answer：Okay, so I have this problem about a regular 26-gon, and I need to prove that if I pick any 9 vertices, there must be three that form an isosceles triangle. Hmm, interesting. Let me try to break this down step by step.First, I know that a regular 26-gon has all its sides equal and all its internal angles equal. So, each vertex is equally spaced around a circle. That means the distance between any two vertices can be measured by the number of edges between them along the circumference. Now, an isosceles triangle has two sides of equal length. In the context of a regular polygon, this would mean that two of the sides of the triangle correspond to the same number of edges between the vertices. So, if I can find three vertices where two of the distances between them are equal, that would form an isosceles triangle.The problem is asking me to prove that no matter which 9 vertices I choose from the 26, there will always be at least one such trio that forms an isosceles triangle. That sounds like a problem where the pigeonhole principle might come into play. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Maybe I can apply this idea here.Let me think about how to model this. Each vertex can be thought of as a point on a circle, equally spaced. If I number the vertices from 0 to 25, then the distance between two vertices can be represented by the difference in their numbers modulo 26. For example, the distance between vertex 0 and vertex 5 is 5, and the distance between vertex 25 and vertex 1 is 2 (since 1 - 25 = -24, which is equivalent to 2 modulo 26).So, if I pick any three vertices, say a, b, and c, the distances between them would be |a - b|, |b - c|, and |c - a|, all modulo 26. For these to form an isosceles triangle, at least two of these distances must be equal.Now, if I have 9 vertices, I need to ensure that among all the possible distances between them, at least two are equal. But wait, how many distances are there? For 9 vertices, the number of pairs is C(9,2) = 36. Each distance can be from 1 to 13 because beyond 13, the distances start repeating in the opposite direction (since 14 is equivalent to 12 modulo 26, 15 equivalent to 11, and so on up to 25 equivalent to 1). So, there are 13 possible distinct distances.If I have 36 distances and only 13 possible distinct values, by the pigeonhole principle, at least one distance must repeat at least ceiling(36/13) = 3 times. Wait, actually, 36 divided by 13 is about 2.769, so ceiling of that is 3. So, at least one distance must occur at least 3 times.But does that necessarily mean that there are three vertices forming an isosceles triangle? Hmm, not exactly. Because having three pairs with the same distance doesn't necessarily mean they share a common vertex. For example, if I have three pairs (a,b), (c,d), (e,f) all with the same distance, but none of these pairs share a common vertex, then they don't form a triangle.So, maybe I need a different approach. Instead of looking at all pairs, perhaps I should consider the distances from a single vertex. Let's fix a vertex, say vertex 0. Then, the distances from vertex 0 to the other 8 vertices can be from 1 to 12 (since 13 would be the vertex directly opposite, but since we have 26 vertices, 13 is the midpoint). Wait, no, 26 is even, so the distance from 0 to 13 is 13, which is the diameter.But in our case, we have 9 vertices, so one of them is vertex 0, and the other 8 are somewhere else. The distances from 0 to these 8 vertices can be from 1 to 12, but since we have 8 distances and 12 possible distances, it's not immediately clear.Wait, maybe I should consider the concept of arithmetic progressions. In a regular polygon, an isosceles triangle can be thought of as three vertices that form an arithmetic progression in terms of their positions. For example, if I have vertices at positions a, a + d, and a + 2d modulo 26, then the distances between a and a + d, and between a + d and a + 2d are both d, forming an isosceles triangle.So, if I can show that among any 9 vertices, there must be three that form an arithmetic progression, then I have shown that there is an isosceles triangle.Now, how do I show that? This seems related to the Erdős–Szemerédi theorem or maybe van der Waerden's theorem, but I'm not sure. Maybe I can use the pigeonhole principle again.Let me think about the residues modulo something. Since 26 is 2 times 13, maybe I can consider the residues modulo 2 and modulo 13 separately.If I take any 9 vertices, by the pigeonhole principle, at least 5 of them must have the same parity, i.e., they are all even or all odd. Because there are only two parity classes, even and odd, so 9 vertices