🧠:[<think>Okay, so I need to find the derivative of u with respect to x, where u is defined as e raised to (z minus 2y), and z is sin x, and y is x squared. Hmm, this looks like a chain rule problem because u is a function of z and y, which are both functions of x. Right? So, I'll have to use the chain rule for multivariable functions here.First, let me write down what I know. The function u is e^(z - 2y). Then z is sin x, so dz/dx should be cos x. And y is x squared, so dy/dx is 2x. Got that part. But how do I connect these to get du/dx?I remember that when you have a function of multiple variables, each of which is a function of another variable, you take the partial derivatives of the outer function with respect to each variable, multiply them by the derivatives of those variables with respect to x, and then add them up. So, in formula terms, du/dx = (∂u/∂z)*(dz/dx) + (∂u/∂y)*(dy/dx). Is that right?Let me check. Suppose u = f(z, y), then the total derivative du/dx is indeed the sum of the partial derivatives multiplied by the derivatives of z and y with respect to x. Yes, that seems correct. So, I need to compute the partial derivatives of u with respect to z and y first.Starting with ∂u/∂z. Since u = e^(z - 2y), the partial derivative with respect to z would be e^(z - 2y) times the derivative of the exponent with respect to z. The exponent is (z - 2y), so derivative with respect to z is 1. So, ∂u/∂z = e^(z - 2y)*1 = e^(z - 2y).Similarly, the partial derivative of u with respect to y is e^(z - 2y) times the derivative of the exponent with respect to y. The exponent's derivative with respect to y is -2. So, ∂u/∂y = e^(z - 2y)*(-2) = -2e^(z - 2y).Okay, so now I have both partial derivatives. Then I need dz/dx and dy/dx, which we already found as cos x and 2x respectively.Putting it all together: du/dx = (e^(z - 2y))*(cos x) + (-2e^(z - 2y))*(2x). Let me write that out:du/dx = e^(z - 2y) * cos x - 4x e^(z - 2y)Hmm, this simplifies to e^(z - 2y)(cos x - 4x). But z and y are functions of x, so maybe I should substitute those back in to express the answer in terms of x only.Since z = sin x and y = x^2, substituting those into the exponent gives:e^(sin x - 2x^2)So, putting it all together, du/dx = e^(sin x - 2x^2) * (cos x - 4x). That seems to be the derivative. Let me check my steps again to make sure I didn't make a mistake.First, the partial derivatives: yes, with respect to z, it's e^(z - 2y), and with respect to y, it's -2e^(z - 2y). Then the derivatives of z and y with respect to x: cos x and 2x. Multiplying the partials by these derivatives gives e^(z - 2y)*cos x and -2e^(z - 2y)*2x, which is -4x e^(z - 2y). Adding them together, factor out e^(z - 2y) to get (cos x - 4x) times that exponential. Then substituting z and y gives the final expression in terms of x. Looks good.Wait, but let me double-check the multiplication for the second term. The partial derivative of u with respect to y is -2e^(z - 2y), and dy/dx is 2x. So multiplying those together: -2e^(z - 2y) * 2x = -4x e^(z - 2y). Yes, that's correct.Another thing to check: when taking the partial derivative with respect to y, make sure that the exponent is treated correctly. The exponent is (z - 2y), so the derivative with respect to y is -2, which we multiply by the derivative of the exponent. Since the derivative of e^f is e^f * f', where f' here is the derivative with respect to the variable. So yes, that term is -2. So that's correct.Also, substituting z = sin x and y = x^2 into the exponent: sin x - 2x². So the exponential term becomes e^(sin x - 2x²). Then the entire expression for du/dx is that exponential multiplied by (cos x - 4x). Is there any other step where I could have made an error? Let's see. Maybe in the chain rule application. For example, if I forgot to add the two terms or miscalculated the signs. Let's see: the first term is positive because it's (∂u/∂z)(dz/dx) = e^(...) * cos x, and the second term is (∂u/∂y)(dy/dx) = (-2e^(...))