🧠:Okay, so I have this problem here where I need to find the largest possible value of the product of logarithms, specifically log base 2 of π(2) multiplied by log base 3 of π(3) and so on up to log base 2012 of π(2012). Here, π is a permutation of the numbers from 2 through 2012. Hmm, permutations... so each number from 2 to 2012 is mapped to another number in that range without repetition. My goal is to maximize the product of these logs. Let me break this down step by step.First, let me restate the problem to make sure I understand it correctly. We have a permutation π of the numbers 2, 3, ..., 2012. For each number k in this range, we compute log base k of π(k), and then multiply all these logarithms together. The task is to find the permutation π that maximizes this product. So, essentially, I need to assign each number from 2 to 2012 to another number in the same set such that when I take each log and multiply them all, the result is as large as possible.Alright, let's think about the properties of logarithms here. Remember that log_b a is equal to ln a / ln b, right? So, maybe expressing each term in terms of natural logarithms could help? Let me see. If I write each log base k of π(k) as ln π(k) / ln k, then the entire product becomes the product from k=2 to 2012 of [ln π(k) / ln k]. Hmm, that simplifies to [product of ln π(k) from k=2 to 2012] divided by [product of ln k from k=2 to 2012]. Wait, but π is a permutation, so the product of ln π(k) from k=2 to 2012 is the same as the product of ln m from m=2 to 2012, since π is just rearranging the numbers. Therefore, the entire expression simplifies to [product of ln m from m=2 to 2012] divided by [product of ln k from k=2 to 2012] which is just 1. Wait, that can't be right. If that's the case, then regardless of the permutation, the product is always 1. But that would mean the maximum is 1. But that seems contradictory because permutations can affect the product. Wait, maybe I made a mistake here.Wait, no. Let's verify this. If π is a permutation, then the product over k=2 to 2012 of π(k) is the same as the product over m=2 to 2012 of m, because permutations just rearrange the terms. But here, we have the product of ln π(k), so indeed, the product of ln π(k) over all k is equal to the product of ln m over all m from 2 to 2012. Similarly, the denominator is the product of ln k over the same range. Therefore, the ratio is indeed 1. But that would mean that no matter how we permute the numbers, the product of log base k of π(k) is always 1. Therefore, the maximum is 1. But this seems counterintuitive because depending on the permutation, the individual terms can be greater or less than 1. If some terms are greater than 1 and others less than 1, their product might be more or less. But according to this calculation, it's always 1. Hmm, maybe I need to check with a smaller case.Let's take a small example, say numbers 2 and 3. So the permutation π could be either [2,3] or [3,2]. Let's compute the product for both permutations.First permutation: π(2)=2, π(3)=3. Then the product is log2(2) * log3(3) = 1 * 1 = 1.Second permutation: π(2)=3, π(3)=2. Then the product is log2(3) * log3(2). Let's compute that: ln3/ln2 * ln2/ln3 = (ln3 * ln2)/(ln2 * ln3) = 1. So again, the product is 1.Interesting! So regardless of the permutation, the product is always 1? So maybe in general, for any permutation, this product is 1? But that seems strange. Wait, let's try with three numbers: 2, 3, 4.Take permutation π(2)=3, π(3)=4, π(4)=2. Then the product is log2(3) * log3(4) * log4(2). Let's compute each term:log2(3) = ln3/ln2log3(4) = ln4/ln3log4(2) = ln2/ln4Multiplying them together: (ln3/ln2) * (ln4/ln3) * (ln2/ln4) = (ln3 * ln4 * ln2)/(ln2 * ln3 * ln4) = 1. So again, 1.Wait a second, so maybe for any permutation, the product is 1? That would mean that the answer is 1, regardless of the permutation. But the question asks for the largest possible value. If all permutations give the same product, 1, then the maximum is 1. But maybe there's a mistake here.Wait, let's think again. If we consider that each term is log_b a, then the product over all terms is product_{k=2}^{2012} log_k π(k). If π is a permutation, then each a=π(k) is unique. So, the product becomes product_{k=2}^{2012} (ln π(k)/ln k). Then, rearranging the terms in the numerator, since π is a permutation, the product in the numerator is product_{m=2}^{2012} ln m, which is the same as the product in the denominator. Therefore, the ratio is indeed 1. Therefore, no matter how you permute the numbers, the product is always 1. Hence, the maximum is 1.But wait, if that's the case, then why is the problem asking for the largest possible value? Because it seems that regardless of the permutation, the product is 1. Therefore, the answer is 1.But let me verify once more with another permutation. Suppose in the case of three numbers: 2, 3, 4. Let's take another permutation, say π(2)=4, π(3)=2, π(4)=3. Then the product is log2(4) * log3(2) * log4(3). Compute each term:log2(4) = 2log3(2) = ln2/ln3log4(3) = ln3/ln4Multiply them together: 2 * (ln2/ln3) * (ln3/ln4) = 2 * (ln2 / ln4) = 2 * (ln2 / (2 ln2)) ) = 2 * (1/2) = 1. So again, 1.Wait, so even if one term is 2, the other terms compensate to bring the product back to 1. So, it's like the product is always 1. Therefore, no permutation can make the product larger than 1.But this seems counterintuitive. For example, if we have log_b a, if a > b, then log_b a > 1, and if a < b, log_b a < 1. So, if we could have more terms greater than 1, maybe the product increases. But according to the earlier calculations, that doesn't happen. Each time, the terms multiply out to 1. So, perhaps there's some mathematical identity here.Let me consider the general case. Let S = {2, 3, ..., n}. Let π be a permutation of S. Then, the product over k in S of log_k π(k) = product_{k in S} (ln π(k)/ln k) = [product_{k in S} ln π(k)] / [product_{k in S} ln k]. But since π is a permutation, the numerator is just the product of ln m for m in S, which is the same as the denominator. Therefore, the product is 1. So, regardless of the permutation, the product is always 1. Therefore, the maximum is 1. Therefore, the answer is 1.But the problem says "Find the largest possible value". So, perhaps the answer is 1.But wait, let me check for another permutation where perhaps a term is 1. For example, if π(k) = k for all k, then each term is log_k k = 1, so the product is 1. If π is not the identity permutation, then some terms are greater than 1, some less than 1. But as we saw, their product is still 1. Therefore, all permutations yield the same product. Therefore, the maximum is 1.But this seems too straightforward. The problem is presented as a competition-style problem, so maybe there's a trick here. Wait, but according to the mathematical reasoning, it's indeed always 1. Let me check with another example. Take numbers 2, 3, 4, 5. Let's take a permutation π such that π(2)=5, π(3)=4, π(4)=3, π(5)=2. Then compute the product:log2(5) * log3(4) * log4(3) * log5(2).Express in natural logs:(ln5/ln2) * (ln4/ln3) * (ln3/ln4) * (ln2/ln5) = (ln5 * ln4 * ln3 * ln2) / (ln2 * ln3 * ln4 * ln5) ) = 1. So again, 1.So no matter how you permute, the product is 1. Therefore, the answer is 1.But the problem states "Find the largest possible value", implying that there might be a way to get a larger value. But according to this, it's fixed. Maybe the problem is in the way I interpret the permutation.Wait, the problem says "a permutation of the numbers from 2 through 2012". So π is a bijection from {2,3,...,2012} to itself. Each element is mapped to another element without repetition. Then, the product over k=2 to 2012 of log_k π(k). So, as per the earlier reasoning, the product is 1. So, is the answer 1? Then why is the problem asking for the largest possible value? Because if all permutations give the same value, then 1 is the only possible value, hence the maximum is 1. Therefore, the answer is 1.But let me check once again. Suppose we have a different kind of permutation where cycles are involved. For example, if you have a cycle like π(a) = b, π(b) = c, π(c) = a. Then, the terms log_a b, log_b c, log_c a. Multiply them: (ln b / ln a) * (ln c / ln b) * (ln a / ln c) = 1. So again, 1. Therefore, regardless of cycle structure, the product is 1.Alternatively, if we have transpositions, swapping two elements. For example, swap a and b. Then, log_a b * log_b a = (ln b / ln a) * (ln a / ln b) = 1. So even swapping two elements, the product of their logs is 1, and the rest are 1s, so total product remains 1.Therefore, no matter how you permute, the product is 1. Therefore, the maximum is 1. Hence, the answer is boxed{1}.But wait, maybe I made a mistake in assuming that all permutations result in 1. Let me think again. Suppose we have a permutation where π(k) = k+1 for k=2 to 2011, and π(2012)=2. Then, the product is log2(3) * log3(4) * ... * log2011(2012) * log2012(2). Each term is log_k (k+1), except the last term which is log2012(2). Let's compute the product:First, write each term as ln(k+1)/ln k. So, the product is [ln3/ln2] * [ln4/ln3] * ... * [ln2012/ln2011] * [ln2/ln2012]. Multiplying all these together, the numerators and denominators cancel out: ln3 cancels with the denominator of the next term, etc., and finally ln2012 cancels with the denominator of the last term, leaving ln2 in the numerator from the last term and ln2 in the denominator from the first term. Wait, no:Wait, the first term is ln3/ln2, the second term is ln4/ln3, ..., the second to last term is ln2012/ln2011, and the last term is ln2/ln2012. So, multiplying all these fractions:(ln3/ln2) * (ln4/ln3) * ... * (ln2012/ln2011) * (ln2/ln2012) = (ln3 * ln4 * ... * ln2012 * ln2) / (ln2 * ln3 * ... * ln2011 * ln2012) ) = 1. Because the numerator is ln2 * ln3 * ... * ln2012 and the denominator is the same. So, again, 1. Therefore, regardless of the permutation, the product is 1.Therefore, the answer must be 1. So, maybe the problem is designed to trick you into thinking that you need to find a clever permutation, but in reality, it's always 1. Therefore, the maximum is 1.But to make sure, let me check one more example with a different permutation. Suppose in the set {2,3,4}, let π(2)=4, π(3)=3, π(4)=2. Then the product is log2(4) * log3(3) * log4(2). That is, 2 * 1 * 0.5 = 1. So again, 1. If π(3)=3, which is allowed in permutations (since permutations can have fixed points). The product is 2 * 1 * 0.5 = 1. So even if some terms are fixed (log_k k =1), but others are not, the product still remains 1.Therefore, regardless of the permutation, the product is always 1. Therefore, the largest possible value is 1.So, conclusion: The product is always 1, so the maximum is boxed{1}.Final Answerboxed{1}

✅:To find the largest possible value of the product (log_{2} pi(2) cdot log_{3} pi(3) cdots log_{2012} pi(2012)) where (pi) is a permutation of the numbers from 2 through 2012, we start by analyzing the properties of logarithms and permutations.First, we note that each term (log_{k} pi(k)) can be expressed in terms of natural logarithms as (frac{ln pi(k)}{ln k}). Therefore, the entire product can be written as:[prod_{k=2}^{2012} frac{ln pi(k)}{ln k}]Since (pi) is a permutation, the set ({pi(k) mid k = 2, 3, ldots, 2012}) is exactly the set ({2, 3, ldots, 2012}). This means the product of the numerators (prod_{k=2}^{2012} ln pi(k)) is the same as the product (prod_{m=2}^{2012} ln m). Similarly, the product of the denominators (prod_{k=2}^{2012} ln k) is also (prod_{m=2}^{2012} ln m). Therefore, the entire expression simplifies to:[frac{prod_{m=2}^{2012} ln m}{prod_{m=2}^{2012} ln m} = 1]This shows that regardless of the permutation (pi), the product of the logarithms is always 1. This result holds because the permutation merely rearranges the terms in the numerator, which does not change the product. To verify, we considered smaller cases and specific permutations, confirming that the product of the logarithms always results in 1. For example, in permutations involving cycles or transpositions, the terms cancel out appropriately, leading to the same result.Thus, the largest possible value of the given product is (boxed{1}).

🔑:1. We start with the product we need to maximize: [ prod_{i=2}^{2012} log_i pi(i) ]2. Recall the change of base formula for logarithms: [ log_i x = frac{log x}{log i} ] 3. Applying this to each term in our product, we have: [ prod_{i=2}^{2012} log_i pi(i) = prod_{i=2}^{2012} frac{log pi(i)}{log i} ]4. We can rewrite this product in a more convenient form: [ prod_{i=2}^{2012} log_i pi(i) = frac{prod_{i=2}^{2012} log pi(i)}{prod_{i=2}^{2012} log i} ]5. Observe that since (pi) is a permutation of the numbers (2) through (2012), (pi(i)) will go through all of these numbers exactly once.6. Consider the numerator (prod_{i=2}^{2012} log pi(i)): [ prod_{i=2}^{2012} log pi(i) = log 2 cdot log 3 cdot log 4 cdots log 2012 ] This simplifies to: [ log(2 cdot 3 cdot 4 cdots 2012) ]7. Similarly, we analyze the denominator (prod_{i=2}^{2012} log i): [ prod_{i=2}^{2012} log i = log 2 cdot log 3 cdot log 4 cdots log 2012 = log(2 cdot 3 cdot 4 cdots 2012) ]8. Upon division, we get: [ frac{log(2 cdot 3 cdot 4 cdots 2012)}{log(2 cdot 3 cdot 4 cdots 2012)} = 1 ]9. Therefore, the product ( prod_{i=2}^{2012} log_i pi(i) ) simplifies to: [ boxed{1} ]

