y are positive.So, from p, we get x < y without any restrictions on x and y, whereas from q, we get x < y with the additional constraints that x and y are positive.Now, let's see how p relates to q. If p is true (x < y), does that necessarily make q true? Well, not always. Because if x and y are negative, p could still be true (since x < y), but q wouldn't make sense because log of a negative number isn't defined. So, p being true doesn't guarantee q is true because q requires x and y to be positive. Therefore, p is not a sufficient condition for q.But is p a necessary condition for q? Let's see. If q is true, meaning log base 2 of x < log base 2 of y, then we know that x < y. So, for q to be true, p must be true. Therefore, p is a necessary condition for q.So, putting it all together, p is a necessary condition for q because q can't be true without p being true. However, p alone isn't enough to ensure q is true because p doesn't account for the positivity of x and y, which is required for q. Therefore, p is a necessary but not sufficient condition for q.Let me double-check to make sure I haven't missed anything. If p is true, does it sometimes lead to q being true? Yes, if x and y are positive, then p being true would make q true. But since p doesn't ensure x and y are positive, it's not always sufficient. On the flip side, if q is true, then p must be true because the logarithm inequality directly implies x < y. So, p is necessary for q.Is there any scenario where q is true but p is false? No, because q implies x < y, so p must be true. Is there a scenario where p is true but q is false? Yes, if x or y are negative, p could be true (x < y) but q wouldn't hold because the logarithm isn't defined. Therefore, p is necessary but not sufficient.I think that makes sense. So, the answer should be option B: A necessary but not sufficient condition.

answer：First, I need to determine how many apples Jack's basket can hold when it is full. Jack's basket is full at 12 apples and currently has space for 4 more apples. This means Jack's basket currently has 12 - 4 = 8 apples.Next, since Jill's basket can hold twice as much as Jack's basket when both are full, Jill's basket can hold 2 * 12 = 24 apples when it is full.Each apple weighs 150 grams, so the total weight of apples in Jill's basket when it is full would be 24 apples * 150 grams/apple = 3600 grams.Therefore, the total weight of apples in Jill's basket when it is full is 3600 grams.

answer：Okay, so I have this problem where there are several points, A1 through An, and each pair of these points is more than 2 units apart. I need to prove that any shape with an area less than π can be moved by a vector of length at most 1 so that it doesn't contain any of these points. Hmm, interesting.First, let me try to visualize this. If the points are all more than 2 units apart, that means they're pretty spread out. If I have a shape with a small area, less than π, which is the area of a circle with radius 1, maybe I can shift this shape so it doesn't cover any of these points.Wait, the area of the shape is less than π. That makes me think of circles because the area of a circle is πr². So if the area is less than π, maybe the shape is smaller than a circle with radius 1? Or maybe not necessarily a circle, but something with a similar area.The problem says I can shift the shape by a vector of length no more than 1. So, moving it by at most 1 unit in some direction. I need to ensure that after this shift, none of the points A1 through An are inside the shape.Let me think about the circles around each point. If each point is more than 2 units apart, then if I draw a circle of radius 1 around each point, these circles won't overlap. Because if two points are more than 2 units apart, their circles of radius 1 won't intersect. That seems right.So, if I have these non-overlapping circles around each point, and my shape has an area less than π, which is the area of one of these circles. Maybe I can use some kind of area argument here.If I consider the intersection of the shape with each of these circles, each intersection would be a part of the shape that's near each point. Since the circles don't overlap, these intersections are all separate.The total area of the shape is less than π, so the sum of the areas of these intersections must also be less than π. But each circle has an area of π, so if the total area of the intersections is less than π, that means that not all of the shape is covering these circles.Wait, maybe I can translate the shape so that none of these intersections are covering the points. Since the shape can be moved by at most 1 unit, maybe I can move it so that each intersection is moved away from the corresponding point.But how exactly? Maybe if I move the shape so that each intersection is moved into a common area, and since the total area is less than π, there must be some space left where none of the intersections are.Let me think step by step.1. **Define the problem and elements:** - Points A1, A2, ..., An with pairwise distances > 2. - Figure Φ with area < π. - Need to show Φ can be shifted by a vector of length ≤ 1 so that it doesn't contain any Ai.2. **Introduce circles around each point:** - Circles Si with radius 1 around each Ai. - Since pairwise distances > 2, circles Si are disjoint.3. **Consider intersections of Φ with each Si:** - Let Vi = Φ ∩ Si. - Since Si are disjoint, Vi are disjoint. - Total area of Vi is less than area of Φ, which is < π.4. **Move each Vi to a common point:** - Choose a point O. - Translate each Vi to O, resulting in regions Wi. - Each Wi is within a circle T of radius 1 around O. - Total area of Wi is less than π.5. **Existence of a point in T not covered by Wi:** - Since total area of Wi < π, there exists a point B in T not in any Wi. - Translate Φ by vector BO, which has length ≤ 1.6. **Conclusion:** - After translation, Φ doesn't contain any Ai.Wait, that seems a bit abstract. Let me try to make it more concrete.Imagine each point Ai has a circle of radius 1 around it. Since the points are more than 2 units apart, these circles don't overlap. Now, the shape Φ has an area less than π, so it's smaller than a circle of radius 1.If I look at where Φ overlaps with each of these circles, each overlap is a part of Φ near each Ai. Since the circles don't overlap, these overlaps are all separate.The total area of Φ is less than π, so the sum of the areas of these overlaps is also less than π. Now, if I move each of these overlaps to a common point O, the total area covered by all these moved overlaps is still less than π. Since the area of a circle of radius 1 is π, there must be some space left in this common circle where none of the overlaps are.So, if I move Φ by a vector that takes each overlap to this empty space, the entire Φ will be shifted by at most 1 unit, and none of the points Ai will be inside Φ anymore.I think that makes sense. The key idea is that since the overlaps are small and the circles are disjoint, we can find a common spot to move the shape so it avoids all the points.But let me double-check if moving each overlap individually is the same as moving the entire shape. Yes, because translating the entire shape by a vector affects all parts of it equally. So, by choosing the right translation, we can ensure that none of the overlaps end up covering the original points.Another thing to consider is whether the translation vector can always be found. Since the total area of the overlaps is less than π, and the target circle has area π, there must be some point in the target circle that isn't covered by any of the translated overlaps. That point gives us the translation vector needed.I think this approach works. It uses the fact that the circles are disjoint and the area of the shape is limited to ensure that a suitable translation exists.**Final Answer**boxed{text{Such a shift is always possible.}}