*(2x) = -4x e^(...). So when you add them, it's positive cos x times e^(...) minus 4x e^(...). That's right. So the final answer is correct.Therefore, I think the derivative du/dx is e raised to (sin x - 2x squared) multiplied by (cos x minus 4x). Final AnswerThe derivative of ( u ) with respect to ( x ) is (boxed{e^{sin x - 2x^2} (cos x - 4x)}).</think>To find (frac{du}{dx}) where (u = e^{z - 2y}), (z = sin x), and (y = x^2), we use the chain rule for multivariable functions. First, we compute the partial derivatives of (u) with respect to (z) and (y):[frac{partial u}{partial z} = e^{z - 2y} quad text{and} quad frac{partial u}{partial y} = -2e^{z - 2y}]Next, we find the derivatives of (z) and (y) with respect to (x):[frac{dz}{dx} = cos x quad text{and} quad frac{dy}{dx} = 2x]Applying the chain rule, we combine these results:[frac{du}{dx} = frac{partial u}{partial z} cdot frac{dz}{dx} + frac{partial u}{partial y} cdot frac{dy}{dx}][frac{du}{dx} = e^{z - 2y} cdot cos x + (-2e^{z - 2y}) cdot 2x]Substituting (z = sin x) and (y = x^2) into the expression, we get:[frac{du}{dx} = e^{sin x - 2x^2} cdot cos x - 4x e^{sin x - 2x^2}]Factoring out the common term (e^{sin x - 2x^2}), we obtain:[frac{du}{dx} = e^{sin x - 2x^2} (cos x - 4x)]Thus, the derivative of (u) with respect to (x) is (boxed{e^{sin x - 2x^2} (cos x - 4x)}).,<think>Okay, let me try to figure out how to find the derivative of u with respect to x here. So, the problem is u equals e raised to the power of z minus 2y, where z is sin x and y is x squared. Alright, first, I need to remember how to take derivatives of composite functions. This seems like a chain rule problem because u is expressed in terms of z and y, which are both functions of x. Let me start by writing down what u is. So, u = e^(z - 2y). And z is sin x, y is x². So, substituting those in, u would be e^(sin x - 2x²). But maybe I don't need to substitute them first? Wait, the problem says to find du/dx, so using the chain rule might be the right approach here. Right, since u depends on z and y, and both z and y depend on x, I need to take the derivative of u with respect to z and y first, and then multiply by the derivatives of z and y with respect to x. Let me recall the chain rule for multiple variables. If u is a function of z and y, then du/dx = (∂u/∂z) * dz/dx + (∂u/∂y) * dy/dx. Yeah, that sounds right. So, let's compute each part step by step. First, let's find ∂u/∂z. Since u = e^(z - 2y), the partial derivative of u with respect to z is e^(z - 2y) times the derivative of (z - 2y) with respect to z, which is 1. So, ∂u/∂z = e^(z - 2y). Then, dz/dx is the derivative of sin x, which is cos x. Next, the partial derivative of u with respect to y. Again, u = e^(z - 2y). The derivative of the exponent with respect to y is -2, so ∂u/∂y = e^(z - 2y) * (-2) = -2e^(z - 2y). Then, dy/dx is the derivative of x², which is 2x. Putting it all together, du/dx = (e^(z - 2y))(cos x) + (-2e^(z - 2y))(2x). Simplify that. Factor out e^(z - 2y), so we have e^(z - 2y)[cos x - 4x]. Then, since z is sin x and y is x², maybe substitute back in to write the answer in terms of x? So, replacing z and y, we get e^(sin x - 2x²)[cos x - 4x]. Wait, let me check if I did that right. Let me go through each step again. Starting with u = e^(z - 2y). Then, ∂u/∂z = e^(z - 2y) * (1) and ∂u/∂y = e^(z - 2y) * (-2). Then dz/dx = cos x and dy/dx = 2x. So multiplying the corresponding terms and adding them: e^(z - 2y)*cos x + (-2e^(z - 2y))*2x. Wait, yes, that's e^(z - 2y) times cos x minus 4x. So, that's correct. Alternatively, if I substitute z and y first