🧠:Okay, so I need to solve this trigonometric equation: sin²x - (2a + 1)cosx - a² = 0, and find the values of real number a for which the equation has real solutions x. The answer choices are given, so I need to figure out which interval is correct. Let me start by recalling some trigonometric identities and methods to solve such equations.First, I know that sin²x can be written in terms of cosx using the Pythagorean identity: sin²x = 1 - cos²x. Maybe substituting that into the equation will help simplify things. Let me try that.So replacing sin²x with 1 - cos²x, the equation becomes:1 - cos²x - (2a + 1)cosx - a² = 0.Let me rewrite this to make it more organized:- cos²x - (2a + 1)cosx + (1 - a²) = 0.Hmm, maybe multiply both sides by -1 to make the coefficient of cos²x positive. That would give:cos²x + (2a + 1)cosx + (a² - 1) = 0.Wait, let's check the signs again. Original substitution:1 - cos²x - (2a +1)cosx -a² = 0So bringing all terms to the left side:- cos²x - (2a +1)cosx + (1 - a²) = 0.Multiply by -1:cos²x + (2a +1)cosx + (a² -1) = 0.Yes, that's correct.So now we have a quadratic equation in terms of cosx. Let me denote y = cosx. Then the equation becomes:y² + (2a +1)y + (a² -1) = 0.Since we're dealing with real solutions for x, the variable y must lie within the range of cosx, which is [-1, 1]. So even if the quadratic equation in y has real roots, these roots must satisfy -1 ≤ y ≤ 1. Therefore, the problem reduces to finding the values of a for which the quadratic equation in y has at least one real root within the interval [-1, 1].To approach this, I can use the quadratic equation theory. For a quadratic equation Ay² + By + C = 0, the roots are real if the discriminant is non-negative. The discriminant D = B² - 4AC.First, let's compute the discriminant for our quadratic equation in y:D = (2a +1)² - 4 * 1 * (a² -1)Let me expand that:D = 4a² + 4a +1 - 4a² +4Simplify:4a² -4a² cancels out, so D = 4a +1 +4 = 4a +5.So discriminant D = 4a +5.For the quadratic to have real roots, we need D ≥ 0. Therefore:4a +5 ≥ 0 ⇒ 4a ≥ -5 ⇒ a ≥ -5/4.So that's the first condition: a must be greater than or equal to -5/4.But even if the roots are real, we need at least one of them to lie within [-1, 1]. So even if a ≥ -5/4, we need to ensure that the quadratic equation has at least one root in [-1, 1].Therefore, the next step is to analyze when the quadratic equation y² + (2a +1)y + (a² -1) = 0 has roots in [-1, 1].To do this, there are a few approaches. One approach is to use the concept that a quadratic function f(y) = y² + (2a +1)y + (a² -1) must cross the y-axis within the interval [-1, 1]. So, either:1. The quadratic has exactly one root in [-1, 1] (i.e., the quadratic touches the x-axis at that root), or2. The quadratic has two roots, at least one of which is in [-1, 1].Alternatively, we can use the following criteria:- Either f(-1) and f(1) have opposite signs (which would imply at least one root in between), or- The quadratic has a root exactly at -1 or 1, or- The quadratic has its vertex within [-1, 1] and the discriminant is non-negative.But perhaps a more systematic approach is to use the method of interval analysis.Let me recall that for a quadratic equation Ay² + By + C = 0, the roots lie in [m, n] if:1. The discriminant is non-negative (already established a ≥ -5/4).2. Either: a. The quadratic evaluated at m and n have opposite signs (f(m) * f(n) ≤ 0), or b. The quadratic has a root exactly at m or n, or c. The vertex is within [m, n] and the function at the vertex has a sign opposite to the leading coefficient.But this might get complicated, so maybe it's better to use the following method.Since we need at least one root in [-1, 1], we can consider two cases:Case 1: Exactly one root in [-1, 1]. This happens when f(-1) and f(1) have opposite signs.Case 2: Both roots are in [-1, 1]. For this, the following must hold:- The discriminant is non-negative (already D ≥ 0, so a ≥ -5/4),- The axis of symmetry is within [-1, 1]. The axis of symmetry for quadratic Ay² + By + C is at y = -B/(2A). Here, A = 1, B = (2a +1), so axis at y = -(2a +1)/2.- Additionally, f(-1) ≥ 0 and f(1) ≥ 0 (since the quadratic opens upwards, A = 1 > 0).Wait, but if the quadratic opens upwards, then if both roots are in [-1, 1], then f(-1) and f(1) must be non-negative. But actually, if both roots are between -1 and 1, then since the parabola opens upwards, the minimum is between -1 and 1, and the function is negative between the roots. So, actually, if the quadratic has two roots within [-1,1], then between those roots the function is negative, and outside it is positive. Therefore, f(-1) and f(1) would be positive, and there's a minimum in between. However, this seems a bit conflicting. Maybe I need to think more carefully.Alternatively, to have both roots in [-1,1], the following must hold:1. The quadratic has two real roots (D ≥ 0, which is a ≥ -5/4).2. The smaller root is ≥ -1.3. The larger root is ≤ 1.So, if I find the roots of the quadratic equation, then set them to be within [-1,1].The roots of the quadratic equation y² + (2a +1)y + (a² -1) = 0 can be found using the quadratic formula:y = [-(2a +1) ± sqrt(D)] / 2But D = 4a +5, so sqrt(D) = sqrt(4a +5). Therefore, the roots are:y = [ - (2a +1) ± sqrt(4a +5) ] / 2.So the two roots are:y1 = [ - (2a +1) + sqrt(4a +5) ] / 2,y2 = [ - (2a +1) - sqrt(4a +5) ] / 2.We need at least one of these roots to be in [-1,1].Alternatively, since y1 is the larger root (because we have +sqrt(4a +5)) and y2 is the smaller root.Therefore, the possible scenarios are:1. The larger root y1 ≤ 1,2. The smaller root y2 ≥ -1,Or both. But actually, for at least one root in [-1,1], it's sufficient that either y1 ≥ -1 and y2 ≤ 1? Wait, no. Wait, perhaps another approach.Wait, the problem states that the equation has real solutions for x. So there exists some x such that cosx is equal to one of the roots of the quadratic. Therefore, as long as at least one of the roots is in [-1,1], there is a real solution x.Therefore, the equation will have real solutions if and only if at least one of the roots y1 or y2 is in [-1,1].So, our goal is to find all a ≥ -5/4 such that either y1 ∈ [-1,1] or y2 ∈ [-1,1].So, perhaps we can analyze the conditions for y1 and y2.Let me first compute y1 and y2.Given that y1 = [ - (2a +1) + sqrt(4a +5) ] / 2,and y2 = [ - (2a +1) - sqrt(4a +5) ] / 2.So let's first analyze y1. Let's see when y1 ≤ 1 and y1 ≥ -1.Similarly, for y2, check when y2 ≥ -1 and y2 ≤1.But since y1 is the larger root, and y2 is the smaller root, perhaps y1 is more likely to be in the upper bound and y2 in the lower.But let's proceed step by step.First, let's find when y1 ≤ 1.Compute:[ - (2a +1) + sqrt(4a +5) ] / 2 ≤ 1Multiply both sides by 2:- (2a +1) + sqrt(4a +5) ≤ 2Bring - (2a +1) to the right:sqrt(4a +5) ≤ 2 + 2a +1sqrt(4a +5) ≤ 2a +3Now, since sqrt(4a +5) is non-negative, the right-hand side must be non-negative as well. So 2a +3 ≥ 0 ⇒ a ≥ -3/2.But we already have a ≥ -5/4 from the discriminant. Since -5/4 ≈ -1.25, which is greater than -3/2 = -1.5, so the condition a ≥ -5/4 is more restrictive. Therefore, in the interval a ≥ -5/4, 2a +3 is:At a = -5/4: 2*(-5/4) +3 = -5/2 +3 = 1/2 ≥0. So yes, 2a +3 ≥0 for all a ≥ -5/4.Therefore, we can square both sides without reversing inequality:sqrt(4a +5) ≤ 2a +3 ⇒ 4a +5 ≤ (2a +3)^2Compute the right-hand side:(2a +3)^2 = 4a² +12a +9Thus:4a +5 ≤4a² +12a +9Bring all terms to the right:0 ≤4a² +12a +9 -4a -5Simplify:4a² +8a +4 ≥0Divide both sides by 4:a² +2a +1 ≥0Which factors as:(a +1)^2 ≥0Since a square is always non-negative, this inequality is always true. Therefore, the inequality sqrt(4a +5) ≤2a +3 holds for all a ≥ -5/4. Therefore, the original inequality y1 ≤1 is always true for a ≥ -5/4. Therefore, y1 is always less than or equal to 1.Wait, but that seems strange. Let me check with a specific value. Let's take a = 0.Then y1 = [ - (0 +1) + sqrt(0 +5) ] / 2 = [ -1 + sqrt(5) ] / 2 ≈ (-1 + 2.236)/2 ≈1.236/2≈0.618, which is ≤1. Correct.Another example, a = 1:y1 = [ -3 + sqrt(9) ] / 2 = (-3 +3)/2 =0 ≤1.Another example, a = -5/4:Then sqrt(4*(-5/4)+5)=sqrt(-5 +5)=sqrt(0)=0.Thus, y1 = [ - (2*(-5/4) +1) +0 ] /2 = [ - ( -10/4 +4/4 ) ] /2 = [ - (-6/4) ] /2 = (6/4)/2= (3/2)/2=3/4 ≤1. Correct.So indeed, y1 ≤1 for all a ≥-5/4.Therefore, the upper bound of y1 is always ≤1. Therefore, the only condition we need is that y1 ≥-1 (since y1 is already ≤1), to ensure that y1 ∈ [-1,1]. Alternatively, if y1 is between -1 and1, then there is a solution. If y1 < -1, then perhaps y2 might be in [-1,1].But since y1 is the larger root, if y1 < -1, then y2 would also be < -1, since it's smaller. Therefore, if the larger root is less than -1, then both roots are less than -1, so no solutions in [-1,1]. Similarly, if the smaller root y2 >1, then both roots are greater than1, so no solutions. But since the quadratic opens upwards, the minimum is between the roots. Wait, no, the quadratic opens upwards, so it tends to infinity as y increases. Wait, actually, the quadratic is y² + (2a +1)y + (a² -1). The leading coefficient is positive, so it opens upwards, meaning it has a minimum between the two roots.Therefore, if both roots are on the left of -1, then the quadratic is positive for y ≥ -1. If both roots are on the right of 1, the quadratic is positive for y ≤1. Therefore, in these cases, there would be no solution. But if at least one root is within [-1,1], then there is a solution. Since we have y1 ≤1 always, we need to check when y1 ≥-1.Therefore, the key condition is y1 ≥ -1. Because if y1 is ≥-1 and ≤1, then there is a root in [-1,1]. If y1 < -1, then even the larger root is less than -1, so both roots are less than -1, so no solution. Similarly, if the smaller root is greater than1, but since y1 is the larger root, which is ≤1, that can't happen.Wait, if the smaller root y2 >1, then the larger root y1 would be greater than y2, which would be impossible because y1 is already ≤1. Wait, maybe this is conflicting. Let me see.If y2 >1, then since y1 is the larger root, it must be y1 ≥ y2 >1, but we already saw that y1 ≤1. Therefore, y2 cannot be greater than1. Therefore, the only possible cases are:1. Both roots are ≤1. Since y1 ≤1 always, then if the smaller root y2 is ≥-1, then both roots are in [-1,1]. If y2 < -1, then only the larger root y1 is in (-∞,1], but if y1 ≥-1, then there is a solution. If y1 < -1, then both roots are < -1, no solution.Therefore, the equation has real solutions when:Either:- Both roots are in [-1,1], which requires y2 ≥-1 and y1 ≤1 (but y1 is always ≤1, so just y2 ≥-1), or- At least one root is in [-1,1]. Since y1 is always ≤1, the condition is y1 ≥-1.Wait, but if y1 ≥-1, then since y1 is the larger root, which is ≤1, that would imply that the larger root is in [-1,1]. Then, since the quadratic opens upwards, if the larger root is in [-1,1], the smaller root can be either in [-1,1] or less than -1. However, if the larger root is in [-1,1], regardless of the smaller root, there is at least one solution (the larger root). Therefore, the key condition is y1 ≥-1.Similarly, even if y2 is in [-1,1], but since y2 is smaller than y1, if y1 is already in [-1,1], then both roots are in [-1,1] if y2 ≥-1. If y2 < -1, but y1 ≥-1, then only y1 is in [-1,1]. Therefore, the overall condition is y1 ≥-1.Therefore, to have at least one root in [-1,1], it's necessary and sufficient that y1 ≥-1.So let's set up the inequality:y1 = [ - (2a +1) + sqrt(4a +5) ] / 2 ≥ -1.Multiply both sides by 2:- (2a +1) + sqrt(4a +5) ≥ -2Bring - (2a +1) to the right:sqrt(4a +5) ≥ -2 + 2a +1 = 2a -1So sqrt(4a +5) ≥2a -1Now, sqrt(4a +5) is non-negative, so the right-hand side 2a -1 must be ≤ sqrt(4a +5). However, we have to consider two cases:Case 1: 2a -1 ≤0. In this case, since the left-hand side (sqrt) is non-negative, the inequality sqrt(4a +5) ≥2a -1 holds automatically.Case 2: 2a -1 >0. In this case, both sides are positive, so we can square both sides:4a +5 ≥ (2a -1)^2Compute the right-hand side:(2a -1)^2 =4a² -4a +1Thus:4a +5 ≥4a² -4a +1Bring all terms to the left:0 ≥4a² -8a -4Multiply both sides by -1 (reverse inequality):4a² -8a -4 ≤0Divide both sides by 4:a² -2a -1 ≤0Solve the quadratic inequality a² -2a -1 ≤0.Find the roots:a = [2 ± sqrt(4 +4)]/2 = [2 ± sqrt(8)]/2 = [2 ± 2*sqrt(2)]/2 = 1 ± sqrt(2)Therefore, the inequality a² -2a -1 ≤0 holds for a ∈ [1 - sqrt(2), 1 + sqrt(2)].But remember, this is under Case 2 where 2a -1 >0 ⇒ a > 1/2.So, in Case 2, we have a >1/2 and a ∈ [1 - sqrt(2),1 + sqrt(2)]. But 1 - sqrt(2) ≈1 -1.414≈-0.414, and 1 + sqrt(2)≈2.414. So the intersection of a >1/2 and a ∈ [1 - sqrt(2),1 + sqrt(2)] is a ∈ (1/2,1 + sqrt(2)].But in Case 1, where 2a -1 ≤0 ⇒ a ≤1/2, the inequality sqrt(4a +5) ≥2a -1 holds automatically. Therefore, combining both cases:- For a ≤1/2, the inequality holds.- For a >1/2, the inequality holds when a ∈ [1 - sqrt(2),1 + sqrt(2)] intersected with a >1/2, which is a ∈ (1/2,1 + sqrt(2)].But wait, this is confusing. Let me clarify:The original inequality sqrt(4a +5) ≥2a -1 is satisfied in two cases:1. When 2a -1 ≤0 (i.e., a ≤1/2), in which case the left-hand side is ≥0 and the right-hand side is ≤0, so inequality holds.2. When 2a -1 >0 (i.e., a >1/2), then we must have 4a +5 ≥(2a -1)^2, leading to a ∈ [1 - sqrt(2),1 + sqrt(2)]. But since in this case a >1/2, the valid interval is a ∈ (1/2,1 + sqrt(2)].But 1 - sqrt(2) is approximately -0.414, which is less than 1/2. So combining both cases, the inequality sqrt(4a +5) ≥2a -1 is satisfied for:a ∈ (-∞,1 + sqrt(2)].But wait, no. Because in Case 1, a ≤1/2, and in Case 2, a ∈ (1/2,1 + sqrt(2)]. Therefore, combining both, the inequality holds for a ∈ (-∞,1 + sqrt(2)].But we have the initial condition from the discriminant that a ≥-5/4. Therefore, combining both conditions, the values of a are:a ∈ [-5/4,1 + sqrt(2)].But wait, this seems to conflict with the answer choices. Let me check the answer choices:(A) [−5/4, 1−√2](B) [−5/4, 1+√2](C) [1−√2, 1+√2](D) [−3/2,1−√2]So option B is [-5/4,1 + sqrt(2)], which is what I just derived. But wait, but in the previous steps, we had the quadratic inequality a² -2a -1 ≤0 leading to a ∈ [1 - sqrt(2),1 + sqrt(2)] when a >1/2, but combined with a ≥-5/4, the total interval is [-5/4,1 + sqrt(2)]. Therefore, the answer should be option B. However, let me verify this with the original problem.But wait, this seems like we might have made a mistake. Because when we considered the inequality sqrt(4a +5) ≥2a -1, and found that this is true for a ∈ (-∞,1 + sqrt(2)] intersected with a ≥-5/4. Therefore, the valid a is [-5/4,1 + sqrt(2)]. But let me test this with some examples.Take a =1 + sqrt(2). Then check if y1 ≥-1.Compute y1:y1 = [ - (2*(1 + sqrt(2)) +1) + sqrt(4*(1 + sqrt(2)) +5) ] /2First, compute 2*(1 + sqrt(2)) +1 =2 +2sqrt(2) +1=3 +2sqrt(2)sqrt(4*(1 + sqrt(2)) +5) =sqrt(4 +4sqrt(2) +5)=sqrt(9 +4sqrt(2))Need to compute sqrt(9 +4sqrt(2)). Let me see if this simplifies. Suppose sqrt(9 +4sqrt(2))=sqrt(a) + sqrt(b), then squaring both sides: a + b + 2sqrt(ab)=9 +4sqrt(2). Therefore, a +b=9 and 2sqrt(ab)=4sqrt(2) ⇒ sqrt(ab)=2sqrt(2) ⇒ ab=8. So solving a +b=9 and ab=8. The solutions are roots of x² -9x +8=0 ⇒ x=(9±sqrt(81-32))/2=(9±sqrt(49))/2=(9±7)/2 ⇒ x=8 or x=1. Therefore, sqrt(9 +4sqrt(2))=sqrt(8)+sqrt(1)=2sqrt(2)+1. Therefore, sqrt(9 +4sqrt(2))=1 +2sqrt(2).Therefore, y1 = [ - (3 +2sqrt(2)) + (1 +2sqrt(2)) ] /2 = [ -3 -2sqrt(2) +1 +2sqrt(2) ] /2 = (-2)/2 = -1.Therefore, when a=1 +sqrt(2), y1=-1, which is within [-1,1]. Therefore, valid.Similarly, at a= -5/4, as before, y1=3/4 which is within [-1,1]. Therefore, valid.Therefore, the interval is from -5/4 to 1 +sqrt(2), which is option B.Wait, but let me check if there is a mistake here. Because when we solved the inequality sqrt(4a +5) ≥2a -1, we concluded that the solution is a ∈ (-∞,1 + sqrt(2)], but combined with a ≥-5/4, it becomes [-5/4,1 + sqrt(2)]. However, the answer options include this as option B. But the original problem's answer options are:(A) [−5/4, 1−√2](B) [−5/4, 1+√2](C) [1−√2, 1+√2](D) [−3/2,1−√2]Therefore, option B is the correct answer. However, in some of the steps, we derived a quadratic inequality leading to a ∈ [1 - sqrt(2),1 + sqrt(2)] when a >1/2, but combined with the rest, but perhaps I missed something.Wait, let's recap:We had the inequality sqrt(4a +5) ≥2a -1, which is equivalent to y1 ≥ -1.We split into cases based on 2a -1 ≤0 or >0.Case 1: a ≤1/2, inequality holds automatically.Case 2: a >1/2, need to satisfy a² -2a -1 ≤0 ⇒ a ∈ [1 - sqrt(2),1 + sqrt(2)].But since a >1/2, the intersection is a ∈ (1/2,1 + sqrt(2)].Therefore, combining both cases:For a ≤1/2: valid.For a >1/2: valid only if a ∈ (1/2,1 + sqrt(2)].Thus, the entire solution set is a ∈ (-∞,1 + sqrt(2)] intersected with a ≥-5/4.But since 1 + sqrt(2) ≈2.414, which is greater than -5/4≈-1.25. Therefore, the intersection is a ∈ [-5/4,1 + sqrt(2)].Therefore, the correct answer is option B.But let me check if there is any restriction I missed.Wait, when a=1 + sqrt(2), we saw that y1=-1, which is acceptable because cosx=-1 has solution x=π. So that's valid.If a is greater than1 + sqrt(2), say a=3. Then check if the equation has real solutions.Compute D=4*3 +5=17>0.Compute the roots:y1 = [ -7 + sqrt(17) ] /2 ≈ (-7 +4.123)/2≈(-2.877)/2≈-1.4385, which is less than -1.y2 = [ -7 - sqrt(17) ] /2 ≈ (-7 -4.123)/2≈-5.5615, also less than -1.Therefore, both roots are less than -1, so cosx cannot be equal to them, hence no real solutions. Therefore, a=3 is invalid, as expected since a=3 >1 + sqrt(2)≈2.414.Another check at a=1 - sqrt(2). Let's compute a=1 - sqrt(2)≈-0.414.Check if a=1 - sqrt(2) is included in the interval.Since 1 - sqrt(2)≈-0.414 which is greater than -5/4≈-1.25, so it's within the interval [-5/4,1 + sqrt(2)].Compute y1 when a=1 - sqrt(2).y1 = [ - (2*(1 - sqrt(2)) +1) + sqrt(4*(1 - sqrt(2)) +5) ] /2First compute 2*(1 - sqrt(2)) +1=2 -2sqrt(2) +1=3 -2sqrt(2)sqrt(4*(1 - sqrt(2)) +5)=sqrt(4 -4sqrt(2) +5)=sqrt(9 -4sqrt(2)).Again, trying to simplify sqrt(9 -4sqrt(2)). Suppose sqrt(9 -4sqrt(2))=sqrt(a)-sqrt(b), then squaring: a + b -2sqrt(ab)=9 -4sqrt(2). Therefore:a + b=9,-2sqrt(ab)= -4sqrt(2) ⇒ sqrt(ab)=2sqrt(2) ⇒ ab=8.Therefore, solving a + b=9 and ab=8. Same as before, roots are 8 and1. Therefore, sqrt(9 -4sqrt(2))=sqrt(8) - sqrt(1)=2sqrt(2)-1.Thus, sqrt(9 -4sqrt(2))=2sqrt(2)-1.Therefore, y1 = [ - (3 -2sqrt(2)) + (2sqrt(2)-1) ] /2 = [ -3 +2sqrt(2) +2sqrt(2) -1 ] /2 = [ -4 +4sqrt(2) ] /2 = (-4 +4sqrt(2))/2= -2 +2sqrt(2).Compute this numerically: sqrt(2)≈1.414, so 2sqrt(2)=2.828. Then -2 +2.828≈0.828. So y1≈0.828, which is within [-1,1]. Therefore, a=1 - sqrt(2) is valid, and y1 is within the interval. Therefore, a=1 - sqrt(2) is indeed part of the solution set. But according to our previous interval, [-5/4,1 + sqrt(2)], a=1 - sqrt(2) is included. However, the answer option B includes it as well. Therefore, all seems consistent.But wait, in our analysis, the solution is a ∈ [-5/4,1 + sqrt(2)], which includes 1 - sqrt(2). So why is there an option A which is [−5/4,1−√2]?Ah, perhaps I need to re-examine the logic. Because when a=1 - sqrt(2), which is approximately -0.414, which is greater than -5/4≈-1.25, so it's within the interval. But when we set the inequality sqrt(4a +5) ≥2a -1, the solution was a ∈ (-∞,1 + sqrt(2)], intersected with a ≥-5/4. So indeed, all a from -5/4 up to1 + sqrt(2) are valid. However, in this case, why does the quadratic equation have a solution?Wait, but perhaps there is a mistake here. Because when a=1 - sqrt(2), which is approximately -0.414, let's check if the equation has real solutions.Take a=1 - sqrt(2). Then the quadratic equation in y is:y² + (2*(1 - sqrt(2)) +1)y + ( (1 - sqrt(2))² -1 )=0Compute coefficients:2*(1 - sqrt(2)) +1=2 -2sqrt(2)+1=3 -2sqrt(2)(1 - sqrt(2))² -1=1 -2sqrt(2)+2 -1=2 -2sqrt(2)Therefore, equation is y² + (3 -2sqrt(2))y + (2 -2sqrt(2))=0Let me compute the discriminant D=(3 -2sqrt(2))² -4*1*(2 -2sqrt(2))Expand (3 -2sqrt(2))²=9 -12sqrt(2) +8=17 -12sqrt(2)4*1*(2 -2sqrt(2))=8 -8sqrt(2)Thus, D=17 -12sqrt(2) -8 +8sqrt(2)=9 -4sqrt(2)Which is positive, as sqrt(2)≈1.414, so 4sqrt(2)≈5.656. Then 9 -5.656≈3.344>0.Therefore, the roots are real. Now, compute y1 and y2.y1 = [ - (3 -2sqrt(2)) + sqrt(9 -4sqrt(2)) ] /2Earlier, we found sqrt(9 -4sqrt(2))=2sqrt(2)-1, so:y1 = [ -3 +2sqrt(2) +2sqrt(2)-1 ] /2 = [ -4 +4sqrt(2) ] /2 = -2 +2sqrt(2)≈-2 +2.828≈0.828∈ [-1,1]Similarly, y2 = [ -3 +2sqrt(2) - (2sqrt(2)-1) ] /2 = [ -3 +2sqrt(2) -2sqrt(2)+1 ] /2 = (-2)/2 = -1.Therefore, y2=-1, which is in [-1,1]. Therefore, when a=1 - sqrt(2), both roots y1≈0.828 and y2=-1 are in [-1,1]. Therefore, a=1 - sqrt(2) is valid. However, according to the interval we found earlier, [-5/4,1 + sqrt(2)], 1 - sqrt(2) is within that interval, so it's included.But looking at the answer choices, option B is [-5/4,1 + sqrt(2)], which includes all values from -5/4 up to1 + sqrt(2). Option A is [-5/4,1 - sqrt(2)], which would be from -1.25 to approximately -0.414. But in our analysis, the valid a goes up to1 + sqrt(2). So unless there is a mistake, option B is correct.But let me check another value between1 - sqrt(2) and1 + sqrt(2), say a=0.5.At a=0.5, check if the equation has solutions.Quadratic equation: y² + (2*0.5 +1)y + (0.25 -1)= y² +2y -0.75=0Discriminant D=4 +3=7>0. Roots: y=(-2 ±sqrt(7))/2≈(-2 ±2.645)/2.Thus, y1≈(0.645)/2≈0.322, y2≈(-4.645)/2≈-2.322. Only y1≈0.322 is in [-1,1]. So there is a solution. Hence, a=0.5 is valid, and since0.5 is between1 - sqrt(2)≈-0.414 and1 + sqrt(2)≈2.414, it's within the interval [-5/4,1 + sqrt(2)], so valid.Another test at a=1. Let's compute:Quadratic equation: y² + (2*1 +1)y + (1 -1)= y² +3y +0=0 ⇒ y(y +3)=0. Roots y=0 and y=-3. So y=0 is within [-1,1], so valid. Hence, a=1 is valid and included in the interval.Therefore, all tests confirm that the interval is [-5/4,1 + sqrt(2)], which is option B. Therefore, the correct answer should be option B.But wait, initially, I thought the answer might involve [1 - sqrt(2),1 + sqrt(2)], but that is not the case. Because the discriminant requires a ≥-5/4, and the other condition expands up to1 + sqrt(2). Therefore, option B is correct.Final Answerboxed{B}

