🧠:[<think>Okay, let's tackle these two problems one by one. I need to make sure I follow the order of operations correctly, which is parentheses, exponents, multiplication and division (from left to right), and then addition and subtraction (from left to right). Let's start with the first problem.Problem (1): -(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)Alright, let's break this down step by step. First, I need to handle the exponents. The first term is -(-1)^4. Hmm, the exponent here is 4, so I need to compute (-1) to the power of 4 first and then apply the negative sign. Wait, but is the negative sign inside the exponentiation or outside? Since there are parentheses around -1, the exponent applies to the entire -1. So, (-1)^4 would be (-1)*(-1)*(-1)*(-1). Multiplying two negatives gives a positive, so (-1)^2 is 1, then (-1)^3 is -1, and (-1)^4 is 1. So, (-1)^4 is 1. Then applying the negative sign outside, it becomes -1. So the first term simplifies to -1.Next, let's look at the second part: -(1 - 0.5) times 1/3 times (2 - 3^2). Let's handle the parentheses first.Starting with (1 - 0.5). That's straightforward: 1 - 0.5 is 0.5. Then, there's a negative sign in front of that parentheses, so it becomes -0.5.Then, we have to multiply that by 1/3 and by (2 - 3^2). Let's compute the exponent in the last parentheses first: 3^2 is 9. So, 2 - 9 is -7. Therefore, the expression inside the parentheses simplifies to -7.Putting it all together: -0.5 * (1/3) * (-7). Let's compute this step by step. First, multiply -0.5 by 1/3. Multiplying a negative by a positive gives a negative. 0.5 * 1/3 is (1/2)*(1/3) = 1/6. So, -0.5 * 1/3 is -1/6.Then, multiply that result by -7. So, (-1/6) * (-7) is positive 7/6. Because a negative times a negative is a positive.So, the entire second part simplifies to 7/6.Now, the original problem is the first term minus the second part, but wait, actually, the entire expression is:First term: -(-1)^4 = -1Minus: the entire second part, which we found to be 7/6. Wait, no, hold on. Let me check again.Wait, the original problem is written as:-(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)So, it's the first term (which is -(-1)^4) minus the product of (1 - 0.5), 1/3, and (2 - 3^2). But when we simplified the product, we considered the negative sign in front of (1 - 0.5) as part of the multiplication. Let me verify that.Wait, the expression is: - (1 - 0.5) * (1/3) * (2 - 3^2). So the negative sign is in front of the entire product. So, first compute (1 - 0.5) = 0.5, then multiply by 1/3 and (2 - 3^2). Then apply the negative sign to that product.Wait, but actually, order of operations: multiplication and division are done left to right. So, let's parse the expression properly.Original expression after the first term: - (1 - 0.5) × (1/3) × (2 - 3²)So, first compute (1 - 0.5) = 0.5Then, the expression becomes -0.5 × (1/3) × (2 - 9) [since 3² is 9, so 2 - 9 = -7]So, compute each multiplication step by step.First, -0.5 × (1/3). Let's do this: 0.5 × (1/3) = (1/2) × (1/3) = 1/6. Then, apply the negative sign: -1/6.Then, multiply that by -7: (-1/6) × (-7) = 7/6. So, the entire second part is +7/6.Therefore, the entire expression is first term minus the second part? Wait, no. Wait, the original expression is:First term: -(-1)^4Then minus: [ (1 - 0.5) × (1/3) × (2 - 3²) ]Wait, but when we computed the second part, we included the negative sign as part of the multiplication. Wait, perhaps I made a mistake here.Let me re-express the original problem:-(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)This is equivalent to:[ -(-1)^4 ] - [ (1 - 0.5) × (1/3) × (2 - 3^2) ]So, the first term is -(-1)^4, which is -1 as we calculated before. The second term is subtracted, so it's minus [the product].But when we calculated the product, we included the negative sign in front of (1 - 0.5). Wait, no. The problem is written as:- (1 - 0.5) × (1/3) × (2 - 3^2)Wait, but actually, the entire second term after the first minus is:- (1 - 0.5) × (1/3) × (2 - 3^2)So, the structure is:First term: -(-1)^4Second term: - [ (1 - 0.5) × (1/3) × (2 - 3^2) ]So, the second term is a subtraction of the product. But in the original problem, the first term is subtracted by the product, which is multiplied by a negative? Wait, no. Let's parse the original expression carefully.Looking at the original problem: -(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)It can be seen as:[ -(-1)^4 ] - [ (1 - 0.5) × (1/3) × (2 - 3^2) ]So, each of these terms is separate. The first term is -(-1)^4, which is -1. The second term is minus the product. So, the total expression is -1 minus [product]. But when we computed the product earlier, we considered the negative sign as part of the product, but actually, the negative is part of the subtraction. Wait, maybe I confused the order.Wait, no. Let's recast it:Original problem:First term: -(-1)^4Second term: - (1 - 0.5) × (1/3) × (2 - 3^2)So, the entire expression is first term plus second term.But the second term is negative of the product. So, the total is:First term: -1Second term: - [ (1 - 0.5) × (1/3) × (2 - 3^2) ]Therefore, we need to compute the product inside the brackets first, then apply the negative sign, and then add it to the first term.So, compute the product:(1 - 0.5) = 0.5(2 - 3^2) = (2 - 9) = -7Therefore, the product is 0.5 × (1/3) × (-7)Let's compute 0.5 × (1/3) first: 0.5 × 1/3 = (1/2) × (1/3) = 1/6Then, 1/6 × (-7) = -7/6Therefore, the product inside the brackets is -7/6Then, applying the negative sign in front: -(-7/6) = +7/6Therefore, the entire expression is:First term (-1) + second term (7/6) = -1 + 7/6Convert -1 to sixths: -6/6 + 7/6 = ( -6 + 7 ) / 6 = 1/6Wait, so the final answer is 1/6?Wait, that contradicts my previous calculation. Let me check again.Hold on, perhaps I messed up the signs. Let me go through this again step by step.Original problem:-(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)First term: -(-1)^4Compute (-1)^4: (-1)*(-1)=1, *(-1)=-1, *(-1)=1. So (-1)^4=1. Therefore, -1^4= -1. Wait, no: the expression is -(-1)^4. Since exponentiation has higher precedence than the negation, it's - [ (-1)^4 ].Since (-1)^4 is 1, then - [1] = -1. So first term is -1.Second term: - (1 - 0.5) × (1/3) × (2 - 3^2)Compute each component:1 - 0.5 = 0.52 - 3^2 = 2 - 9 = -7So, substituting back:-0.5 × (1/3) × (-7)Compute left to right:First, -0.5 × (1/3) = (-0.5) × (1/3) = -0.5/3 = -1/6Then, -1/6 × (-7) = 7/6Therefore, the second term is 7/6.Therefore, the entire expression is:First term (-1) minus (something)? Wait, no. Wait, the original expression is:First term: -(-1)^4Then minus the product: - (1 - 0.5) × ... which becomes +7/6.Wait, no. Wait, the structure is:[ -(-1)^4 ] - [ (1 - 0.5) × (1/3) × (2 - 3^2) ]Therefore, first term is -1, then subtract the product. But when I computed the product, I had -0.5 × 1/3 × -7 = 7/6. Therefore, the entire expression is:-1 - 7/6 ?Wait, no. Wait, no, the second term is the product [ (1 - 0.5) × (1/3) × (2 - 3^2) ] which is 0.5 × 1/3 × (-7) = (0.5 × 1/3) = 1/6, then × (-7) = -7/6. Then, since the entire second term is subtracted, it becomes - (-7/6) = +7/6.Wait, now I'm confused. Let me clarify.The original expression is:-(-1)^4 - (1 - 0.5) × (1/3) × (2 - 3^2)This can be broken down into two parts:1. The first part is -(-1)^4.2. The second part is - (1 - 0.5) × (1/3) × (2 - 3^2)So, both parts are combined by subtraction. Wait, no, actually, the entire expression is:Start with -(-1)^4, then subtract the product of (1 - 0.5), 1/3, and (2 - 3^2). So:Total = -(-1)^4 - [(1 - 0.5) × (1/3) × (2 - 3^2)]So, compute each part:First term: -(-1)^4 = -1Second term: - [0.5 × (1/3) × (-7)] = - [ (0.5 × 1/3) × (-7) ] = - [ (1/6) × (-7) ] = - [ -7/6 ] = +7/6Therefore, total = -1 + 7/6Convert -1 to sixths: -6/6 + 7/6 = 1/6So the final answer is 1/6. Wait, that seems correct now. So the confusion was whether the negative sign was part of the multiplication or part of the subtraction. But in reality, since there's a minus sign before the product, it's equivalent to subtracting the entire product. However, when we computed the product, we noticed that the product itself is negative, so subtracting a negative becomes addition.Let me verify once again:First term: -(-1)^4 = -1Second term: (1 - 0.5) = 0.5; (2 - 3^2) = -7. So the product is 0.5 * 1/3 * (-7) = 0.5 * (-7/3) = -7/6. Then, since the entire product is subtracted, we have - (-7/6) = +7/6. Therefore, total is -1 + 7/6 = (-6/6 + 7/6) = 1/6.Yes, that's correct. So problem (1) is 1/6.Problem (2): frac{5}{13} times (-3frac{1}{4}) - 0.5 div |{-3 -1}|Alright, let's parse this one step by step. First, convert the mixed number to an improper fraction. Also, handle the absolute value.First term: (5/13) multiplied by (-3 1/4)Second term: 0.5 divided by the absolute value of (-3 -1)Let's compute each part separately.First term: 5/13 × (-3 1/4)Convert -3 1/4 to an improper fraction. 3 1/4 is 13/4, so -3 1/4 is -13/4.Therefore, 5/13 × (-13/4). Let's compute this. Multiply numerator: 5 × (-13) = -65. Denominator: 13 × 4 = 52. So, -65/52. Simplify this fraction.Divide numerator and denominator by 13: -65 ÷13 = -5; 52 ÷13=4. So simplifies to -5/4.First term simplifies to -5/4.Second term: 0.5 ÷ | -3 -1 |. Compute the absolute value first.Inside the absolute value: -3 -1 = -4. The absolute value of -4 is 4.Therefore, the second term is 0.5 ÷ 4. Convert 0.5 to a fraction: 0.5 = 1/2. Therefore, 1/2 ÷4 = (1/2) × (1/4) = 1/8.So, the second term is 1/8.Now, the entire expression is first term minus second term:-5/4 - 1/8To subtract these fractions, convert them to a common denominator. The denominators are 4 and 8. The common denominator is 8.Convert -5/4 to eighths: -5/4 = -10/8So, -10/8 - 1/8 = (-10 -1)/8 = -11/8Therefore, the result is -11/8.Wait, but let me confirm step by step.Original problem:(5/13) × (-3 1/4) = (5/13) × (-13/4) = (5 × -13)/(13 ×4) = (-65)/52 = simplifies to (-5)/4 (since 65 ÷13=5 and 52 ÷13=4). Correct.Second term: 0.5 ÷ | -3 -1 |. Compute inside the absolute value: -3 -1 = -4, absolute value is 4. Then 0.5 ÷4 = 0.125, which is 1/8. Correct.Then, combining the two terms: (-5/4) - (1/8). Convert to eighths: -10/8 -1/8 = -11/8. Yes, that's correct. So the final answer is -11/8, which can also be written as -1 3/8, but since the question doesn't specify the form, probably best to leave it as an improper fraction, which is -11/8.But let me check if there's any miscalculation.Wait, 0.5 divided by 4: 0.5 is a half. Half divided by 4 is the same as half multiplied by 1/4, which is 1/8. Correct. And 5/13 times -13/4: the 13s cancel out, leaving 5/1 times -1/4, which is -5/4. Correct. Then, -5/4 minus 1/8 is indeed -10/8 -1/8 = -11/8. Yes, so problem (2) is -11/8.Final Answer(1) boxed{dfrac{1}{6}}(2) boxed{-dfrac{11}{8}}</think>To solve the given problems, we follow the order of operations (parentheses, exponents, multiplication and division, addition and subtraction).Problem (1):[-(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)]1. Calculate the exponent: (-(-1)^4): [ (-1)^4 = 1 quad text{so} quad -(-1)^4 = -1 ]2. Simplify the expression inside the parentheses: [ 1 - 0.5 = 0.5 ] [ 2 - 3^2 = 2 - 9 = -7 ]3. Compute the product: [ 0.5 times frac{1}{3} times (-7) = 0.5 times frac{1}{3} times (-7) = left(frac{1}{2} times frac{1}{3}right) times (-7) = frac{1}{6} times (-7) = -frac{7}{6} ]4. Apply the negative sign to the product: [ -left(-frac{7}{6}right) = frac{7}{6} ]5. Combine the results: [ -1 + frac{7}{6} = -frac{6}{6} + frac{7}{6} = frac{1}{6} ]Final Answer (1):[boxed{dfrac{1}{6}}]Problem (2):[frac{5}{13} times left(-3frac{1}{4}right) - 0.5 div |{-3-1}|]1. Convert the mixed number to an improper fraction: [ -3frac{1}{4} = -frac{13}{4} ]2. Compute the first term: [ frac{5}{13} times left(-frac{13}{4}right) = frac{5 times (-13)}{13 times 4} = -frac{65}{52} = -frac{5}{4} ]3. Simplify the absolute value and compute the second term: [ |-3 - 1| = |-4| = 4 ] [ 0.5 div 4 = frac{1}{2} div 4 = frac{1}{2} times frac{1}{4} = frac{1}{8} ]4. Combine the results: [ -frac{5}{4} - frac{1}{8} = -frac{10}{8} - frac{1}{8} = -frac{11}{8} ]Final Answer (2):[boxed{-dfrac{11}{8}}],<think>Okay, let me try to solve these two problems step by step. I need to be careful with the order of operations, parentheses, exponents, multiplication/division, and addition/subtraction. Let's start with the first one.Problem (1): Calculate: -(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2)First, I need to break this down into parts. Let me identify the components:1. The first term is -(-1)^4. 2. The second part is subtracting the product of three factors: - (1 - 0.5) - frac{1}{3} - (2 - 3^2) Starting with the first term: -(-1)^4. I remember that exponents come before multiplication and the negative sign is like multiplying by -1. So, according to PEMDAS, exponentiation is first. Let's compute (-1)^4 first. Since the exponent is 4 (even), the result will be positive. (-1)^4 = 1. Then, apply the negative sign in front: -1. So the first term simplifies to -1. Now, moving on to the second part: - (1 - 0.5) = 0.5 - The next term is just frac{1}{3} - The last term is (2 - 3^2). Let's compute 3^2 first: 9. Then, 2 - 9 = -7. Now, multiply these three results together: 0.5 * (1/3) * (-7) Let's compute step by step. First, 0.5 * (1/3) = (1/2) * (1/3) = 1/6 Then, 1/6 * (-7) = -7/6 So the second part simplifies to -7/6. Now, putting it all together: First term - second part Which is: -1 - (-7/6) Subtracting a negative is like adding a positive. So this becomes: -1 + 7/6 Convert -1 to a fraction with denominator 6: -6/6 + 7/6 = ( -6 + 7 ) / 6 = 1/6 Wait, that seems like the result is 1/6? Let me verify to make sure I didn't make a mistake. First term: -(-1)^4 (-1)^4 is 1, so negative of that is -1. Correct. Second term: (1 - 0.5) = 0.5 frac{1}{3} stays the same. (2 - 3^2) = 2 - 9 = -7. Correct. Multiplying them: 0.5 * 1/3 = 1/6, then 1/6 * (-7) = -7/6. Correct. So the entire expression is: -1 - (-7/6) = -1 + 7/6. Convert -1 to sixths: -6/6 + 7/6 = 1/6. So yes, 1/6. But wait, 1/6 is approximately 0.1667. Let me check again. Alternatively, maybe I messed up the signs? Let's go step by step again. Original problem: -(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2) Compute each part: 1. -(-1)^4 Exponent first: (-1)^4 = 1 Then apply the negative: -1 2. (1 - 0.5) = 0.5 3. frac{1}{3} remains 4. (2 - 3^2) = 2 - 9 = -7 Multiply 0.5 * (1/3) = 1/6, then 1/6 * (-7) = -7/6 So the entire expression is: -1 - (-7/6) = -1 + 7/6 = (-6/6 + 7/6) = 1/6 Yes, that seems correct. So problem (1) answer is 1/6. Problem (2): Calculate: frac{5}{13} times (-3frac{1}{4}) - 0.5 div |{-3 - 1}| Again, let's break this down. First, convert mixed numbers to improper fractions. -3 1/4 is equal to - (3 + 1/4) = -13/4. So the first term is (5/13) * (-13/4) The second term is 0.5 divided by the absolute value of (-3 -1). Let's compute the absolute value first: |-3 -1| = |-4| = 4. Then 0.5 / 4. Let me compute each part step by step. First term: (5/13) * (-13/4) Notice that 13 in the denominator and numerator will cancel out. 5/13 * (-13/4) = 5 * (-1/4) = -5/4 Second term: 0.5 / 4 0.5 is 1/2, so (1/2) / 4 = 1/2 * 1/4 = 1/8 Now, the entire expression is: -5/4 - 1/8 To subtract these fractions, find a common denominator, which is 8. Convert -5/4 to eighths: -10/8 So, -10/8 - 1/8 = (-10 -1)/8 = -11/8 So the result is -11/8. Let me check again. First term: 5/13 multiplied by -13/4. 13 cancels out, 5 remains in numerator, 4 in denominator, with a negative sign: -5/4. Correct. Second term: 0.5 divided by | -3 -1 | Compute inside absolute value: -3 -1 = -4, absolute value is 4. 0.5 / 4 = 0.125, which is 1/8. Correct. So combining: -5/4 - 1/8 Convert to eighths: -5/4 = -10/8 -10/8 - 1/8 = -11/8. Correct. Therefore, problem (2) answer is -11/8. But let me verify once more to be thorough. Alternatively, converting fractions to decimals to cross-validate. First term: 5/13 * (-13/4) 5 divided by 13 is approximately 0.3846 Multiply by -13/4: 0.3846 * (-13) ≈ -5, then divided by 4 ≈ -1.25, which is -5/4. Correct. Second term: 0.5 / 4 = 0.125. Correct. Total: -1.25 - 0.125 = -1.375. Convert -11/8 to decimal: 8 goes into 11 once (8), remainder 3. 3/8 is 0.375. So total is -1.375. Which matches. Therefore, problem (2) is indeed -11/8. So summarizing: Problem (1) answer is 1/6 Problem (2) answer is -11/8 Final Answer(1) boxed{dfrac{1}{6}} (2) boxed{-dfrac{11}{8}}</think>To solve the given problems, we follow the order of operations (PEMDAS) carefully.Problem (1): Calculate: [ -(-1)^4 - (1 - 0.5) times frac{1}{3} times (2 - 3^2) ]1. Compute the first term: [ -(-1)^4 = -1 ] (Since ((-1)^4 = 1), applying the negative sign gives (-1)).2. Compute the second part step by step: - (1 - 0.5 = 0.5) - (2 - 3^2 = 2 - 9 = -7) - Multiply the results: [ 0.5 times frac{1}{3} times (-7) = left(frac{1}{2} times frac{1}{3}right) times (-7) = frac{1}{6} times (-7) = -frac{7}{6} ]3. Combine the terms: [ -1 - left(-frac{7}{6}right) = -1 + frac{7}{6} = -frac{6}{6} + frac{7}{6} = frac{1}{6} ]Final Answer (1): [boxed{dfrac{1}{6}}]Problem (2): Calculate: [ frac{5}{13} times left(-3frac{1}{4}right) - 0.5 div |{-3-1}| ]1. Convert the mixed number to an improper fraction: [ -3frac{1}{4} = -frac{13}{4} ]2. Compute the first term: [ frac{5}{13} times left(-frac{13}{4}right) = -frac{5}{4} ] (The 13 in the numerator and denominator cancels out).3. Compute the absolute value and the second term: [ |-3 - 1| = |-4| = 4 ] [ 0.5 div 4 = frac{1}{2} div 4 = frac{1}{8} ]4. Combine the terms: [ -frac{5}{4} - frac{1}{8} = -frac{10}{8} - frac{1}{8} = -frac{11}{8} ]Final Answer (2): [boxed{-dfrac{11}{8}}]]

🔑:Problem 1:Calculate -(-1)^4 - (1-0.5) times frac{1}{3} times (2 - 3^2).Step-by-Step Solution:First, we simplify the expression inside the parentheses and the exponent:-(-1)^4 - (1-0.5) times frac{1}{3} times (2 - 3^2)= -1 - frac{1}{2} times frac{1}{3} times (2 - 9)Here, we calculate 2 - 9 = -7, and simplify the fractions:= -1 - frac{1}{2} times frac{1}{3} times (-7)= -1 + frac{1}{2} times frac{1}{3} times 7= -1 + frac{7}{6}Finally, we find a common denominator and add the fractions:= frac{-6}{6} + frac{7}{6}= frac{1}{6}So, the final answer is boxed{frac{1}{6}}.Problem 2:Calculate frac{5}{13} times (-3frac{1}{4}) - 0.5 div |{-3-1}|.Step-by-Step Solution:First, we convert the mixed number to an improper fraction and simplify the absolute value:frac{5}{13} times (-3frac{1}{4}) - 0.5 div |{-3-1}|= frac{5}{13} times (-frac{13}{4}) - frac{1}{2} div 4Next, we multiply the fractions and divide by 4:= -frac{5}{4} - frac{1}{2} div 4= -frac{5}{4} - frac{1}{8}Finally, we find a common denominator and add the fractions:= -frac{10}{8} - frac{1}{8}= -frac{11}{8}So, the final answer is boxed{-frac{11}{8}}.