into u, u = e^(sin x - 2x²). Then, taking the derivative directly with respect to x, using the chain rule: the derivative of e^something is e^something times the derivative of the exponent. The exponent is sin x - 2x², so its derivative is cos x - 4x. So, du/dx = e^(sin x - 2x²)(cos x - 4x). Which is the same result. So both methods give the same answer. That's reassuring. Maybe the first method using the multivariable chain rule was a bit more involved, but either way, the result is the same. So, the final answer should be e raised to the sin x minus 2x squared times (cos x minus 4x). Let me just make sure I didn't mix up any signs. The derivative of -2y with respect to y is -2, so when taking partial derivative with respect to y, we get -2e^(z - 2y). Then dy/dx is 2x, so multiplying those gives -2e^(z - 2y)*2x = -4x e^(z - 2y). Then the other term is e^(z - 2y)*cos x. So combined, it's e^(z - 2y)(cos x - 4x). Which, substituting z and y, is e^(sin x - 2x²)(cos x - 4x). Yes, that seems right. I don't think I missed any negative signs or coefficients. So I think that's the correct derivative.Final Answerboxed{e^{sin x - 2x^2} (cos x - 4x)}</think>To find (frac{du}{dx}) where (u = e^{z - 2y}), (z = sin x), and (y = x^2), we use the chain rule for multiple variables. First, we compute the partial derivatives of (u) with respect to (z) and (y):[frac{partial u}{partial z} = e^{z - 2y} quad text{and} quad frac{partial u}{partial y} = -2e^{z - 2y}]Next, we find the derivatives of (z) and (y) with respect to (x):[frac{dz}{dx} = cos x quad text{and} quad frac{dy}{dx} = 2x]Applying the chain rule, we combine these results:[frac{du}{dx} = frac{partial u}{partial z} cdot frac{dz}{dx} + frac{partial u}{partial y} cdot frac{dy}{dx}][frac{du}{dx} = e^{z - 2y} cdot cos x + (-2e^{z - 2y}) cdot 2x]Factoring out (e^{z - 2y}) and substituting (z = sin x) and (y = x^2), we get:[frac{du}{dx} = e^{sin x - 2x^2} (cos x - 4x)]Thus, the final answer is:[boxed{e^{sin x - 2x^2} (cos x - 4x)}]]

🔑:Solution. The function u=e^{z-2 y} depends on two intermediate arguments v_{1}=y, v_{2}=z. According to formula (7.21), which in this case takes the formfrac{d u}{d x}=frac{partial u}{partial y} cdot frac{d y}{d x}+frac{partial u}{partial z} cdot frac{d z}{d x}we obtainbegin{gathered}frac{d u}{d x}=e^{z-2 y}(-2) 2 x+e^{z-2 y} cos x=e^{z-2 y}(cos x-4 x)= =e^{sin x-2 x^{2}}(cos x-4 x)end{gathered}sincefrac{partial u}{partial y}=e^{z-2 y}(-2), frac{partial u}{partial z}=e^{z-2 y}, frac{d y}{d x}=2 x, frac{d z}{d x}=cos xRemark. The same result can be obtained in another way, by expressing u as an explicit function of xu=e^{sin x-2 x^{2}}and differentiating the latter:frac{d u}{d x}=left(e^{sin x-2 x^{2}}right)^{prime}left(sin x-2 x^{2}right)^{prime}=e^{sin x-2 x^{2}}(cos x-4 x)

🧠:[<think>Okay, so I need to find the value of S₁₅ divided by S₅ for an arithmetic sequence where the 8th term is twice the 3rd term. Let me start by recalling some formulas related to arithmetic sequences.First, the nth term of an arithmetic sequence can be expressed as aₙ = a₁ + (n - 1)d, where a₁ is the first term and d is the common difference. The sum of the first n terms, Sₙ, is given by Sₙ = n/2 [2a₁ + (n - 1)d] or alternatively, Sₙ = n(a₁ + aₙ)/2. Both formulas are equivalent, so I can use whichever is more convenient.The problem states that a₈ = 2a₃. Let me write down those terms using the nth term formula.For the 3rd term, a₃ = a₁ + (3 - 1)d = a₁ + 2d.For the 8th term, a₈ = a₁ + (8 - 1)d = a₁ + 7d.According to the problem, a₈ = 2a₃. So substituting the expressions I found:a₁ + 7d = 2(a₁ + 2d)Let me expand the right side:a₁ + 7d = 2a₁ + 4dNow, subtract a₁ and 4d from both sides to solve for the relationship between a₁ and d:7d - 4d = 2a₁ - a₁3d = a₁So, a₁ = 3d. That's interesting. So