✅:Given the equation (sin^{2} x - (2a + 1) cos x - a^2 = 0), we need to determine the set of values for the real number (a) such that the equation has real solutions for (x).1. Convert to Quadratic Equation: Using the identity (sin^2 x = 1 - cos^2 x), substitute (sin^2 x) in the equation: [ 1 - cos^2 x - (2a + 1) cos x - a^2 = 0 ] Rearrange to form a quadratic equation in (cos x): [ cos^2 x + (2a + 1) cos x + (a^2 - 1) = 0 ]2. Quadratic Equation Analysis: Let (y = cos x). The quadratic equation becomes: [ y^2 + (2a + 1)y + (a^2 - 1) = 0 ] For real solutions, the discriminant must be non-negative: [ (2a + 1)^2 - 4(a^2 - 1) geq 0 ] Simplify the discriminant: [ 4a + 5 geq 0 implies a geq -frac{5}{4} ]3. Roots within Interval ([-1, 1]): The roots of the quadratic equation are: [ y = frac{-(2a + 1) pm sqrt{4a + 5}}{2} ] We need at least one root within ([-1, 1]). The larger root (y_1) is: [ y_1 = frac{-(2a + 1) + sqrt{4a + 5}}{2} ] The smaller root (y_2) is: [ y_2 = frac{-(2a + 1) - sqrt{4a + 5}}{2} ]4. Condition for Roots: - For (y_1 geq -1): [ sqrt{4a + 5} geq 2a - 1 ] - If (2a - 1 leq 0) (i.e., (a leq frac{1}{2})), the inequality holds. - If (2a - 1 > 0) (i.e., (a > frac{1}{2})), square both sides: [ 4a + 5 geq (2a - 1)^2 implies a^2 - 2a - 1 leq 0 implies a in [1 - sqrt{2}, 1 + sqrt{2}] ] Combining these results, (a) must be in ([- frac{5}{4}, 1 + sqrt{2}]).Thus, the set of values for (a) is (left[-frac{5}{4}, 1 + sqrt{2}right]).[boxed{B}]

🔑:1. Let's start by transforming the given equation (sin^2 x - (2a + 1) cos x - a^2 = 0). Using the Pythagorean identity (sin^2 x = 1 - cos^2 x), we can rewrite the equation: [ 1 - cos^2 x - (2a + 1) cos x - a^2 = 0 ] Simplifying, we get: [ cos^2 x + (2a + 1) cos x + a^2 - 1 = 0 ]2. Let ( t = cos x ). Therefore, the equation becomes: [ t^2 + (2a + 1)t + a^2 - 1 = 0 ]3. Define ( f(t) = t^2 + (2a + 1)t + a^2 - 1 ) where ( t in [-1, 1] ).4. For ( f(t) ) to have real roots in the interval ([-1, 1]), the following conditions must be met: [ begin{cases} (2a + 1)^2 - 4(a^2 - 1) geq 0 & text{(discriminant condition)} f(1) geq 0 & text{(value at (t=1) condition)} f(-1) geq 0 & text{(value at (t=-1) condition)} -1 leq -frac{2a + 1}{2} leq 1 & text{(root location condition)} end{cases} ] 5. Solve the discriminant condition: [ (2a + 1)^2 - 4(a^2 - 1) = 4a^2 + 4a + 1 - 4a^2 + 4 = 4a + 5 geq 0 implies a geq -frac{5}{4} ]6. Evaluate ( f(1) ): [ f(1) = 1 + (2a + 1) + a^2 - 1 = 2a + 1 + a^2 geq 0 ]7. Evaluate ( f(-1) ): [ f(-1) = 1 - (2a + 1) + a^2 - 1 = a^2 - 2a - 2 geq 0 ]8. Solve the root location condition: [ -1 leq -frac{2a + 1}{2} leq 1 implies -2 leq -(2a + 1) leq 2 implies 1 - sqrt{2} leq a leq 1 + sqrt{2} ]Combining all the conditions, we find that:[-frac{5}{4} leq a leq 1 + sqrt{2}] Conclusion:The real number (a) that satisfies all conditions is in the interval (left[-frac{5}{4}, 1 + sqrt{2}right]).(boxed{text{B}})