🧠:[<think>Okay, so I need to figure out what fraction of the cake Postman Pechkin ate. Let's see, there are four people: Uncle Fyodor, Cat Matroskin, Sharik, and Postman Pechkin. They all ate the entire cake. The problem gives me some relationships between how much each person ate, and I know Sharik ate one-tenth of the cake. I need to translate these relationships into equations and solve for Pechkin's portion.First, let's denote the fraction of the cake each person ate as variables:- Let P be the fraction Pechkin ate.- Let U be the fraction Uncle Fyodor ate.- Let M be the fraction Cat Matroskin ate.- Let S be the fraction Sharik ate.From the problem, I know that Sharik ate one-tenth, so S = 0.1.The total cake eaten by all four is 1 (the whole cake), so:U + M + S + P = 1Now, the problem states:1. "Uncle Fyodor ate half as much as Pechkin." So, U = (1/2)P.2. "Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat." Hmm, this one is a bit trickier. Let's parse that."The part of the cake that Pechkin did not eat" would be 1 - P, since Pechkin ate P. So, Matroskin ate half of that? Wait, the problem says "half as much as the part of the cake that Pechkin did not eat." So, M = (1/2)(1 - P). Is that correct? Let me make sure.Wait, the problem says "Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat." So if Pechkin did not eat (1 - P), then Matroskin ate half of that quantity? So yes, M = (1/2)(1 - P). That seems right.So summarizing:U = (1/2)PM = (1/2)(1 - P)S = 0.1And total:U + M + S + P = 1Now substitute U, M, and S in terms of P into the total equation.So:(1/2)P + (1/2)(1 - P) + 0.1 + P = 1Let me compute each term step by step.First, expand (1/2)P + (1/2)(1 - P):(1/2)P + (1/2)(1) - (1/2)P = (1/2)P - (1/2)P + 1/2 = 0 + 1/2 = 1/2Wait, hold on, that's interesting. The first two terms when combined cancel each other out? Let me check that again.Wait, (1/2)P + (1/2)(1 - P) = (1/2)P + 1/2 - (1/2)P = 1/2. Yes, the P terms cancel. So that simplifies to 1/2. Then add the other terms:1/2 + 0.1 + P = 1So combining 1/2 and 0.1: 0.5 + 0.1 = 0.6So the equation becomes:0.6 + P = 1Therefore, P = 1 - 0.6 = 0.4Wait, so according to this, Pechkin ate 0.4 of the cake? Let me verify this again because that seems straightforward, but maybe I missed something.Let me check each step again.Given:U = (1/2)PM = (1/2)(1 - P)S = 0.1Total equation:U + M + S + P = 1Substituting U and M:(1/2)P + (1/2)(1 - P) + 0.1 + P = 1First, let's compute (1/2)P + (1/2)(1 - P):= (1/2)P + (1/2) - (1/2)P= (1/2)P - (1/2)P + 1/2= 0 + 1/2= 1/2So that part is correct, the first two terms add up to 1/2.Then adding S (0.1) and P:1/2 + 0.1 + P = 1Convert 1/2 to 0.5:0.5 + 0.1 + P = 10.6 + P = 1Therefore, P = 1 - 0.6 = 0.4So P = 0.4, so Pechkin ate 0.4, which is 40% of the cake.Wait, but let me check if that satisfies all conditions.If Pechkin ate 0.4:Uncle Fyodor ate half as much as Pechkin: U = 0.4 / 2 = 0.2Cat Matroskin ate half as much as the part Pechkin did not eat. Pechkin did not eat 1 - 0.4 = 0.6, so half of that is 0.3. So M = 0.3Sharik ate 0.1Adding them up: 0.2 + 0.3 + 0.1 + 0.4 = 1.0Which is correct. So that checks out. So Pechkin ate 0.4 of the cake.But wait, let me just make sure that the interpretation of Matroskin's portion is correct.The problem says: "Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat."So "the part of the cake that Pechkin did not eat" is 1 - P. Then "half as much as" that part. So half of (1 - P). So M = (1/2)(1 - P). Which in this case, with P = 0.4, would be 0.6 / 2 = 0.3, which is correct.Alternatively, is there another interpretation?Like, maybe "half as much as Pechkin did not eat" could be referring to Pechkin's uneaten portion relative to something else? Hmm. But "the part of the cake that Pechkin did not eat" is clearly 1 - P. So half as much as that part. So M = 0.5*(1 - P). So that seems correct.Alternatively, sometimes "half as much as" can be confusing. For instance, if someone says A ate half as much as B, that means A = (1/2)B. So in this case, Matroskin ate half as much as (the part Pechkin did not eat). Since the part Pechkin did not eat is (1 - P), then M = (1/2)*(1 - P). So yes, that's correct.Therefore, the calculation seems correct, and all the portions add up to 1. So the answer is 0.4. So Pechkin ate 0.4 of the cake, which is 2/5, but as a decimal, it's 0.4.Therefore, the answer is 0.4.Final Answerboxed{0.4}</think>Let ( P ) be the fraction of the cake Postman Pechkin ate. We denote the fractions eaten by Uncle Fyodor, Cat Matroskin, and Sharik as ( U ), ( M ), and ( S ) respectively. We know the following:1. Sharik ate one-tenth of the cake: ( S = 0.1 ).2. Uncle Fyodor ate half as much as Pechkin: ( U = frac{1}{2}P ).3. Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat: ( M = frac{1}{2}(1 - P) ).The total fraction of the cake eaten by all four is 1:[U + M + S + P = 1]Substituting the expressions for ( U ), ( M ), and ( S ) into the equation:[frac{1}{2}P + frac{1}{2}(1 - P) + 0.1 + P = 1]Simplifying the terms step by step:1. Combine (frac{1}{2}P) and (frac{1}{2}(1 - P)): [ frac{1}{2}P + frac{1}{2} - frac{1}{2}P = frac{1}{2} ]2. Add the remaining terms: [ frac{1}{2} + 0.1 + P = 1 ]3. Convert (frac{1}{2}) to 0.5 and combine with 0.1: [ 0.5 + 0.1 + P = 1 ]4. Solve for ( P ): [ 0.6 + P = 1 implies P = 1 - 0.6 implies P = 0.4 ]Thus, the fraction of the cake that Postman Pechkin ate is (boxed{0.4}).,<think>Okay, let's see. So there's this cake that Uncle Fyodor, Cat Matroskin, Sharik, and Postman Pechkin all ate. The whole cake was eaten, so the fractions they each ate should add up to 1. The problem is asking for the fraction that Postman Pechkin ate, right? And we know that Sharik ate one-tenth of the cake. So first, let's list out what we know.First, let me assign variables to each person to make it easier. Let me let P be the fraction of the cake that Postman Pechkin ate. Then Uncle Fyodor ate half as much as Pechkin, so that would be (1/2)P. Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat. Hmm, that part might be a bit tricky. The part Pechkin did not eat is (1 - P), right? Because the whole cake is 1. So Matroskin ate half of that? So Matroskin ate (1/2)(1 - P). And Sharik ate 1/10. So all these should add up to 1.Let me write that out as an equation. So:Uncle Fyodor + Cat Matroskin + Sharik + Pechkin = 1Substituting the variables:(1/2)P + (1/2)(1 - P) + 1/10 + P = 1Okay, let's check if that's correct. Uncle Fyodor is half of Pechkin's amount, so (1/2)P. Matroskin is half of what Pechkin did not eat, which is half of (1 - P). Sharik is 1/10, and Pechkin is P. Yeah, that seems right.Now, let's simplify this equation step by step. Let's first combine the terms with P.First, expand (1/2)(1 - P). That's (1/2) - (1/2)P.So now the equation becomes:(1/2)P + (1/2 - (1/2)P) + 1/10 + P = 1Now, let's combine like terms. Let's handle the P terms first.(1/2)P - (1/2)P + P. The first two terms cancel each other out: (1/2)P - (1/2)P = 0. Then we just have P left.Then the constants are (1/2) + 1/10.So now the equation simplifies to:P + (1/2 + 1/10) = 1Let's compute 1/2 + 1/10. Convert them to tenths: 5/10 + 1/10 = 6/10 = 3/5.So now the equation is:P + 3/5 = 1Subtract 3/5 from both sides:P = 1 - 3/5 = 2/5Wait, so P is 2/5? Let me check that again because this seems straightforward, but let me verify.So, if Pechkin ate 2/5, then Uncle Fyodor ate half of that, which is 1/5. Matroskin ate half of what Pechkin did not eat. Pechkin did not eat 1 - 2/5 = 3/5. Half of that is 3/10. Sharik ate 1/10. Let's add them all up:1/5 (Uncle Fyodor) + 3/10 (Matroskin) + 1/10 (Sharik) + 2/5 (Pechkin)Convert everything to tenths:1/5 = 2/10, 3/10 stays, 1/10 stays, 2/5 = 4/10.Adding them: 2/10 + 3/10 + 1/10 + 4/10 = (2 + 3 + 1 + 4)/10 = 10/10 = 1.Okay, that checks out. So Pechkin ate 2/5 of the cake, which is 0.4 as a decimal. So the answer should be 0.4.Wait, but let me make sure I didn't misinterpret the problem. Let me read the problem again to check."Uncle Fyodor ate half as much as Pechkin," so that's correct, (1/2)P."Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat." So Pechkin didn't eat (1 - P), and Matroskin ate half of that, so (1/2)(1 - P). That seems right.Sharik ate 1/10. Total is 1. So all the substitutions and equations are correct.So the calculation seems correct. Therefore, the answer should be 2/5, which is 0.4 in decimal. So I think that's the answer.Final AnswerThe fraction of the cake that Postman Pechkin ate is boxed{0.4}.</think>Let ( P ) be the fraction of the cake that Postman Pechkin ate. According to the problem:- Uncle Fyodor ate half as much as Pechkin: ( frac{1}{2}P ).- Cat Matroskin ate half as much as the part of the cake that Pechkin did not eat: ( frac{1}{2}(1 - P) ).- Sharik ate ( frac{1}{10} ) of the cake.The total fraction of the cake eaten by all four should sum to 1. Therefore, we set up the equation:[frac{1}{2}P + frac{1}{2}(1 - P) + frac{1}{10} + P = 1]Expanding and simplifying the equation:[frac{1}{2}P + frac{1}{2} - frac{1}{2}P + frac{1}{10} + P = 1]Combining like terms:[P + left( frac{1}{2} + frac{1}{10} right) = 1]Converting the constants to a common denominator:[frac{1}{2} + frac{1}{10} = frac{5}{10} + frac{1}{10} = frac{6}{10} = frac{3}{5}]Thus, the equation simplifies to:[P + frac{3}{5} = 1]Solving for ( P ):[P = 1 - frac{3}{5} = frac{2}{5}]Converting ( frac{2}{5} ) to a decimal gives ( 0.4 ).The fraction of the cake that Postman Pechkin ate is boxed{0.4}.]

🔑:Answer: 0.4.Solution. Let x be the portion of the cake eaten by Pechkin. Then Fyodor ate frac{x}{2}, and Matroskin the cat ate -frac{1-x}{2}. That is, they together ate half of the cake. This means that Sharik and Pechkin the postman together ate the other half of the cake.We can set up the equation: x (Pechkin's portion) +0.1 (Sharik's portion) =0.5 ; x=0.4.

🧠:[<think>Okay, so I need to figure out this problem about the even function f(x) that's increasing on the interval (-∞, -1]. The question is asking which of the options A, B, C, or D is true regarding the order of f(2), f(-1.5), and f(-1). Let me start by recalling what an even function is. An even function satisfies the condition f(x) = f(-x) for all x in its domain. So, the graph of the function is symmetric with respect to the y-axis. That means if I know the value of the function at some positive x, I also know it at -x, and vice versa.Now, the problem states that f(x) is increasing on (-∞, -1]. When a function is increasing on an interval, as x increases, the value of the function also increases. So, in this case, as we move from left to right on the interval (-∞, -1], the function's value is going up. But wait, since the interval is from negative infinity up to -1, moving left to right there means x is getting larger (less negative) approaching -1. So, for example, at x = -2 (which is further left), the function value would be less than at x = -1.5, which is less than at x = -1. So, f(-2) < f(-1.5) < f(-1).But since the function is even, f(2) is equal to f(-2), right? Because f(2) = f(-2) by the definition of an even function. So, substituting that in, f(2) = f(-2) < f(-1.5) < f(-1). Therefore, f(2) < f(-1.5) < f(-1). That would correspond to option A: f(2) < f(-1.5) < f(-1). Hmm, but let me make sure I'm not making a mistake here.Wait, hold on. The function is increasing on (-∞, -1], so as x increases from -∞ to -1, f(x) increases. Therefore, at x = -2, which is less than x = -1.5, which is less than x = -1, so f(-2) < f(-1.5) < f(-1). But since f is even, f(2) = f(-2). So, substituting, f(2) = f(-2) < f(-1.5) < f(-1). So yes, that would mean f(2) < f(-1.5) < f(-1), which is option A.But wait, let me think again. Is there a possibility that the function's behavior on the other side of the y-axis (for positive x) affects this? Since the function is even, the right side (positive x) is a mirror image of the left side. So, if the function is increasing on (-∞, -1], then on the interval [1, ∞), which is the mirror image, the function would be decreasing. Because if you reflect an increasing function over the y-axis, it becomes decreasing. Let me confirm that.Suppose we have a point x = a on the left side (negative), so f(a) is some value. On the right side, x = -a (positive), and f(-a) = f(a). If the function is increasing as we move from left to right on the left side (i.e., as x increases from -∞ to -1), then on the right side, moving from right to left (from ∞ to 1), the function would be increasing, but since we usually consider moving from left to right on the number line, the function on the right side [1, ∞) would be decreasing as x increases. So, as x increases beyond 1, f(x) decreases.So, if I consider f(2), since 2 is in [1, ∞), and the function is decreasing there, then the larger x is, the smaller f(x) becomes. Therefore, f(2) would be less than f(1), because at x=1, which is the start of the interval [1, ∞), the function is higher and then decreases. But wait, how do we relate this to the values at -1.5 and -1?Hold on, maybe I need to directly compare all these values. Let's list out the key points:1. The function is increasing on (-∞, -1], so for any a < b ≤ -1, f(a) < f(b).2. The function is even, so f(-x) = f(x) for all x.Therefore:- f(-1.5) is in the interval (-∞, -1], and since the function is increasing there, as x increases from -∞ to -1, f(x) increases. Therefore, comparing f(-1.5) and f(-1): since -1.5 < -1, and the function is increasing, f(-1.5) < f(-1).Similarly, f(-2) < f(-1.5) because -2 < -1.5, and since it's increasing, f(-2) < f(-1.5). But f(-2) = f(2) because the function is even. So, f(2) = f(-2) < f(-1.5) < f(-1). Therefore, ordering is f(2) < f(-1.5) < f(-1), which is option A.But wait, earlier I thought about the function decreasing on [1, ∞). Let me check if that affects f(2). Since 2 is greater than 1, and if the function is decreasing on [1, ∞), then f(2) < f(1). But f(1) is equal to f(-1) because the function is even. Therefore, f(1) = f(-1). So, combining this, f(2) < f(1) = f(-1). Therefore, f(2) < f(-1). But we also have f(-1.5) < f(-1) from the increasing on the left side. So, how does f(-1.5) compare to f(2)?Wait, since f(-1.5) is greater than f(-2) (because the function is increasing on (-∞, -1], and -2 < -1.5 < -1). Then, since f(-2) = f(2), so f(2) = f(-2) < f(-1.5) < f(-1). Therefore, f(2) < f(-1.5) < f(-1). So that's the order. Therefore, option A is correct.But let me check again. If the function is even, then the right side is a mirror image. So, if it's increasing on the left side (from -∞ to -1), then on the right side (from 1 to ∞), it's decreasing. Therefore, on the right side, as x increases, f(x) decreases. So, x=1 corresponds to x=-1, and x=2 corresponds to x=-2. So, since on the left side, as x increases from -∞ to -1, f(x) increases, so f(-2) < f(-1.5) < f(-1). On the right side, as x increases from 1 to ∞, f(x) decreases, so f(1) > f(2) > f(3) > ... But f(1) = f(-1), f(2) = f(-2). Therefore, since on the left side, f(-2) < f(-1.5) < f(-1), on the right side, that corresponds to f(2) = f(-2) < f(-1.5) = f(1.5) < f(-1) = f(1). Wait, but hold on. Wait, f(-1.5) is equal to f(1.5), right? Because f(-x) = f(x). So, f(-1.5) = f(1.5). But where is 1.5 located? 1.5 is between 1 and 2. But since on the right side [1, ∞), the function is decreasing, then at x=1.5, which is greater than 1, f(1.5) would be less than f(1) but greater than f(2). Because as x increases from 1 to ∞, f(x) decreases. So, f(1) > f(1.5) > f(2). Therefore, f(1) = f(-1) > f(1.5) = f(-1.5) > f(2) = f(-2). Therefore, ordering would be f(2) < f(-1.5) < f(-1), which is the same as before, so that's still option A. Wait, but in this case, f(1.5) is between f(1) and f(2). But since f(-1.5) = f(1.5), then f(-1.5) is between f(-1) and f(2). But according to the previous reasoning, since on the left side, f(-2) < f(-1.5) < f(-1), and f(2) = f(-2), that gives f(2) < f(-1.5) < f(-1). However, if we use the right side, since f(1.5) is between f(1) and f(2), but decreasing, then f(1) > f(1.5) > f(2). Since f(1) = f(-1), then f(-1) > f(-1.5) > f(2). Wait, that contradicts the previous conclusion. Now I'm confused.Wait, so which is correct? If f is decreasing on [1, ∞), then as x increases from 1 to infinity, f(x) decreases. So, x=1.5 is greater than 1, so f(1.5) < f(1). But x=1.5 is less than 2, so f(1.5) > f(2). Therefore, f(1) > f(1.5) > f(2). Therefore, since f(-1.5) = f(1.5), then f(-1.5) = f(1.5) which is between f(1) and f(2). But f(1) = f(-1), so f(-1) > f(-1.5) > f(2). Therefore, this would imply that f(-1) > f(-1.5) > f(2), so the order is f(2) < f(-1.5) < f(-1). Wait, but according to the right side, it's f(2) < f(1.5) < f(1). Translating that to the left side, f(2) = f(-2), f(1.5) = f(-1.5), f(1) = f(-1). Therefore, f(-2) < f(-1.5) < f(-1). Therefore, this matches the original conclusion.So, both approaches give the same ordering. So, why was there a confusion? Because when we think about the right side decreasing, we have to remember that as x increases, f(x) decreases, so larger x on the right gives smaller f(x). Therefore, 1.5 is between 1 and 2, so f(1.5) is between f(1) and f(2), but since f is decreasing, f(1) > f(1.5) > f(2). Translating to the left side, f(-1) > f(-1.5) > f(-2) = f(2). So, the order is f(2) < f(-1.5) < f(-1), which is option A. Therefore, the correct answer should be A.Wait, but let me check once more with actual numbers. Let's assume a specific even function that is increasing on (-∞, -1]. For example, let's take f(x) = - (x + 1)^2 for x <= -1, but since it's even, f(x) = f(-x). Wait, but if it's increasing on (-∞, -1], then the function must be increasing as x approaches -1 from the left. Let's pick a simple even function that is increasing on (-∞, -1]. For example, consider f(x) = -1/x for x < 0 and f(x) = -1/x for x > 0, but this isn't even. Wait, no. To make it even, f(x) should equal f(-x). Let's think of another function. Maybe f(x) = |x|, but that's increasing on [0, ∞), not (-∞, -1]. Wait, but |x| is actually decreasing on (-∞, 0]. So that's not the case.Alternatively, take f(x) = -x^2. Wait, that's an even function because (-x)^2 = x^2. But the function f(x) = -x^2 is decreasing on (-∞, 0], which is the opposite of what we want. So maybe f(x) = - (x + 2)^2 for x <= -1 and then even. Wait, this is getting complicated. Let me try a piecewise function.Suppose we define f(x) as follows: For x <= -1, f(x) = x + 2. Since it's increasing on (-∞, -1]. Then, since it's even, for x >= 1, f(x) = f(-x) = (-x) + 2 = -x + 2. For -1 < x < 1, we can define it as any even function, but maybe a constant? Let's not worry about that part since the problem doesn't mention behavior there.So, with this setup, f(-2) = -2 + 2 = 0, f(-1.5) = -1.5 + 2 = 0.5, f(-1) = -1 + 2 = 1. Then, on the right side, f(1) = -1 + 2 = 1, f(1.5) = -1.5 + 2 = 0.5, f(2) = -2 + 2 = 0. So, in this case, f(2) = 0, f(-1.5) = 0.5, f(-1) = 1. So, f(2) < f(-1.5) < f(-1). Which is 0 < 0.5 < 1. So, this supports option A.Alternatively, if I take another example, say f(x) = e^x for x <= -1, but since it's even, for x >= 1, f(x) = e^{-x}. Then, on (-∞, -1], f(x) = e^x, which is increasing because the derivative is e^x, which is positive. Wait, but e^x is actually increasing everywhere, but since x is going to -infinity, e^x approaches 0. Wait, at x = -infinity, e^x approaches 0, and at x = -1, e^x is e^{-1}. So, as x increases from -infty to -1, e^x increases from 0 to e^{-1}. So, that's increasing. Then, on the right side, for x >= 1, f(x) = e^{-x}, which is decreasing. So, f(2) = e^{-2}, f(1.5) = e^{-1.5}, f(1) = e^{-1}. So, e^{-2} < e^{-1.5} < e^{-1}, so f(2) < f(1.5) < f(1). But since f(1.5) = f(-1.5) and f(1) = f(-1), this translates to f(2) < f(-1.5) < f(-1), again option A.Therefore, in both concrete examples, the order is A. Therefore, I think the answer is A. However, let me check the answer choices again to make sure I didn't misread.Options:A. f(2) < f(-1.5) < f(-1)B. f(-1) < f(-1.5) < f(2)C. f(2) < f(-1) < f(-1.5)D. f(-1.5) < f(-1) < f(2)So, my conclusion is A. But wait, in my second example, with the exponential function, f(-1.5) is e^{-1.5} and f(-1) is e^{-1}, which are approximately 0.223 and 0.368, so indeed 0.223 < 0.368. And f(2) is e^{-2} ≈ 0.135, so 0.135 < 0.223 < 0.368, so yes, f(2) < f(-1.5) < f(-1). Similarly, in the first example with the linear function, 0 < 0.5 < 1. So, A is correct.Therefore, the correct answer should be option A. Therefore, I think the answer is A.Final Answerboxed{A}</think>Given that ( f(x) ) is an even function and is increasing on the interval ((-infty, -1]), we need to determine the correct order of ( f(2) ), ( f(-1.5) ), and ( f(-1) ).1. Even Function Property: Since ( f(x) ) is even, ( f(x) = f(-x) ) for all ( x ). This means ( f(2) = f(-2) ), ( f(-1.5) = f(1.5) ), and ( f(-1) = f(1) ).2. Increasing Interval: The function ( f(x) ) is increasing on ((-infty, -1]). Therefore, for any ( a < b leq -1 ), we have ( f(a) < f(b) ). Specifically, since (-2 < -1.5 < -1), it follows that ( f(-2) < f(-1.5) < f(-1) ).3. Applying the Even Function Property: - ( f(2) = f(-2) ) - From the increasing property, ( f(-2) < f(-1.5) < f(-1) ) - Therefore, ( f(2) < f(-1.5) < f(-1) )4. Behavior on the Right Side: Since ( f(x) ) is even, the behavior on the right side ([1, infty)) is a mirror image of the left side. As ( f(x) ) is decreasing on ([1, infty)), ( f(1) > f(1.5) > f(2) ). Translating this to the left side using the even property confirms ( f(-1) > f(-1.5) > f(2) ).Thus, the correct order is ( f(2) < f(-1.5) < f(-1) ), which corresponds to option A.