would mean at least ceiling(9/2) = 5 in one class.So, suppose I have 5 vertices that are all even. Then, if I divide their positions by 2, they become positions in a regular 13-gon. Similarly, if they are all odd, subtracting 1 and dividing by 2 would also give positions in a regular 13-gon.Therefore, the problem reduces to showing that in any 5 vertices of a regular 13-gon, there must be three that form an arithmetic progression. If I can prove that, then I can extend it back to the 26-gon.So, let's focus on the 13-gon. I need to show that any 5 vertices contain an arithmetic progression of length 3. Hmm, this seems more manageable.I recall that in additive combinatorics, there's a result called Szemerédi's theorem, which states that any subset of the integers with positive density contains arbitrarily long arithmetic progressions. But 13 is a small number, so maybe I can use a more elementary approach.Alternatively, maybe I can use the pigeonhole principle again. Let's consider the differences between the selected vertices. If I have 5 vertices, the number of differences is C(5,2) = 10. The possible differences in a 13-gon are from 1 to 6 (since beyond 6, the differences start repeating in the opposite direction). So, there are 6 possible differences.By the pigeonhole principle, with 10 differences and 6 possible values, at least one difference must occur at least ceiling(10/6) = 2 times. So, there are at least two pairs of vertices with the same difference.But again, similar to before, having two pairs with the same difference doesn't necessarily mean they form an arithmetic progression. They could be overlapping or non-overlapping.Wait, if I have two pairs with the same difference, say (a, a + d) and (b, b + d), then if a + d = b, then we have an arithmetic progression a, b, b + d. Similarly, if b + d = a, then we have b, a, a + d. Otherwise, if a, a + d, b, b + d are all distinct, then we might not have an arithmetic progression.Hmm, so maybe this approach isn't sufficient. Perhaps I need a different strategy.Let me think about the specific structure of the 13-gon. Since 13 is a prime number, it has nice properties in modular arithmetic. Maybe I can use this to my advantage.Suppose I have 5 vertices in the 13-gon. Let me label them as x1, x2, x3, x4, x5. Without loss of generality, I can assume they are in increasing order. Now, consider the differences between consecutive vertices: x2 - x1, x3 - x2, x4 - x3, x5 - x4. These differences must sum up to x5 - x1.Since we're working modulo 13, the total sum of differences is x5 - x1 modulo 13. If any of these differences is repeated, then we might be able to find an arithmetic progression.Alternatively, maybe I can use the concept of graph theory. If I represent each vertex as a node and connect them if their difference is the same, then finding an arithmetic progression is equivalent to finding a path of length 2 with the same edge labels.But I'm not sure if this helps directly. Maybe I need to think about it differently.Wait, another idea: in a 13-gon, the number of possible arithmetic progressions of length 3 is quite large. Maybe I can count the number of possible arithmetic progressions and compare it to the number of possible triples in the selected 5 vertices.The number of arithmetic progressions in a 13-gon is 13 * 6 = 78, since for each starting point, there are 6 possible common differences (1 to 6). The number of triples in 5 vertices is C(5,3) = 10. So, if I can show that these 10 triples must intersect with the 78 arithmetic progressions, then perhaps one of them must be an arithmetic progression.But this seems too vague. I need a more concrete approach.Let me try a different angle. Suppose I have 5 vertices in the 13-gon. I want to show that among them, three form an arithmetic progression. Let's assume the contrary: that no three form an arithmetic progression. Then, for any two vertices, the midpoint (which would complete the arithmetic progression) is not among the selected vertices.So, if I pick any two vertices, the point that would make an arithmetic progression with them is excluded. How many such excluded points would there be?For each pair of vertices, there is exactly one midpoint. So, with 5 vertices, the number of pairs is C(5,2) = 10. Therefore, there are 10 midpoints that must be excluded. But since we're working in a 13-gon, there are only 13 points in total. If we have 5 selected points, the number of excluded points is 13 - 5 = 8. But we have 10 midpoints that need to be excluded, which is more than the number of available excluded points. This is a contradiction because we can't exclude more points than we