🧠:Okay, so I need to figure out if a 10x10 checkerboard can be tiled completely with 1x4 tiles. Hmm, let me start by understanding the problem. Tiling means covering the entire board without overlapping or leaving any gaps, right? Each tile is 1 unit by 4 units, so they can be placed either horizontally or vertically. First, let me check the basic area requirement. The area of the checkerboard is 10x10, which is 100 squares. Each 1x4 tile covers 4 squares, so the total number of tiles needed would be 100 divided by 4, which is 25. Since 25 is an integer, the area condition is satisfied. That means it's possible in terms of area, but that's just the first step. There might be other constraints that make it impossible.Next, I remember that tiling problems often involve coloring arguments. Maybe a checkerboard coloring could help here. Let me think. A standard checkerboard uses two colors alternating, but with 1x4 tiles, maybe a different coloring scheme is needed. Let me try that.If I color the 10x10 board in the usual checkerboard pattern with black and white squares, each 1x4 tile will always cover either 2 black and 2 white squares, regardless of its orientation. Because if you place it horizontally, it alternates colors every square, so starting on black, it would be black, white, black, white. Similarly, vertical placement would also cover two of each. So each tile covers 2 black and 2 white squares. Since there are 25 tiles, that would cover 50 black and 50 white squares. But in a 10x10 checkerboard, the number of black and white squares is equal, right? Since 10x10 is even, there are 50 black and 50 white squares. So the coloring argument here doesn't rule out the tiling, because the counts match. So maybe that's not the issue.Hmm, maybe another coloring scheme. I've heard of using four colors in some tiling problems. Let me try that. If I divide the board into 4x4 sections and color them in a repeating pattern, but since the tile is 1x4, maybe a four-coloring where each row cycles through four colors. Wait, let me think. If I color each row in a cycle of four colors, say, red, blue, green, yellow, repeating every four columns. Then a horizontal 1x4 tile would cover one of each color. A vertical tile, however, would cover four squares of the same color, because each row in the same column would have the same color if we cycle every four columns. Wait, no, if each row is colored in a repeating four-color pattern, then in a column, the color would cycle every four rows. Wait, maybe this is getting complicated. Let me try to visualize.Alternatively, maybe use a checkerboard with four colors instead of two. For example, a 4-coloring where each 2x2 block has all four colors. Then a 1x4 tile would cover different numbers of each color depending on its placement. Let me try to formalize this. If the four-color scheme is such that every 2x2 square has all four colors, then a horizontal 1x4 tile would cover two squares of two different colors each. Wait, in a 2x2 block, colors are arranged in a cycle, so a horizontal tile spanning four squares would cover two of each color in two different 2x2 blocks. Hmm, maybe not. Alternatively, if the coloring is such that each row is colored in a repeating four-color pattern, like color 1,2,3,4,1,2,3,4,... then in each row, every fourth square is the same color. Then a horizontal tile would cover one of each color, since it spans four different columns, each with a different color. A vertical tile would cover four squares in the same column, but since the coloring cycles every four rows, a vertical tile of length 4 would cover each color once. Wait, if the rows are colored with a 4-coloring repeating every four columns, then columns would have colors that cycle every four rows. Let me check that.Suppose each row is colored as 1,2,3,4,1,2,3,4,1,2. Then in column 1, the first row is color 1, second row is color 1 (if the next row starts with color 1 again?), wait no, maybe I need to define the coloring properly. Let's use a toroidal coloring where each cell (i,j) is colored (i + j) mod 4. Then each cell's color is determined by the sum of its row and column indices modulo 4. Let's see. For example, cell (0,0) is color 0, (0,1) color 1, (0,2) color 2, (0,3) color 3, (0,4) color 0 again, etc. Similarly, cell (1,0) is 1+0=1 mod4, cell (1,1)=2, etc. In this coloring, a horizontal tile from (i,j) to (i,j+3) would cover colors (i+j), (i+j+1), (i+j+2), (i+j+3) mod4, which are four consecutive colors, so 0,1,2,3 if starting at 0. Similarly, a vertical tile from (i,j) to (i+3,j) would cover colors (i+j), (i+1+j), (i+2+j), (i+3+j) mod4, which is (i+j), (i+j+1), (i+j+2), (i+j+3) mod4, again all four colors. So in this 4-coloring, every 1x4 tile, regardless of orientation, covers exactly one of each color. Therefore, if the entire board is colored this way, each color would need to be covered exactly 25 times, since 100 squares divided by 4 colors is 25 each. But let's check how many squares of each color there are in a 10x10 grid with this coloring.The total number of cells of each color: For each cell (i,j), color is (i+j) mod4. Since both i and j range from 0 to 9 (assuming 0-based indexing), the sum (i+j) can range from 0 to 18. However, modulo4, the possible residues are 0,1,2,3. Let's count how many times each residue occurs. For a 10x10 grid, there are 100 cells. The number of cells where (i+j) ≡k mod4 for k=0,1,2,3.This can be calculated as follows: For each row i, the residues of (i+j) mod4 as j goes from 0 to9. For a fixed i, j varies, so (i+j) mod4 cycles through (i mod4), (i+1) mod4, ..., (i+9) mod4. Since 10=4*2 +2, each residue class will appear either 2 or 3 times per row. Wait, maybe a better approach is to note that for each i, the number of times each residue occurs in j from 0 to9. For example, if i is fixed, then j runs through 0 to9, so i+j runs through i to i+9. The residues mod4 here will cycle every 4. So over 10 columns, each residue will appear either 2 or 3 times. Specifically, for each row, each residue occurs floor(10/4)=2 times, with a remainder of 2, so two residues will have an extra count. Therefore, in each row, two residues will have 3 occurrences and two will have 2. Therefore, over 10 rows, each residue will have 10*(2 or 3). Wait, maybe this is getting too detailed. Alternatively, note that in total, for all cells, (i+j) mod4. The total count for each residue can be found by noting that for each i from 0 to9, and j from 0 to9, the sum i+j. The number of times each residue occurs is (number of pairs (i,j) with (i+j) ≡k mod4). Because i and j are independent, this is equivalent to convolution of the distributions of i mod4 and j mod4. Since both i and j run from 0 to9, let's compute how many i ≡a mod4 and how many j≡b mod4.For i from 0 to9: Each residue a=0,1,2,3 occurs floor(10/4)=2 times with a remainder of 2. So residues 0,1,2,3 occur 3,3,2,2 times respectively. Wait: Let's check. 10 divided by4 is 2 with remainder 2. So the first two residues (0 and1) get an extra count. Wait, no, starting from i=0: i=0,1,2,3,4,5,6,7,8,9. i mod4: 0,1,2,3,0,1,2,3,0,1. So residue counts: 0 occurs 3 times (i=0,4,8), 1 occurs 3 times (i=1,5,9), 2 occurs 2 times (i=2,6), 3 occurs 2 times (i=3,7). Similarly for j from0 to9, same counts. So the number of pairs (i,j) where i≡a and j≡b is the product of counts for a and b. To get the number of pairs where (i+j)≡k mod4, we need to sum over all a and b such that a+b ≡k mod4. For example, for k=0: sum over a=0 to3, and b=(0 -a) mod4. So for each a, b=( -a) mod4. So the total count for k=0 is sum_{a=0}^3 count_i(a) * count_j((-a) mod4). Since count_i and count_j are same for both i and j.So count_i(0)=3, count_i(1)=3, count_i(2)=2, count_i(3)=2.Similarly count_j(0)=3, count_j(1)=3, count_j(2)=2, count_j(3)=2.For k=0:sum_{a=0}^3 count_i(a) * count_j((-a) mod4)= count_i(0)*count_j(0) + count_i(1)*count_j(3) + count_i(2)*count_j(2) + count_i(3)*count_j(1)=3*3 +3*2 +2*2 +2*3 =9 +6 +4 +6=25Similarly for k=1:sum_{a=0}^3 count_i(a)*count_j((1 -a) mod4)= count_i(0)*count_j(1) + count_i(1)*count_j(0) + count_i(2)*count_j(3) + count_i(3)*count_j(2)=3*3 +3*3 +2*2 +2*2=9+9+4+4=26For k=2:sum_{a=0}^3 count_i(a)*count_j((2 -a) mod4)= count_i(0)*count_j(2) + count_i(1)*count_j(1) + count_i(2)*count_j(0) + count_i(3)*count_j(3)=3*2 +3*3 +2*3 +2*2=6+9+6+4=25For k=3:sum_{a=0}^3 count_i(a)*count_j((3 -a) mod4)= count_i(0)*count_j(3) + count_i(1)*count_j(2) + count_i(2)*count_j(1) + count_i(3)*count_j(0)=3*2 +3*2 +2*3 +2*3=6+6+6+6=24Wait, so total counts are 25,26,25,24 for k=0,1,2,3. That sums to 25+26+25+24=100, which checks out. So in this 4-coloring, the counts are 25,26,25,24. But each 1x4 tile covers exactly one of each color. So if we need to cover 25 of each color, but here we have 26 of color1 and 24 of color3. Therefore, it's impossible because the tiles require equal numbers of each color, but the board doesn't have equal counts. Therefore, such a tiling is impossible.Wait, that seems like a solid argument. So even though the area is compatible and the two-coloring didn't rule it out, the four-coloring shows that the color counts are unequal, making it impossible. Therefore, the answer is no, it cannot be tiled with 1x4 tiles.But let me just check if my 4-coloring counts are correct. Let's recount. For the rows and columns each having counts 3,3,2,2 for residues 0,1,2,3. Then convolution for k=0: 0+0,1+3,2+2,3+1. So 3*3 (i=0,j=0) +3*2 (i=1,j=3) +2*2 (i=2,j=2) +2*3 (i=3,j=1). Wait, wait, when k=0, we need j ≡ (-a) mod4. So for a=0, j≡0; a=1, j≡3; a=2, j≡2; a=3, j≡1. So counts are count_i(0)*count_j(0) + count_i(1)*count_j(3) + count_i(2)*count_j(2) + count_i(3)*count_j(1). Which is 3*3 +3*2 +2*2 +2*3 =9+6+4+6=25. Similarly for k=1: j≡1 -a. For a=0, j≡1; a=1, j≡0; a=2, j≡3; a=3, j≡2. So counts: count_i(0)*count_j(1) + count_i(1)*count_j(0) + count_i(2)*count_j(3) + count_i(3)*count_j(2) =3*3 +3*3 +2*2 +2*2=9+9+4+4=26. For k=2: j≡2 -a. a=0:j≡2; a=1:j≡1; a=2:j≡0; a=3:j≡3. So counts: count_i(0)*count_j(2) + count_i(1)*count_j(1) + count_i(2)*count_j(0) + count_i(3)*count_j(3)=3*2 +3*3 +2*3 +2*2=6+9+6+4=25. For k=3: j≡3 -a. a=0:j≡3; a=1:j≡2; a=2:j≡1; a=3:j≡0. Counts: count_i(0)*count_j(3) + count_i(1)*count_j(2) + count_i(2)*count_j(1) + count_i(3)*count_j(0)=3*2 +3*2 +2*3 +2*3=6+6+6+6=24. Yes, that adds up. So the 4-coloring shows that colors 0,1,2,3 have 25,26,25,24 squares respectively. Therefore, since each tile must cover one of each color, you would need exactly 25 of each color, but color1 has 26 and color3 has 24, making it impossible. Therefore, the tiling is impossible.But let me think if there's another way to approach this. Maybe parity or something else. Alternatively, divide the board into 2x2 blocks. Each 2x2 block has 4 squares. A 1x4 tile covers 4 squares, so it could cover exactly one 2x2 block if placed appropriately. But wait, a 1x4 tile can't fit into a 2x2 block. Alternatively, maybe considering that each 1x4 tile will cover parts of multiple 2x2 blocks. Not sure. Alternatively, think about the number of tiles in horizontal vs vertical orientation. But I don't see an immediate way to use that.Alternatively, think about the positions of the tiles. Since the board is 10x10, and 10 is not a multiple of 4, except when considering that 10=4*2 +2. So if we try to tile the board, maybe there's a problem with the leftover spaces. But I think the coloring argument is more solid here.Wait, but let's try to actually attempt a tiling. Suppose we try to place tiles horizontally. If we place them in rows, each row is 10 units long. 10 divided by4 is 2 with a remainder of 2. So two tiles of length4, covering 8 units, leaving 2 units. Similarly, vertically, each column is 10 units, which also leaves a remainder of2 when divided by4. So if we try to tile the board with horizontal tiles, we end up with leftover 2-unit spaces in each row, which can't be filled with vertical tiles. Similarly, if we use vertical tiles, leftover spaces in columns. But if we mix horizontal and vertical tiles, maybe we can cover the leftovers. But this is getting vague. Let me think.Suppose we divide the board into 4x4 sections. Since 10 isn't a multiple of4, this would leave a border of 2 units around. For example, the board would have a 8x8 section (which is divisible by4) and then 2-unit borders on the right and bottom. But 8x8 can be tiled with 1x4 tiles. Then the borders would be 8x2 and 2x8, which can also be tiled with 1x4 tiles. Wait, 8x2 can be tiled with horizontal 1x4 tiles (since 8 is divisible by4). Similarly, 2x8 can be tiled with vertical 1x4 tiles (since 8 is divisible by4). Then the remaining 2x2 section? Wait, no. If the board is 10x10, subtracting 8x8 leaves a border of 2 around all sides, but actually, 10-8=2, so it's an L-shaped border of width 2. But an L-shaped 2x10 + 2x8 - 2x2 (to avoid overlapping). Not sure. Alternatively, maybe it's possible to partition the 10x10 into regions that are each 4xN or Nx4. But 10 is not a multiple of4, so that might not work. Alternatively, think of the 10x10 as a combination of 4x10 and 6x10. But 6 is also not a multiple of4. This approach seems messy.Alternatively, maybe use induction or some other method. But I think the four-coloring argument is the key here. Since the counts of colors are unequal, and each tile must cover exactly one of each color, it's impossible. Therefore, the conclusion is that a 10x10 checkerboard cannot be tiled with 1x4 tiles.

✅:A 10 times 10 checkerboard cannot be tiled with 1 times 4 tiles. Here's the systematic reasoning:1. Area Check: The total area is 10 times 10 = 100, and each 1 times 4 tile covers 4 squares. Since 100 div 4 = 25 is an integer, the area condition is satisfied. However, this alone does not guarantee a valid tiling.2. Two-Coloring Argument: Using a standard checkerboard pattern, each 1 times 4 tile covers 2 black and 2 white squares. The board has 50 black and 50 white squares, so parity is preserved. This does not rule out tiling.3. Four-Coloring Argument: - Assign colors using (i + j) mod 4 for cell (i, j). This results in four colors (0, 1, 2, 3). - Counts of each color: Calculations show colors 0, 1, 2, 3 occur 25, 26, 25, and 24 times, respectively. - Each 1 times 4 tile must cover one of each color. For a valid tiling, all color counts must match (25 each). However, color 1 appears 26 times and color 3 appears 24 times, violating this requirement.4. Conclusion: The imbalance in color counts under the four-coloring scheme makes it impossible to tile the board with 1 times 4 tiles. The tiling is not possible.boxed{text{No}}

🔑:In order to determine whether a 10 times 10 chessboard can be tiled using 1 times 4 tiles, we will describe three different approaches. Each approach demonstrates why it is impossible to completely cover the chessboard with such tiles.# First Approach:1. Coloring the Chessboard: - Color part of the 10 times 10 board with a black and white checkerboard pattern such that there is an equal number of black and white cells but starting arbitrarily without consideration for a perfect 1-to-1 balance within each tile 1 times 4. - Each 1 times 4 tile can only cover either 0 or 2 black cells since it is 1 cell wide and 4 cells long. Therefore, it always covers an even number of black cells (either 0 or 2). - If the entire board is to be covered by such tiles, the total number of black cells should be an even number.2. Counting Black Cells: - On a 10 times 10 board, with the described arbitrarily initial coloring, there could be 25 black cells and 25 white cells, thus the board does not maintain the required even number of black cells for tiling.3. Conclusion: - Since tiles can't perfectly cover the board under the described conditions, it follows that the tiling is impossible given any arbitrary coloring leading to initial imbalance holding 25 black cells. 4. Conclusion: [ boxed{text{Impossible to tile a } 10 times 10 text{ board with } 1 times 4 text{ tiles.}} ]# Second Approach:1. Coloring in Four Colors: - Color the 10 times 10 board with four colors so that each group of rows gets a different color. Let's say the colors used rotate after each 4 rows. - Specifically, check each cell assignment in sequence, ensuring the first block of 4 cells is one color, the next four cells the second color, etc.2. Counting Cells of Each Color: - Each color will occur 25 times (since 10/4 = 2.5 iteration sets meaning each color localizes its frequency out to sustain an average 26.5).3. Observing the Tile Placement: - A tile 1 times 4 intersects one cell of each color. - Given any hypothetical tiling completion would imply each segment covered makes achieving even set cell of similar 1 times 4 value impossible.4. Conclusion: - Conclusively, as we can't impose such even 4 variations over to applicability mismatch setting balanced yet local 2.5 frequency means, implies placement issues thus impossible tiling. 5. Conclusion: [ boxed{text{Impossible to tile a } 10 times 10 text{ board with } 1 times 4 text{ tiles.}} ]# Third Approach:1. Further Board Division: - Sub-divide the 10 times 10 board into 2 times 2 mini-squares. - Color cells in mini-squares in a checkerboard pattern, ensuring each 2 times 2 has 2 black and 2 white.2. Analyzing Tile Placement: - Note, each 1 times 4 tile intersection constraints thus within any tiling attempt will cover 2 cells (let's assume equal i.e., black/white split mandates within 2times 2 checker condition).3. Observation: - Given if 2 times 2 overall completion would mean perfectly balanced split of entire board etc, but re-arranging poses significant local balancing thus impossible tiling.4. Conclusion: - The impossibility lane noted mandates establishing valuation incompatible 25 big/black-white mismatch issue. 5. Conclusion: [ boxed{text{Impossible to tile a } 10 times 10 text{ board with } 1 times 4 text{ tiles.}} ]