[boxed{A}],<think>Alright, so I have this problem here about an even function f(x) that's increasing on the interval (-∞, -1]. I need to figure out which of the options A, B, C, or D is true based on that information. Let me start by recalling what an even function is. An even function satisfies the condition f(-x) = f(x) for all x in its domain. That means the graph of the function is symmetric with respect to the y-axis. So whatever happens on the right side of the y-axis is mirrored on the left side, and vice versa.The problem states that f(x) is increasing on (-∞, -1]. If a function is increasing on an interval, its value increases as x increases. But since we're dealing with negative x-values here, increasing on (-∞, -1] means that as x becomes more negative (i.e., moves leftwards on the number line), the function's value increases. Wait, actually, in terms of calculus, if a function is increasing on an interval, then for any two points a < b in that interval, f(a) ≤ f(b). But since we're dealing with negative numbers here, maybe I need to think carefully about how the function behaves.So, for example, take two points in (-∞, -1], say x1 = -3 and x2 = -2. Since -3 < -2 and the function is increasing on this interval, we must have f(-3) ≤ f(-2). But because these are negative numbers, moving from left to right (i.e., from more negative to less negative) on the interval (-∞, -1], the function is increasing. So as x increases (becomes less negative), f(x) increases. Therefore, closer to -1 (which is the right end of the interval), the function has higher values.Now, since the function is even, f(-x) = f(x). So the behavior of the function on the positive side (right of the y-axis) is a mirror image of the left side. Specifically, the interval [1, ∞) would correspond to the interval (-∞, -1]. Because the function is even, the increasing nature on (-∞, -1] should translate to decreasing on [1, ∞). Let me verify that. Suppose we take two points on [1, ∞), say x1 = 1 and x2 = 2. Then their counterparts on the left side are -1 and -2. Since the function is increasing on the left, f(-2) ≤ f(-1). But by the evenness, f(-2) = f(2) and f(-1) = f(1). So f(2) ≤ f(1). Therefore, on the right side, as x increases from 1 to 2, the function's value decreases. So f(x) is decreasing on [1, ∞).So, putting this together: On the left side, from -∞ to -1, the function is increasing. On the right side, from 1 to ∞, the function is decreasing. So between these intervals, how do the function values compare?Now, let's look at the points given in the options: 2, -1.5, and -1. Let's note their positions. -1.5 is in (-∞, -1], so on the left interval where the function is increasing. 2 is in [1, ∞), so on the right interval where the function is decreasing. -1 is the endpoint of the interval where the function is increasing on the left.First, let's analyze the left side. Since f(x) is increasing on (-∞, -1], then for any two points a < b ≤ -1, f(a) ≤ f(b). So, for example, take a = -2 and b = -1.5. Since -2 < -1.5 < -1, then f(-2) ≤ f(-1.5) ≤ f(-1). But since the function is even, f(-2) = f(2) and f(-1.5) = f(1.5), and f(-1) = f(1).But wait, maybe we don't need to convert them to positive numbers. Let me see. Let's consider the points given: -1.5 is in (-∞, -1], so f(-1.5) is part of the increasing interval. Since the function is increasing there, as x increases (i.e., becomes less negative), the function's value increases. So, moving from -∞ towards -1, the function's value goes from lower to higher. Therefore, the closer x is to -1, the larger f(x) is. So f(-1.5) is less than f(-1) because -1.5 is less than -1 (wait, but hold on: in terms of numbers, -1.5 is less than -1, but in terms of position on the interval (-∞, -1], -1.5 is to the left of -1. So since the function is increasing on that interval, moving from left to right (from -∞ to -1), f(x) increases. Therefore, since -1.5 < -1, then f(-1.5) < f(-1). So that's one relationship: f(-1.5) < f(-1).Now, what about f(2)? Since 2 is in [1, ∞), and we established that on [1, ∞), the function is decreasing. So for points a < b in [1, ∞), f(a) ≥ f(b). So 1 < 2, so f(1) ≥ f(2). But f(1) is equal to f(-1) because the function is even. Therefore, f(-1) = f(1) ≥ f(2). So f(2) ≤ f(-1). But we need to compare f(2) with f(-1.5) and f(-1).So far, we know:1. f(-1.5) < f(-1) (from the increasing on the left interval)2. f(2) ≤ f(1) = f(-1) (from the decreasing on the right interval)But where does f(2) stand in relation to f(-1.5)?To figure that out, let's consider that f(2) = f(-2) (since it's even). Now, since -2 is in (-∞, -1], and the function is increasing there, and -2 < -1.5 < -1, so f(-2) < f(-1.5) < f(-1). Therefore, f(2) = f(-2) < f(-1.5) < f(-1). So putting that together: f(2) < f(-1.5) < f(-1). Therefore, the order from smallest to largest is f(2) < f(-1.5) < f(-1). Which corresponds to option A.But let me verify this again step by step to make sure I haven't confused anything.First, f(x) is even, so f(-x) = f(x). So f(2) = f(-2). Since the function is increasing on (-∞, -1], and -2 < -1.5 < -1, then f(-2) < f(-1.5) < f(-1). Therefore, substituting f(2) for f(-2), we get f(2) < f(-1.5) < f(-1). So that's option A. Therefore, the answer should be A.Wait, but let me check if there's another perspective. Maybe I made a mistake in the direction of the inequality.Wait, if the function is increasing on (-∞, -1], that means as x increases (from left to right, i.e., from more negative to less negative), f(x) increases. So, for example, x1 = -2 and x2 = -1.5. Then x1 < x2, so f(x1) < f(x2). So indeed, f(-2) < f(-1.5). Since f(-2) = f(2), we have f(2) < f(-1.5). Similarly, since -1.5 < -1, f(-1.5) < f(-1). So combining these, f(2) < f(-1.5) < f(-1). Therefore, option A is correct.But wait, let me check the options again. The options are:A. f(2) < f(-1.5) < f(-1)B. f(-1) < f(-1.5) < f(2)C. f(2) < f(-1) < f(-1.5)D. f(-1.5) < f(-1) < f(2)So yes, according to our reasoning, A is the correct one. But let me just cross-verify once again.Alternatively, consider specific examples. Let's take a simple even function that is increasing on (-∞, -1]. For instance, consider a function that is symmetric about the y-axis, increasing from -∞ up to x = -1, and then decreasing from x = 1 onwards. For example, let's take f(x) = -x^2. Wait, but that's a parabola opening downward, which is increasing on (-∞, 0) and decreasing on (0, ∞). But that's not increasing on (-∞, -1]; after x = -1, which is part of the interval (-∞, -1], but if the vertex is at 0, then from -∞ to 0, the function is increasing. Wait, maybe that's not the best example.Alternatively, maybe take a piecewise function. Let's define f(x) as follows: For x ≤ -1, f(x) = x + 2, which is increasing (since the slope is positive). For x ≥ 1, since it's even, f(x) = f(-x) = (-x) + 2 = -x + 2. Then between -1 and 1, maybe define it as a constant or something. Wait, but even functions need to satisfy f(-x) = f(x) for all x. So if we define it as x + 2 for x ≤ -1, then for x ≥ 1, f(x) must be -x + 2. Then, for example, f(2) = -2 + 2 = 0. f(-1.5) = -1.5 + 2 = 0.5. f(-1) = -1 + 2 = 1. So f(2) = 0, f(-1.5) = 0.5, f(-1) = 1. So in this case, f(2) < f(-1.5) < f(-1), which matches option A. So that example supports the conclusion.Alternatively, consider another even function. Let's take f(x) = |x|. But |x| is decreasing on (-∞, 0], which is not our case. So that's not a good example.Wait, but if the function is increasing on (-∞, -1], then in that interval, the function's slope is positive. For example, f(x) = -1/x for x ≤ -1. Then for x ≤ -1, f(x) = -1/x, which is increasing because the derivative f'(x) = 1/x², which is positive, so the function is increasing. Then for x ≥ 1, f(x) = f(-x) = -1/(-x) = 1/x. So f(2) = 1/2, f(-1.5) = -1/(-1.5) = 2/3, f(-1) = -1/(-1) = 1. Then f(2) = 0.5, f(-1.5) ≈ 0.666..., f(-1) = 1. So again, 0.5 < 0.666... < 1, which is f(2) < f(-1.5) < f(-1). So that's also option A.Therefore, based on both the theoretical reasoning and concrete examples, option A is correct. So I think the answer is A.Final Answerboxed{A}</think>Given that the even function ( f(x) ) is increasing on the interval ((-infty, -1]), we need to determine the correct order of the function values at points (2), (-1.5), and (-1).1. Understanding Even Functions: An even function satisfies ( f(-x) = f(x) ), meaning the function is symmetric about the y-axis. Therefore, the behavior of the function on the right side of the y-axis (positive (x)) is a mirror image of the left side (negative (x)).2. Behavior on Intervals: - Since ( f(x) ) is increasing on ((-infty, -1]), as (x) increases (becomes less negative), ( f(x) ) increases. - Due to the even nature, the function on the interval ([1, infty)) will be decreasing. This is because if ( f(x) ) is increasing on ((-infty, -1]), then ( f(-x) = f(x) ) implies ( f(x) ) is decreasing on ([1, infty)).3. Comparing Function Values: - For ( x = -1.5 ) in ((-infty, -1]), since the function is increasing, ( f(-1.5) < f(-1) ). - For ( x = 2 ) in ([1, infty)), since the function is decreasing, ( f(2) < f(1) ). Given ( f(1) = f(-1) ), we have ( f(2) < f(-1) ). - To compare ( f(2) ) and ( f(-1.5) ), note ( f(2) = f(-2) ). Since ( f ) is increasing on ((-infty, -1]), ( f(-2) < f(-1.5) ). Thus, ( f(2) < f(-1.5) ).Combining these results, we get the order ( f(2) < f(-1.5) < f(-1) ).The correct answer is (boxed{A}).]

🔑:The correct answer is A.Firstly, let's clarify the properties of even functions. An even function satisfies the property that for all values of x, f(x) = f(-x). Since f(x) is an even function, we can say that f(2) = f(-2).Next, since f(x) is increasing on the interval (-infty, -1], we have that for x_1 < x_2 leq -1, f(x_1) < f(x_2). The given interval (-infty, -1] corresponds to the left side of the graph where x is negative.Now, let's compare the given values:- Since -2 < -1.5 < -1, and since the function is increasing on (-infty, -1], f(-2) < f(-1.5).- Also from the increasing property, f(-1.5) < f(-1).- Combining both above points, f(-2) < f(-1.5) < f(-1).- Utilizing the even property of the function, f(2) = f(-2).Putting it all together, we get f(2) < f(-1.5) < f(-1).Therefore, the correct answer is boxed{A}.

🧠:[<think>Okay, so I need to solve this inequality: 2x² - 5x - 3 ≥ 0. The question is asking for a sufficient and necessary condition, and there are four options to choose from. Let me try to figure this out step by step.First, I remember that quadratic inequalities can be solved by finding the roots of the corresponding quadratic equation and then determining the intervals where the quadratic expression is positive or negative. The general form is ax² + bx + c, so here, a is 2, b is -5, and c is -3.So, first step, find the roots of the equation 2x² -5x -3 = 0. To do that, I can use the quadratic formula: x = [-b ± √(b² - 4ac)]/(2a). Plugging in the values, that becomes x = [5 ± √(25 + 24)]/4 because b² is (-5)² = 25, 4ac is 4*2*(-3) = -24, so the discriminant is 25 + 24 = 49. The square root of 49 is 7. Therefore, the roots are [5 + 7]/4 and [5 - 7]/4.Calculating those: [5 + 7]/4 = 12/4 = 3, and [5 - 7]/4 = (-2)/4 = -0.5. So the roots are x = 3 and x = -0.5. Now, since the quadratic coefficient a is 2, which is positive, the parabola opens upwards. That means the graph of the quadratic function will be a U-shaped curve. The inequality 2x² -5x -3 ≥ 0 is asking where the function is above or equal to zero. Since the parabola opens upwards, the graph will be above the x-axis outside the interval between the two roots. So the solution should be x ≤ -0.5 or x ≥ 3. Wait, let me confirm that. If the parabola opens upwards, then it's positive outside the roots and negative between them. So yes, the inequality 2x² -5x -3 ≥ 0 holds when x is less than or equal to the smaller root or greater than or equal to the larger root. So in interval notation, that would be (-∞, -0.5] ∪ [3, ∞).Now, let's check the options given:A: x ≥ 0. Hmm, that can't be right because from x = -0.5 to x = 3, the inequality doesn't hold. So if x is between 0 and 3, which is part of option A, the inequality is actually negative. So A is not correct.B: x < 0 or x > 2. Wait, x < 0 would include values less than 0, but our critical points are -0.5 and 3. So from -0.5 to 3, the inequality is negative. So x < 0 is partially correct, but between -0.5 and 0, x is still negative but the inequality is not holding. Wait, actually, if x is less than -0.5, which is part of x < 0, then the inequality does hold. But from -0.5 to 0, the inequality doesn't hold. So x < 0 is not entirely correct. Also, x > 2: our critical point is 3, so from 3 onwards, the inequality holds. But between 2 and 3, x > 2 but less than 3, the inequality is still negative. So option B says x < 0 or x > 2, but in reality, the correct intervals are x ≤ -0.5 or x ≥ 3. So B includes some incorrect regions. Therefore, B is not the correct answer.C: x ∈ {-1, 3, 5}. Let's test these points. For x = -1: 2(-1)^2 -5(-1) -3 = 2*1 +5 -3 = 2 +5 -3 = 4, which is ≥ 0. For x = 3: 2(9) -5(3) -3 = 18 -15 -3 = 0, which is okay. For x = 5: 2(25) -5(5) -3 = 50 -25 -3 = 22 ≥0. So those three points do satisfy the inequality. However, the question is asking for a sufficient and necessary condition. Option C is giving specific points, but the inequality holds for entire intervals, not just individual points. So even though those points are solutions, the necessary and sufficient condition should cover all solutions, not just some specific ones. So C is not the right answer.D: x < -1/2 or x > 3. Let's check. Our solution was x ≤ -0.5 or x ≥ 3. The option D is x < -0.5 or x > 3. So it's almost the same, except that it excludes the points x = -0.5 and x = 3. But in the original inequality, it's "≥ 0", which includes equality. So at x = -0.5 and x = 3, the expression equals zero, which satisfies the inequality. Therefore, the correct solution should include those points. So option D is incorrect because it uses strict inequalities instead of including the endpoints.Wait, but let me double-check. The options given: D says x < -1/2 or x > 3. But our solution is x ≤ -1/2 or x ≥ 3. So D is excluding the points -1/2 and 3. However, when x = -1/2, plugging into the inequality: 2*(-1/2)^2 -5*(-1/2) -3. Let's compute that: 2*(1/4) + 5/2 -3 = 0.5 + 2.5 -3 = 0. So 0 ≥ 0 is true. Similarly, at x = 3: 2*(9) -5*(3) -3 = 18 -15 -3 = 0, which is also true. Therefore, the correct solution should include those endpoints. So D is wrong because it uses strict inequalities. But none of the other options seem correct either.Wait, the options are A, B, C, D. Maybe I made a mistake here. Let me check again.Original inequality: 2x² -5x -3 ≥ 0. Roots at x = -0.5 and x =3. So solution is x ≤ -0.5 or x ≥3. So the correct intervals are (-∞, -0.5] ∪ [3, ∞). Now, check the options again.Option D is x < -1/2 or x > 3. So this is (-∞, -0.5) ∪ (3, ∞). This excludes the points -0.5 and 3. But at those points, the inequality is equal to zero, so they should be included. So the correct answer should include those. However, none of the options mention x ≤ -0.5 or x ≥3. Wait, maybe I need to check the options again.Looking at the options:A: x ≥ 0 – No, because from 0 to3, it's negative.B: x <0 or x >2 – No, because between -0.5 and 0, it's still negative, and between 2 and 3, also negative.C: x ∈ {-1,3,5} – These are specific points that work, but the inequality is true for intervals, so C is not a necessary and sufficient condition.D: x < -1/2 or x >3 – This is almost the correct interval except it's missing the equality at -0.5 and 3.Wait, but perhaps the original problem had a typo? Let me check the original problem again.The inequality is 2x² -5x -3 ≥0. The options:A: x ≥0B: x <0 or x >2C: x ∈ {-1,3,5}D: x＜−1/2 or x＞3Wait, in the original problem, is option D using strict inequalities? If so, then since at x=-1/2 and x=3, the expression is zero, which satisfies the inequality. So the correct answer should include those points. But option D does not. So maybe the options are incorrect? But the user is giving this problem, so perhaps there's a mistake here.Alternatively, maybe the person who made the problem considered that the roots are excluded, but that's not correct because the inequality is ≥0, which includes the roots. Therefore, unless there was a typo in the problem, perhaps the intended answer is D, but with a note that it should include the endpoints. However, given the options provided, D is the closest, even though technically incorrect. Wait, but maybe there is a mistake in my calculation.Wait, let me check the roots again. The quadratic equation: 2x² -5x -3 =0. Using quadratic formula:x = [5 ± √(25 +24)] /4 = [5 ±7]/4. So 5+7=12, 12/4=3; 5-7=-2, -2/4=-0.5. So roots at x=3 and x=-0.5. Correct. So the solution is x ≤ -0.5 or x ≥3. So the correct answer is intervals including those endpoints. But none of the options have that. Option D is x < -0.5 or x >3. So it's excluding the exact points where the expression equals zero. That's odd.Wait, maybe I made a mistake in interpreting the inequality. Let me test with some numbers. Let's take x = -0.5. Plug into 2x² -5x -3:2*(0.25) -5*(-0.5) -3 = 0.5 +2.5 -3 = 0. So 0 ≥0, which is true. Similarly, x=3: 2*9 -5*3 -3 =18-15-3=0, which is 0≥0, true. So those points should be included.But option D excludes them, so D is incorrect. However, none of the other options are correct either. Wait, maybe the answer is C? But C is a set of specific points. If the problem had asked "which of the following is a solution", then C would be a correct answer. But the problem says "a sufficient and necessary condition", which means the condition has to cover all solutions and only the solutions. So C is not a necessary and sufficient condition because there are infinitely many solutions (all x ≤ -0.5 or x ≥3), not just those three points.Hmm. Maybe the problem is written incorrectly? Or perhaps I made a mistake in the roots. Wait, let me check the quadratic equation again.2x² -5x -3 =0. a=2, b=-5, c=-3.Discriminant: b² -4ac = (-5)^2 -4*2*(-3) =25 +24=49. Correct. Square root of 49 is 7. So x=(5 ±7)/4. So x=(12)/4=3 and x=(-2)/4=-0.5. Correct. So the roots are correct.Therefore, the solution set is x ≤ -0.5 or x ≥3, but none of the options exactly match. However, option D is x < -0.5 or x >3, which is very close but excludes the endpoints. But the original inequality includes equality. So unless there is a typo in the options, maybe the answer is D? Because perhaps in the original problem, the inequality was strict (>) instead of ≥. But the user wrote the inequality as "≥".Alternatively, maybe the answer expects the user to ignore the equality? But that's not right. Wait, maybe the options have a typo. For example, if option D was "x ≤ -1/2 or x ≥3", that would be correct. But given the options as stated, D is incorrect. So this is confusing.Looking back at the options:A: x ≥0B: x <0 or x>2C: x ∈ {-1,3,5}D: x＜−1/2 or x＞3If none of the options are correct, but the closest is D, even though it's technically incorrect. But maybe in the original problem, there was a mistake. Alternatively, perhaps when they say "sufficient and necessary condition", they might consider the closure? But no, necessary and sufficient condition should exactly match the solution set.Alternatively, maybe there's a mistake in the problem statement. Let me check again.Original problem: "A sufficient and necessary condition for the inequality 2x²−5x−3⩾0 to hold is ( )."Given that the user has written the options as:A: x≥0B: x <0 or x >2C: x∈{−1,3,5}D: x＜−1/2 or x＞3Hmm. If the answer is D, then the endpoints are excluded. But we know at x=-1/2 and x=3, the inequality holds. Therefore, D is not a necessary condition because it excludes points where the inequality is true, but it is a sufficient condition because all points in D satisfy the inequality. However, the question asks for a sufficient and necessary condition, which means it needs to include all solutions and only the solutions. Therefore, D is not necessary because there are solutions not in D (like x=-0.5 and x=3), and D is not necessary because even if a point is in D, it's sufficient, but there are points not in D that also satisfy the inequality.Wait, no. Wait, if a necessary and sufficient condition must exactly describe all solutions. So if the answer is D, but D excludes some solutions (x=-0.5 and x=3), then it's not a necessary condition because those points are also solutions, so the condition is not necessary (since you don't need to be in D to be a solution; being at x=-0.5 or x=3 also works). Therefore, D is insufficient as a necessary condition.Similarly, if a student chose D, they would be excluding valid solutions. Therefore, none of the options are correct? But the problem must have an answer. Wait, maybe there was a miscalculation.Wait, let's check if x= -1/2 is included in any option. Option C includes x=-1, which is less than -0.5, so in D, x=-1 would satisfy x < -0.5, so in D. But x=-0.5 is not in any option. Similarly, x=3 is not included in D. So perhaps the answer is D, assuming that the original problem had a strict inequality. But according to the user's problem statement, it's "≥".Alternatively, maybe the original quadratic was supposed to be 2x² -5x -3 >0, which would make D correct. But as written, it's "≥".Alternatively, perhaps the options have a typo, and D was supposed to have "≤ -1/2" and "≥3", but written as strict inequalities. If that's the case, then maybe the answer is D. But given the options as presented, strictly speaking, none of them are correct. However, since the options are given, and D is the closest, maybe the answer is D. The problem might have a typo. Alternatively, maybe in Chinese notation, the symbols are different? Wait, the user wrote "x＜−1/2", which is correct for strict inequality. So perhaps in the original problem, the inequality was non-strict, but the answer options are using strict inequalities. In that case, maybe it's a mistake, but given the options, D is the only one that's close.Alternatively, maybe I made a mistake in the direction of the inequality. Wait, the quadratic opens upwards, so it's positive outside the roots. So for 2x² -5x -3 ≥0, x ≤ -0.5 or x ≥3. So correct. So if D was x ≤ -1/2 or x ≥3, it would be correct. But D is x < -1/2 or x >3.Wait, perhaps in the problem statement, the inequality is written as 2x² -5x -3 >0 (strictly greater), in which case D is correct. But the user wrote "≥". Hmm. This is confusing.Alternatively, maybe the options are correct in another way. Let's check option C: x ∈ {-1,3,5}. We tested x=-1, which is in the solution set, x=3 is a boundary point (which works), and x=5 is in the solution set. So all three points satisfy the inequality. However, the problem is asking for a necessary and sufficient condition. A necessary and sufficient condition must include all solutions. Since C only lists three specific points, it's not including all solutions. Therefore, C is not a correct answer.Option B: x <0 or x >2. Let's check some points in B. For example, x= -1 is in B and satisfies the inequality. x= -0.5 is not in B (since B is x <0 or x >2), but x=-0.5 is a solution. Wait, x=-0.5 is not in B because B is x <0 or x >2. So x=-0.5 is actually equal to -0.5, which is less than 0, so it is included in B. Wait, hold on: x=-0.5 is less than 0, so it is included in B. But at x=-0.5, the inequality holds (it's equal to zero). Wait, but B says x <0 or x>2. So x=-0.5 is included in x <0. However, according to our solution, between x=-0.5 and x=3, the inequality does not hold. So in B, when x is between -0.5 and0, even though x is <0, the inequality is not holding. Therefore, B includes some x's where the inequality does not hold. So B is not a correct answer.Option A: x ≥0. Well, x=0 is in A. Plugging into the inequality: 2*0 -5*0 -3 = -3, which is not ≥0. So A is not correct.So all options except C and D have clear mistakes. Option C is only specific points, so it's not a condition. Option D is the closest but excludes the endpoints. But maybe in the answer options, they considered the endpoints negligible or there's a typo. Given that, maybe D is the intended answer.Alternatively, since the original inequality is 2x² -5x -3 ≥0, and option D is x < -1/2 or x >3, which corresponds to the regions where the quadratic is positive (not including zero). But since the inequality includes zero, technically D is incorrect. However, if the question had been 2x² -5x -3 >0, then D would be correct. Maybe the problem was mistyped. Given that, perhaps the answer is D.Given that the options are A, B, C, D, and from the analysis, none are perfectly correct, but D is the closest. Maybe the original problem intended the inequality to be strict, so the answer is D. Therefore, I will go with D.Final Answerboxed{D}</think>To solve the inequality (2x^2 - 5x - 3 geq 0), we first find the roots of the quadratic equation (2x^2 - 5x - 3 = 0) using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, (a = 2), (b = -5), and (c = -3). Plugging in these values, we get:[x = frac{5 pm sqrt{25 + 24}}{4} = frac{5 pm 7}{4}]This gives the roots:[x = frac{12}{4} = 3 quad text{and} quad x = frac{-2}{4} = -0.5]The quadratic (2x^2 - 5x - 3) opens upwards (since the coefficient of (x^2) is positive), so the inequality holds outside the interval between the roots. Therefore, the solution to the inequality is:[x leq -0.5 quad text{or} quad x geq 3]Next, we check the given options:- Option A: (x geq 0) is incorrect because the inequality does not hold between (0) and (3).- Option B: (x < 0) or (x > 2) is incorrect because the inequality does not hold between (-0.5) and (0) or between (2) and (3).- Option C: (x in {-1, 3, 5}) is incorrect because it only lists specific points, not intervals.- Option D: (x < -frac{1}{2}) or (x > 3) is almost correct but excludes the endpoints (-0.5) and (3). However, since these endpoints satisfy the inequality, the closest correct answer is D, considering possible typographical errors in the options.Thus, the final answer is:[boxed{D}],<think>Okay, so I need to solve this inequality: 2x² - 5x - 3 ≥ 0. And then choose the correct option from A to D. Let me think step by step.First, I remember that to solve quadratic inequalities, it's useful to find the roots of the corresponding quadratic equation and then determine the intervals where the quadratic expression is positive or negative. The quadratic here is 2x² -5x -3. Let me try to factor it or find its roots using the quadratic formula.The quadratic formula is x = [ -b ± sqrt(b² - 4ac) ] / 2a. Here, a = 2, b = -5, c = -3. Let's compute the discriminant first: b² - 4ac = (-5)² - 4*2*(-3) = 25 + 24 = 49. Oh, that's a perfect square, so the roots should be nice numbers.Calculating the roots:x = [5 ± sqrt(49)] / (2*2) = [5 ± 7]/4.So the two roots are:x = (5 + 7)/4 = 12/4 = 3,andx = (5 - 7)/4 = (-2)/4 = -0.5.So the roots are x = 3 and x = -0.5. That means the quadratic can be factored as 2x² -5x -3 = 2(x - 3)(x + 0.5). Wait, let me check that. Hmm, if we factor out the leading coefficient, maybe. Let me verify:Suppose we factor it as (2x + 1)(x - 3). Let's multiply out: 2x*x = 2x², 2x*(-3) = -6x, 1*x = x, 1*(-3) = -3. So combining terms: 2x² -6x +x -3 = 2x² -5x -3. Yes, that works. So the quadratic factors to (2x + 1)(x - 3). Good, so the roots are x = -1/2 and x = 3.Now, to solve the inequality (2x + 1)(x - 3) ≥ 0. The critical points are at x = -1/2 and x = 3. These divide the real line into three intervals:1. x < -1/22. -1/2 < x < 33. x > 3We need to test each interval to see the sign of the quadratic in that interval.First interval: x < -1/2. Let's pick a test point, say x = -1.Plugging into (2x + 1): 2*(-1) + 1 = -2 +1 = -1(x - 3): (-1) -3 = -4Multiply them: (-1)*(-4) = 4, which is positive. So the quadratic is positive here.Second interval: -1/2 < x < 3. Let's choose x = 0.(2*0 +1) = 1(0 -3) = -3Multiply: 1*(-3) = -3, which is negative. So the quadratic is negative here.Third interval: x > 3. Let's take x = 4.(2*4 +1) = 9(4 -3) = 1Multiply: 9*1 = 9, positive. So quadratic is positive here.Now, the inequality is ≥ 0, so we include the intervals where the quadratic is positive, which are x ≤ -1/2 or x ≥ 3. But also, we need to check the equality at the critical points. At x = -1/2 and x = 3, the quadratic equals zero. So these points are included in the solution.Therefore, the solution set is x ≤ -1/2 or x ≥ 3.Looking at the options:A: x ≥ 0. That's only part of the solution (from x ≥3 is included in x ≥0, but x ≤ -1/2 is not covered). So A is not correct.B: x <0 or x >2. Hmm. x <0 would include x < -1/2 and between -1/2 and 0. But between -1/2 and 0, the quadratic is negative, so those points don't satisfy the inequality. Similarly, x >2 includes x >3 where it's positive, but between 2 and 3, the quadratic is still negative. So B is incorrect because it includes parts where the inequality doesn't hold.C: x ∈ {-1,3,5}. Let's check these points. x = -1: plug into quadratic: 2*(-1)^2 -5*(-1) -3 = 2*1 +5 -3 = 2 +5 -3 =4 ≥0. Okay. x=3: 2*9 -5*3 -3= 18 -15 -3=0, which is equal to zero, so satisfies. x=5: 2*25 -25 -3=50-25-3=22≥0. So all three points satisfy the inequality. However, the problem is asking for a necessary and sufficient condition. The correct solution is all x ≤ -1/2 or x ≥3. Option C only includes three specific points. So even though those points are in the solution set, the condition is not equivalent because there are infinitely many other solutions. So C is not correct.D: x < -1/2 or x >3. Wait, but in our solution, we have x ≤ -1/2 or x ≥3. So D is x < -1/2 or x >3, which excludes the points x = -1/2 and x =3. But in our solution, those points are included because the inequality is ≥0. So D is not exactly correct. But wait, let's check the original inequality. The question says "a sufficient and necessary condition". Hmm. Wait, maybe there's a mistake here. Let me double-check the original inequality: 2x² -5x -3 ≥0.We found the roots at x = -1/2 and x =3. At x = -1/2: 2*(-1/2)^2 -5*(-1/2) -3 = 2*(1/4) + 5/2 -3 = 1/2 + 5/2 -3 = 3 -3 =0. Similarly, at x =3, it's zero. So those points satisfy the inequality, so the solution set includes those points. So the correct intervals should be x ≤ -1/2 or x ≥3. However, option D says x < -1/2 or x >3, which excludes the endpoints. So D is incorrect. But none of the options include the endpoints? Wait, let me check the options again.Option A: x ≥0. No. B: x <0 or x >2. No. C: specific points. D: x < -1/2 or x >3. So none of the options actually include the endpoints. Wait, but maybe the original problem has a typo? Or perhaps I made a mistake in considering the endpoints?Wait, the quadratic factors as (2x +1)(x -3) ≥0. So the critical points are x = -1/2 and x=3. To determine where the product is non-negative, we need to check the intervals. Since the parabola opens upwards (since the coefficient of x² is positive), the graph is a U-shaped parabola. So it's positive outside the roots and negative between the roots. So the solution is indeed x ≤ -1/2 or x ≥3. So including the endpoints.But the options given don't have an option that includes the endpoints. Wait, let me check the options again:A: x≥0B: x<0 or x>2C: x∈{-1,3,5}D: x＜−1/2 or x＞3Wait, so D is x < -1/2 or x >3. But in the correct solution, it's x ≤ -1/2 or x ≥3. So D is almost correct, except it's missing the equality at the critical points. But the problem states "a sufficient and necessary condition". So technically, the correct solution set includes x = -1/2 and x=3, but none of the options include those. However, if we look at the options, D is the closest, but missing the equality. Hmm.Wait, maybe there was a mistake in my factoring. Let's confirm. The quadratic is 2x² -5x -3. Factored as (2x +1)(x -3). Let's multiply back: 2x*x =2x², 2x*(-3)= -6x, 1*x =x, 1*(-3)= -3. So combining terms: 2x² -6x +x -3 =2x² -5x -3. Correct. So the roots are x=-1/2 and x=3.So the inequality (2x +1)(x -3) ≥0. The critical points are at x=-1/2 and x=3. Since the coefficient of x² is positive, the parabola opens upwards. Therefore, the expression is positive when x ≤ -1/2 or x ≥3, and negative in between. Therefore, including the endpoints.But looking at the answer choices, D says x < -1/2 or x >3, which excludes the endpoints. However, in the original inequality, it's "≥0", so the endpoints are included. Therefore, D is not exactly correct. However, perhaps the problem allows approximate answers or considers that in some contexts, the endpoints might not be included? Wait, but the problem says "sufficient and necessary condition". So the answer must exactly match the solution set. Since none of the options exactly include the endpoints except perhaps C, which has specific points.But option C includes x=3 and x=5 (and x=-1). x=3 is a boundary point, x=5 is in the solution set, and x=-1 is also in the solution set (since x=-1 is less than -1/2). However, the solution set is all x ≤ -1/2 or x ≥3. So option C only lists three specific points, which are all valid solutions, but it's not a complete solution. So C is not a necessary and sufficient condition because there are other solutions. For example, x=-2 is also a solution, but it's not in option C.Therefore, none of the options A, B, C, D exactly match the solution set x ≤ -1/2 or x ≥3. However, option D is the closest, but it's missing the equality. But perhaps in the original problem's options, maybe there was a typo, and the inequality was supposed to be strict? Let me check again.The original problem states: "A sufficient and necessary condition for the inequality 2x²−5x−3⩾0 to hold is ( )." So the inequality is non-strict (≥0). Therefore, the endpoints must be included. But since none of the options include the endpoints, perhaps there's an error in the options. Alternatively, maybe I made a mistake in thinking the endpoints are included.Wait, let's check at x=-1/2. Plugging into the inequality: 2*(-1/2)² -5*(-1/2) -3 = 2*(1/4) + 5/2 -3 = 1/2 +5/2 -3 = 3 -3 =0. So it's equal to zero, which satisfies the inequality. Similarly, x=3: 2*(9) -5*3 -3 =18 -15 -3=0. So they do satisfy the inequality.Therefore, the correct solution must include those points. But none of the options do. Wait, looking back at the options:A: x≥0. This includes x=0 to x=3, which is part of the negative interval, so wrong.B: x <0 or x >2. This includes x <0 (which includes x < -1/2 and between -1/2 and 0, which is part of the negative interval) and x>2 (which includes between 2 and 3, which is negative, and x>3 which is positive). So B is incorrect.C: x∈{-1,3,5}. These are specific points. As discussed, they are solutions, but not the entire solution set.D: x＜−1/2 or x＞3. As discussed, this is almost correct but excludes the endpoints. However, in some contexts, maybe the answer expects D because the original inequality is written with ≥, but perhaps in the problem's options, the endpoints are excluded. Alternatively, maybe the problem had a typo and the inequality was supposed to be >0 instead of ≥0. Let me check:If the inequality was 2x² -5x -3 >0, then the solution would be x < -1/2 or x >3, which matches option D. But the problem states ≥0. So unless there's a mistake in the problem, option D is incorrect. But given the options, D is the closest, and maybe there's a typo.Alternatively, perhaps the answer is D, considering that in the options, there is no option with ≤ or ≥, but perhaps in Chinese notation, the symbols might be different. Wait, no, the options are given as:A: x≥0B: x <0 or x >2C: x∈{-1,3,5}D: x＜−1/2 or x＞3So in the options, only A uses ≥, others use < or >. So perhaps the problem expects D as the answer, assuming that maybe the original inequality was strict. But the user has written the problem as "2x²−5x−3⩾0", so with the greater than or equal sign.Alternatively, maybe the problem considers that at x=-1/2 and x=3, the expression is zero, which is technically satisfying the inequality, but perhaps in the answer choices, they are not considered. But in standard math, they should be included.Wait, but let's check the original question again. It says: "A sufficient and necessary condition for the inequality 2x²−5x−3⩾0 to hold is ( )."A sufficient and necessary condition must exactly match the solution set. If none of the options exactly match, then there's a problem. However, since option D is the closest and perhaps in the answer choices, the endpoints are not considered, maybe due to a mistake. Alternatively, maybe I made a mistake in calculation.Wait, double-check the factoring. The quadratic 2x² -5x -3. Let me try factoring again. Looking for two numbers that multiply to (2)*(-3) = -6 and add to -5. Hmm, -6 and +1. Yes: 2x² -6x +x -3 = (2x² -6x) + (x -3) = 2x(x -3) +1(x -3) = (2x +1)(x -3). So correct. So roots at x=-1/2 and x=3.Then, as per the inequality, including the roots. So the solution is (-∞, -1/2] ∪ [3, ∞). But none of the options have these intervals. Option D is (-∞, -1/2) ∪ (3, ∞). So if the problem had strict inequality, D would be correct. But since it's non-strict, perhaps the answer is not among the options. But since the user provides these options, and likely expects one of them, maybe there's a mistake.Wait, maybe the test maker made an error, but given the options, D is the closest. Also, looking back at the options:C is a set of three points. Since the inequality holds for all x ≤ -1/2 or x ≥3, which includes infinitely many points, C cannot be a necessary and sufficient condition because it's only three points. So C is out.A and B are incorrect. So D is the only possible answer, even though it's technically incorrect due to the exclusion of the endpoints. However, in some contexts, people might consider the critical points where the expression is zero as separate, but since the question asks for a sufficient and necessary condition, D is not exactly correct. But maybe in the options given, D is the intended answer. Alternatively, maybe I made a mistake in considering the direction of the inequality.Wait, let me verify with a test point in D. For example, x= -1 is in D's solution (x < -1/2). Plugging into the quadratic: 2*(-1)^2 -5*(-1) -3 = 2 +5 -3 =4 ≥0. Good. x=4 is in D's solution: 2*16 -20 -3=32-20-3=9≥0. Good. If we take x=-1/2, which is excluded by D, the value is 0, which should be included. Similarly, x=3 is excluded but gives zero. However, the problem says "sufficient and necessary condition". If the condition is D, then all x in D satisfy the inequality (sufficient), and all x satisfying the inequality are in D (necessary). But since x=-1/2 and x=3 satisfy the inequality but are not in D, the condition D is not necessary. So D is insufficient.But the problem states "sufficient and necessary condition". Therefore, none of the options are correct. However, given the options provided, D is the best possible answer, assuming that perhaps the question had a typo with the inequality sign. Since the user might expect an answer from the given options, and D is the closest, albeit technically incorrect, I might have to go with D.Alternatively, maybe the original problem in Chinese had the correct option, but in translation, the symbols got messed up. Alternatively, maybe my analysis is wrong.Wait, another way to check: the quadratic 2x² -5x -3. The graph is a parabola opening upwards, with vertex at x = -b/(2a) = 5/(4). The vertex is at x=1.25. Plugging x=1.25 into the quadratic: 2*(1.25)^2 -5*(1.25) -3 = 2*(1.5625) -6.25 -3 = 3.125 -6.25 -3 = -6.125. So the vertex is at (1.25, -6.125), which is a minimum point. Therefore, the parabola is below zero between the roots and above zero outside. Therefore, the solution set is x ≤ -1/2 or x ≥3, including the endpoints.Since none of the options include the endpoints, but D is the only one that covers the correct intervals (excluding endpoints), while the other options are clearly wrong. Therefore, perhaps the answer is D, assuming that the problem expects the strict inequality solution. Alternatively, there might be a mistake in the problem statement.But given the options provided, D is the only one that is close. Therefore, I think the answer is D.Final Answerboxed{D}</think>To solve the inequality (2x^2 - 5x - 3 geq 0), we first find the roots of the quadratic equation (2x^2 - 5x - 3 = 0) using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, (a = 2), (b = -5), and (c = -3). Calculating the discriminant:[b^2 - 4ac = (-5)^2 - 4 cdot 2 cdot (-3) = 25 + 24 = 49]The roots are:[x = frac{5 pm 7}{4}]This gives the roots (x = 3) and (x = -frac{1}{2}). The quadratic can be factored as ((2x + 1)(x - 3)).Next, we determine the intervals where the quadratic expression is non-negative by testing values in each interval divided by the roots (x = -frac{1}{2}) and (x = 3):1. For (x < -frac{1}{2}), the expression is positive.2. For (-frac{1}{2} < x < 3), the expression is negative.3. For (x > 3), the expression is positive.Since the inequality is (geq 0), we include the roots (x = -frac{1}{2}) and (x = 3). Thus, the solution set is (x leq -frac{1}{2}) or (x geq 3).However, the given options do not include the endpoints (-frac{1}{2}) and (3). Option D: (x < -frac{1}{2}) or (x > 3) is the closest, excluding the endpoints. Given the provided options, D is the best choice as it covers the correct intervals, albeit excluding the endpoints which might be a typo or oversight in the problem statement.[boxed{D}]]

🔑:To solve the inequality 2x^{2}-5x-3geqslant 0, we first find the roots of the quadratic equation 2x^{2}-5x-3=0. 1. Finding the Roots: The roots of the equation can be found using the quadratic formula, x = frac{-b pm sqrt{b^2-4ac}}{2a}, where a=2, b=-5, and c=-3. [ x = frac{-(-5) pm sqrt{(-5)^2-4(2)(-3)}}{2(2)} ] Simplifying inside the square root: [ x = frac{5 pm sqrt{25+24}}{4} ] Further simplification gives: [ x = frac{5 pm sqrt{49}}{4} ] Therefore, the roots are: [ x_1 = frac{5 + 7}{4} = 3, quad x_2 = frac{5 - 7}{4} = -frac{1}{2} ]2. Analyzing the Inequality: The inequality 2x^{2}-5x-3geqslant 0 changes sign at the roots x=3 and x=-frac{1}{2}. To determine the intervals where the inequality holds, we test values in the intervals (-infty, -frac{1}{2}), (-frac{1}{2}, 3), and (3, infty). - For x 3, the inequality holds because the value at x > 3 is above the x-axis.Therefore, the sufficient and necessary condition for the inequality 2x^{2}-5x-3geqslant 0 to hold is xleq-frac{1}{2} or xgeqslant 3.Considering the options, only options D correctly matches the condition derived from the analysis. Hence, the correct answer is encapsulated as boxed{D}.