have.Therefore, our assumption that no three vertices form an arithmetic progression must be false. Hence, among any 5 vertices in a 13-gon, there must be three that form an arithmetic progression.Okay, so that works for the 13-gon. Now, going back to the original problem with the 26-gon. We had 9 vertices, which by the pigeonhole principle, at least 5 must be of the same parity. These 5 can be mapped to a 13-gon, and as we just proved, among these 5, there must be three forming an arithmetic progression. Translating back to the 26-gon, these three vertices will form an isosceles triangle.Therefore, we've shown that in any selection of 9 vertices from a regular 26-gon, there must be three that form an isosceles triangle.I think that makes sense. Let me just recap to make sure I didn't miss anything.1. Start with 9 vertices in a 26-gon.2. By pigeonhole, at least 5 have the same parity.3. These 5 correspond to 5 vertices in a 13-gon.4. In the 13-gon, any 5 vertices must contain an arithmetic progression of length 3.5. This arithmetic progression translates back to an isosceles triangle in the 26-gon.Yes, that seems solid. I don't see any gaps in the logic. The key was reducing the problem to the 13-gon and using the pigeonhole principle there to force an arithmetic progression, which then gives the isosceles triangle.**Final Answer**boxed{9}

answer：Okay, so I have this complex number z, which is equal to (1 - i) divided by (1 + i). I need to find the conjugate of z. Hmm, complex numbers can sometimes be tricky, especially when they're in the form of fractions like this. Let me think about how to approach this step by step.First, I remember that to simplify expressions like this, where there's a complex number in the denominator, it's helpful to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of a complex number a + bi is a - bi, right? So, in this case, the denominator is (1 + i), so its conjugate should be (1 - i). That makes sense.So, if I multiply both the numerator and the denominator by (1 - i), I can eliminate the imaginary unit from the denominator. Let me write that out:z = (1 - i)/(1 + i) * (1 - i)/(1 - i)Wait, so that's multiplying both the top and the bottom by (1 - i). Okay, so the numerator becomes (1 - i) squared, and the denominator becomes (1 + i)(1 - i). Let me compute each part separately.Starting with the numerator: (1 - i)^2. I know that (a - b)^2 is a^2 - 2ab + b^2, so applying that here:(1)^2 - 2*(1)*(i) + (i)^2 = 1 - 2i + i^2But wait, i^2 is equal to -1, right? So substituting that in:1 - 2i + (-1) = 1 - 2i - 1 = (1 - 1) - 2i = 0 - 2i = -2iOkay, so the numerator simplifies to -2i. That wasn't too bad.Now, the denominator: (1 + i)(1 - i). I recall that when you multiply a binomial by its conjugate, it's a difference of squares, so (a + b)(a - b) = a^2 - b^2. Applying that here:(1)^2 - (i)^2 = 1 - i^2Again, i^2 is -1, so:1 - (-1) = 1 + 1 = 2So the denominator simplifies to 2.Putting it all together, z = (-2i)/2 = -i.Wait, so z is equal to -i? That seems straightforward. So z is a purely imaginary number with no real part. Now, I need to find the conjugate of z.I remember that the conjugate of a complex number a + bi is a - bi. But in this case, z is -i, which can be written as 0 - i. So, applying the conjugate formula:Conjugate of z = 0 + i = iSo, the conjugate of z is i.Let me just double-check my steps to make sure I didn't make any mistakes. Starting with z = (1 - i)/(1 + i), I multiplied numerator and denominator by (1 - i), which gave me (-2i)/2 = -i. Then, taking the conjugate of -i, which is i. That seems correct.Alternatively, I could have found the conjugate before simplifying z. Let me try that approach to see if I get the same result.The conjugate of z is the conjugate of (1 - i)/(1 + i). I know that the conjugate of a quotient is the quotient of the conjugates, so:Conjugate(z) = conjugate(1 - i)/conjugate(1 + i) = (1 + i)/(1 - i)Now, let's simplify this expression. Again, I'll multiply numerator and denominator by the conjugate of the denominator, which is (1 + i):(1 + i)/(1 - i) * (1 + i)/(1 + i) = (1 + i)^2 / (1 - i)(1 + i)Calculating the numerator: (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2iDenominator: (1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 2So, Conjugate(z) = 2i/2 = iOkay, that's the same result as before. So whether I simplify z first and then take the conjugate, or take the conjugate first and then simplify, I end up with the same answer, which is i. That gives me more confidence that i is indeed the correct conjugate of z.Just to recap, the steps I took were:1. Recognize that z is a complex number in the form of a fraction.2. Multiply