🧠:Alright, so I need to figure out the highest number of successful free throws Andrew and Beatrice could have made together, given that they attempted a total of 105 free throws, each took at least one, Andrew made 1/3 of his attempts, and Beatrice made 3/5 of hers. Let's break this down step by step.First, let me denote the number of free throws Andrew attempted as A and Beatrice as B. According to the problem, A + B = 105, and both A and B are at least 1. Andrew's made free throws would be (1/3)A, and Beatrice's made free throws would be (3/5)B. The total successful free throws would be the sum of these two: (1/3)A + (3/5)B. Our goal is to maximize this total.Since A and B are integers (you can't attempt a fraction of a free throw), we need to find integer values of A and B such that A + B = 105, A ≥ 1, B ≥ 1, and (1/3)A + (3/5)B is as large as possible.Let me express one variable in terms of the other using the equation A + B = 105. Let's solve for A: A = 105 - B. Then substitute this into the total made free throws equation:Total made = (1/3)(105 - B) + (3/5)BLet me simplify this expression:First, distribute the 1/3:= (1/3)*105 - (1/3)B + (3/5)B= 35 - (1/3)B + (3/5)BNow, combine the terms with B. To combine - (1/3)B and + (3/5)B, find a common denominator, which is 15.Convert fractions:- (1/3)B = - (5/15)B+ (3/5)B = + (9/15)BSo adding these together: (-5/15 + 9/15)B = (4/15)BTherefore, total made = 35 + (4/15)BSo the total made free throws depend linearly on B with a positive coefficient (4/15). That means, to maximize the total made, we need to maximize B. Because as B increases, the total made increases. Therefore, the larger B is, the higher the total made free throws. But we have constraints: A and B must each be at least 1, so B can be as large as 105 - 1 = 104. However, there's another constraint: Beatrice's made free throws are (3/5)B, which requires B to be a multiple of 5 for her made free throws to be an integer. Similarly, Andrew's made free throws are (1/3)A, which requires A = 105 - B to be a multiple of 3. Therefore, both A and B must satisfy these divisibility conditions.So, to recap, B must be such that:1. B is an integer between 1 and 104 inclusive.2. A = 105 - B must be a multiple of 3.3. B must be a multiple of 5.Therefore, B has to be a multiple of 5, and 105 - B has to be a multiple of 3.Let me check the divisibility conditions:First, since B is a multiple of 5, we can write B = 5k, where k is a positive integer (since B ≥1, k ≥1). Then, A = 105 - 5k must be a multiple of 3.So, 105 - 5k ≡ 0 mod 3.105 divided by 3 is 35, so 105 ≡ 0 mod 3.Therefore, -5k ≡ 0 mod 3 ⇒ 5k ≡ 0 mod 3.Since 5 ≡ 2 mod 3, so 2k ≡ 0 mod 3 ⇒ 2k is divisible by 3 ⇒ k must be divisible by 3.Therefore, k must be a multiple of 3. Let k = 3m, where m is a positive integer. Then, B = 5k = 15m.But B must also satisfy B ≤ 104 (since A ≥1). Therefore, 15m ≤ 104 ⇒ m ≤ 104/15 ≈6.933, so m ≤6 (since m is integer).Therefore, possible values of m are 1, 2, 3, 4, 5, 6.Therefore, possible values of B are 15, 30, 45, 60, 75, 90.Corresponding A values are 105 -15=90, 105-30=75, 105-45=60, 105-60=45, 105-75=30, 105-90=15.Now, check if these values satisfy A being a multiple of 3:90 is multiple of 3 (90/3=30), yes.75: 75/3=25, yes.60: 60/3=20, yes.45: 45/3=15, yes.30:30/3=10, yes.15:15/3=5, yes. All these A values are multiples of 3.So the possible B values are 15, 30, 45, 60, 75, 90, and A correspondingly 90, 75, 60, 45, 30, 15.Now, since the total made free throws is 35 + (4/15)B, and since we need to maximize this, and since (4/15)B increases as B increases, the maximum total made free throws would occur at the largest possible B, which is 90 (when m=6). Let's check:If B=90, then A=15.Andrew made (1/3)*15=5 free throws.Beatrice made (3/5)*90=54 free throws.Total made=5+54=59.But wait, let's verify this because 35 + (4/15)*90=35 + (4*6)=35 +24=59. Correct.But wait, let's check if B=90 is allowed. Since A=15, which is ≥1, yes. B=90 is allowed. So that's the maximum.But wait, let me check the next lower B value just to be thorough. For example, if B=75, A=30.Andrew made 10, Beatrice made 45. Total 55. Which is less than 59. Similarly, B=60: Andrew made 20, Beatrice made 36. Total 56. B=45: Andrew made 20, Beatrice made 27. Wait, wait, wait, hold on. Wait, if A=60, Andrew made (1/3)*60=20. Beatrice made (3/5)*45=27. Total 20+27=47? Wait, but according to the equation, 35 + (4/15)*45=35 +12=47. Correct. So yes, as B decreases, the total made decreases. So 59 is the maximum.But wait, let me check another thing. What if B is not a multiple of 15? Wait, but according to the earlier analysis, B must be a multiple of 15. Wait, no: B must be a multiple of 5, and A=105-B must be a multiple of 3, so the conclusion was that k must be a multiple of 3, hence B=15m. So B has to be multiples of 15. Therefore, possible B values are 15,30,45,60,75,90. Therefore, the next possible value above 90 would be 105, but that's not allowed because A would be zero, which is invalid. So 90 is the maximum allowed B.But let me think again: perhaps the original reasoning is correct, but maybe there's a way to have B not a multiple of 15 but still satisfy that A is a multiple of 3 and B a multiple of 5. Wait, for example, suppose B=5, then A=100. A=100 is not a multiple of 3 (100/3≈33.333). So Andrew's made free throws would be 100/3, which is not an integer. Therefore, invalid.Similarly, B=10: A=95. 95 is not a multiple of 3 (95/3≈31.666). Not integer. B=15: A=90. 90 is multiple of 3. So works.So only multiples of 15 work for B because only then does 105 - B become a multiple of 3 when B is multiple of 5. Because 105 is a multiple of both 3 and 5 (since 105=3*5*7). Therefore, 105 - B is multiple of 3 if and only if B is multiple of 3. But since B must be multiple of 5, then B must be multiple of LCM(3,5)=15. Therefore, B must be multiples of 15. Hence, the possible values are indeed 15,30,...,90.Therefore, the maximum B is 90, leading to total made free throws of 59. So the answer is 59.But wait, let me check this again. Suppose we try B=95. Then A=10. But A=10 is not a multiple of 3. So Andrew's made free throws would be 10/3≈3.333, which is not an integer. Therefore, invalid. Similarly, B=85: A=20. A=20 is not a multiple of 3. B=80: A=25, not multiple of 3. B=75: A=30, which is a multiple of 3. So yes, B=75 is allowed. So our previous reasoning holds.Therefore, the maximum possible B is 90, leading to 59 made free throws. So the answer is 59. But let me check once more with B=90:Andrew attempts 15, makes 5. Beatrice attempts 90, makes 54. Total 59. Correct. If I try to see if there's any other combination where B is higher than 90 but still satisfies A being a multiple of 3. For example, B=93. Then A=12. A=12 is a multiple of 3, but B=93 is not a multiple of 5. So Beatrice's made free throws would be (3/5)*93=55.8, which is not an integer. Therefore, invalid. Similarly, B=90 is the maximum that satisfies both conditions. So yes, 59 is the correct answer.But wait, another angle: maybe instead of requiring both A and B to be integers such that A is multiple of 3 and B multiple of 5, perhaps if we relax this, but since the problem states that each made exactly 1/3 and 3/5 of their attempts, then the number of made free throws must be integers. Therefore, A must be divisible by 3 and B must be divisible by 5. Therefore, our initial constraints are correct. Therefore, the possible B's are multiples of 15 up to 90.Therefore, the maximum total made is 59. Hence, the answer is boxed{59}.

✅:To determine the highest number of successful free throws Andrew and Beatrice could have made together, we start by defining the number of attempts as ( A ) for Andrew and ( B ) for Beatrice, with ( A + B = 105 ). Andrew made ( frac{1}{3}A ) free throws, and Beatrice made ( frac{3}{5}B ). We need to maximize ( frac{1}{3}A + frac{3}{5}B ).Expressing ( A ) in terms of ( B ), we get ( A = 105 - B ). Substituting this into the total made free throws equation:[text{Total made} = frac{1}{3}(105 - B) + frac{3}{5}B]Simplifying:[text{Total made} = 35 - frac{1}{3}B + frac{3}{5}B = 35 + frac{4}{15}B]To maximize this, ( B ) should be as large as possible. However, ( A ) must be a multiple of 3 and ( B ) a multiple of 5. Since ( A = 105 - B ), ( 105 - B ) must be a multiple of 3. Given ( 105 ) is a multiple of 3, ( B ) must also be a multiple of 3. Thus, ( B ) must be a multiple of both 3 and 5, i.e., a multiple of 15.Possible values of ( B ) are 15, 30, 45, 60, 75, and 90. The corresponding ( A ) values are 90, 75, 60, 45, 30, and 15. The largest ( B ) is 90, leading to:- Andrew's made free throws: ( frac{1}{3} times 15 = 5 )- Beatrice's made free throws: ( frac{3}{5} times 90 = 54 )Total made free throws: ( 5 + 54 = 59 ).Thus, the highest number of successful free throws they could have made is (boxed{59}).

🔑:To determine the highest number of successful free throws that Andrew and Beatrice could make between them, let’s follow the structured steps and employ some simple arithmetic and algebra.# 1:1. Identify the constraints and initial multiples: Since Andrew makes frac{1}{3} of his free throw attempts and Beatrice makes frac{3}{5} of hers, we conclude that both must have attempted a number of free throws that are multiples of these denominators. Specifically, Andrew's number of throws, denoted as (a), must be a multiple of 3, and Beatrice's number of throws, denoted as (b), must be a multiple of 5. The total number of throws is 105: [ a + b = 105 ] Since 15 is the least common multiple of 3 and 5, (a) and (b) must both be multiples of 15.2. Set a strategy for maximization: To maximize the total number of successful free throws, note that Beatrice's success rate ( frac{3}{5} ) is higher than Andrew's success rate ( frac{1}{3} ). Hence, we should maximize Beatrice’s throw attempts:3. Minimize Andrew's attempts and maximize Beatrice’s: Minimize Andrew’s attempts by setting (a = 15): [ a = 15 quad Rightarrow quad b = 105 - 15 = 90 ]4. Calculate successful free throws: Calculate the successful free throws for both: [ text{Successful throws by Andrew} = frac{1}{3} times 15 = 5 ] [ text{Successful throws by Beatrice} = frac{3}{5} times 90 = 54 ] Therefore, the total number of successful free throws made by both: [ 5 + 54 = 59 ]Thus, the maximum number of successful free throws they could have made together is:[boxed{59}]# 2:1. Introduce variables and set up initial equations: Let Andrew attempt (a) throws and Beatrice (b) throws, with the constraint: [ a + b = 105 quad text{with} quad a > 0 quad text{and} quad b > 0 ]2. Express the total number of successful throws (M): Since Andrew makes (frac{1}{3}) of his throws and Beatrice makes (frac{3}{5}) of hers: [ M = frac{a}{3} + frac{3b}{5} ]3. Substitute (b) with (105 - a): [ M = frac{a}{3} + frac{3 (105 - a)}{5} ]4. Simplify the expression for (M): [ M = frac{a}{3} + frac{3 times 105 - 3a}{5} = frac{a}{3} + frac{315 - 3a}{5} ]5. Combine terms into a common denominator of 15: [ M = frac{5a}{15} + frac{945 - 9a}{15} = frac{5a + 945 - 9a}{15} = frac{945 - 4a}{15} ]6. Determine conditions for maximization: Since (945 - 4a) achieves its maximum when (4a) is minimized. Therefore, set (a) to the smallest multiple of 15: [ frac{4a}{15} rightarrow min text{ when } a = 15 ]7. Final calculation for (M): [ M = frac{945 - 4 times 15}{15} = frac{945 - 60}{15} = frac{885}{15} = 59 ]Thus, the maximum number of successful free throws they could have made together is:[boxed{59}]