🧠:[<think>Okay, so I have this problem here where I need to find the maximum value of the expression (2a³ + 27c - 9ab)/λ³, given some conditions on the cubic polynomial f(x) = x³ + a x² + b x + c. The cubic has three real roots x₁, x₂, x₃, with the difference between x₂ and x₁ being λ, and x₃ is greater than the average of x₁ and x₂. Alright, let's break this down step by step.First, since the polynomial has three real roots, it must be expressible in its factored form as f(x) = (x - x₁)(x - x₂)(x - x₃). Expanding this should give me the coefficients a, b, c in terms of the roots. Let me do that.Expanding (x - x₁)(x - x₂)(x - x₃):First, multiply the first two factors: (x - x₁)(x - x₂) = x² - (x₁ + x₂)x + x₁x₂.Then multiply by (x - x₃):x³ - (x₁ + x₂ + x₃)x² + (x₁x₂ + x₁x₃ + x₂x₃)x - x₁x₂x₃.Comparing this with f(x) = x³ + a x² + b x + c, so the coefficients are:a = -(x₁ + x₂ + x₃),b = x₁x₂ + x₁x₃ + x₂x₃,c = -x₁x₂x₃.Got that. So a is the negative sum of the roots, b is the sum of products of roots two at a time, and c is the negative product of the roots.Now, the problem states that x₂ - x₁ = λ, which is a given positive number. Also, x₃ > (x₁ + x₂)/2. So, the third root is greater than the average of the first two roots. That probably means that x₃ is to the right of the midpoint of x₁ and x₂. So, if I imagine x₁ and x₂ on the number line with a distance λ between them, x₃ is somewhere to the right of their midpoint. Interesting.Now, the expression we need to maximize is (2a³ + 27c - 9ab)/λ³. Let's substitute a, b, c in terms of the roots.First, let's compute 2a³:2a³ = 2[-(x₁ + x₂ + x₃)]³ = -2(x₁ + x₂ + x₃)³.Then 27c: 27*(-x₁x₂x₃) = -27x₁x₂x₃.Then -9ab: -9*(-(x₁ + x₂ + x₃))*(x₁x₂ + x₁x₃ + x₂x₃) = 9(x₁ + x₂ + x₃)(x₁x₂ + x₁x₃ + x₂x₃).So putting all together:2a³ + 27c - 9ab = -2(x₁ + x₂ + x₃)³ -27x₁x₂x₃ + 9(x₁ + x₂ + x₃)(x₁x₂ + x₁x₃ + x₂x₃).Hmm, that's a bit complicated. Let's see if we can simplify this expression. Maybe it's related to some symmetric functions or known identities.Wait, in the theory of cubic equations, there's a discriminant which is 18abcd - 4b³d + ... but maybe that's not directly helpful here. Alternatively, maybe this expression can be related to the discriminant or another invariant.Alternatively, let's try to compute this expression step by step.Let me denote S = x₁ + x₂ + x₃, P = x₁x₂ + x₁x₃ + x₂x₃, Q = x₁x₂x₃.Then, we have:2a³ + 27c - 9ab = 2(-S)^3 + 27(-Q) - 9*(-S)(P) = -2S³ -27Q + 9S P.So the expression is -2S³ -27Q + 9S P.Wait, but maybe there's an identity that connects these terms. Let's recall that for cubic polynomials, the discriminant D is given by D = 18S Q -4S³ P + S² P² - 4P³ - 27Q². Hmm, not sure if that's helpful.Alternatively, maybe we can relate this expression to the derivative of the polynomial? Because the problem involves the distance between roots. Let's see.The polynomial f(x) has three real roots, so its derivative f'(x) = 3x² + 2a x + b must have two real critical points, which are a local maximum and a local minimum. The distance between the roots x₁ and x₂ is λ, so maybe we can use Rolle's theorem here? Between x₁ and x₂, there must be a critical point. Similarly, between x₂ and x₃, another critical point.Wait, but actually, between any two roots of a differentiable function, there's at least one critical point. Since f(x) has three roots, x₁ < x₂ < x₃, then between x₁ and x₂, there is a critical point, say m, and between x₂ and x₃, another critical point, say M. So, the derivative f'(x) has two real roots m and M.But how is this helpful? Maybe we can relate the critical points to the roots? Alternatively, since we have the distance between x₁ and x₂ given as λ, and x₃ is greater than (x₁ + x₂)/2, perhaps we can parametrize the roots in terms of variables that capture these conditions.Let me try to parametrize the roots. Let’s set x₁ = t - λ/2, x₂ = t + λ/2, so that x₂ - x₁ = λ. Then the average of x₁ and x₂ is t, so x₃ > t. Let’s denote x₃ = t + s, where s > 0. So, the three roots are t - λ/2, t + λ/2, and t + s, with s > 0.Then, the sum S = x₁ + x₂ + x₃ = (t - λ/2) + (t + λ/2) + (t + s) = 3t + s.The sum of products P = x₁x₂ + x₁x₃ + x₂x₃.Calculating each term:x₁x₂ = (t - λ/2)(t + λ/2) = t² - (λ/2)² = t² - λ²/4.x₁x₃ = (t - λ/2)(t + s) = t(t + s) - (λ/2)(t + s) = t² + t s - (λ/2)t - (λ/2)s.Similarly, x₂x₃ = (t + λ/2)(t + s) = t(t + s) + (λ/2)(t + s) = t² + t s + (λ/2)t + (λ/2)s.Adding them together:P = [t² - λ²/4] + [t² + t s - (λ/2)t - (λ/2)s] + [t² + t s + (λ/2)t + (λ/2)s].Let me compute term by term:First term: t² - λ²/4.Second term: t² + t s - (λ/2)t - (λ/2)s.Third term: t² + t s + (λ/2)t + (λ/2)s.Adding all three terms:t² - λ²/4 + t² + t s - (λ/2)t - (λ/2)s + t² + t s + (λ/2)t + (λ/2)s.Let's combine like terms:t² + t² + t² = 3t².t s + t s = 2t s.-λ²/4.Then the terms with λ/2 t: - (λ/2)t + (λ/2)t = 0.Similarly, terms with λ/2 s: - (λ/2)s + (λ/2)s = 0.So overall, P = 3t² + 2t s - λ²/4.Now, the product Q = x₁x₂x₃ = (t - λ/2)(t + λ/2)(t + s) = [t² - λ²/4](t + s).So Q = (t² - λ²/4)(t + s) = t³ + t² s - (λ²/4)t - (λ²/4)s.Now, let's recall that the expression we need is:-2S³ -27Q +9S P.Let me compute each part.First, S = 3t + s.So S³ = (3t + s)³ = 27t³ + 27t² s + 9t s² + s³.Multiply by -2: -2S³ = -54t³ -54t² s -18t s² -2s³.Next, Q = t³ + t² s - (λ²/4)t - (λ²/4)s.Multiply by -27: -27Q = -27t³ -27t² s + (27λ²/4)t + (27λ²/4)s.Then, S P: (3t + s)(3t² + 2t s - λ²/4).Let me compute that:First, expand (3t)(3t²) = 9t³.3t*(2t s) = 6t² s.3t*(-λ²/4) = - (3λ²/4) t.s*(3t²) = 3t² s.s*(2t s) = 2t s².s*(-λ²/4) = - (λ²/4) s.So adding all terms:9t³ + 6t² s - (3λ²/4) t + 3t² s + 2t s² - (λ²/4)s.Combine like terms:9t³.6t² s + 3t² s = 9t² s.- (3λ²/4) t.2t s².- (λ²/4)s.Therefore, S P = 9t³ +9t² s +2t s² - (3λ²/4)t - (λ²/4)s.Multiply by 9: 9S P = 81t³ +81t² s +18t s² - (27λ²/4)t - (9λ²/4)s.Now, combining all three parts:Total expression = (-54t³ -54t² s -18t s² -2s³) + (-27t³ -27t² s + (27λ²/4)t + (27λ²/4)s) + (81t³ +81t² s +18t s² - (27λ²/4)t - (9λ²/4)s).Let me compute term by term.First, the t³ terms:-54t³ -27t³ +81t³ = (-54 -27 +81)t³ = 0. So t³ terms cancel out.Next, t² s terms:-54t² s -27t² s +81t² s = (-54 -27 +81)t² s = 0. These also cancel.t s² terms:-18t s² +18t s² = 0.s³ terms:-2s³. No other s³ terms.Now, terms with λ² t:(27λ²/4)t - (27λ²/4)t = 0.Terms with λ² s:(27λ²/4)s - (9λ²/4)s = (18λ²/4)s = (9λ²/2)s.So the total expression simplifies to:-2s³ + (9λ²/2)s.So the entire expression 2a³ + 27c -9ab simplifies to -2s³ + (9λ²/2)s.Therefore, the expression we need to maximize is:(-2s³ + (9λ²/2)s)/λ³ = (-2s³ + (9/2)λ² s)/λ³ = (-2s³)/λ³ + (9/2)λ² s /λ³ = -2(s/λ)^3 + (9/2)(s/λ).Let’s denote k = s/λ. Since s > 0 and λ > 0, k > 0.So the expression becomes: -2k³ + (9/2)k.We need to find the maximum value of this expression with respect to k > 0. Wait, but is there any restriction on k?Wait, originally, we had x₃ = t + s, and x₃ > (x₁ + x₂)/2 = t. So, s > 0. But is there another constraint?From the fact that all roots are real, maybe there's a relation between s and λ. Let me check.The cubic polynomial must have three real roots, so the discriminant must be non-negative. The discriminant of a cubic x³ + a x² + b x + c is given by D = 18abc - 4a³c + a²b² - 4b³ - 27c².But maybe it's complicated to compute. Alternatively, since we have expressed everything in terms of t and s, perhaps we can find if there's a relation.But wait, in our parametrization, x₁ = t - λ/2, x₂ = t + λ/2, x₃ = t + s. Since the roots are real, t and s can be any real numbers such that x₃ > t (i.e., s > 0). But is there any other constraint? For example, maybe the cubic's critical points must be real, but since it's a cubic, the derivative is a quadratic, which always has two real critical points (since discriminant is positive). So perhaps the only constraint is s > 0. Therefore, k = s/λ > 0.Therefore, the problem reduces to maximizing the function g(k) = -2k³ + (9/2)k over k > 0. Let's do that.To find the maximum, take derivative of g(k):g’(k) = -6k² + 9/2.Set derivative equal to zero:-6k² + 9/2 = 0 → 6k² = 9/2 → k² = (9/2)/6 = 3/4 → k = sqrt(3)/2 ≈ 0.866.Since k > 0, we take the positive root. Now, check second derivative to confirm it's a maximum:g''(k) = -12k. At k = sqrt(3)/2, g''(k) = -12*(sqrt(3)/2) = -6sqrt(3) < 0. So yes, maximum at k = sqrt(3)/2.Compute g(sqrt(3)/2):g(k) = -2*( (sqrt(3)/2)^3 ) + (9/2)*(sqrt(3)/2).First compute (sqrt(3)/2)^3 = (3^(1/2))/2)^3 = 3^(3/2)/8 = (3*sqrt(3))/8.Thus, -2*(3*sqrt(3)/8) = - (6*sqrt(3))/8 = - (3*sqrt(3))/4.Then, (9/2)*(sqrt(3)/2) = (9 sqrt(3))/4.Adding these together: (-3 sqrt(3)/4) + (9 sqrt(3)/4) = (6 sqrt(3))/4 = (3 sqrt(3))/2 ≈ 2.598.Therefore, the maximum value of the expression is (3 sqrt(3))/2. However, let me confirm that there are no constraints on k beyond k > 0.Wait, but in our parametrization, we set x₁ = t - λ/2, x₂ = t + λ/2, x₃ = t + s. So, s can be any positive number, but we must ensure that the cubic actually has three real roots. However, in our analysis, we didn't consider the ordering of the roots or possible multiplicities. Wait, but if x₁, x₂, x₃ are distinct, then the discriminant is positive. However, even if two roots coincide, the discriminant is zero, but in our problem statement, it just says three real roots, so they could be multiple roots. But the given difference x₂ - x₁ = λ implies that x₂ and x₁ are distinct (since λ is positive). So x₁ and x₂ are distinct, but x₃ could coincide with one of them? Wait, but x₃ > (x₁ + x₂)/2. If x₃ coincided with x₂, then x₃ = x₂, but x₂ = t + λ/2, and x₃ = t + s. So x₃ = x₂ would mean s = λ/2. But since x₃ must be greater than t, which is the average of x₁ and x₂. If s = λ/2, then x₃ = t + λ/2 = x₂, so x₃ coincides with x₂, making it a double root. But the problem says three real roots, so perhaps allowing multiple roots. Hmm.But in our parametrization, we can have s approaching zero, but s > 0. However, the discriminant of the cubic must be non-negative. Let's check when s approaches zero.If s approaches zero, then x₃ approaches t, which is the average of x₁ and x₂. But the problem states x₃ > (x₁ + x₂)/2, so s must be greater than zero, not equal. But as s approaches zero, we approach the case where x₃ is a root approaching t, which would make the cubic have a multiple root at x₂ (if s approaches zero from above, x₃ approaches t, but x₂ is at t + λ/2, so it doesn't coincide). Wait, if x₃ approaches t, then the roots are t - λ/2, t + λ/2, and t. So they are distinct as long as s > 0. Wait, even if s is very small, as long as s > 0, x₃ is distinct. So maybe the discriminant remains positive. Therefore, perhaps our only constraint is s > 0, hence k > 0.Therefore, the maximum of the expression occurs at k = sqrt(3)/2, leading to a maximum value of (3 sqrt(3))/2. Therefore, the maximum value of (2a³ + 27c -9ab)/λ³ is (3 sqrt(3))/2. So I should check my steps again to ensure I didn't make a mistake.Wait, let's recap:1. Expressed roots as t - λ/2, t + λ/2, t + s with s > 0.2. Calculated S, P, Q in terms of t and s.3. Plugged into the expression and found it simplifies to -2s³ + (9/2)λ² s, leading to expression in terms of k = s/λ as -2k³ + (9/2)k.4. Took derivative, found critical point at k = sqrt(3)/2, confirmed it's a maximum.5. Calculated maximum value as (3 sqrt(3))/2.This seems correct. Let me just verify with an example. Suppose we take λ = 2, then s = sqrt(3)/2 * 2 = sqrt(3). Then the roots are t - 1, t + 1, t + sqrt(3). Let's pick t = 0 for simplicity. Then roots are -1, 1, sqrt(3). Wait, sqrt(3) ≈ 1.732, which is greater than ( -1 + 1)/2 = 0. So x₃ = sqrt(3) > 0, which satisfies the condition. Then, forming the polynomial f(x) = (x +1)(x -1)(x - sqrt(3)).Compute coefficients:First, (x +1)(x -1) = x² -1.Multiply by (x - sqrt(3)): x³ - sqrt(3)x² - x + sqrt(3).Therefore, a = -sqrt(3), b = -1, c = sqrt(3).Compute the expression (2a³ + 27c -9ab)/λ³.First, compute each term:2a³ = 2*(-sqrt(3))³ = 2*(-3*sqrt(3)) = -6 sqrt(3).27c = 27*sqrt(3).-9ab = -9*(-sqrt(3))*(-1) = -9*sqrt(3).Therefore, total expression:-6 sqrt(3) +27 sqrt(3) -9 sqrt(3) = ( -6 +27 -9 ) sqrt(3) = 12 sqrt(3).Divide by λ³ = 2³ =8:12 sqrt(3)/8 = (3 sqrt(3))/2 ≈ 2.598, which matches our earlier calculation. Therefore, this example works.Therefore, the maximum value is indeed (3 sqrt(3))/2.Final AnswerThe maximum value is boxed{dfrac{3sqrt{3}}{2}}.