numerator and denominator by the conjugate of the denominator to simplify z.3. Simplify both the numerator and denominator separately.4. Find the conjugate of the simplified z.5. Alternatively, take the conjugate of the original expression and simplify it to verify the result.Both methods led me to the same conclusion, so I feel confident that the conjugate of z is i.I also remember that another way to find the conjugate is to reflect the complex number over the real axis on the complex plane. Since z is -i, which is located on the negative imaginary axis, its conjugate should be its mirror image over the real axis, which would be i. That visual makes sense too.Another thing I can do to verify is to represent z in polar form and then find its conjugate. Let's see.First, z = -i. In polar form, a complex number is represented as r*(cosθ + i sinθ), where r is the modulus and θ is the argument.For z = -i, the modulus r is sqrt(0^2 + (-1)^2) = 1.The argument θ is the angle from the positive real axis to the point. Since -i is on the negative imaginary axis, the angle is -π/2 or 3π/2 radians.So, z can be written as 1*(cos(-π/2) + i sin(-π/2)).The conjugate of a complex number in polar form is r*(cosθ - i sinθ). So, substituting θ = -π/2:Conjugate(z) = 1*(cos(-π/2) - i sin(-π/2)) = cos(-π/2) - i sin(-π/2)But cos(-π/2) is 0, and sin(-π/2) is -1. So:Conjugate(z) = 0 - i*(-1) = 0 + i = iAgain, the same result. This method also confirms that the conjugate is i.I think I'm pretty sure now that the conjugate of z is i. It's consistent across multiple methods: simplifying first, taking the conjugate first, and using polar form. All approaches lead to the same conclusion.Just to ensure I haven't overlooked anything, let me consider if there's another perspective or method to confirm this. Maybe using Euler's formula?Euler's formula states that e^(iθ) = cosθ + i sinθ. For z = -i, which is e^(-iπ/2), the conjugate would be e^(iπ/2), which is i. So, yes, that also matches.Therefore, through algebraic manipulation, polar form, and Euler's formula, I consistently find that the conjugate of z is i. I think that's thorough enough to be confident in the answer.Another way to think about it is that if z = -i, then its conjugate should have the opposite sign for the imaginary part. Since z is purely imaginary, the conjugate is just the negative of z. So, conjugate(z) = -(-i) = i. That's a simpler way to see it once you've simplified z.In summary, whether I approach it by simplifying first, taking the conjugate first, using polar coordinates, or applying Euler's formula, all methods confirm that the conjugate of z is i. I don't see any errors in my reasoning, and all the cross-checks align. So, I'm pretty certain that the answer is correct.

answer：Alright, so I have this problem here with three parts. Let me try to tackle each part step by step. I need to make sure I understand each part before moving on to the next.**Problem 1: Find the expression for ( f(x) ).**Okay, so the function is given as ( f(log_{2}x) = ax^{2} - 2x + 1 - a ). I need to express ( f(x) ) in terms of ( x ), not in terms of ( log_{2}x ).Hmm, substitution seems like the way to go here. If I let ( t = log_{2}x ), then by definition, ( x = 2^{t} ). So, substituting ( x ) with ( 2^{t} ) in the given function should give me ( f(t) ).Let me write that out:( f(t) = a(2^{t})^{2} - 2(2^{t}) + 1 - a ).Simplify that:( f(t) = a cdot 2^{2t} - 2 cdot 2^{t} + 1 - a ).Since ( t ) is just a placeholder variable, I can replace ( t ) with ( x ) to get ( f(x) ):( f(x) = a cdot 2^{2x} - 2 cdot 2^{x} + 1 - a ).Wait, that looks good. So, ( f(x) = a cdot 4^{x} - 2 cdot 2^{x} + 1 - a ). Yeah, since ( 2^{2x} = 4^{x} ).So, I think that's the expression for ( f(x) ).**Problem 2: Solve the equation ( f(x) = (a - 1) cdot 4^{x} ) with respect to ( x ).**Alright, so I have ( f(x) = (a - 1) cdot 4^{x} ). From part 1, I know that ( f(x) = a cdot 4^{x} - 2 cdot 2^{x} + 1 - a ). So, setting them equal:( a cdot 4^{x} - 2 cdot 2^{x} + 1 - a = (a - 1) cdot 4^{x} ).Let me bring all terms to one side:( a cdot 4^{x} - 2 cdot 2^{x} + 1 - a - (a - 1) cdot 4^{x} = 0 ).Simplify the terms with ( 4^{x} ):( [a - (a - 1)] cdot 4^{x} - 2 cdot 2^{x} + 1 - a = 0 ).Calculating ( a - (a - 1) ):( a - a + 1 = 1 ).So, the equation becomes:( 4^{x} - 2 cdot 2^{x} + 1 - a = 0 ).Hmm, this looks like a quadratic in terms of ( 2^{x} ). Let me set ( y = 2^{x} ). Then, ( 4^{x} = (2^{2})^{x} = 2^{2x} = y^{2} ).Substituting into the equation:( y^{2} - 2y + 1 - a = 0 ).So, ( y^{2} - 2y + (1 - a) = 0 ).This is a quadratic equation in ( y ). Let's solve for ( y ):Using the quadratic formula:( y = frac{2 pm sqrt{(2)^2 - 4 cdot 