🧠:Okay, let's tackle this problem step by step. The question is about the relationship between the values of a gold coin and a silver coin, expressed in percentages. We need to find how many possible integer values for x exist, given that x% is how much more a gold coin is worth compared to a silver coin, and y% is how much less a silver coin is worth compared to a gold coin. Both x and y are positive integers.First, I need to make sure I understand the problem correctly. Let me paraphrase it: The gold coin's value is x% higher than the silver coin's. Conversely, the silver coin's value is y% lower than the gold coin's. Both percentages, x and y, are positive integers. We have to determine the number of possible integer values for x.Hmm. So, we have two different percentages here. When something is x% more than another, and then the reverse is y% less, the percentages aren't the same. For example, if something increases by 50%, then to go back to the original value, it needs to decrease by 33.33...%. So, they aren't reciprocal. Therefore, x and y are different. But in this problem, both x and y must be integers. So, we need to find all x (positive integers) such that there exists a positive integer y where these percentage relationships hold.Let me set up some variables. Let's denote the value of the silver coin as S. Then, the value of the gold coin G is x% more than S. So, G = S + (x/100)*S = S*(1 + x/100).On the other hand, the silver coin is y% less than the gold coin. So, S = G - (y/100)*G = G*(1 - y/100).So, substituting the first equation into the second, we can relate S and G. Let's do that. From the first equation, G = S*(1 + x/100). Plugging that into the second equation, we get:S = [S*(1 + x/100)] * (1 - y/100)We can divide both sides by S (assuming S ≠ 0, which makes sense because coins have value), so:1 = (1 + x/100)*(1 - y/100)That's the key equation. Let's write that out:1 = (1 + x/100)(1 - y/100)We need to find positive integers x and y such that this equation holds. Then, we have to count the possible x values.Let me rearrange the equation. Let's expand the right-hand side:1 = 1*(1 - y/100) + (x/100)*(1 - y/100)But maybe expanding directly is better:1 = (1 + x/100)(1 - y/100) = 1 - y/100 + x/100 - (xy)/10000Subtracting 1 from both sides:0 = - y/100 + x/100 - (xy)/10000Multiply both sides by 10000 to eliminate denominators:0 = -100y + 100x - xyLet's rearrange the terms:xy + 100y = 100xFactor out y on the left side:y(x + 100) = 100xTherefore, solving for y:y = (100x)/(x + 100)Since y must be a positive integer, this fraction must simplify to an integer. Therefore, (100x) must be divisible by (x + 100). So, x + 100 divides 100x. Let's denote d = x + 100. Then, d divides 100x. But d = x + 100, so x = d - 100. Therefore, substituting back, d divides 100*(d - 100). So:d | 100(d - 100)Which means that d divides 100d - 10000Therefore, d divides (100d - 10000) - 100*(d) = -10000So, d divides 10000Therefore, d is a positive divisor of 10000. But d = x + 100, and x is a positive integer, so d must be greater than 100 (since x is positive, d = x + 100 > 100). Also, since x is positive, d = x + 100 ≥ 101.Therefore, the possible values of d are the positive divisors of 10000 that are greater than 100. Then, x = d - 100, so each such d gives a corresponding x. But we also need to check that y is a positive integer. Since y = (100x)/d, and d divides 100x (from earlier), as d divides 10000 and d divides 100x, but since d divides 100x, but we have already set d = x + 100. Wait, maybe since d divides 10000, and we have y = 100x / d, which must be an integer.But since d divides 100x, and d divides 10000, perhaps we can just focus on d being a divisor of 10000 greater than 100.Let me confirm. So, if d is a divisor of 10000, then 100x = y*d. But since x = d - 100, substituting:100*(d - 100) = y*dTherefore, y = [100*(d - 100)] / dSince y must be a positive integer, [100*(d - 100)] must be divisible by d. So, d divides 100*(d - 100). But as we saw earlier, d divides 10000. So, perhaps this is redundant? Because if d divides 10000, then 100*(d - 100) = 100d - 10000, and since d divides 100d and d divides 10000, then d divides 100d - 10000. Therefore, y is an integer.Therefore, all divisors d of 10000 greater than 100 will yield an integer y. Hence, the values of d are the divisors of 10000 greater than 100, and x = d - 100. Therefore, the number of possible x is equal to the number of divisors of 10000 that are greater than 100.So, now, we need to find all divisors of 10000 greater than 100 and subtract 100 from each to get x. Then, the number of such divisors is the number of possible x.First, let's factorize 10000. 10000 = 10^4 = (2*5)^4 = 2^4 * 5^4. Therefore, the number of divisors is (4 + 1)(4 + 1) = 25. So, 10000 has 25 divisors. Now, we need to find how many of these divisors are greater than 100.But note that divisors come in pairs multiplying to 10000. For each divisor d less than sqrt(10000), there is a corresponding divisor 10000/d greater than sqrt(10000). Since sqrt(10000) is 100, so divisors less than 100 pair with divisors greater than 100. Divisors equal to 100 pair with themselves (since 100*100=10000). But 100 is a divisor here, right? Since 100 divides 10000.Therefore, total divisors: 25. Number of divisors less than 100: Let's compute. Then, the number of divisors greater than 100 would be equal to the number of divisors less than 100. But since 100 is a perfect square, there's an odd number of divisors, so one of them is equal to 100. Therefore, the number of divisors less than 100 is equal to the number of divisors greater than 100, and then there's the divisor 100 itself. So, total 25 divisors: 12 pairs (each pair multiplies to 10000) and 1 divisor at 100. Therefore, number of divisors greater than 100 is 12, and number less than 100 is 12, and one equal to 100. Therefore, the number of divisors greater than 100 is 12. Therefore, there are 12 possible values of d, each greater than 100, leading to 12 possible x = d - 100. Wait, but wait, is that correct?Wait, actually, the number of divisors greater than 100 would be equal to the number of divisors less than 100, which is 12 each, and one divisor equal to 100. But let's confirm this by enumerating the divisors.But maybe that's not entirely accurate because 10000 has 25 divisors, but the number of divisors less than 100 might not be exactly 12. Let's compute all divisors of 10000.Given the prime factorization 2^4 * 5^4, the divisors are of the form 2^a * 5^b where 0 ≤ a ≤ 4 and 0 ≤ b ≤ 4. Let's list all divisors:Start with exponents of 2:a=0:5^0 =15^1=55^2=255^3=1255^4=625a=1:2*1=22*5=102*25=502*125=2502*625=1250a=2:4*1=44*5=204*25=1004*125=5004*625=2500a=3:8*1=88*5=408*25=2008*125=10008*625=5000a=4:16*1=1616*5=8016*25=40016*125=200016*625=10000So, let's list all divisors:1, 5, 25, 125, 625,2, 10, 50, 250, 1250,4, 20, 100, 500, 2500,8, 40, 200, 1000, 5000,16, 80, 400, 2000, 10000.Now, let's count the divisors greater than 100.Looking at the list:From a=0:125, 625a=1:250, 1250a=2:500, 2500a=3:200, 1000, 5000a=4:400, 2000, 10000Wait, let's check each row:First row (a=0): 1,5,25,125,625. The ones greater than 100 are 125, 625. So two.Second row (a=1): 2,10,50,250,1250. Greater than 100: 250, 1250. Two.Third row (a=2):4,20,100,500,2500. Greater than 100:500,2500. Two.Fourth row (a=3):8,40,200,1000,5000. Greater than 100:200,1000,5000. Three.Fifth row (a=4):16,80,400,2000,10000. Greater than 100:400,2000,10000. Three.Total greater than 100: 2+2+2+3+3=12. Correct, as expected. Plus the divisor 100 itself. So total 12 divisors greater than 100. Therefore, the possible d values are these 12 divisors: 125, 625, 250, 1250, 500, 2500, 200, 1000, 5000, 400, 2000, 10000. Wait, let's confirm from the list:List of divisors greater than 100:From a=0:125,625a=1:250,1250a=2:500,2500a=3:200,1000,5000a=4:400,2000,10000Yes, that's 12 divisors. So, 12 possible values of d. Therefore, x = d - 100. Let's compute each x:For each d:125: x = 125 - 100 =25625:625 -100=525250:250-100=1501250:1250-100=1150500:500-100=4002500:2500-100=2400200:200-100=1001000:1000-100=9005000:5000-100=4900400:400-100=3002000:2000-100=190010000:10000-100=9900Therefore, the x values are:25,525,150,1150,400,2400,100,900,4900,300,1900,9900.So, these are 12 different x values.But wait, let me check if all these x values are distinct. Let's list them:25, 525, 150, 1150, 400, 2400, 100, 900, 4900, 300, 1900, 9900.Yes, all are distinct. Therefore, there are 12 possible integer values for x.Wait a minute, but hold on. Let me check if there's a mistake here. Because when d=200, x=100. But x is supposed to be a positive integer. 100 is a positive integer, so that's okay. Similarly, others are positive integers as well.But the problem states "Both x and y are positive integers." So, we need to confirm that y is also a positive integer. But earlier reasoning says that if d is a divisor of 10000 greater than 100, then y=100x/d. Since x=d-100, then y=100*(d-100)/d. Since d divides 100*(d -100), which we have already established because d divides 10000. Wait, but let's check with an example.Take d=125. Then x=25. Then y=100*25 /125=2500/125=20. That's an integer. Good.Take d=625, x=525. y=100*525 /625=52500/625=84. Integer. Good.d=250, x=150. y=100*150/250=15000/250=60. Integer.d=1250, x=1150. y=100*1150 /1250=115000/1250=92. Integer.d=500, x=400. y=100*400 /500=40000/500=80. Integer.d=2500, x=2400. y=100*2400 /2500=240000/2500=96. Integer.d=200, x=100. y=100*100 /200=10000/200=50. Integer.d=1000, x=900. y=100*900 /1000=90000/1000=90. Integer.d=5000, x=4900. y=100*4900 /5000=490000/5000=98. Integer.d=400, x=300. y=100*300 /400=30000/400=75. Integer.d=2000, x=1900. y=100*1900 /2000=190000/2000=95. Integer.d=10000, x=9900. y=100*9900 /10000=990000/10000=99. Integer.So, in all cases, y is an integer. Therefore, all 12 x values are valid. Therefore, the answer should be 12 possible values for x.Wait, but the problem says "how many possible values for x are there?" So, 12? But let's check again if there's any mistake in the process.Wait, but in the list of divisors greater than 100, we had 12 divisors, leading to 12 x's, all distinct. Each gives a unique x. So, 12 possible values.But wait, let's check if perhaps some different d's could lead to the same x. For example, if two different d's result in the same x. But x is d -100. Since d's are distinct and greater than 100, x's will be distinct. For example, if d1 ≠ d2, then x1 = d1 -100 ≠ d2 -100 =x2. Therefore, all x's are distinct. So, 12 different x's.Therefore, the answer is 12.But wait, let me cross-verify this with another approach. Let's consider the equation y = (100x)/(x + 100). For y to be an integer, x + 100 must divide 100x. Let’s denote k = x + 100. Then, x = k - 100. So, y = 100(k - 100)/k. Then, y must be an integer. Therefore, 100(k - 100)/k must be integer. Which implies that k divides 100(k - 100). Therefore, k divides 100k - 10000, which implies k divides 10000. Therefore, k must be a divisor of 10000. Therefore, k is a positive divisor of 10000, and since x = k - 100 > 0 (because x is positive), then k > 100. Therefore, k is a divisor of 10000 greater than 100, which is exactly what we had before. So, same result. Therefore, the number of possible x's is equal to the number of divisors of 10000 greater than 100, which is 12. Therefore, 12.But wait, is there a chance that some divisors of 10000 greater than 100 might result in the same x? For example, two different divisors d1 and d2 such that d1 - 100 = d2 - 100. But that would mean d1 = d2, so they can't be different. Therefore, each divisor greater than 100 gives a unique x. So, 12 divisors give 12 x's.Therefore, the answer is 12. Hmm, but let's check if the problem says "possible values for x", and all these x's are positive integers, which they are. So, 12.But wait, the problem states "the silver coin is worth y% less than the gold coin". So, y must be a positive integer less than 100, right? Because if something is worth y% less than another, y must be less than 100, otherwise, it would imply a negative value. Wait, is that a consideration here?Wait, let's check. If y were 100%, then the silver coin would be worth 0. If y is more than 100%, then the silver coin would have a negative value, which is impossible. Therefore, y must be less than 100. So, in our previous calculations, do we have any y that is 100 or more?Looking back at our examples:For d=10000, x=9900, y=99. So, 99, which is less than 100.Similarly, for d=2000, x=1900, y=95. Also less than 100.All other y values calculated were 20,84,60,92,80,96,50,90,98,75. All less than 100. So, all y values are indeed less than 100, which is necessary. Therefore, no problem here.Therefore, all 12 pairs are valid. So, the answer is 12.But just to make sure, let me check if 10000 has exactly 25 divisors, which we listed. Let's count them again:First row: 5Second row:5Third row:5Fourth row:5Fifth row:5Total 25. Correct. Divisors greater than 100:12 as listed. Correct.Therefore, the answer is 12. So, the number of possible integer values for x is 12.Final AnswerThe number of possible values for ( x ) is boxed{12}.

✅:Given that a gold coin is worth ( x% ) more than a silver coin and a silver coin is worth ( y% ) less than a gold coin, where both ( x ) and ( y ) are positive integers, we need to determine the number of possible values for ( x ).1. Let the value of the silver coin be ( S ). The value of the gold coin ( G ) can be expressed as: [ G = S left(1 + frac{x}{100}right) ]2. Conversely, the value of the silver coin can be expressed in terms of the gold coin: [ S = G left(1 - frac{y}{100}right) ]3. Substituting ( G ) from the first equation into the second equation: [ S = S left(1 + frac{x}{100}right) left(1 - frac{y}{100}right) ]4. Simplifying this, we get: [ 1 = left(1 + frac{x}{100}right) left(1 - frac{y}{100}right) ]5. Expanding and rearranging terms: [ 0 = -frac{y}{100} + frac{x}{100} - frac{xy}{10000} ] [ 0 = -100y + 100x - xy ] [ y(x + 100) = 100x ] [ y = frac{100x}{x + 100} ]6. For ( y ) to be an integer, ( x + 100 ) must divide ( 100x ). Let ( d = x + 100 ). Then ( d ) must be a divisor of 10000. The divisors of 10000 are found from its prime factorization ( 2^4 times 5^4 ), giving 25 divisors.7. We need divisors ( d ) greater than 100. The divisors of 10000 greater than 100 are: [ 125, 625, 250, 1250, 500, 2500, 200, 1000, 5000, 400, 2000, 10000 ]8. Each divisor ( d ) corresponds to ( x = d - 100 ). These values of ( x ) are: [ 25, 525, 150, 1150, 400, 2400, 100, 900, 4900, 300, 1900, 9900 ]9. All these values are distinct positive integers, confirming there are 12 possible values for ( x ).Thus, the number of possible values for ( x ) is (boxed{12}).

🔑:: 1. Let's denote the cost of a gold coin as (pounds g) and the cost of a silver coin as (pounds s).2. According to the problem, a gold coin is worth (x%) more than a silver coin. This can be represented mathematically as: [ g = left(1 + frac{x}{100}right) s ]3. Similarly, a silver coin is worth (y%) less than a gold coin, which can be represented as: [ s = left(1 - frac{y}{100}right) g ]4. We can express (g/s) and (s/g) from these equations: [ frac{g}{s} = 1 + frac{x}{100} = frac{100 + x}{100} ] [ frac{s}{g} = 1 - frac{y}{100} = frac{100 - y}{100} ]5. By multiplying the expressions for (frac{g}{s}) and (frac{s}{g}), we get: [ left(frac{100 + x}{100}right) left(frac{100 - y}{100}right) = 1 ]6. Simplify the equation: [ frac{(100 + x)(100 - y)}{100^2} = 1 ]7. Multiply both sides by (100^2): [ (100 + x)(100 - y) = 10000 ]8. Solve for (x): [ 100 + x = frac{10000}{100 - y} ] [ x = frac{10000}{100 - y} - 100 ]9. To find possible values for (x), we need to consider the values of (100 - y) that are factors of 10000 and also positive integers less than 100.10. The factors of 10000 that are less than 100 are: (1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80).11. Now, we compute the corresponding values of (x) for each factor: [ begin{aligned} & text{For } 100 - y = 1, & x &= frac{10000}{1} - 100 = 9900, & text{For } 100 - y = 2, & x &= frac{10000}{2} - 100 = 4900, & text{For } 100 - y = 4, & x &= frac{10000}{4} - 100 = 2400, & text{For } 100 - y = 5, & x &= frac{10000}{5} - 100 = 1900, & text{For } 100 - y = 8, & x &= frac{10000}{8} - 100 = 1150, & text{For } 100 - y = 10, & x &= frac{10000}{10} - 100 = 900, & text{For } 100 - y = 16, & x &= frac{10000}{16} - 100 = 525, & text{For } 100 - y = 20, & x &= frac{10000}{20} - 100 = 400, & text{For } 100 - y = 25, & x &= frac{10000}{25} - 100 = 300, & text{For } 100 - y = 40, & x &= frac{10000}{40} - 100 = 150, & text{For } 100 - y = 50, & x &= frac{10000}{50} - 100 = 100, & text{For } 100 - y = 80, & x &= frac{10000}{80} - 100 = 25. end{aligned} ]12. Thus, there are twelve possible values of (x).Conclusion:[boxed{12}]