</think>Given the cubic polynomial ( f(x) = x^3 + ax^2 + bx + c ) with three real roots ( x_1, x_2, x_3 ) satisfying ( x_2 - x_1 = lambda ) and ( x_3 > frac{1}{2}(x_1 + x_2) ), we need to find the maximum value of ( frac{2a^3 + 27c - 9ab}{lambda^3} ).First, we express the coefficients ( a, b, c ) in terms of the roots ( x_1, x_2, x_3 ):- ( a = -(x_1 + x_2 + x_3) )- ( b = x_1x_2 + x_1x_3 + x_2x_3 )- ( c = -x_1x_2x_3 )We parametrize the roots as ( x_1 = t - frac{lambda}{2} ), ( x_2 = t + frac{lambda}{2} ), and ( x_3 = t + s ) where ( s > 0 ). This ensures ( x_2 - x_1 = lambda ) and ( x_3 > frac{1}{2}(x_1 + x_2) ).Substituting these roots into the expressions for ( a, b, c ):- ( a = -(3t + s) )- ( b = 3t^2 + 2ts - frac{lambda^2}{4} )- ( c = -((t^2 - frac{lambda^2}{4})(t + s)) )We then compute the expression ( 2a^3 + 27c - 9ab ):- ( 2a^3 = -2(3t + s)^3 )- ( 27c = -27(t^2 - frac{lambda^2}{4})(t + s) )- ( 9ab = 9(3t + s)(3t^2 + 2ts - frac{lambda^2}{4}) )After expanding and simplifying, we find that the expression simplifies to:[ -2s^3 + frac{9}{2}lambda^2s ]Introducing ( k = frac{s}{lambda} ), the expression becomes:[ -2k^3 + frac{9}{2}k ]To find the maximum value, we take the derivative with respect to ( k ):[ frac{d}{dk} left( -2k^3 + frac{9}{2}k right) = -6k^2 + frac{9}{2} ]Setting the derivative to zero, we solve for ( k ):[ -6k^2 + frac{9}{2} = 0 implies k = frac{sqrt{3}}{2} ]Evaluating the expression at ( k = frac{sqrt{3}}{2} ):[ -2 left( frac{sqrt{3}}{2} right)^3 + frac{9}{2} left( frac{sqrt{3}}{2} right) = frac{3sqrt{3}}{2} ]Thus, the maximum value of ( frac{2a^3 + 27c - 9ab}{lambda^3} ) is (boxed{dfrac{3sqrt{3}}{2}}).,<think>Okay, let me try to work through this problem step by step. So, we have a cubic polynomial f(x) = x³ + a x² + b x + c with three real roots x₁, x₂, x₃. The conditions given are:1. The difference between two roots is x₂ - x₁ = λ, where λ is a positive number.2. The third root x₃ is greater than the average of x₁ and x₂, i.e., x₃ > (x₁ + x₂)/2.We need to find the maximum value of the expression (2a³ + 27c - 9ab)/λ³.First, since f(x) is a cubic polynomial with roots x₁, x₂, x₃, we can express it in factored form as:f(x) = (x - x₁)(x - x₂)(x - x₃)Expanding this, we can relate the coefficients a, b, c to the roots. Let me recall that for a cubic polynomial (x - r)(x - s)(x - t), the expansion is x³ - (r + s + t)x² + (rs + rt + st)x - rst. Comparing this to our polynomial, which is x³ + a x² + b x + c, we need to adjust the signs. So:f(x) = (x - x₁)(x - x₂)(x - x₃) = x³ - (x₁ + x₂ + x₃)x² + (x₁x₂ + x₁x₃ + x₂x₃)x - x₁x₂x₃But our polynomial is x³ + a x² + b x + c. Therefore, equating coefficients:a = - (x₁ + x₂ + x₃)b = x₁x₂ + x₁x₃ + x₂x₃c = - x₁x₂x₃So, we can express a, b, c in terms of the roots. Our target expression is (2a³ + 27c - 9ab)/λ³. Let's substitute a, b, c with the expressions above.First, let's compute each term:2a³ = 2[- (x₁ + x₂ + x₃)]³ = -2(x₁ + x₂ + x₃)³27c = 27(-x₁x₂x₃) = -27x₁x₂x₃-9ab = -9[- (x₁ + x₂ + x₃)][x₁x₂ + x₁x₃ + x₂x₃] = 9(x₁ + x₂ + x₃)(x₁x₂ + x₁x₃ + x₂x₃)So, combining all three terms:2a³ + 27c - 9ab = -2(x₁ + x₂ + x₃)³ -27x₁x₂x₃ + 9(x₁ + x₂ + x₃)(x₁x₂ + x₁x₃ + x₂x₃)Hmm, that seems complicated. Maybe there's a way to simplify this expression. Let me see.Alternatively, maybe there's a known identity or formula that relates these terms. Let me recall that for cubic polynomials, there's the identity related to the discriminant. However, the discriminant of a cubic is given by Δ = 18abcd - 4b³d + b²c² - 4ac³ - 27a²d² for a general cubic ax³ + bx² + cx + d. But in our case, the leading coefficient is 1, so a=1, and our polynomial is x³ + a x² + b x + c. Wait, that might be confusing with the standard notation. Let me check.Wait, hold on, the problem statement uses a, b, c as coefficients: f(x) = x³ + a x² + b x + c. So in standard notation, that would correspond to cubic term coefficient 1, quadratic term a, linear term b, and constant term c. Therefore, the discriminant Δ of this cubic would be:Δ = 18(1)(a)(b)(c) - 4(a)³c + (a)²(b)² - 4(1)(b)³ - 27(1)²(c)²But I'm not sure if the discriminant is directly useful here. The problem mentions that all roots are real, so the discriminant is non-negative. But perhaps instead of working directly with the discriminant, we can use Vieta's formulas and the given conditions.Given the roots x₁, x₂, x₃, with x₂ - x₁ = λ and x₃ > (x₁ + x₂)/2. We need to maximize the expression (2a³ + 27c - 9ab)/λ³.Let me try to express everything in terms of the roots. Let me denote the roots as x₁, x₂ = x₁ + λ, and x₃. Then, since x₂ - x₁ = λ, we can set x₂ = x₁ + λ. Then, the third root x₃ must satisfy x₃ > (x₁ + x₂)/2 = (x₁ + x₁ + λ)/2 = x₁ + λ/2.So, x₃ > x₁ + λ/2. Let me set variables in terms of x₁ and λ. Let's let x₁ be some real number, x₂ = x₁ + λ, and x₃ > x₁ + λ/2.But since the problem involves a ratio divided by λ³, perhaps we can normalize variables by λ. Let me consider a substitution to make things dimensionless. Let’s set t = x₁ / λ, then x₁ = tλ, x₂ = tλ + λ = λ(t + 1), and x₃ = x₁ + λ/2 + s, where s > 0. But maybe another substitution would be better.Alternatively, since the expression we need to maximize is (2a³ + 27c - 9ab)/λ³, it's a homogeneous expression in terms of the roots scaled by λ. Let me check the scaling.Suppose we scale all roots by a factor of k: x_i → k x_i. Then λ = x₂ - x₁ scales as k λ, and the expression (2a³ + 27c - 9ab) would scale as follows:a = - (x₁ + x₂ + x₃) → -k(x₁ + x₂ + x₃) = k ab = x₁x₂ + x₁x₃ + x₂x₃ → k²(x₁x₂ + x₁x₃ + x₂x₃) = k² bc = -x₁x₂x₃ → -k³ x₁x₂x₃ = k³ cTherefore, 2a³ + 27c -9ab becomes 2(k a)³ + 27(k³ c) - 9(k a)(k² b) = 2k³ a³ + 27k³ c - 9k³ ab = k³(2a³ + 27c -9ab). Then, the ratio (2a³ + 27c -9ab)/λ³ would become (k³(2a³ +27c -9ab))/(k³ λ³) = (2a³ +27c -9ab)/λ³. So the expression is invariant under scaling of roots. Therefore, without loss of generality, we can set λ = 1 to simplify the problem, find the maximum value, and it would be the same for any λ. So, let's assume λ = 1. Then, the problem reduces to finding the maximum of (2a³ +27c -9ab) under the conditions x₂ - x₁ =1 and x₃ > (x₁ +x₂)/2.After solving it for λ=1, the maximum value would be the same for any λ, due to the homogeneity. So, let's proceed with λ=1.Therefore, let me set x₂ = x₁ +1. Then, x₃ > x₁ + 0.5.So, the roots are x₁, x₁ +1, x₃, with x₃ > x₁ + 0.5.Express a, b, c in terms of x₁ and x₃.a = - (x₁ + x₂ + x₃) = - (x₁ + (x₁ +1) + x₃) = - (2x₁ +1 + x₃)b = x₁x₂ + x₁x₃ + x₂x₃ = x₁(x₁ +1) + x₁x₃ + (x₁ +1)x₃ = x₁² +x₁ +x₁x₃ +x₁x₃ +x₃ = x₁² +x₁ + 2x₁x₃ +x₃c = -x₁x₂x₃ = -x₁(x₁ +1)x₃Now, let's compute the expression E = 2a³ +27c -9ab.First, compute a³:a = - (2x₁ +1 +x₃) = - (2x₁ +x₃ +1)a³ = - (2x₁ +x₃ +1)³So, 2a³ = -2(2x₁ +x₃ +1)³27c = 27(-x₁(x₁ +1)x₃) = -27x₁(x₁ +1)x₃-9ab = -9[- (2x₁ +x₃ +1)][x₁² +x₁ +2x₁x₃ +x₃] = 9(2x₁ +x₃ +1)(x₁² +x₁ +2x₁x₃ +x₃)So, E = -2(2x₁ +x₃ +1)³ -27x₁(x₁ +1)x₃ +9(2x₁ +x₃ +1)(x₁² +x₁ +2x₁x₃ +x₃)This seems very complicated. Maybe there's a smarter substitution or coordinate system.Alternatively, since the problem is symmetric in some way, maybe we can shift coordinates to make x₁ and x₂ symmetric around some point. Since x₂ = x₁ +1, if we let the midpoint of x₁ and x₂ be m = x₁ +0.5, then x₁ = m -0.5, x₂ = m +0.5. Then, x₃ > m.So, let's set m = x₁ +0.5, so x₁ = m -0.5, x₂ = m +0.5, and x₃ = m + t, where t >0 (since x₃ > m). Then, the roots are m -0.5, m +0.5, and m + t.Now, express a, b, c in terms of m and t.a = - [ (m -0.5) + (m +0.5) + (m + t) ] = - [3m + t]b = (m -0.5)(m +0.5) + (m -0.5)(m + t) + (m +0.5)(m + t)Compute each term:First term: (m -0.5)(m +0.5) = m² -0.25Second term: (m -0.5)(m + t) = m² + m t -0.5 m -0.5 tThird term: (m +0.5)(m + t) = m² + m t +0.5 m +0.5 tAdd them together:First term: m² -0.25Second term: m² + m t -0.5 m -0.5 tThird term: m² + m t +0.5 m +0.5 tSum:3m² + 2m t -0.25 + (-0.5 m +0.5 m) + (-0.5 t +0.5 t) = 3m² + 2m t -0.25So, b = 3m² + 2m t -0.25c = - (m -0.5)(m +0.5)(m + t) = - [ (m² -0.25)(m + t) ] = - [ m³ + m² t -0.25 m -0.25 t ]So, c = -m³ -m² t +0.25 m +0.25 tNow, compute E = 2a³ +27c -9ab.First, compute a:a = - (3m + t)So, a = -3m -tCompute a³:a³ = (-3m -t)^3 = - (3m + t)^3 = - [27m³ +27m² t +9m t² + t³]Therefore, 2a³ = -2(27m³ +27m² t +9m t² +t³) = -54m³ -54m² t -18m t² -2t³27c =27(-m³ -m² t +0.25 m +0.25 t) = -27m³ -27m² t +6.75 m +6.75 t-9ab = -9*(-3m -t)*(3m² +2m t -0.25)First compute (-3m -t)*(3m² +2m t -0.25):Multiply term by term:-3m*(3m²) = -9m³-3m*(2m t) = -6m² t-3m*(-0.25) = 0.75m-t*(3m²) = -3m² t-t*(2m t) = -2m t²-t*(-0.25) = 0.25tSum all terms:-9m³ -6m² t +0.75m -3m² t -2m t² +0.25tCombine like terms:-9m³ -9m² t -2m t² +0.75m +0.25tTherefore, -9ab = -9*( -9m³ -9m² t -2m t² +0.75m +0.25t ) = 81m³ +81m² t +18m t² -6.75m -2.25tNow, combine all three terms for E:E = ( -54m³ -54m² t -18m t² -2t³ ) + ( -27m³ -27m² t +6.75 m +6.75 t ) + ( 81m³ +81m² t +18m t² -6.75m -2.25t )Let's compute term by term:For m³:-54m³ -27m³ +81m³ = 0For m² t:-54m² t -27m² t +81m² t = 0For m t²:-18m t² +18m t² =0For t³:-2t³For m terms:6.75m -6.75m = 0For t terms:6.75t -2.25t =4.5tSo, combining all terms, E = -2t³ +4.5tSo, E = -2t³ + 4.5tTherefore, the expression simplifies to E = -2t³ + (9/2)tBut t >0 (since x₃ > m, and x₃ = m + t, so t >0).So, we need to maximize E = -2t³ + (9/2)t over t >0.This is a single-variable calculus problem. Let's find the critical points by differentiating E with respect to t:dE/dt = -6t² + 9/2Set derivative equal to zero:-6t² + 9/2 =0=> 6t² =9/2=> t²= (9/2)/6 = (9)/(12) = 3/4=> t= sqrt(3)/2 ≈0.866Since t >0, we take the positive root.Check second derivative to confirm if it's a maximum:d²E/dt² = -12tAt t = sqrt(3)/2 >0, the second derivative is -12*(sqrt(3)/2) = -6*sqrt(3) <0, so it's a local maximum.Therefore, the maximum occurs at t= sqrt(3)/2.Compute E at t= sqrt(3)/2:E = -2*( (sqrt(3)/2)^3 ) + (9/2)*(sqrt(3)/2 )First, compute (sqrt(3)/2)^3:(sqrt(3))^3 = (3)^(3/2) =3*sqrt(3)So, (sqrt(3)/2)^3= 3*sqrt(3)/8Then, -2*(3*sqrt(3)/8 )= -6*sqrt(3)/8 = -3*sqrt(3)/4Next, (9/2)*(sqrt(3)/2 )=9*sqrt(3)/4Therefore, E= (-3*sqrt(3)/4) + (9*sqrt(3)/4 )= (6*sqrt(3))/4= (3*sqrt(3))/2Thus, the maximum value of E when λ=1 is 3√3/2.Since we normalized λ to 1 earlier, and due to homogeneity, the original expression (2a³ +27c -9ab)/λ³ will have the maximum value 3√3/2 regardless of λ. Therefore, the maximum value is 3√3/2.But let me verify this result again. Because when we substituted variables, we set λ=1 and found the maximum E =3√3/2. Since the original expression is E/λ³, which when λ=1 is 3√3/2, but when scaling back to any λ, the expression remains the same. Therefore, the maximum value is 3√3/2.Wait, but let me check the substitution again. When we set λ=1, we found that (2a³ +27c -9ab) for that case is 3√3/2. Therefore, (2a³ +27c -9ab)/λ³ = (3√3/2)/1³=3√3/2. For a general λ, due to homogeneity, scaling all variables by λ would lead to (2a³ +27c -9ab) scaling as λ³, so the ratio remains the same. Hence, the maximum value is indeed 3√3/2.Therefore, the answer is 3√3/2, which can be written as (3√3)/2. In boxed form, that's boxed{dfrac{3sqrt{3}}{2}}.Final AnswerThe maximum value is boxed{dfrac{3sqrt{3}}{2}}.</think>Given a cubic polynomial ( f(x) = x^3 + ax^2 + bx + c ) with three real roots ( x_1, x_2, x_3 ) satisfying ( x_2 - x_1 = lambda ) and ( x_3 > frac{1}{2}(x_1 + x_2) ), we need to find the maximum value of the expression ( frac{2a^3 + 27c - 9ab}{lambda^3} ).First, we express the polynomial in terms of its roots:[ f(x) = (x - x_1)(x - x_2)(x - x_3) ]Expanding this, we relate the coefficients ( a, b, c ) to the roots:[ a = -(x_1 + x_2 + x_3), quad b = x_1x_2 + x_1x_3 + x_2x_3, quad c = -x_1x_2x_3 ]We introduce variables ( m ) and ( t ) such that ( x_1 = m - frac{lambda}{2} ), ( x_2 = m + frac{lambda}{2} ), and ( x_3 = m + t ) where ( t > 0 ). Substituting these into the expressions for ( a, b, c ):[ a = -(3m + t), quad b = 3m^2 + 2mt - frac{lambda^2}{4}, quad c = -m^3 - m^2t + frac{lambda^2}{4}m + frac{lambda^2}{4}t ]We then compute the expression ( 2a^3 + 27c - 9ab ). By normalizing ( lambda = 1 ) and solving, we find that the expression simplifies to:[ -2t^3 + frac{9}{2}t ]Maximizing this expression using calculus, we find the critical point at ( t = frac{sqrt{3}}{2} ), leading to the maximum value:[ frac{3sqrt{3}}{2} ]Due to the homogeneity of the problem, this result holds for any ( lambda ). Thus, the maximum value of ( frac{2a^3 + 27c - 9ab}{lambda^3} ) is:[ boxed{dfrac{3sqrt{3}}{2}} ]]

🔑:Let S=frac{2 a^{3}+27 c-9 a b}{lambda^{3}}, thenbegin{aligned}S & =frac{27left(frac{2}{27} a^{3}-frac{1}{3} a b+cright)}{lambda^{3}}=frac{27 fleft(-frac{1}{3} aright)}{lambda^{3}} & =frac{27left(-frac{1}{3} a-x_{1}right)left(-frac{1}{3} a-x_{2}right)left(-frac{1}{3} a-x_{3}right)}{left(x_{2}-x_{1}right)^{3}} & =frac{-27left(x_{1}+frac{1}{3} aright)left(x_{2}+frac{1}{3} aright)left(x_{3}+frac{1}{3} aright)}{left(x_{2}-x_{1}right)^{3}} .end{aligned}Let hat{imath} u_{i}=x_{i}+frac{a}{3}(i=1,2,3), then u_{2}-u_{1}=x_{2}-x_{1}=lambda, stackrel{1}{.}left(u_{1}+u_{2}right), and u_{1}+u_{2}+u_{3}=x_{1}+x_{2}+x_{3}+a=0. Therefore, u_{1}, u_{2}, u_{3} still satisfy the corresponding conditions, andS=frac{-27 u_{1} u_{2} u_{3}}{left(u_{2}-u_{1}right)^{3}} .From u_{3}=-left(u_{1}+u_{2}right)>frac{u_{1}+u_{2}}{2}, we get u_{1}+u_{2}<0, and letv_{1}=frac{-u_{1}}{u_{2}-u_{1}}, v=v_{2}=frac{u_{2}}{u_{2}-u_{1}},then v_{1}+v_{2}=1, v_{1}, v_{2} are both greater than 0, v_{1}-v_{2}=frac{u_{3}}{u_{2}-u_{1}}>0,begin{aligned}S & =27 v_{1} v_{2}left(v_{1}-v_{2}right)=27 v(1-v)(1-2 v) & =27 sqrt{left(v-v^{2}right)^{2}(1-2 v)^{2}} & =27 sqrt{2left(v-v^{2}right)left(v-v^{2}right)left(frac{1}{2}-2 v+2 v^{2}right)} & leqslant 27 sqrt{2 timesleft(frac{1}{6}right)^{3}} & =frac{3}{2} sqrt{3} .end{aligned}The equality holds when v=frac{1}{2}left(1-frac{sqrt{3}}{3}right), and the corresponding cubic equation is x^{3}-frac{1}{2} x+frac{sqrt{3}}{18}=0, lambda=1left(x_{1}=-frac{1}{2}left(1+frac{sqrt{3}}{3}right), x_{2}=frac{1}{2}left(1-frac{sqrt{3}}{3}right), x_{3}=frac{sqrt{3}}{3}right).In summary, the maximum value sought is frac{3}{2} sqrt{3}.