1 cdot (1 - a)}}{2 cdot 1} ).Simplify the discriminant:( sqrt{4 - 4(1 - a)} = sqrt{4 - 4 + 4a} = sqrt{4a} = 2sqrt{a} ).So, the solutions are:( y = frac{2 pm 2sqrt{a}}{2} = 1 pm sqrt{a} ).So, ( y = 1 + sqrt{a} ) or ( y = 1 - sqrt{a} ).But remember, ( y = 2^{x} ), which is always positive. So, we need ( 1 + sqrt{a} > 0 ) and ( 1 - sqrt{a} > 0 ).Since ( sqrt{a} ) is non-negative, ( 1 + sqrt{a} ) is always positive. For ( 1 - sqrt{a} ), we need ( 1 - sqrt{a} > 0 ), which implies ( sqrt{a} < 1 ), so ( a < 1 ).So, depending on the value of ( a ), we have different solutions.Case 1: ( a < 0 ).Wait, if ( a < 0 ), then ( sqrt{a} ) is not real. So, in this case, the equation ( y^{2} - 2y + 1 - a = 0 ) would have no real solutions because the discriminant would be negative. Therefore, no solution for ( x ) when ( a < 0 ).Case 2: ( a = 0 ).Then, the equation becomes ( y^{2} - 2y + 1 = 0 ), which factors to ( (y - 1)^2 = 0 ). So, ( y = 1 ), which means ( 2^{x} = 1 ), so ( x = 0 ).Case 3: ( 0 < a < 1 ).Here, ( sqrt{a} ) is real and less than 1. So, both ( y = 1 + sqrt{a} ) and ( y = 1 - sqrt{a} ) are positive. Thus, we have two solutions:( 2^{x} = 1 + sqrt{a} ) => ( x = log_{2}(1 + sqrt{a}) ).( 2^{x} = 1 - sqrt{a} ) => ( x = log_{2}(1 - sqrt{a}) ).Case 4: ( a = 1 ).Then, the equation becomes ( y^{2} - 2y = 0 ), which factors to ( y(y - 2) = 0 ). So, ( y = 0 ) or ( y = 2 ). But ( y = 2^{x} ) can't be zero, so only ( y = 2 ). Thus, ( x = log_{2}2 = 1 ).Case 5: ( a > 1 ).Here, ( sqrt{a} > 1 ), so ( 1 - sqrt{a} ) is negative. Therefore, only ( y = 1 + sqrt{a} ) is valid. So, ( x = log_{2}(1 + sqrt{a}) ).So, summarizing:- If ( a < 0 ): No solution.- If ( a = 0 ): ( x = 0 ).- If ( 0 < a < 1 ): Two solutions, ( x = log_{2}(1 + sqrt{a}) ) and ( x = log_{2}(1 - sqrt{a}) ).- If ( a = 1 ): One solution, ( x = 1 ).- If ( a > 1 ): One solution, ( x = log_{2}(1 + sqrt{a}) ).I think that covers all cases.**Problem 3: Let ( h(x) = 2^{-x}f(x) ). If ( a geq frac{1}{2} ), then for any ( x_{1}, x_{2} in [-1, 1] ), it always holds that ( |h(x_{1}) - h(x_{2})| leq frac{a + 1}{2} ). Find the range of ( a ).**Alright, this seems more involved. Let me break it down.First, ( h(x) = 2^{-x}f(x) ). From part 1, ( f(x) = a cdot 4^{x} - 2 cdot 2^{x} + 1 - a ). So, let's compute ( h(x) ):( h(x) = 2^{-x} cdot (a cdot 4^{x} - 2 cdot 2^{x} + 1 - a) ).Simplify each term:- ( 2^{-x} cdot a cdot 4^{x} = a cdot 2^{-x} cdot 2^{2x} = a cdot 2^{x} ).- ( 2^{-x} cdot (-2 cdot 2^{x}) = -2 cdot 2^{-x} cdot 2^{x} = -2 cdot 1 = -2 ).- ( 2^{-x} cdot (1 - a) = (1 - a) cdot 2^{-x} ).So, putting it all together:( h(x) = a cdot 2^{x} - 2 + (1 - a) cdot 2^{-x} ).Simplify further:( h(x) = a cdot 2^{x} + (1 - a) cdot 2^{-x} - 2 ).So, ( h(x) = a cdot 2^{x} + (1 - a) cdot 2^{-x} - 2 ).Now, the problem states that for any ( x_{1}, x_{2} in [-1, 1] ), the difference ( |h(x_{1}) - h(x_{2})| leq frac{a + 1}{2} ).This is essentially saying that the maximum variation of ( h(x) ) on the interval ( [-1, 1] ) is bounded by ( frac{a + 1}{2} ). So, the maximum value of ( h(x) ) minus the minimum value of ( h(x) ) on this interval should be less than or equal to ( frac{a + 1}{2} ).Therefore, to solve this, I need to find the maximum and minimum of ( h(x) ) on ( x in [-1, 1] ), compute their difference, and set that difference less than or equal to ( frac{a + 1}{2} ). Then, solve for ( a ).Let me denote ( t = 2^{x} ). Since ( x in [-1, 1] ), ( t ) will range from ( 2^{-1} = frac{1}{2} ) to ( 2^{1} = 2 ). So, ( t in [frac{1}{2}, 2] ).Express ( h(x) ) in terms of ( t ):( h(x) = a cdot t + (1 - a) cdot frac{1}{t} - 2 ).So, let me define ( g(t) = a cdot t + frac{1 - a}{t} - 2 ), where ( t in [frac{1}{2}, 2] ).Now, I need to find the maximum and minimum of ( g(t) ) on ( t in [frac{1}{2}, 2] ).To find extrema, I can take the derivative of ( g(t) ) with respect to ( t ) and set it to zero.Compute ( g'(t) ):( g'(t) = a - frac{1 - a}{t^{2}} ).Set ( g'(t) = 0 ):( a - frac{1 - a}{t^{2}} = 0 ).Solve for ( t ):( a = frac{1 - a}{t^{2}} ).Multiply both sides by ( t^{2} ):( a t^{2} = 1 - a ).Bring all terms to one side:( a t^{2} + a - 1 = 0 ).Factor:( a(t^{2} + 1) - 1 = 0 ).Wait, that might not be helpful. Alternatively, solve for ( t^{2} ):( t^{2} = frac{1 - a}{a} ).So, ( t = sqrt{frac{1 - a}{a}} ).But ( t ) must be positive, so we take the positive square root.Now, this critical point ( t = sqrt{frac{1 - a}{a}} ) must lie within ( [frac{1}{2}, 2] ) for it to be considered.So, we have to consider two cases:1. The critical point ( t = sqrt{frac{1 - a}{a}} ) lies within ( [frac{1}{2}, 2] ).2. The critical point lies outside this interval, in which case the extrema occur at the endpoints.But before that, let's analyze the expression ( sqrt{frac{1 - a}{a}} ).First, the expression inside the square root must be non-negative:( frac{1 - a}{a} geq 0 ).This implies that ( (1 - a) ) and ( a ) have the same sign.Given that ( a geq frac{1}{2} ), as per the problem statement, so ( a > 0 ).Therefore, ( 1 - a geq 0 ) => ( a leq 1 ).So, the critical point exists only when ( a leq 1 ). If ( a > 1 ), then ( frac{1 - a}{a} ) is negative, so no real critical point.Therefore, for ( a in [frac{1}{2}, 1] ), there is a critical point at ( t = sqrt{frac{1 - a}{a}} ). For ( a > 1 ), no critical point within the domain.So, let's break it down into cases.