🧠:Okay, so I need to solve this problem where we have a function f(x) = ax - b, and for all x in the interval [-1,1], the value of f(x) is between 0 and 1 inclusive. Then, we have to find the range of the expression (3a + b + 1)/(a + 2b - 2). The options are given as A through D. Let me try to figure this out step by step.First, since f(x) is between 0 and 1 for all x in [-1,1], that imposes some constraints on the values of a and b. My goal is to translate those constraints into inequalities involving a and b, and then use those to determine the possible values of the given expression.Let me start by considering the function f(x) = ax - b. For x in [-1,1], 0 ≤ ax - b ≤ 1. So, this inequality must hold for all x in that interval. To find the constraints on a and b, I need to ensure that the minimum and maximum values of f(x) on the interval [-1,1] are within [0,1].Since f(x) is a linear function in x, its extrema on the interval [-1,1] will occur at the endpoints. Therefore, I can just check the values of f(-1) and f(1) to ensure they lie within [0,1]. Wait, but actually, if a is positive, then f(x) is increasing, so the minimum is at x = -1 and the maximum at x = 1. If a is negative, then f(x) is decreasing, so the minimum is at x = 1 and maximum at x = -1. If a is zero, then f(x) is constant -b, so in that case, we need 0 ≤ -b ≤ 1, which would mean -1 ≤ b ≤ 0. But since a is a real number, we need to consider all possibilities.So, to cover all cases, let's compute f(-1) and f(1) and set up inequalities for both.For x = -1: f(-1) = -a - b. For x = 1: f(1) = a - b.But since the function is linear, depending on the slope a, the maximum and minimum will be at these endpoints. So regardless of the sign of a, the maximum and minimum of f(x) on [-1,1] will be at x = 1 or x = -1. Therefore, to ensure that 0 ≤ f(x) ≤ 1 for all x in [-1,1], it is necessary and sufficient that both f(-1) and f(1) are within [0,1], and if a ≠ 0, also check that the function doesn't exceed the bounds in between. Wait, no, since it's linear, if the endpoints are within [0,1], then the entire interval will be within [0,1] because linear functions don't have local extrema except at endpoints. So, actually, if both f(-1) and f(1) are between 0 and 1, then all values in between will be between those two endpoints. Therefore, since f(x) is linear, the entire function on [-1,1] is between f(-1) and f(1). So, if we ensure both f(-1) and f(1) are in [0,1], then the entire interval is covered. However, this is only true if the function is increasing or decreasing. If a = 0, then f(x) is constant, so just need that constant to be in [0,1].Therefore, the constraints are:Case 1: a > 0 (increasing function)Then f(-1) = -a - b ≥ 0 and f(1) = a - b ≤ 1But wait, if a > 0, the function is increasing, so the minimum is at x = -1 and maximum at x = 1. So to have f(x) ≥ 0 everywhere, we need f(-1) ≥ 0, and to have f(x) ≤ 1 everywhere, we need f(1) ≤ 1. So:- a - b ≥ 0 (from f(-1) ≥ 0)a - b ≤ 1 (from f(1) ≤ 1)Additionally, since the function is increasing, the other endpoint conditions (f(-1) ≤ 1 and f(1) ≥ 0) would automatically hold if the above are satisfied? Wait, no. Wait, if a > 0, then f(-1) is the minimum and f(1) is the maximum. So if the minimum is ≥ 0 and the maximum is ≤1, then all values are between 0 and 1. Therefore, only these two inequalities are needed for a > 0:1) -a - b ≥ 02) a - b ≤ 1Similarly, for a < 0 (decreasing function), then the maximum is at x = -1 and the minimum at x = 1. So, we need f(-1) ≤ 1 and f(1) ≥ 0:3) -a - b ≤ 14) a - b ≥ 0And if a = 0, then f(x) = -b, so 0 ≤ -b ≤ 1 ⇒ -1 ≤ b ≤ 0.So, combining all cases:If a > 0:- a - b ≥ 0 ⇒ b ≤ -aanda - b ≤ 1 ⇒ b ≥ a - 1Also, since a > 0, these two inequalities must hold: b ≤ -a and b ≥ a -1. Therefore, for a > 0, the possible b is between a -1 and -a. But since a > 0, a -1 can be less than -a or not. Let's see:Suppose a > 0. Then, a -1 must be ≤ -a. So, a -1 ≤ -a ⇒ 2a ≤ 1 ⇒ a ≤ 1/2. Therefore, if a > 0, then for the inequalities b ≥ a -1 and b ≤ -a to have a solution, we need a -1 ≤ -a ⇒ a ≤ 1/2. Therefore, for a > 0, possible a is in (0, 1/2], and for each such a, b must be between a -1 and -a.Similarly, for a < 0:From the decreasing case, we need:f(-1) = -a - b ≤ 1 ⇒ -a - b ≤ 1 ⇒ b ≥ -a -1andf(1) = a - b ≥ 0 ⇒ a - b ≥ 0 ⇒ b ≤ aSince a < 0, and b must satisfy both b ≥ -a -1 and b ≤ a. But since a < 0, -a is positive, so -a -1 = (-1)(a + 1). Let's check if these intervals overlap.We need b ≤ a and b ≥ -a -1. Since a is negative, let's take a = -k where k > 0. Then:b ≤ -kb ≥ k -1But since a < 0, and k = |a|, so for these two inequalities to have a solution, k -1 ≤ -k ⇒ 2k ≤ 1 ⇒ k ≤ 1/2 ⇒ a ≥ -1/2.Therefore, for a < 0, possible a is in [-1/2, 0), and b must be between -a -1 and a.Additionally, when a = 0, as mentioned earlier, b ∈ [-1, 0].So, summarizing all constraints:Case 1: a > 0Then 0 < a ≤ 1/2, and b ∈ [a -1, -a]Case 2: a < 0Then -1/2 ≤ a < 0, and b ∈ [-a -1, a]Case 3: a = 0Then b ∈ [-1, 0]Now, we need to find the range of the expression (3a + b + 1)/(a + 2b - 2). Let's denote this expression as E = (3a + b + 1)/(a + 2b - 2). Our goal is to find the maximum and minimum values of E given the constraints on a and b.This seems like an optimization problem with variables a and b subject to linear inequalities. The expression E is a linear fractional function. To find its range, perhaps we can express E in terms of a single variable by substituting the constraints, or use Lagrange multipliers or other optimization techniques. But since the constraints are linear and the objective is a linear fractional function, the extrema should occur at the vertices of the feasible region.First, let's consider each case separately.Case 1: a ∈ (0, 1/2], b ∈ [a -1, -a]Case 2: a ∈ [-1/2, 0), b ∈ [-a -1, a]Case 3: a = 0, b ∈ [-1, 0]So, let's analyze each case.Starting with Case 1: a ∈ (0, 1/2], b ∈ [a -1, -a]In this case, for each a, b can vary between a -1 and -a. So, perhaps for each a, we can express E as a function of b, find its extremum over b, then optimize over a.Alternatively, since E is linear in b for fixed a, the extrema in b will occur at the endpoints.Similarly, for fixed a, E is linear in b, so let's check the endpoints for b.So, for Case 1, let's fix a in (0, 1/2], then b can be either b = a -1 or b = -a. Let's compute E at these two endpoints.First, when b = a -1:E = (3a + (a -1) + 1)/(a + 2(a -1) - 2) = (4a -1 + 1)/(a + 2a - 2 - 2) = (4a)/(3a -4)Second, when b = -a:E = (3a + (-a) + 1)/(a + 2(-a) -2) = (2a +1)/( -a -2 )So, for each a in (0, 1/2], E can take values between these two expressions. However, since E is continuous in a and b, and the coefficients are linear, the maximum and minimum over the region will occur at these endpoints.Therefore, we can parametrize E in terms of a for each endpoint.Let me compute E for b = a -1: E1(a) = 4a / (3a -4)Similarly, E for b = -a: E2(a) = (2a +1)/(-a -2) = -(2a +1)/(a +2)Now, we can analyze E1(a) and E2(a) over a ∈ (0, 1/2]First, E1(a) = 4a / (3a -4). Let's compute the derivative to see if it's increasing or decreasing.Let’s compute dE1/da:Numerator derivative: 4*(3a -4) - 4a*3 = 12a -16 -12a = -16Denominator: (3a -4)^2Therefore, dE1/da = -16 / (3a -4)^2, which is negative since denominator is squared, so derivative is negative. Thus, E1(a) is decreasing in a on (0, 1/2]So, maximum at a =0, but a ∈ (0, 1/2], so approaching a→0+:E1 approaches 0 / (-4) = 0. But at a =1/2:E1(1/2) = 4*(1/2) / (3*(1/2) -4) = 2 / (1.5 -4) = 2 / (-2.5) = -0.8 = -4/5.So, E1(a) goes from approaching 0 (as a approaches 0) to -4/5 at a=1/2. Since it's decreasing, the range of E1(a) in this case is [-4/5, 0).Now, for E2(a) = -(2a +1)/(a +2). Let's simplify:E2(a) = -(2a +1)/(a +2). Let's compute derivative:dE2/da = [ -2(a +2) - ( - (2a +1) ) ] / (a +2)^2Wait, better to compute directly:Let E2(a) = (-2a -1)/(a +2). Then derivative:Numerator derivative: -2*(a +2) - (-2a -1)*1 = -2a -4 +2a +1 = (-4 +1) = -3Denominator: (a +2)^2Therefore, dE2/da = -3 / (a +2)^2 < 0. Therefore, E2(a) is decreasing in a.Thus, as a increases from 0 to 1/2, E2(a) decreases from E2(0) = -1 / 2 = -0.5 to E2(1/2) = -(2*(1/2) +1)/( (1/2) +2 ) = -(1 +1)/(2.5) = -2 / 2.5 = -0.8 = -4/5.Wait, that's interesting. So E2(a) starts at -0.5 when a approaches 0 and decreases to -4/5 at a=1/2. But since E2(a) is decreasing, the maximum is at a=0 (approaching from the right) and minimum at a=1/2.Wait, but in Case 1, a ∈ (0, 1/2], so approaching a=0, E2(a) approaches -0.5, and at a=1/2, it's -4/5. So the range of E2(a) here is [-4/5, -0.5).Wait, but in Case 1, when a is approaching 0+, b is approaching -0 from above? Wait, no: when a approaches 0+, in Case 1, b is in [a -1, -a]. So as a approaches 0+, a -1 approaches -1, and -a approaches 0. So b is approaching the interval [-1, 0]. But when a is 0, we have Case 3.But in this case, when we take the endpoints for b in Case 1, when a approaches 0+, then E1(a) approaches 0 (from 4a/(3a -4) as a→0+ is 0/-4 = 0), and E2(a) approaches -0.5.Therefore, in Case 1, the possible values of E are from -4/5 (at a=1/2, both E1 and E2 reach -4/5) up to approaching 0 (from E1) and approaching -0.5 (from E2). Wait, but since E1(a) is decreasing from 0 to -4/5, and E2(a) is decreasing from -0.5 to -4/5, so the union of these ranges would be from -4/5 to 0 (since E1 can approach 0) and from -4/5 to -0.5. But actually, when a is in (0, 1/2], for each a, E can be between E1(a) and E2(a). Wait, no: when we fix a, and vary b between a -1 and -a, since E is linear in b, then depending on whether the coefficient of b in E is positive or negative, the maximum and minimum of E will be at the endpoints.Let me check the expression E = (3a + b +1)/(a + 2b -2). Let's write it as:E = [ (3a +1) + b ] / [ (a -2) + 2b ]To see how E changes with b, let's compute derivative of E with respect to b:dE/db = [1*(a + 2b -2) - (3a + b +1)*2 ] / (a + 2b -2)^2Simplify numerator:(a + 2b -2) - 2*(3a + b +1) = a + 2b -2 -6a -2b -2 = (-5a -4)Therefore, dE/db = (-5a -4)/(a + 2b -2)^2The denominator is squared, so the sign of dE/db depends on the numerator: -5a -4.In Case 1: a ∈ (0, 1/2]. So, -5a -4 is always negative (since a is positive, -5a is negative, and -4 is negative, so total is negative). Therefore, derivative dE/db is negative, meaning E is decreasing in b. Therefore, for fixed a in (0,1/2], maximum E occurs at minimal b (which is a -1) and minimal E occurs at maximal b (which is -a).Therefore, in Case 1, the range of E is from E1(a) (at b = a -1) to E2(a) (at b = -a). But since E is decreasing in b, and for each a, E1(a) > E2(a). Wait, since when b increases, E decreases. So for each a, the maximum E is at b = a -1 (the lower b) and the minimum E is at b = -a (the higher b). Therefore, over the interval of a, the maximum possible E in Case 1 is the maximum of E1(a), which is approaching 0 as a approaches 0, and the minimum possible E is the minimum of E2(a), which is -4/5 at a=1/2. But wait, actually, for each a, E varies between E1(a) and E2(a). But since E1(a) is decreasing from 0 to -4/5, and E2(a) is decreasing from -0.5 to -4/5, then the overall maximum in Case 1 is 0 (approached as a→0+), and the overall minimum is -4/5 (attained at a=1/2). But for a fixed a, E ranges from E1(a) down to E2(a). But since both E1(a) and E2(a) are decreasing functions, the maximum of E over Case 1 is the upper limit of E1(a) as a approaches 0, which is 0, and the minimum is the lower limit of E2(a) as a approaches 1/2, which is -4/5. However, for a given a, the E ranges between E1(a) and E2(a). Since E1(a) is always above E2(a) (since E1(a) = 4a/(3a-4) and E2(a) = -(2a+1)/(a +2). Let's check at a=1/2:E1(1/2) = 4*(1/2)/(3*(1/2)-4) = 2/(1.5 -4) = 2/(-2.5) = -0.8 = -4/5E2(1/2) = -(2*(1/2) +1)/(1/2 +2) = -(2)/(2.5) = -0.8 = -4/5. So at a=1/2, both endpoints give E=-4/5. For a approaching 0:E1(a) approaches 0, E2(a) approaches -0.5. So for a approaching 0, E can approach 0 from below (since E1(a) approaches 0 but is negative because denominator 3a -4 is negative, numerator 4a is positive, so overall E1(a) approaches 0 from the negative side). Wait, 4a/(3a -4) as a approaches 0 from the right: numerator approaches 0+, denominator approaches -4, so E1(a) approaches 0 from below (i.e., approaches 0 from the negative side). Similarly, E2(a) approaches -0.5 from above (since as a approaches 0+, E2(a) = -(2a +1)/(a +2) approaches -1/2, which is -0.5). Therefore, in Case 1, the possible values of E are between E1(a) and E2(a), which when a approaches 0, is from near 0 (but slightly below) down to -0.5. But as a increases, both E1(a) and E2(a) decrease until they reach -4/5 at a=1/2. Therefore, the range in Case 1 is from -4/5 up to just below 0, and down to -0.5? Wait, this is confusing.Wait, no: For each a in (0,1/2], since E is decreasing in b, and for each a, E can take any value between E1(a) and E2(a). Since E1(a) is decreasing from 0 to -4/5 and E2(a) is decreasing from -0.5 to -4/5, the set of all possible E in Case 1 is the union over a ∈ (0,1/2] of [E1(a), E2(a)]. But because E1(a) starts at 0 (approaching) and decreases to -4/5, and E2(a) starts at -0.5 and decreases to -4/5, the union of all these intervals would be from -4/5 up to 0. Wait, how?Imagine when a is near 0, E1(a) is near 0 (but negative), and E2(a) is near -0.5. So for a near 0, E can range from near 0 down to -0.5. Then, as a increases, the upper bound of E (E1(a)) decreases to -4/5, and the lower bound (E2(a)) also decreases to -4/5. Therefore, the overall range contributed by Case 1 is from -4/5 up to 0, but between a=0 and a=1/2, the upper part near 0 is only approached as a→0. However, since a cannot be zero in Case 1 (it's handled in Case 3), but in Case 3, a=0, so perhaps we need to consider Case 3 separately.Case 3: a=0, b ∈ [-1,0]. Then, E = (0 + b +1)/(0 + 2b -2) = (b +1)/(2b -2). Let's compute this for b ∈ [-1,0].Let’s denote E3(b) = (b +1)/(2b -2). Let's compute E3(b):At b = -1: E3(-1) = (0)/(-4) = 0At b = 0: E3(0) = (1)/(-2) = -0.5Also, let's check if E3(b) is increasing or decreasing. Compute derivative:dE3/db = [1*(2b -2) - (b +1)*2]/(2b -2)^2 = [2b -2 -2b -2]/(2b -2)^2 = (-4)/(2b -2)^2Since the derivative is -4/(denominator)^2, which is always negative (denominator is squared, numerator is -4). Therefore, E3(b) is decreasing in b. Therefore, when b increases from -1 to 0, E3(b) decreases from 0 to -0.5. So in Case 3, E ranges from -0.5 to 0.Therefore, combining Case 1 and Case 3:Case 1 gives E from -4/5 up to just below 0 (approaching 0 as a→0+), and down to -4/5.Case 3 gives E from -0.5 to 0.Therefore, overlapping ranges? Wait, but Case 1 includes E values down to -4/5, which is -0.8, which is less than -0.5. But in Case 3, the minimum is -0.5, so the combined range from Case 1 and Case 3 would be from -4/5 to 0. However, when a=0, the maximum E is 0 (attained when b=-1), and the minimum is -0.5 (attained at b=0). But in Case 1, when a approaches 0+, E can approach 0 from below (since in Case 1, when a→0+, b approaches -1, so E approaches (0 + (-1) +1)/(0 + 2*(-1) -2) = 0 / (-4) = 0. But actually, when a approaches 0+, in Case 1, b can approach -1 from above (since in Case 1, b ∈ [a -1, -a], so when a approaches 0+, a -1 approaches -1, so b can approach -1 from above). Therefore, in Case 1, as a→0+ and b approaches -1 from above, E approaches (0 + (-1) +1)/(0 + 2*(-1) -2) = 0/-4 = 0. But since in Case 3, when a=0 and b=-1, E=0. So the upper bound of E is 0, which is attainable in Case 3.Similarly, the minimum in Case 1 is -4/5, which is less than -0.5, so the overall minimum so far is -4/5.Now, let's check Case 2: a ∈ [-1/2, 0), b ∈ [-a -1, a]Again, we can analyze E = (3a + b +1)/(a + 2b -2). For fixed a, E is linear in b, so extremum at endpoints.Compute E at b = -a -1 and b = a.First, when b = -a -1:E = (3a + (-a -1) +1)/(a + 2*(-a -1) -2) = (2a -1 +1)/(a -2a -2 -2) = (2a)/(-a -4) = -2a/(a +4)Second, when b = a:E = (3a + a +1)/(a + 2a -2) = (4a +1)/(3a -2)Now, let's compute these expressions for a ∈ [-1/2, 0)First, E at b = -a -1: E4(a) = -2a/(a +4). Let's simplify:E4(a) = -2a/(a +4). Let's analyze this for a ∈ [-1/2,0)At a = -1/2: E4(-1/2) = -2*(-1/2)/(-1/2 +4) = 1/(3.5) ≈ 0.2857 = 2/7At a approaching 0 from below: E4(a) approaches 0/(0 +4) = 0Now, check if E4(a) is increasing or decreasing. Compute derivative:dE4/da = [ -2*(a +4) - (-2a)*1 ] / (a +4)^2 = [ -2a -8 +2a ] / (a +4)^2 = (-8)/(a +4)^2 < 0Therefore, E4(a) is decreasing in a over [-1/2,0). Therefore, maximum at a=-1/2 (which is 2/7) and minimum approaching 0 as a→0-.Second, E at b = a: E5(a) = (4a +1)/(3a -2). Let's compute this for a ∈ [-1/2,0)At a = -1/2: E5(-1/2) = (4*(-1/2) +1)/(3*(-1/2)-2) = (-2 +1)/(-1.5 -2) = (-1)/(-3.5) ≈ 0.2857 = 2/7At a approaching 0 from below: E5(a) approaches (0 +1)/(0 -2) = 1/(-2) = -0.5Compute derivative of E5(a):dE5/da = [4*(3a -2) - (4a +1)*3 ] / (3a -2)^2 = [12a -8 -12a -3]/(3a -2)^2 = (-11)/(3a -2)^2 < 0Therefore, E5(a) is decreasing in a. Hence, maximum at a=-1/2 (2/7) and minimum approaching -0.5 as a→0-.Now, similar to Case 1, we need to check for each a ∈ [-1/2,0), the range of E is between E4(a) and E5(a). But since E is linear in b, we need to check the coefficient of b in E.Let’s compute dE/db again for general case:Earlier, we found dE/db = (-5a -4)/(a + 2b -2)^2In Case 2: a ∈ [-1/2,0). So, -5a -4. Since a is negative, -5a is positive. So, -5a -4 = positive -4. When a is in [-1/2,0), let's see:At a = -1/2: -5*(-1/2) -4 = 2.5 -4 = -1.5 <0At a approaching 0 from below: -5a -4 approaches -4 <0Therefore, in Case 2, -5a -4 is negative for all a ∈ [-1/2,0). Therefore, dE/db is negative, so E is decreasing in b. Therefore, for fixed a in [-1/2,0), maximum E occurs at minimal b, which is b = -a -1, and minimum E occurs at maximal b, which is b = a.Thus, for each a in [-1/2,0), E ranges from E4(a) to E5(a), decreasing from E4(a) to E5(a). Since E4(a) is decreasing from 2/7 to 0, and E5(a) is decreasing from 2/7 to -0.5. Therefore, for each a, E varies between E4(a) and E5(a). The maximum value of E in Case 2 is 2/7 (attained at a=-1/2), and the minimum approaches -0.5 as a approaches 0-.Therefore, the range contributed by Case 2 is from -0.5 up to 2/7.But wait, in Case 2, when a=-1/2, both E4 and E5 give 2/7, so at a=-1/2, E=2/7. As a increases from -1/2 to 0, E4(a) decreases from 2/7 to 0, and E5(a) decreases from 2/7 to -0.5. Therefore, the range in Case 2 is from -0.5 to 2/7.Now, combining all cases:Case 1: E ∈ [-4/5, 0)Case 2: E ∈ [-0.5, 2/7]Case 3: E ∈ [-0.5, 0]But we need to consider the union of these ranges. So:- From Case 1: E can go as low as -4/5 and up to just below 0.- From Case 2: E can go as low as -0.5 and up to 2/7.- From Case 3: E can go from -0.5 to 0.So, combining these, the overall range is from -4/5 (approx -0.8) up to 2/7 (approx 0.2857), but need to check if there's any overlap or missing parts.But wait, in Case 1, the lower bound is -4/5, which is -0.8, less than -0.5. In Case 2 and 3, the lower bound is -0.5. So the overall minimum is -4/5. The upper bound is the maximum of 0 (from Case 1 and 3) and 2/7 (from Case 2). 2/7 is approximately 0.2857, which is higher than 0, so the upper bound is 2/7.Therefore, the range of E is from -4/5 to 2/7. However, need to verify if all values between -4/5 and 2/7 are attainable.Wait, but in Case 1, we have E from -4/5 to 0, in Case 2 from -0.5 to 2/7, and in Case 3 from -0.5 to 0. So the combined range would be the union of these intervals:[-4/5, 0) ∪ [-0.5, 2/7]. But overlapping regions?From -0.5 to 0, covered by both Case 1 and Case 3. So overall, union is [-4/5, 2/7].But need to confirm that there are no gaps. For example, between -4/5 (-0.8) and -0.5, is that interval covered? In Case 1, E can go down to -4/5 and up to 0, but does it cover the entire interval [-4/5, 0]?Wait, in Case 1, when a varies from 0 to 1/2, and for each a, E varies from E1(a) to E2(a). When a=1/4, for example:E1(1/4) = 4*(1/4)/(3*(1/4) -4) = 1 / (0.75 -4) = 1 / (-3.25) ≈ -0.3077E2(1/4) = -(2*(1/4)+1)/( (1/4)+2 ) = -(0.5 +1)/(2.25) = -1.5 / 2.25 = -0.666...So between E1(a) ≈ -0.3077 and E2(a) ≈ -0.666..., E can take any value in between because E is continuous in b. So for a=1/4, E can be anywhere between -0.666... and -0.3077. Similarly, for other a values, the interval between E1(a) and E2(a) fills in the gap between -4/5 and 0. So the entire interval from -4/5 to 0 is covered by Case 1.Similarly, Case 2 covers from -0.5 to 2/7. But since -0.5 is higher than -4/5, the overall range is from -4/5 to 2/7, with no gaps.Additionally, when a=-1/2 in Case 2, E=2/7, which is the upper bound. When a=1/2 in Case 1, E=-4/5, which is the lower bound. Therefore, the overall range of E is [-4/5, 2/7].Looking back at the options:(A) [-1/2, 0] ≈ [-0.5, 0](B) [-4/5, 0] ≈ [-0.8, 0](C) [-1/2, 2/7] ≈ [-0.5, 0.2857](D) [-4/5, 2/7] ≈ [-0.8, 0.2857]Our conclusion is that the range is [-4/5, 2/7], which corresponds to option D.But let me double-check if 2/7 is indeed attainable. In Case 2, when a=-1/2, b is either -a -1 = 0.5 -1 = -0.5 or b=a=-0.5. Wait, no:Wait, in Case 2, a ∈ [-1/2,0), and b ∈ [-a -1, a]. For a=-1/2, b ∈ [-(-1/2) -1, -1/2] = [0.5 -1, -0.5] = [-0.5, -0.5]. Wait, that can't be. Wait, a=-1/2:b ∈ [-a -1, a] = [-(-1/2) -1, -1/2] = [ (1/2 -1), -1/2 ] = [ -1/2, -1/2 ]. So when a=-1/2, b must be -1/2. Therefore, E when a=-1/2 and b=-1/2 is:E = (3*(-1/2) + (-1/2) +1)/( (-1/2) + 2*(-1/2) -2 ) = ( -1.5 -0.5 +1 )/( -0.5 -1 -2 ) = (-1)/(-3.5) = 1/3.5 = 2/7. So yes, attainable.Similarly, when a=-1/2, in Case 2, b must be -1/2, so E=2/7. Therefore, the upper bound is 2/7.Similarly, the lower bound of -4/5 is attained in Case 1 when a=1/2 and b=-1/2:Wait, in Case 1, a=1/2, b ∈ [a -1, -a] = [1/2 -1, -1/2] = [-1/2, -1/2]. So b=-1/2. Then, E=(3*(1/2) + (-1/2) +1)/(1/2 + 2*(-1/2) -2) = (1.5 -0.5 +1)/(0.5 -1 -2) = (2)/(-2.5) = -0.8 = -4/5. So yes, attainable.Therefore, the range is from -4/5 to 2/7, which is option D.Final Answerboxed{D}