🧠:[<think>Okay, let's try to solve this problem step by step. So, we have a quadratic function f(x) that satisfies three conditions: f(-2) = 1, f(2) = 2, and for all real numbers x, the inequality x ≤ f(x) ≤ 1/4(x² + 4) holds. We need to find the value of f(4) from the given options.First, since f(x) is a quadratic function, it has the general form f(x) = ax² + bx + c, where a, b, and c are constants we need to determine. The problem gives us two specific values: when x = -2, f(-2) = 1, and when x = 2, f(2) = 2. So, let's write those equations down.For x = -2:a*(-2)² + b*(-2) + c = 1Simplify: 4a - 2b + c = 1 ...(1)For x = 2:a*(2)² + b*(2) + c = 2Simplify: 4a + 2b + c = 2 ...(2)Now, we have two equations here, but three unknowns (a, b, c). So, we need another equation to solve for all three variables. The third condition is that for all real x, x ≤ f(x) ≤ (1/4)(x² + 4). Maybe we can use this inequality to find more constraints on a, b, c.Let me think. Since f(x) is a quadratic function, and it's bounded below by x and above by (1/4)(x² + 4). That means for all x, the quadratic function f(x) must lie between the line y = x and the quadratic function y = (1/4)(x² + 4).First, let's analyze the upper bound: f(x) ≤ (1/4)(x² + 4). Let's write that as:ax² + bx + c ≤ (1/4)x² + 1Subtracting the right-hand side from both sides:ax² + bx + c - (1/4)x² - 1 ≤ 0Simplify:(a - 1/4)x² + bx + (c - 1) ≤ 0 for all x.Since this inequality must hold for all real x, the quadratic expression (a - 1/4)x² + bx + (c - 1) must be ≤ 0 for all x. For a quadratic to be non-positive for all x, two conditions must be satisfied:1. The leading coefficient must be ≤ 0.2. The quadratic must have at most one real root (i.e., discriminant ≤ 0).Similarly, the lower bound is x ≤ f(x), which can be written as:x ≤ ax² + bx + cRearranged:ax² + bx + c - x ≥ 0Simplify:ax² + (b - 1)x + c ≥ 0 for all x.Again, since this must hold for all real x, the quadratic ax² + (b - 1)x + c must be ≥ 0 for all x. For this quadratic to be non-negative for all x:1. The leading coefficient a must be ≥ 0.2. The discriminant must be ≤ 0.So, let's note these conditions.From the upper bound:1. a - 1/4 ≤ 0 ⇒ a ≤ 1/42. Discriminant of (a - 1/4)x² + bx + (c - 1) must be ≤ 0. The discriminant D = b² - 4*(a - 1/4)*(c - 1) ≤ 0.From the lower bound:1. a ≥ 02. Discriminant of ax² + (b - 1)x + c must be ≤ 0. The discriminant D' = (b - 1)^2 - 4*a*c ≤ 0.Also, from the upper bound, since the quadratic must be ≤ 0 everywhere, if a - 1/4 is negative (which it is, from the first condition), the parabola opens downward. But since it's ≤ 0 for all x, there can be no real roots or exactly one real root. Similarly, the lower bound quadratic opens upward (since a ≥ 0) and must be ≥ 0 everywhere, so again no real roots or exactly one.So, combining the leading coefficients:From upper bound: a ≤ 1/4From lower bound: a ≥ 0Thus, 0 ≤ a ≤ 1/4.Now, let's see if we can use the two equations we have from f(-2) and f(2):Equation (1): 4a - 2b + c = 1Equation (2): 4a + 2b + c = 2If we subtract equation (1) from equation (2), we get:(4a + 2b + c) - (4a - 2b + c) = 2 - 1Simplify:4b = 1 ⇒ b = 1/4So, we have found b = 1/4.Now, substitute b = 1/4 into equations (1) and (2):Equation (1): 4a - 2*(1/4) + c = 1 ⇒ 4a - 1/2 + c = 1 ⇒ 4a + c = 1 + 1/2 ⇒ 4a + c = 3/2 ...(1a)Equation (2): 4a + 2*(1/4) + c = 2 ⇒ 4a + 1/2 + c = 2 ⇒ 4a + c = 2 - 1/2 ⇒ 4a + c = 3/2 ...(2a)Wait, so both equations (1a) and (2a) are the same: 4a + c = 3/2. That means that with b = 1/4, we only have one equation for a and c. So, we need another condition to find a and c. The other conditions come from the inequalities.Let's recall the upper bound discriminant and lower bound discriminant conditions.From upper bound: discriminant D = b² - 4*(a - 1/4)*(c - 1) ≤ 0We know b = 1/4, so:(1/4)^2 - 4*(a - 1/4)*(c - 1) ≤ 0Calculate 1/16 - 4*(a - 1/4)*(c - 1) ≤ 0Similarly, from lower bound: discriminant D' = (b - 1)^2 - 4ac ≤ 0With b = 1/4:(1/4 - 1)^2 - 4ac ≤ 0 ⇒ (-3/4)^2 - 4ac ≤ 0 ⇒ 9/16 - 4ac ≤ 0 ⇒ -4ac ≤ -9/16 ⇒ 4ac ≥ 9/16 ⇒ ac ≥ 9/64So, two inequalities:1. 1/16 - 4*(a - 1/4)*(c - 1) ≤ 02. ac ≥ 9/64Also, from 4a + c = 3/2. Let's express c in terms of a:c = 3/2 - 4aSo, substitute c into the inequalities.First inequality:1/16 - 4*(a - 1/4)*( (3/2 - 4a) - 1 ) ≤ 0Simplify inside the second term:(3/2 - 4a) - 1 = (3/2 - 1) - 4a = 1/2 - 4aThus, the inequality becomes:1/16 - 4*(a - 1/4)*(1/2 - 4a) ≤ 0Let's compute the product (a - 1/4)(1/2 - 4a):Multiply out:(a)(1/2) + a*(-4a) - (1/4)(1/2) + (1/4)(4a)= (1/2)a - 4a² - 1/8 + a= (1/2a + a) - 4a² - 1/8= (3/2)a - 4a² - 1/8So, the product is -4a² + (3/2)a - 1/8Therefore, the inequality:1/16 - 4*(-4a² + (3/2)a - 1/8) ≤ 0Wait, hold on. Wait, the term we're multiplying is (a - 1/4)(1/2 - 4a). Let me check that multiplication again.Wait, let's do it step by step.(a - 1/4)(1/2 - 4a) = a*(1/2) + a*(-4a) - (1/4)*(1/2) + (1/4)*(4a)= (a/2 - 4a²) - (1/8) + a= a/2 - 4a² - 1/8 + a= (a/2 + a) - 4a² - 1/8= (3a/2) - 4a² - 1/8So, correct. Then multiply by 4:4*(3a/2 - 4a² - 1/8) = 4*(3a/2) - 4*(4a²) - 4*(1/8)= 6a - 16a² - 0.5So, the inequality is:1/16 - (6a - 16a² - 0.5) ≤ 0Wait, but originally, the term was:1/16 - 4*(a - 1/4)*(1/2 - 4a) ≤ 0Which is 1/16 - [4*(a - 1/4)*(1/2 - 4a)] ≤ 0So, substituting:1/16 - [6a - 16a² - 0.5] ≤ 0Which becomes:1/16 -6a +16a² +0.5 ≤0Convert 0.5 to 8/16:1/16 + 8/16 =9/16So:9/16 -6a +16a² ≤0Multiply all terms by 16 to eliminate denominators:9 -96a +256a² ≤0So, 256a² -96a +9 ≤0Let me check if this quadratic can be simplified.256a² -96a +9 =0We can compute the discriminant:D = (-96)^2 -4*256*9 = 9216 - 9216 =0So, discriminant is 0, meaning there's one real root. Thus, the quadratic is a perfect square.Let me factor it.256a² -96a +9 = (16a)^2 - 2*16a*3 + 3^2 = (16a -3)^2Yes, so:(16a -3)^2 ≤0The square of a real number is always ≥0, so the inequality (16a -3)^2 ≤0 holds only when 16a -3 =0 ⇒ a= 3/16Therefore, the first inequality gives a = 3/16Now, since a must be 3/16, we can find c from c=3/2 -4aSo, c = 3/2 -4*(3/16) = 3/2 -12/16 = 3/2 -3/4 = (6/4 -3/4)= 3/4Therefore, c = 3/4So, we have a=3/16, b=1/4, c=3/4Therefore, the quadratic function is f(x) = (3/16)x² + (1/4)x + 3/4Now, we need to check if this satisfies the lower bound condition.From the lower bound discriminant condition, we had ac ≥9/64Let's compute a*c:(3/16)*(3/4) =9/64, which is exactly 9/64. So, since ac=9/64 ≥9/64, the inequality holds. Also, the discriminant D' was (b -1)^2 -4ac = ( -3/4 )^2 -4*(3/16)*(3/4)= 9/16 -4*(9/64)=9/16 -9/16=0, so discriminant is zero. Therefore, the quadratic ax² + (b -1)x +c touches the x-axis at one point, which is okay since it's allowed to be tangent (equality holds there). Thus, the lower bound condition is satisfied.Similarly, the upper bound discriminant was zero, so the upper quadratic touches the line at one point. So, both inequalities are satisfied as equalities at certain points, but overall, the function f(x) lies between x and (1/4)(x² +4) for all x.Therefore, the quadratic function is f(x) = (3/16)x² + (1/4)x + 3/4Now, we need to compute f(4):f(4) = (3/16)*(4)^2 + (1/4)*4 + 3/4Compute each term:(3/16)*(16) = 3(1/4)*4 =13/4 remains.So, total f(4) =3 +1 +3/4=4 +3/4=19/4=4.75Looking at the options:(A) 17/4=4.25(B)9/2=4.5(C)19/4=4.75(D)5=5So, 19/4 is option C. Therefore, the answer should be C.But let me just double-check my calculations to make sure I didn't make an arithmetic error.First, a=3/16, b=1/4, c=3/4.Compute f(4):3/16*(4)^2 =3/16*16=31/4*4=13/4=0.75Adding together: 3 +1 +0.75=4.75=19/4. Correct.Therefore, the answer is C) 19/4.Final Answerboxed{C}</think>Given the quadratic function ( f(x) ) satisfies ( f(-2) = 1 ), ( f(2) = 2 ), and ( x leqslant f(x) leqslant frac{1}{4}(x^2 + 4) ) for all real numbers ( x ). We need to find the value of ( f(4) ).First, we express ( f(x) ) in the general quadratic form ( f(x) = ax^2 + bx + c ). Using the given conditions ( f(-2) = 1 ) and ( f(2) = 2 ), we set up the following equations:[4a - 2b + c = 1 quad text{(1)}][4a + 2b + c = 2 quad text{(2)}]Subtracting equation (1) from equation (2) gives ( 4b = 1 ), so ( b = frac{1}{4} ).Substituting ( b = frac{1}{4} ) into equations (1) and (2), we get:[4a + c = frac{3}{2} quad text{(3)}]Next, we use the inequalities ( x leqslant f(x) leqslant frac{1}{4}(x^2 + 4) ). For the upper bound, we derive the quadratic inequality:[(a - frac{1}{4})x^2 + bx + (c - 1) leqslant 0]For this to hold for all ( x ), the discriminant must be non-positive. Solving this, we find ( a = frac{3}{16} ).For the lower bound, we derive the quadratic inequality:[ax^2 + (b - 1)x + c geqslant 0]For this to hold for all ( x ), the discriminant must be non-positive. Solving this confirms ( ac = frac{9}{64} ).Using ( a = frac{3}{16} ) in equation (3), we find ( c = frac{3}{4} ).Thus, the quadratic function is ( f(x) = frac{3}{16}x^2 + frac{1}{4}x + frac{3}{4} ).Calculating ( f(4) ):[f(4) = frac{3}{16}(4^2) + frac{1}{4}(4) + frac{3}{4} = 3 + 1 + frac{3}{4} = frac{19}{4}]The value of ( f(4) ) is (boxed{C}).,<think>Okay, so I need to find the value of f(4) for a quadratic function f(x) that satisfies three conditions: f(-2) = 1, f(2) = 2, and for all real numbers x, x ≤ f(x) ≤ (1/4)(x² + 4). The answer choices are given, so maybe after figuring it out, I can check which one it matches. Let me start by recalling that a quadratic function has the form f(x) = ax² + bx + c, where a, b, c are constants. Since it's a quadratic, its graph is a parabola, which either opens upwards or downwards depending on the coefficient a.First, let's note the given conditions:1. f(-2) = 12. f(2) = 23. For all real x, x ≤ f(x) ≤ (1/4)(x² + 4)So, I need to determine the coefficients a, b, c such that these conditions are satisfied. Then, once I have the function f(x), I can compute f(4).Let me start by using the first two conditions to set up equations. Let's plug in x = -2 and x = 2 into the quadratic function.For x = -2:f(-2) = a(-2)² + b(-2) + c = 4a - 2b + c = 1. Let's call this equation (1): 4a - 2b + c = 1.For x = 2:f(2) = a(2)² + b(2) + c = 4a + 2b + c = 2. Let's call this equation (2): 4a + 2b + c = 2.Now, if I subtract equation (1) from equation (2), I can eliminate the terms with a and c:(4a + 2b + c) - (4a - 2b + c) = 2 - 14a + 2b + c - 4a + 2b - c = 1(0a) + 4b + 0c = 1So, 4b = 1 => b = 1/4. Okay, so we found that b is 1/4.Now, let's add equations (1) and (2) to see if that helps:(4a - 2b + c) + (4a + 2b + c) = 1 + 28a + 2c = 3. Let's call this equation (3): 8a + 2c = 3.So, we have equation (3): 8a + 2c = 3. Let me solve for c in terms of a.Divide both sides by 2: 4a + c = 3/2 => c = 3/2 - 4a. Let's keep that in mind.So, now we have expressions for b and c in terms of a. But we need another equation to solve for a. The third condition is that for all real x, x ≤ f(x) ≤ (1/4)(x² + 4). So, the quadratic function f(x) must lie between the line y = x and the function y = (1/4)(x² + 4) for all real numbers x.Therefore, for all x, x ≤ ax² + (1/4)x + c ≤ (1/4)(x² + 4). Let me write that down:x ≤ ax² + (1/4)x + c ≤ (x² + 4)/4.This inequality must hold for every real number x. So, let's first analyze the upper bound and the lower bound separately.First, the upper bound:ax² + (1/4)x + c ≤ (x² + 4)/4.Let me rearrange this inequality:ax² + (1/4)x + c - (x² + 4)/4 ≤ 0.Simplify the left side:ax² + (1/4)x + c - (x²)/4 - 1 ≤ 0.Combine like terms:( a - 1/4 )x² + (1/4)x + (c - 1) ≤ 0.This inequality must hold for all real x. Since this is a quadratic in x, the only way this can be true for all x is if the quadratic is always non-positive (≤ 0), which requires two things:1. The leading coefficient (a - 1/4) must be ≤ 0. So, a ≤ 1/4.2. The quadratic must have no real roots, which means its discriminant must be negative.Similarly, the lower bound inequality:ax² + (1/4)x + c ≥ x.Subtract x from both sides:ax² + (1/4)x + c - x ≥ 0.Simplify:ax² - (3/4)x + c ≥ 0.Again, this is a quadratic in x, and this must hold for all real x. Therefore, the quadratic must be non-negative for all x, which requires:1. The leading coefficient a must be ≥ 0.2. The discriminant must be ≤ 0.So, let's handle both inequalities step by step.First, the upper bound:( a - 1/4 )x² + (1/4)x + (c - 1) ≤ 0 for all x.Since this is a quadratic, for it to be ≤ 0 for all x, it must open downward (leading coefficient < 0) or be a downward opening parabola that touches the x-axis at most once (i.e., discriminant ≤ 0). However, if the leading coefficient is zero, then the quadratic becomes linear, which cannot be ≤ 0 for all x unless it's a constant function. But if a - 1/4 = 0, then a = 1/4, and the quadratic reduces to (1/4)x + (c - 1) ≤ 0 for all x. However, a linear function can't be ≤ 0 for all x unless the coefficient of x is 0 and the constant term is ≤ 0. But (1/4)x + (c - 1) can't be ≤ 0 for all x unless 1/4 = 0, which it's not, so a must be < 1/4, and the quadratic must open downward and have discriminant ≤ 0.Similarly, for the lower bound:ax² - (3/4)x + c ≥ 0 for all x.Here, since the quadratic must be ≥ 0 for all x, it must open upward (a > 0) and have discriminant ≤ 0.Wait, but from the upper bound, we have a ≤ 1/4, and from the lower bound, we have a ≥ 0. So, combining these, 0 ≤ a ≤ 1/4.So, a is between 0 and 1/4, inclusive. Also, from the lower bound quadratic discriminant:Discriminant D = (-3/4)^2 - 4*a*c ≤ 0.Which is 9/16 - 4ac ≤ 0 => 4ac ≥ 9/16 => ac ≥ 9/(64).Similarly, for the upper bound quadratic:Quadratic: (a - 1/4)x² + (1/4)x + (c - 1) ≤ 0 for all x.Since the leading coefficient (a - 1/4) is ≤ 0 (as a ≤ 1/4), and for the quadratic to be ≤ 0 for all x, the discriminant must be ≤ 0.Compute discriminant D for the upper bound quadratic:D = (1/4)^2 - 4*(a - 1/4)*(c - 1) ≤ 0.So, 1/16 - 4*(a - 1/4)*(c - 1) ≤ 0.Let me compute that:1/16 ≤ 4*(a - 1/4)*(c - 1)Divide both sides by 4:1/64 ≤ (a - 1/4)*(c - 1)So, (a - 1/4)*(c - 1) ≥ 1/64.Hmm, that's another inequality. Let's note that.So, we have several conditions here:1. 0 ≤ a ≤ 1/42. c = 3/2 - 4a (from equation (3))3. From lower bound discriminant: ac ≥ 9/644. From upper bound discriminant: (a - 1/4)*(c - 1) ≥ 1/64So, let's substitute c from equation (3) into these inequalities.First, compute ac:ac = a*(3/2 - 4a) = (3a/2) - 4a².This must be ≥ 9/64. So:(3a/2) - 4a² ≥ 9/64.Multiply both sides by 64 to eliminate denominators:64*(3a/2) - 64*4a² ≥ 64*(9/64)Simplify:32*3a - 256a² ≥ 9Which is:96a - 256a² - 9 ≥ 0Let me rearrange this:-256a² + 96a - 9 ≥ 0Multiply both sides by -1 (which reverses the inequality):256a² - 96a + 9 ≤ 0This is a quadratic in a. Let's compute its discriminant:D = (-96)^2 - 4*256*9 = 9216 - 9216 = 0.So, discriminant D is zero, meaning the quadratic has a repeated root. Therefore, the quadratic 256a² -96a +9 = 0 has one solution:a = [96 ± sqrt(0)]/(2*256) = 96/(512) = 96/512 = 12/64 = 3/16.Therefore, the inequality 256a² -96a +9 ≤ 0 holds only when a = 3/16.So, the inequality ac ≥ 9/64 holds only when a = 3/16.Therefore, a must equal 3/16. Then, since c = 3/2 -4a, substitute a = 3/16:c = 3/2 - 4*(3/16) = 3/2 - 12/16 = 3/2 - 3/4 = (6/4 - 3/4) = 3/4.So, c = 3/4.So, with a = 3/16, b = 1/4, c = 3/4, let's check if this satisfies the other inequality from the upper bound discriminant.Compute (a - 1/4)*(c - 1):a -1/4 = 3/16 - 4/16 = (-1)/16c -1 = 3/4 - 4/4 = (-1)/4Therefore, (a -1/4)*(c -1) = (-1/16)*(-1/4) = 1/64So, (a -1/4)*(c -1) = 1/64, which is equal to 1/64. Therefore, the inequality (a -1/4)*(c -1) ≥ 1/64 is satisfied since 1/64 ≥ 1/64.Therefore, equality holds here. So, the upper bound quadratic will touch the x-axis at one point, meaning the original quadratic f(x) touches the upper bound function (1/4)(x² +4) at one point. Similarly, the lower bound quadratic inequality becomes equality when discriminant is zero, so f(x) touches the line y = x at one point.Therefore, the quadratic function f(x) is tangent to both y = x and y = (1/4)(x² +4). Thus, this must be the unique quadratic that satisfies all the given conditions.Therefore, with a = 3/16, b = 1/4, c = 3/4, the quadratic function is:f(x) = (3/16)x² + (1/4)x + 3/4.Now, let's verify if this function satisfies f(-2)=1 and f(2)=2.Compute f(-2):(3/16)(4) + (1/4)(-2) + 3/4 = (12/16) - (2/4) + 3/4 = (3/4) - (1/2) + (3/4) = (3/4 + 3/4) - 1/2 = (6/4) - 2/4 = 4/4 = 1. Correct.Compute f(2):(3/16)(4) + (1/4)(2) + 3/4 = (12/16) + (2/4) + 3/4 = (3/4) + (1/2) + (3/4) = (3/4 + 3/4) + 1/2 = 6/4 + 2/4 = 8/4 = 2. Correct.So, the quadratic satisfies the given points. Now, let's check if it lies between y = x and y = (1/4)(x² +4) for all x.First, check that f(x) ≥ x for all x. Let's compute f(x) - x:(3/16)x² + (1/4)x + 3/4 - x = (3/16)x² - (3/4)x + 3/4.This quadratic must be ≥ 0 for all x. Let's check its discriminant:D = (-3/4)^2 - 4*(3/16)*(3/4) = 9/16 - 4*(9/64) = 9/16 - 9/16 = 0.So discriminant is zero, which means the quadratic touches the x-axis at one point. Therefore, f(x) -x is always non-negative (since the leading coefficient is 3/16 >0). So, f(x) ≥ x for all x.Similarly, check the upper bound: f(x) ≤ (1/4)(x² +4).Compute (1/4)(x² +4) - f(x):= (1/4)x² +1 - [ (3/16)x² + (1/4)x + 3/4 ]= (1/4 - 3/16)x² - (1/4)x + (1 - 3/4)= (4/16 - 3/16)x² - (1/4)x + 1/4= (1/16)x² - (1/4)x + 1/4Again, check if this quadratic is ≥0 for all x. Compute discriminant:D = (-1/4)^2 - 4*(1/16)*(1/4) = 1/16 - 4*(1/64) = 1/16 - 1/16 = 0.Discriminant is zero, so the quadratic touches the x-axis at one point, and since the leading coefficient is 1/16 >0, the quadratic is always non-negative. Therefore, (1/4)(x² +4) - f(x) ≥0 for all x, which means f(x) ≤ (1/4)(x² +4) for all x.Therefore, this quadratic function satisfies all the given conditions. Now, we can compute f(4):f(4) = (3/16)(16) + (1/4)(4) + 3/4 = (3/16 *16) + (1/4*4) + 3/4 = 3 + 1 + 3/4 = 4 + 3/4 = 19/4.So, 19/4 is the value of f(4). Let's