**Case 1: ( frac{1}{2} leq a leq 1 ).**Here, the critical point ( t = sqrt{frac{1 - a}{a}} ) is real and positive. We need to check if it lies within ( [frac{1}{2}, 2] ).Compute ( t = sqrt{frac{1 - a}{a}} ).Let me denote ( t_c = sqrt{frac{1 - a}{a}} ).We need ( frac{1}{2} leq t_c leq 2 ).So, let's find the range of ( a ) such that ( t_c in [frac{1}{2}, 2] ).Compute ( t_c geq frac{1}{2} ):( sqrt{frac{1 - a}{a}} geq frac{1}{2} ).Square both sides:( frac{1 - a}{a} geq frac{1}{4} ).Multiply both sides by ( a ) (positive, so inequality remains same):( 1 - a geq frac{a}{4} ).Bring ( a ) terms to one side:( 1 geq frac{5a}{4} ).Multiply both sides by ( frac{4}{5} ):( frac{4}{5} geq a ).Similarly, compute ( t_c leq 2 ):( sqrt{frac{1 - a}{a}} leq 2 ).Square both sides:( frac{1 - a}{a} leq 4 ).Multiply both sides by ( a ):( 1 - a leq 4a ).Bring ( a ) terms to one side:( 1 leq 5a ).So, ( a geq frac{1}{5} ).But since ( a geq frac{1}{2} ), this condition is automatically satisfied.Therefore, ( t_c in [frac{1}{2}, 2] ) when ( a leq frac{4}{5} ).So, for ( frac{1}{2} leq a leq frac{4}{5} ), the critical point ( t_c ) lies within ( [frac{1}{2}, 2] ).For ( frac{4}{5} < a leq 1 ), ( t_c < frac{1}{2} ), so the critical point is outside the interval on the left.So, within ( frac{1}{2} leq a leq frac{4}{5} ), the function ( g(t) ) has a critical point inside the interval, so we need to evaluate ( g(t) ) at ( t = frac{1}{2} ), ( t = 2 ), and ( t = t_c ).For ( frac{4}{5} < a leq 1 ), the critical point is outside the interval, so the extrema occur at the endpoints ( t = frac{1}{2} ) and ( t = 2 ).Additionally, for ( a > 1 ), as mentioned earlier, there is no critical point, so again, extrema occur at the endpoints.So, let's handle each subcase.**Subcase 1.1: ( frac{1}{2} leq a leq frac{4}{5} ).**Here, ( t_c in [frac{1}{2}, 2] ).Compute ( g(t) ) at ( t = frac{1}{2} ), ( t = 2 ), and ( t = t_c ).First, ( g(frac{1}{2}) ):( g(frac{1}{2}) = a cdot frac{1}{2} + frac{1 - a}{frac{1}{2}} - 2 = frac{a}{2} + 2(1 - a) - 2 ).Simplify:( frac{a}{2} + 2 - 2a - 2 = frac{a}{2} - 2a = -frac{3a}{2} ).Next, ( g(2) ):( g(2) = a cdot 2 + frac{1 - a}{2} - 2 = 2a + frac{1 - a}{2} - 2 ).Simplify:( 2a + frac{1}{2} - frac{a}{2} - 2 = frac{4a - a}{2} + frac{1}{2} - 2 = frac{3a}{2} - frac{3}{2} = frac{3(a - 1)}{2} ).Now, ( g(t_c) ):Since ( t_c ) is a critical point, it's a local minimum or maximum. Let's compute ( g(t_c) ).We have ( t_c = sqrt{frac{1 - a}{a}} ).Compute ( g(t_c) = a cdot t_c + frac{1 - a}{t_c} - 2 ).Substitute ( t_c = sqrt{frac{1 - a}{a}} ):( g(t_c) = a cdot sqrt{frac{1 - a}{a}} + frac{1 - a}{sqrt{frac{1 - a}{a}}} - 2 ).Simplify each term:First term: ( a cdot sqrt{frac{1 - a}{a}} = sqrt{a^{2} cdot frac{1 - a}{a}} = sqrt{a(1 - a)} ).Second term: ( frac{1 - a}{sqrt{frac{1 - a}{a}}} = sqrt{(1 - a) cdot a} ).So, both terms are ( sqrt{a(1 - a)} ).Thus, ( g(t_c) = sqrt{a(1 - a)} + sqrt{a(1 - a)} - 2 = 2sqrt{a(1 - a)} - 2 ).So, ( g(t_c) = 2sqrt{a(1 - a)} - 2 ).Now, to determine whether this is a maximum or minimum, let's look at the second derivative or analyze the behavior.Alternatively, since ( g(t) ) is a function that tends to infinity as ( t ) approaches 0 or infinity, but within our interval, the critical point is a minimum or maximum.Wait, let's compute the second derivative.Compute ( g''(t) ):( g''(t) = frac{2(1 - a)}{t^{3}} ).Since ( a geq frac{1}{2} ), ( 1 - a leq frac{1}{2} ), but ( t > 0 ), so ( g''(t) ) is positive if ( 1 - a > 0 ), which is when ( a < 1 ). So, for ( a < 1 ), ( g''(t) > 0 ), so the critical point is a local minimum.For ( a = 1 ), ( g''(t) = 0 ), but that's a boundary case.So, for ( frac{1}{2} leq a < 1 ), ( t_c ) is a local minimum.Therefore, in this subcase, the maximum of ( g(t) ) occurs at one of the endpoints, and the minimum occurs at ( t_c ).So, compute ( g(frac{1}{2}) = -frac{3a}{2} ), ( g(2) = frac{3(a - 1)}{2} ), and ( g(t_c) = 2sqrt{a(1 - a)} - 2 ).We need to find which of ( g(frac{1}{2}) ) and ( g(2) ) is larger.Compute ( g(frac{1}{2}) = -frac{3a}{2} ), ( g(2) = frac{3(a - 1)}{2} ).Compare ( g(frac{1}{2}) ) and ( g(2) ):( -frac{3a}{2} ) vs. ( frac{3(a - 1)}{2} ).Which one is larger?Let's compute ( g(2) - g(frac{1}{2}) = frac{3(a - 1)}{2} - (-frac{3a}{2}) = frac{3a - 3 + 3a}{2} = frac{6a - 3}{2} ).So, if ( 6a - 3 > 0 ), then ( g(2) > g(frac{1}{2}) ).