✅:Given the function ( f(x) = ax - b ) where ( a, b in mathbf{R} ) and ( 0 leq f(x) leq 1 ) for all ( x in [-1, 1] ), we need to determine the range of the expression ( frac{3a + b + 1}{a + 2b - 2} ).First, we establish the constraints on ( a ) and ( b ) by ensuring ( f(x) ) lies within [0, 1] for ( x in [-1, 1] ). This results in the following conditions:- For ( a > 0 ): ( -a - b geq 0 ) and ( a - b leq 1 )- For ( a < 0 ): ( -a - b leq 1 ) and ( a - b geq 0 )- For ( a = 0 ): ( -1 leq b leq 0 )We analyze each case to find the range of the expression ( frac{3a + b + 1}{a + 2b - 2} ):1. Case 1: ( a in (0, 1/2] ) - ( b in [a - 1, -a] ) - Evaluate at endpoints ( b = a - 1 ) and ( b = -a ): - ( frac{4a}{3a - 4} ) (decreasing from 0 to -4/5) - ( frac{2a + 1}{-a - 2} ) (decreasing from -0.5 to -4/5) - Range: ([-4/5, 0))2. Case 2: ( a in [-1/2, 0) ) - ( b in [-a - 1, a] ) - Evaluate at endpoints ( b = -a - 1 ) and ( b = a ): - ( frac{-2a}{a + 4} ) (decreasing from 2/7 to 0) - ( frac{4a + 1}{3a - 2} ) (decreasing from 2/7 to -0.5) - Range: ([-0.5, 2/7])3. Case 3: ( a = 0 ) - ( b in [-1, 0] ) - Evaluate at endpoints ( b = -1 ) and ( b = 0 ): - 0 and -0.5 - Range: ([-0.5, 0])Combining all cases, the overall range is the union of intervals:- Case 1: ([-4/5, 0))- Case 2: ([-0.5, 2/7])- Case 3: ([-0.5, 0])The union of these intervals results in the range ([-4/5, 2/7]), which corresponds to option D.boxed{D}

🔑:Given (a in mathbf{R}, b in mathbf{R}), the function is defined as (f(x) = ax - b). We know for any (x in [-1, 1]), (0 leqslant f(x) leqslant 1). We need to find the range of the expression (frac{3a + b + 1}{a + 2b - 2}).1. From the problem, we derive the inequalities: [ begin{cases} 0 leqslant f(1) leqslant 1, 0 leqslant f(-1) leqslant 1. end{cases} ]2. Rewrite the above values in terms of (f(x)): [ begin{cases} 0 leqslant a - b leqslant 1, -1 leqslant a + b leqslant 0. end{cases} ] Let's call (u = a + b) and (v = a - b).3. The inequalities become: [ begin{cases} 0 leqslant v leqslant 1, -1 leqslant u leqslant 0. end{cases} ]4. Solve for (a) and (b) in terms of (u) and (v): [ a = frac{u + v}{2}, quad b = frac{u - v}{2}. ]5. Substitute these expressions into (frac{3a + b + 1}{a + 2b - 2}): [ begin{aligned} frac{3a + b + 1}{a + 2b - 2} &= frac{3left( frac{u + v}{2} right) + frac{u - v}{2} + 1}{frac{u + v}{2} + 2 left( frac{u - v}{2} right) - 2} &= frac{frac{3(u+v) + (u-v)}{2} + 1}{frac{u + v + 2u - 2v}{2} - 2} &= frac{frac{3u + 3v + u - v}{2} + 1}{frac{3u - v}{2} - 2} &= frac{frac{4u + 2v}{2} + 1}{frac{3u - v}{2} - 2}. end{aligned} ]6. Simplify the fraction further: [ begin{aligned} &= frac{2u + v + 1}{frac{3u - v}{2} - 2} &= frac{2u + v + 1}{frac{3u - v - 4}{2}} &= frac{2(2u + v + 1)}{3u - v - 4}. end{aligned} ]7. We note that there is a transformation for simplification: [ -2 + frac{10}{3 - left(frac{5v - 11}{5u - 3}right)} ]8. After exploring values of (u) and (v) within the bounds: [ -frac{4}{5} leqslant frac{3a + b + 1}{a + 2b - 2} leqslant frac{2}{7} ]Conclusion:[boxed{D}]