check the answer choices: (C) 19/4. So the answer should be C.Wait, but let me verify once again the computation of f(4):(3/16)(4)^2 + (1/4)(4) + 3/4= (3/16)(16) + 1 + 3/4= 3 + 1 + 0.75= 4.75, which is 19/4. Yes, correct.Therefore, the answer is C.Final Answerboxed{C}</think>Given the quadratic function ( f(x) ) satisfies ( f(-2) = 1 ), ( f(2) = 2 ), and ( x leqslant f(x) leqslant frac{1}{4}(x^2 + 4) ) for all real numbers ( x ). We need to find the value of ( f(4) ).First, we assume the quadratic function has the form ( f(x) = ax^2 + bx + c ). Using the given conditions ( f(-2) = 1 ) and ( f(2) = 2 ), we set up the following equations:1. ( 4a - 2b + c = 1 )2. ( 4a + 2b + c = 2 )Subtracting these equations, we find ( 4b = 1 ), so ( b = frac{1}{4} ). Adding these equations, we get ( 8a + 2c = 3 ), leading to ( c = frac{3}{2} - 4a ).Next, we consider the inequalities ( x leqslant f(x) leqslant frac{1}{4}(x^2 + 4) ). For the lower bound ( x leqslant f(x) ), we derive the quadratic inequality ( ax^2 - frac{3}{4}x + c geq 0 ). For the upper bound ( f(x) leqslant frac{1}{4}(x^2 + 4) ), we derive the quadratic inequality ( (a - frac{1}{4})x^2 + frac{1}{4}x + (c - 1) leq 0 ).By analyzing the discriminants and coefficients of these quadratics, we find that ( a = frac{3}{16} ), ( b = frac{1}{4} ), and ( c = frac{3}{4} ). This quadratic function satisfies all conditions, including touching the bounds at one point.Finally, we compute ( f(4) ):[f(4) = frac{3}{16}(4^2) + frac{1}{4}(4) + frac{3}{4} = frac{3}{16}(16) + 1 + frac{3}{4} = 3 + 1 + frac{3}{4} = frac{19}{4}]Thus, the value of ( f(4) ) is (boxed{C}).,<think>Okay, let's try to solve this problem. We have a quadratic function f(x) that satisfies three conditions: f(-2) = 1, f(2) = 2, and for all real numbers x, x ≤ f(x) ≤ (1/4)(x² + 4). We need to find the value of f(4) from the given options.First, let's recall that a quadratic function has the general form f(x) = ax² + bx + c, where a, b, c are constants. Since it's quadratic, we can use the given points to set up equations and solve for a, b, c. But there's also the inequality condition that x ≤ f(x) ≤ (1/4)(x² + 4) for all x. So we need to ensure that whatever quadratic we find satisfies these inequalities everywhere.Let's start by using the two given points. When x = -2, f(-2) = 1:a(-2)² + b(-2) + c = 14a - 2b + c = 1 --- Equation 1When x = 2, f(2) = 2:a(2)² + b(2) + c = 24a + 2b + c = 2 --- Equation 2Now, subtract Equation 1 from Equation 2 to eliminate the a and c terms:(4a + 2b + c) - (4a - 2b + c) = 2 - 14a + 2b + c - 4a + 2b - c = 14b = 1So, b = 1/4Now we can plug b back into either Equation 1 or 2 to find a relationship between a and c. Let's use Equation 1:4a - 2*(1/4) + c = 14a - 1/2 + c = 14a + c = 1 + 1/24a + c = 3/2 --- Equation 3Similarly, using Equation 2:4a + 2*(1/4) + c = 24a + 1/2 + c = 24a + c = 2 - 1/24a + c = 3/2Wait, so both equations give 4a + c = 3/2. That means we only have two equations but three variables, but since both points give the same result after substituting b, we still need another equation. However, we have the inequality constraints which might help us find the specific quadratic.Alternatively, maybe we can express c in terms of a: c = 3/2 - 4a. Then the quadratic function is f(x) = ax² + (1/4)x + (3/2 - 4a).Now, let's consider the inequalities: x ≤ f(x) ≤ (1/4)(x² + 4) for all x.First, let's analyze the upper bound: f(x) ≤ (1/4)(x² + 4)Substitute f(x):ax² + (1/4)x + (3/2 - 4a) ≤ (1/4)x² + 1Subtract (1/4)x² + 1 from both sides:ax² - (1/4)x² + (1/4)x + (3/2 - 4a - 1) ≤ 0Simplify:(a - 1/4)x² + (1/4)x + (1/2 - 4a) ≤ 0This inequality must hold for all real x. Since a quadratic is ≤ 0 for all x only if it opens downward (coefficient of x² < 0) and its discriminant is ≤ 0 (so it doesn't cross the x-axis).Therefore, two conditions:1. Coefficient of x²: a - 1/4 < 0 => a < 1/42. Discriminant ≤ 0:Discriminant D = (1/4)^2 - 4*(a - 1/4)*(1/2 - 4a) ≤ 0Compute D:D = 1/16 - 4*(a - 1/4)*(1/2 - 4a)First, compute the second term:4*(a - 1/4)*(1/2 - 4a)Let me compute (a - 1/4)(1/2 - 4a):= a*(1/2) - a*4a - (1/4)*(1/2) + (1/4)*4a= (a/2) - 4a² - 1/8 + a= (a/2 + a) - 4a² - 1/8= (3a/2) - 4a² - 1/8Multiply by 4:4*(3a/2 - 4a² - 1/8) = 6a - 16a² - 0.5Therefore, D = 1/16 - (6a - 16a² - 0.5)Wait, is that right? Wait:Wait, original discriminant:D = (1/4)^2 - 4*(a - 1/4)*(1/2 - 4a) = 1/16 - 4*(a - 1/4)*(1/2 - 4a)We computed 4*(a - 1/4)*(1/2 - 4a) as 6a - 16a² - 0.5Therefore, D = 1/16 - (6a - 16a² - 0.5)But let's check:Wait, first expand (a - 1/4)(1/2 - 4a):= a*1/2 - a*4a - 1/4*1/2 + 1/4*4a= (a/2) - 4a² - 1/8 + aCombine like terms:a/2 + a = (3a/2)-4a²-1/8So (a - 1/4)(1/2 - 4a) = -4a² + (3a/2) - 1/8Multiply by 4:4*(-4a² + 3a/2 - 1/8) = -16a² + 6a - 0.5Therefore, D = 1/16 - (-16a² + 6a - 0.5) = 1/16 + 16a² - 6a + 0.5Convert 0.5 to 1/2, which is 8/16:D = (1 + 8)/16 + 16a² - 6a = 9/16 + 16a² - 6aTherefore, D = 16a² - 6a + 9/16 ≤ 0So, 16a² - 6a + 9/16 ≤ 0Multiply both sides by 16 to eliminate denominators:256a² - 96a + 9 ≤ 0This is a quadratic in a. Let's compute its discriminant:D' = (-96)^2 - 4*256*9 = 9216 - 9216 = 0So discriminant is zero, which means 256a² -96a +9 =0 has a repeated root.Solving 256a² -96a +9=0:a = [96 ± sqrt(0)]/(2*256) = 96/(512) = 12/64 = 3/16Thus, the quadratic 256a² -96a +9 factors as (16a - 3)^2, since (16a -3)^2 = 256a² - 96a +9.Therefore, 256a² -96a +9 = (16a -3)^2 ≤0But a square is always non-negative, so the only solution is when (16a -3)^2 =0 => 16a -3=0 => a=3/16Therefore, the upper bound inequality holds only when a=3/16So from the upper bound condition, we have a=3/16Wait, but earlier we had a condition that a <1/4 (since coefficient of x² in the upper bound inequality must be negative). Since 3/16 is 0.1875, which is less than 1/4 (0.25), so that's okay.Therefore, a=3/16 is the only value that satisfies the upper bound inequality for all x.Now, let's check the lower bound inequality: x ≤ f(x) for all x.Given that f(x) = ax² + (1/4)x + (3/2 -4a) with a=3/16Substitute a=3/16:f(x) = (3/16)x² + (1/4)x + (3/2 - 4*(3/16)) = (3/16)x² + (1/4)x + (3/2 - 12/16)Simplify 3/2 - 12/16:Convert 3/2 to 24/16, so 24/16 -12/16 =12/16=3/4Therefore, f(x) = (3/16)x² + (1/4)x + 3/4Now, we need to check if x ≤ f(x) for all x.So, f(x) - x ≥0 for all x.Compute f(x) -x:(3/16)x² + (1/4)x + 3/4 -x = (3/16)x² - (3/4)x + 3/4We need to verify that this quadratic is always non-negative.Let's denote g(x) = (3/16)x² - (3/4)x + 3/4Check the leading coefficient: 3/16 >0, so the parabola opens upwards. If the minimum of this quadratic is ≥0, then g(x) ≥0 for all x.Find the vertex of g(x):The x-coordinate is at -B/(2A) where A=3/16, B= -3/4x = -(-3/4)/(2*(3/16)) = (3/4)/(3/8) = (3/4)*(8/3) = 2So the vertex is at x=2.Compute g(2):(3/16)(4) - (3/4)(2) + 3/4 = (12/16) - (6/4) + 3/4Convert to sixteenths:12/16 - 24/16 + 12/16 = (12 -24 +12)/16 = 0/16 =0Therefore, the minimum value of g(x) is 0 at x=2. Thus, g(x) ≥0 for all x, so f(x) -x ≥0 for all x. Hence, x ≤ f(x) for all x.Therefore, with a=3/16, both inequalities are satisfied.Therefore, the quadratic function is f(x) = (3/16)x² + (1/4)x + 3/4Now, compute f(4):f(4) = (3/16)(16) + (1/4)(4) + 3/4Simplify each term:(3/16)(16) = 3(1/4)(4) =13/4 remains.So f(4) =3 +1 +3/4 =4 +3/4= 19/4Which is 4.75, corresponding to option C.Wait, but let's confirm once again:Compute (3/16)*16 = 3, correct.(1/4)*4 =1, correct.3/4 remains. So 3+1=4, 4 +3/4=19/4. Yes.Therefore, the answer is C, 19/4.But let me check if the quadratic we found actually passes through the points (-2,1) and (2,2). Let's verify f(-2):f(-2) = (3/16)(4) + (1/4)(-2) +3/4 = (12/16) - (2/4) +3/4Convert to sixteenths:12/16 -8/16 +12/16 = (12 -8 +12)/16=16/16=1. Correct.f(2)= (3/16)(4) + (1/4)(2) +3/4 =12/16 + 2/4 +3/412/16=3/4, 2/4=1/2, 3/4 remains. So 3/4 +1/2 +3/4= (3/4 +3/4) +1/2=6/4 +2/4=8/4=2. Correct.So everything checks out. Therefore, f(4)=19/4.Final Answerboxed{C}</think>Given the quadratic function ( f(x) ) satisfies ( f(-2) = 1 ), ( f(2) = 2 ), and ( x leqslant f(x) leqslant frac{1}{4}left(x^{2} + 4right) ) for all real numbers ( x ). We need to find the value of ( f(4) ).First, we express the quadratic function in the general form ( f(x) = ax^2 + bx + c ). Using the given points, we set up the following equations:1. ( 4a - 2b + c = 1 )2. ( 4a + 2b + c = 2 )Subtracting these equations, we find ( b = frac{1}{4} ). Substituting ( b ) back into the equations, we get ( 4a + c = frac{3}{2} ).Next, we consider the inequality constraints. For the upper bound ( f(x) leqslant frac{1}{4}(x^2 + 4) ), substituting ( f(x) ) and simplifying, we find the quadratic expression must be non-positive for all ( x ). Solving this, we determine ( a = frac{3}{16} ).For the lower bound ( x leqslant f(x) ), substituting ( a = frac{3}{16} ) into ( f(x) ) and simplifying, we verify that the quadratic expression is non-negative for all ( x ).Thus, the quadratic function is ( f(x) = frac{3}{16}x^2 + frac{1}{4}x + frac{3}{4} ). Evaluating ( f(4) ):[f(4) = frac{3}{16}(16) + frac{1}{4}(4) + frac{3}{4} = 3 + 1 + frac{3}{4} = frac{19}{4}]The value of ( f(4) ) is (boxed{C}).,<think>Okay, let's try to figure out this problem. We have a quadratic function f(x) that satisfies three conditions: f(-2) = 1, f(2) = 2, and for all real numbers x, the inequality x ≤ f(x) ≤ (1/4)(x² + 4). We need to find the value of f(4) from the given options.First, since it's a quadratic function, we know it has the general form f(x) = ax² + bx + c, where a, b, c are constants. Our goal is to determine a, b, and c, and then compute f(4).Given the two points (-2, 1) and (2, 2), we can set up two equations. Let's do that.For x = -2:f(-2) = a(-2)² + b(-2) + c = 4a - 2b + c = 1. --> Equation 1For x = 2:f(2) = a(2)² + b(2) + c = 4a + 2b + c = 2. --> Equation 2So we have two equations:1) 4a - 2b + c = 12) 4a + 2b + c = 2If we subtract Equation 1 from Equation 2, we can eliminate a and c:(4a + 2b + c) - (4a - 2b + c) = 2 - 14a + 2b + c - 4a + 2b - c = 14b = 1So, b = 1/4.Now that we have b, we can substitute back into either equation to find a relation between a and c. Let's use Equation 1:4a - 2*(1/4) + c = 14a - 0.5 + c = 14a + c = 1.5 --> Equation 3Similarly, Equation 2:4a + 2*(1/4) + c = 24a + 0.5 + c = 24a + c = 1.5 --> Wait, same as Equation 3.Hmm, so both equations reduce to 4a + c = 1.5. That means we have two equations but they are dependent, so we need another condition to solve for a and c. The third condition is the inequality x ≤ f(x) ≤ (1/4)(x² + 4) for all real x. So, this must hold true for every x.Therefore, we need to use this inequality to find the possible values of a and c.Let me write the inequality again:x ≤ ax² + (1/4)x + c ≤ (1/4)(x² + 4)So, for all real x, this must hold. Let's split this into two inequalities:1) ax² + (1/4)x + c ≥ x2) ax² + (1/4)x + c ≤ (1/4)(x² + 4)Let's handle the first inequality:ax² + (1/4)x + c - x ≥ 0Simplify:ax² - (3/4)x + c ≥ 0This is a quadratic in x. For this quadratic to be non-negative for all real x, two conditions must hold:a) The leading coefficient a must be positive.b) The discriminant of the quadratic must be ≤ 0.Similarly, for the second inequality:ax² + (1/4)x + c - (1/4)x² - 1 ≤ 0Simplify:(a - 1/4)x² + (1/4 - 0)x + (c - 1) ≤ 0So, (a - 1/4)x² + (1/4)x + (c - 1) ≤ 0Again, for this to hold for all real x, the quadratic must be non-positive everywhere. Therefore, two conditions:a) The leading coefficient (a - 1/4) must be negative or zero.b) The discriminant must be ≤ 0.So, let's process the first inequality:Quadratic: ax² - (3/4)x + c ≥ 0 for all x.Conditions:1. a > 02. Discriminant D = [(-3/4)^2] - 4*a*c ≤ 0Compute D:D = 9/16 - 4ac ≤ 0So, 9/16 ≤ 4acThus, ac ≥ 9/64Similarly, the second inequality:Quadratic: (a - 1/4)x² + (1/4)x + (c - 1) ≤ 0 for all x.Conditions:1. (a - 1/4) ≤ 0 → a ≤ 1/42. Discriminant D' = (1/4)^2 - 4*(a - 1/4)*(c - 1) ≤ 0Compute D':D' = 1/16 - 4*(a - 1/4)*(c - 1) ≤ 0So, 1/16 ≤ 4*(a - 1/4)*(c - 1)Which simplifies to:(a - 1/4)*(c - 1) ≥ 1/64Now, we have from earlier:From the two points, we found that b = 1/4, and 4a + c = 1.5, so c = 1.5 - 4a.So, c = 3/2 - 4aTherefore, we can express c in terms of a. Let's substitute c into the inequalities.First, from the first inequality:ac ≥ 9/64But c = 3/2 - 4a, so:a*(3/2 - 4a) ≥ 9/64Multiply both sides by 64 to eliminate denominators:64a*(3/2 - 4a) ≥ 9Compute 64a*(3/2) = 96a64a*(-4a) = -256a²So:96a - 256a² ≥ 9Rearranged:-256a² + 96a - 9 ≥ 0Multiply both sides by -1 (which reverses the inequality):256a² - 96a + 9 ≤ 0Let me compute discriminant for this quadratic in a:D'' = (-96)^2 - 4*256*9= 9216 - 9216 = 0So, discriminant is zero, which means this quadratic touches the x-axis at one point. So, equality holds when:a = 96/(2*256) = 96/512 = 12/64 = 3/16Therefore, the quadratic 256a² -96a +9 is a perfect square: (16a - 3)^2 = 256a² -96a +9Hence, (16a - 3)^2 ≤ 0But since a square is always non-negative, the only solution is 16a - 3 = 0 → a = 3/16So, from the first inequality, a must be 3/16.Therefore, a = 3/16.Now, since c = 3/2 - 4a = 3/2 - 4*(3/16) = 3/2 - 12/16 = 3/2 - 3/4 = (6/4 - 3/4) = 3/4So, c = 3/4Now, check the second inequality conditions:We already know a = 3/16, which is less than 1/4 (since 1/4 = 4/16, 3/16 < 4/16), so the leading coefficient (a - 1/4) = 3/16 - 4/16 = -1/16 < 0, which satisfies the first condition for the second inequality.Now, compute the discriminant D':D' = 1/16 - 4*(a - 1/4)*(c - 1)We have a = 3/16, c = 3/4So, (a - 1/4) = 3/16 - 4/16 = -1/16(c - 1) = 3/4 - 1 = -1/4Thus, 4*(a - 1/4)*(c -1) = 4*(-1/16)*(-1/4) = 4*(1/64) = 1/16Hence, D' = 1/16 - 1/16 = 0 ≤ 0, which satisfies the discriminant condition.Therefore, all conditions are satisfied when a = 3/16, b = 1/4, c = 3/4.Therefore, the quadratic function is f(x) = (3/16)x² + (1/4)x + 3/4Now, compute f(4):f(4) = (3/16)*(16) + (1/4)*4 + 3/4Compute each term:(3/16)*16 = 3(1/4)*4 = 13/4 remains as is.So, f(4) = 3 + 1 + 3/4 = 4 + 3/4 = 19/4Which is option (C).Wait, but let me double-check in case of any calculation errors.First, a = 3/16, c = 3/4. Then f(x) = (3/16)x² + (1/4)x + 3/4At x=4:(3/16)*16 = 3, correct.(1/4)*4 = 1, correct.3/4 is 0.75.Adding up: 3 + 1 + 0.75 = 4.75, which is 19/4. Since 19 divided by 4 is 4.75, that's correct.Check if this function satisfies the original conditions f(-2)=1 and f(2)=2.Compute f(-2):(3/16)*(4) + (1/4)*(-2) + 3/4 = (12/16) - (2/4) + 3/4 = (3/4) - (1/2) + (3/4) = (3/4 + 3/4) - 1/2 = (6/4 - 2/4) = 4/4 = 1. Correct.Compute f(2):(3/16)*(4) + (1/4)*(2) + 3/4 = (12/16) + (2/4) + 3/4 = (3/4) + (1/2) + (3/4) = (3/4 + 3/4) + 1/2 = 6/4 + 2/4 = 8/4 = 2. Correct.Now, check the inequality x ≤ f(x) ≤ (1/4)(x² +4) for all x.First, check x ≤ f(x):For f(x) - x = (3/16)x² + (1/4)x + 3/4 - x = (3/16)x² - (3/4)x + 3/4This quadratic must be ≥ 0 for all x. As earlier, we found that a = 3/16 > 0, and discriminant D was 0, so it's a perfect square. So, this quadratic touches the x-axis at one point, hence it's always non-negative. So that's satisfied.Then, check f(x) ≤ (1/4)(x² +4):(3/16)x² + (1/4)x + 3/4 ≤ (1/4)x² + 1Subtract (1/4)x² +1 from both sides:(3/16 - 4/16)x² + (1/4)x + (3/4 - 1) ≤ 0Which is (-1/16)x² + (1/4)x -1/4 ≤ 0Multiply both sides by -16 (remember to reverse inequality):x² -4x +4 ≥ 0Which is (x -2)^2 ≥ 0, which is always true. So equality holds when x=2. So, indeed, the inequality holds for all x. Therefore, all conditions are satisfied.Therefore, f(4) = 19/4, which is option C.Final Answerboxed{C}</think>Given the quadratic function ( f(x) ) satisfies ( f(-2) = 1 ), ( f(2) = 2 ), and ( x leq f(x) leq frac{1}{4}(x^2 + 4) ) for all real numbers ( x ). We need to find the value of ( f(4) ).First, we assume the quadratic function has the form ( f(x) = ax^2 + bx + c ). Using the given points ((-2, 1)) and ((2, 2)), we set up the following equations:1. ( 4a - 2b + c = 1 )2. ( 4a + 2b + c = 2 )Subtracting these equations, we find ( b = frac{1}{4} ). Substituting ( b = frac{1}{4} ) back into the equations, we get ( 4a + c = frac{3}{2} ).Next, we use the inequality conditions ( x leq f(x) leq frac{1}{4}(x^2 + 4) ). For ( f(x) geq x ), we derive the quadratic inequality ( ax^2 - frac{3}{4}x + c geq 0 ). For this to hold for all ( x ), the discriminant must be non-positive, leading to ( ac geq frac{9}{64} ).For ( f(x) leq frac{1}{4}(x^2 + 4) ), we derive the quadratic inequality ( (a - frac{1}{4})x^2 + frac{1}{4}x + (c - 1) leq 0 ). The discriminant must also be non-positive, leading to ( (a - frac{1}{4})(c - 1) geq frac{1}{64} ).Substituting ( c = frac{3}{2} - 4a ) into these conditions, we solve for ( a ) and ( c ). We find ( a = frac{3}{16} ) and ( c = frac{3}{4} ). Thus, the quadratic function is ( f(x) = frac{3}{16}x^2 + frac{1}{4}x + frac{3}{4} ).Finally, we compute ( f(4) ):[f(4) = frac{3}{16}(16) + frac{1}{4}(4) + frac{3}{4} = 3 + 1 + frac{3}{4} = frac{19}{4}]The value of ( f(4) ) is (boxed{C}).]

🔑:Let f(x)=a x^{2}+b x+c(a neq 0).From the given information, we haveleft{begin{array}{l}f(-2)=4 a-2 b+c=1, f(2)=4 a+2 b+c=2 .end{array}right.Solving these, we get b=frac{1}{4}, c=frac{3}{2}-4 a.Also, x leqslant f(x) leqslant frac{1}{4}left(x^{2}+4right) holds for all real numbers x, which impliesa x^{2}+(b-1) x+c geqslant 0and left(a-frac{1}{4}right) x^{2}+b x+(c-1) leqslant 0hold for all real numbers x.If a=frac{1}{4}, then c=frac{1}{2}, in which case, frac{1}{4} x-frac{1}{2} leqslant 0 does not hold for all real numbers x. Therefore, a neq frac{1}{4}.Thus, we haveleft{begin{array}{l}a>0, (b-1)^{2}-4 a c leqslant 0, a-frac{1}{4}<0, b^{2}-4left(a-frac{1}{4}right)(c-1) leqslant 0 .end{array}right.Considering b=frac{1}{4}, c=frac{3}{2}-4 a, solving the system of inequalities yields a=frac{3}{16}.When b=frac{1}{4}, a=frac{3}{16}, we have c=frac{3}{4}.Thus, f(x)=frac{3}{16} x^{2}+frac{1}{4} x+frac{3}{4}.Therefore, f(4)=frac{19}{4}.