( 6a - 3 > 0 ) => ( a > frac{1}{2} ).Since ( a geq frac{1}{2} ), for ( a > frac{1}{2} ), ( g(2) > g(frac{1}{2}) ).At ( a = frac{1}{2} ), ( g(2) - g(frac{1}{2}) = frac{6 cdot frac{1}{2} - 3}{2} = frac{3 - 3}{2} = 0 ). So, both are equal.Therefore, for ( frac{1}{2} leq a leq frac{4}{5} ):- Maximum of ( g(t) ) is ( g(2) = frac{3(a - 1)}{2} ).- Minimum of ( g(t) ) is ( g(t_c) = 2sqrt{a(1 - a)} - 2 ).So, the difference ( g_{max} - g_{min} = frac{3(a - 1)}{2} - (2sqrt{a(1 - a)} - 2) ).Simplify:( frac{3(a - 1)}{2} - 2sqrt{a(1 - a)} + 2 ).Combine constants:( frac{3(a - 1)}{2} + 2 - 2sqrt{a(1 - a)} ).Simplify ( frac{3(a - 1)}{2} + 2 ):( frac{3a - 3}{2} + frac{4}{2} = frac{3a + 1}{2} ).So, the difference becomes:( frac{3a + 1}{2} - 2sqrt{a(1 - a)} ).We need this difference to be less than or equal to ( frac{a + 1}{2} ):( frac{3a + 1}{2} - 2sqrt{a(1 - a)} leq frac{a + 1}{2} ).Subtract ( frac{a + 1}{2} ) from both sides:( frac{3a + 1}{2} - frac{a + 1}{2} - 2sqrt{a(1 - a)} leq 0 ).Simplify:( frac{2a}{2} - 2sqrt{a(1 - a)} leq 0 ).Which is:( a - 2sqrt{a(1 - a)} leq 0 ).Bring ( a ) to the other side:( -2sqrt{a(1 - a)} leq -a ).Multiply both sides by -1 (inequality sign reverses):( 2sqrt{a(1 - a)} geq a ).Divide both sides by 2:( sqrt{a(1 - a)} geq frac{a}{2} ).Square both sides (since both sides are non-negative):( a(1 - a) geq frac{a^{2}}{4} ).Simplify:( a - a^{2} geq frac{a^{2}}{4} ).Bring all terms to one side:( a - a^{2} - frac{a^{2}}{4} geq 0 ).Combine like terms:( a - frac{5a^{2}}{4} geq 0 ).Factor:( aleft(1 - frac{5a}{4}right) geq 0 ).So, the inequality ( aleft(1 - frac{5a}{4}right) geq 0 ) holds when:Either both factors are positive or both are negative.But ( a geq frac{1}{2} ), so ( a > 0 ). Therefore, ( 1 - frac{5a}{4} geq 0 ).Thus:( 1 - frac{5a}{4} geq 0 ) => ( frac{5a}{4} leq 1 ) => ( a leq frac{4}{5} ).Therefore, the inequality holds when ( frac{1}{2} leq a leq frac{4}{5} ).So, in this subcase, the condition ( |h(x_{1}) - h(x_{2})| leq frac{a + 1}{2} ) is satisfied when ( frac{1}{2} leq a leq frac{4}{5} ).**Subcase 1.2: ( frac{4}{5} < a leq 1 ).**Here, the critical point ( t_c = sqrt{frac{1 - a}{a}} ) is less than ( frac{1}{2} ), so it's outside the interval ( [frac{1}{2}, 2] ).Therefore, the extrema occur at the endpoints ( t = frac{1}{2} ) and ( t = 2 ).Compute ( g(frac{1}{2}) = -frac{3a}{2} ) and ( g(2) = frac{3(a - 1)}{2} ).We need to find the maximum and minimum of ( g(t) ) on ( [frac{1}{2}, 2] ).Since ( a > frac{4}{5} ), let's see which of ( g(frac{1}{2}) ) and ( g(2) ) is larger.Compute ( g(2) - g(frac{1}{2}) = frac{3(a - 1)}{2} - (-frac{3a}{2}) = frac{3a - 3 + 3a}{2} = frac{6a - 3}{2} ).Since ( a > frac{4}{5} ), ( 6a - 3 > 6 cdot frac{4}{5} - 3 = frac{24}{5} - 3 = frac{24}{5} - frac{15}{5} = frac{9}{5} > 0 ). So, ( g(2) > g(frac{1}{2}) ).Therefore, the maximum is ( g(2) = frac{3(a - 1)}{2} ) and the minimum is ( g(frac{1}{2}) = -frac{3a}{2} ).Thus, the difference ( g_{max} - g_{min} = frac{3(a - 1)}{2} - (-frac{3a}{2}) = frac{3a - 3 + 3a}{2} = frac{6a - 3}{2} ).We need this difference to be less than or equal to ( frac{a + 1}{2} ):( frac{6a - 3}{2} leq frac{a + 1}{2} ).Multiply both sides by 2:( 6a - 3 leq a + 1 ).Subtract ( a ) from both sides:( 5a - 3 leq 1 ).Add 3 to both sides:( 5a leq 4 ).Divide by 5:( a leq frac{4}{5} ).But in this subcase, ( a > frac{4}{5} ). So, there's no solution here. Therefore, for ( frac{4}{5} < a leq 1 ), the condition is not satisfied.**Case 2: ( a > 1 ).**Here, the critical point ( t_c = sqrt{frac{1 - a}{a}} ) is not real, so the extrema occur at the endpoints ( t = frac{1}{2} ) and ( t = 2 ).Compute ( g(frac{1}{2}) = -frac{3a}{2} ) and ( g(2) = frac{3(a - 1)}{2} ).Again, since ( a > 1 ), ( g(2) ) is positive and ( g(frac{1}{2}) ) is negative. So, the maximum is ( g(2) ) and the minimum is ( g(frac{1}{2}) ).The difference ( g_{max} - g_{min} = frac{3(a - 1)}{2} - (-frac{3a}{2}) = frac{6a - 3}{2} ).Set this less than or equal to ( frac{a + 1}{2} ):( frac{6a - 3}{2} leq frac{a + 1}{2} ).Multiply both sides by 2:( 6a - 3 leq a + 1 ).Subtract ( a ):( 5a - 3 leq 1 ).Add 3:( 5a leq 4 ).Divide by 5:( a leq frac{4}{5} ).But in this case, ( a > 1 ), which contradicts ( a leq frac{4}{5} ). So, no solution here either.**Conclusion:**From the above cases, the only valid range for ( a ) is ( frac{1}{2} leq a leq frac{4}{5} ).Therefore, the range of ( a ) is ( boxed{frac{1}{2} leq a leq frac{4}{5}} ).

answer：Alright, so I've got this problem here with two vectors, a and b, and I need to figure out which of the four statements (A, B, C, D) are correct. Let me take it step by step.First, let's write down the vectors:- Vector a is (-2, 1)- Vector b is (-1, 3)Okay, now let's look at each option one by one.**Option A: a and b can be a set of basis vectors.**Hmm, basis vectors. I remember that for two vectors to form a basis in 2D space, they need to be linearly independent. That means neither vector can be a scalar multiple of the other. So, I need to check if there's a scalar λ such that a = λb.Let's see:If a = λb, then:-2 = λ*(-1) => λ = 21 = λ*3 => λ = 1/3But 2 is not equal to 1/3, so there's no such λ. Therefore, a and b are linearly independent and can form a basis. So, Option A seems correct.**Option B: The coordinates of a unit vector in the same direction as a + b are (3/5, -4/5).**Alright, first, let's compute a + b.a + b = (-2 + (-1), 1 + 3) = (-3, 4)Now, to find a unit vector in the same direction, I need to divide this vector by its magnitude.The magnitude of a + b is sqrt((-3)^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.So, the unit vector is ( (-3)/5, 4/5 ) = (-3/5, 4/5)But the option says (3/5, -4/5). That's not the same. It looks like they might have taken the negative of the vector. Maybe they thought the direction was opposite? But the question specifies "in the same direction as a + b," so the unit vector should be (-3/5, 4/5). Therefore, Option B is incorrect.**Option C: The sine value of the angle between a and b is sqrt(2)/2.**Okay, to find the sine of the angle between two vectors, I can use the cross product formula. But wait, in 2D, the magnitude of the cross product is |a x b| = |a1*b2 - a2*b1|, and this equals |a||b|sin(theta), where theta is the angle between them.First, let's compute the cross product:a x b = (-2)*3 - 1*(-1) = -6 + 1 = -5So, |a x b| = 5Now, let's find |a| and |b|:|a| = sqrt((-2)^2 + 1^2) = sqrt(4 + 1) = sqrt(5)|b| = sqrt((-1)^2 + 3^2) = sqrt(1 + 9) = sqrt(10)So, |a||b| = sqrt(5)*sqrt(10) = sqrt(50) = 5*sqrt(2)Therefore, sin(theta) = |a x b| / (|a||b|) = 5 / (5*sqrt(2)) = 1/sqrt(2) = sqrt(2)/2So, yes, sin(theta) is sqrt(2)/2. Therefore, Option C is correct.**Option D: If c = (x, x) satisfies |a + c| = |b - c|, then x = 5/2.**Alright, let's parse this. We have vector c = (x, x). We need to find x such that the magnitude of a + c equals the magnitude of b - c.First, let's write out a + c and b - c:a + c = (-2 + x, 1 + x)b - c = (-1 - x, 3 - x)Now, the magnitudes squared:|a + c|^2 = (-2 + x)^2 + (1 + x)^2|b - c|^2 = (-1 - x)^2 + (3 - x)^2Since the magnitudes are equal, their squares are equal:(-2 + x)^2 + (1 + x)^2 = (-1 - x)^2 + (3 - x)^2Let's expand both sides:Left side:(-2 + x)^2 = (x - 2)^2 = x^2 - 4x + 4(1 + x)^2 = x^2 + 2x + 1Total left side: x^2 - 4x + 4 + x^2 + 2x + 1 = 2x^2 - 2x + 5Right side:(-1 - x)^2 = (x + 1)^2 = x^2 + 2x + 1(3 - x)^2 = x^2 - 6x + 9Total right side: x^2 + 2x + 1 + x^2 - 6x + 9 = 2x^2 - 4x + 10Set left side equal to right side:2x^2 - 2x + 5 = 2x^2 - 4x + 10Subtract 2x^2 from both sides:-2x + 5 = -4x + 10Add 4x to both sides:2x + 5 = 10Subtract 5:2x = 5Divide by 2:x = 5/2So, yes, x is indeed 5/2. Therefore, Option D is correct.**Summary:**- A: Correct- B: Incorrect- C: Correct- D: CorrectSo, the correct options are A, C, and D.

answer：First, calculate the total amount of flour Jessica uses daily by multiplying the number of pie crusts by the flour used per crust:36 crusts × (1/8) cup/crust = 4.5 cups.Next, determine the amount of flour needed for each of the 24 larger pie crusts by dividing the total flour by the new number of crusts:4.5 cups ÷ 24 crusts = 0.1875 cups per crust.Finally, convert the decimal to a fraction:0.1875 cups = 3/16 cups.

answer：Okay, so I have this problem where a pedestrian arrives at a crosswalk, and the traffic light is either red or green. The red light lasts for 45 seconds. If the pedestrian arrives when the light is red, I need to find the probability that they'll have to wait at least 20 seconds before it turns green. The options are A: 4/9, B: 5/9, C: 2/5, D: 3/5.Alright, let me think about this. So, the light is red for 45 seconds. If the pedestrian arrives during this red light, how long do they have to wait? Well, it depends on when exactly they arrive during the red light cycle.If the light is red for 45 seconds, and the pedestrian arrives at some random time during that red light, the waiting time until the light turns green would be somewhere between 0 seconds (if they arrive just as the light is turning red) and 45 seconds (if they arrive just as the light is about to turn green).But the question is asking for the probability that they have to wait at least 20 seconds. So, that means we're looking for the probability that their arrival time is such that there's at least 20 seconds left in the red light when they arrive.Wait, actually, no. If they arrive and the light is red, the waiting time is the remaining time until the light turns green. So, if they arrive at time t, where t is between 0 and 45 seconds, then their waiting time is 45 - t. So, if they arrive at t = 0, they have to wait 45 seconds. If they arrive at t = 45, they have to wait 0 seconds.But the question is about waiting at least 20 seconds. So, we need the probability that 45 - t >= 20. That simplifies to t <= 25.So, the pedestrian arrives during the red light, which lasts 45 seconds. The probability that they arrive in the first 25 seconds of the red light is the probability that they have to wait at least 20 seconds.Since the arrival time is uniformly distributed over the red light period, the probability is just the ratio of the favorable time (first 25 seconds) to the total red light duration (45 seconds).So, probability = 25 / 45 = 5/9.Wait, let me double-check that. So, if the red light is 45 seconds, and we want the pedestrian to arrive in the first 25 seconds, that would mean they have to wait at least 20 seconds. Because if they arrive at 25 seconds, they have 20 seconds left. If they arrive before that, they have more than 20 seconds to wait.So, yes, the favorable time is 25 seconds, and the total possible time is 45 seconds. So, 25/45 simplifies to 5/9.Looking at the options, B is 5/9, so that should be the answer.But let me think again to make sure I didn't make a mistake. Sometimes with probability problems, it's easy to mix up the favorable and unfavorable cases.So, the total red light duration is 45 seconds. The pedestrian arrives at a random time during this period. We want the probability that their waiting time is at least 20 seconds. As I thought earlier, that translates to arriving in the first 25 seconds of the red light.So, the probability is the length of the favorable interval (25 seconds) divided by the total interval (45 seconds), which is 25/45 = 5/9.Yes, that seems correct. So, I think the answer is B: 5/9.