the first term is three times the common difference. I can use this relationship to express all terms in terms of d, which might help in calculating the sums S₁₅ and S₅.Now, I need to find S₁₅ / S₅. Let's compute each sum separately.Starting with S₁₅. Using the sum formula:S₁₅ = 15/2 [2a₁ + (15 - 1)d] = 15/2 [2a₁ + 14d]Similarly, S₅ = 5/2 [2a₁ + (5 - 1)d] = 5/2 [2a₁ + 4d]But since we know a₁ = 3d, we can substitute that into the expressions for S₁₅ and S₅.Let's substitute a₁ = 3d into S₁₅:S₁₅ = 15/2 [2*(3d) + 14d] = 15/2 [6d + 14d] = 15/2 [20d] = 15/2 * 20dSimplify that:15/2 * 20d = (15 * 20d)/2 = (300d)/2 = 150dNow, do the same for S₅:S₅ = 5/2 [2*(3d) + 4d] = 5/2 [6d + 4d] = 5/2 [10d] = 5/2 * 10d = (50d)/2 = 25dTherefore, S₁₅ = 150d and S₅ = 25d. So the ratio S₁₅/S₅ is 150d / 25d. The d cancels out:150d / 25d = 150/25 = 6Wait, so the ratio is 6? Let me double-check my calculations to make sure I didn't make a mistake.First, check the substitution of a₁ = 3d into S₁₅:2a₁ = 2*(3d) = 6d14d is correct because (15 - 1) = 14, so 14d. So 6d + 14d = 20d. Then 15/2 * 20d is indeed 15*10d = 150d. That's correct.For S₅, 2a₁ = 6d, 4d from (5 -1)d. So 6d + 4d = 10d. Then 5/2 * 10d = 5*5d = 25d. That's correct.So the ratio is 150d / 25d = 6. So the answer should be 6. Let me verify this another way just to be thorough.Alternative approach: Maybe use the other sum formula, Sₙ = n(a₁ + aₙ)/2.For S₁₅, a₁₅ = a₁ + 14d. Since a₁ = 3d, then a₁₅ = 3d + 14d = 17d. So S₁₅ = 15*(a₁ + a₁₅)/2 = 15*(3d + 17d)/2 = 15*(20d)/2 = 15*10d = 150d.For S₅, a₅ = a₁ + 4d = 3d + 4d = 7d. So S₅ = 5*(a₁ + a₅)/2 = 5*(3d + 7d)/2 = 5*(10d)/2 = 5*5d = 25d.Same results, so 150d /25d = 6. Seems correct.Another sanity check: If the ratio is 6, then S₁₅ is six times S₅. Let me think about arithmetic sequences. If the first term is 3d and the common difference is d, then the sequence is 3d, 4d, 5d, 6d, 7d, 8d, 9d, 10d, 11d, 12d, 13d, 14d, 15d, 16d, 17d,...The sum of the first 5 terms: 3d +4d +5d +6d +7d = (3+4+5+6+7)d = 25d. Correct.Sum of first 15 terms: Let's add them up. The terms up to 17d. Since it's arithmetic, average of first and last term times number of terms. (3d +17d)/2 *15 = 20d/2 *15 =10d*15=150d. Correct. So 150d /25d =6. All checks out.Therefore, the value of S₁₅/S₅ is 6.Final AnswerThe value of frac{S_{15}}{S_{5}} is boxed{6}.</think>Given an arithmetic sequence ({a_n}), where the sum of the first (n) terms is (S_n), and (a_8 = 2a_3), we need to find the value of (frac{S_{15}}{S_{5}}).First, recall the formula for the (n)-th term of an arithmetic sequence:[ a_n = a_1 + (n - 1)d ]Given (a_8 = 2a_3), we substitute the expressions for the terms:[ a_3 = a_1 + 2d ][ a_8 = a_1 + 7d ]According to the problem:[ a_1 + 7d = 2(a_1 + 2d) ]Solving this equation:[ a_1 + 7d = 2a_1 + 4d ][ 3d = a_1 ]Thus, (a_1 = 3d).Next, we calculate the sums (S_{15}) and (S_5) using the sum formula for an arithmetic sequence:[ S_n = frac{n}{2} [2a_1 + (n - 1)d] ]For (S_{15}):[ S_{15} = frac{15}{2} [2a_1 + 14d] ]Substituting (a_1 = 3d):[ S_{15} = frac{15}{2} [2(3d) + 14d] = frac{15}{2} [6d + 14d] = frac{15}{2} cdot 20d = 15 cdot 10d = 150d ]For (S_5):[ S_5 = frac{5}{2} [2a_1 + 4d] ]Substituting (a_1 = 3d):[ S_5 = frac{5}{2} [2(3d) + 4d] = frac{5}{2} [6d + 4d] = frac{5}{2} cdot 10d = 5 cdot 5d = 25d ]Thus, the ratio (frac{S_{15}}{S_5}) is:[ frac{S_{15}}{S_5} = frac{150d}{25d} = frac{150}{25} = 6 ]The value of (frac{S_{15}}{S_{5}}) is (boxed{6}).,<think>Okay, let's see. I have an arithmetic sequence {a_n}, and the sum of the first n terms is S_n. The problem says that a_8 = 2a_3, and we need to find the value of S_{15}/S_5. Hmm. Alright, let's start by recalling some formulas related to arithmetic sequences.First, the general term of an arithmetic sequence. If the first term is a_1 and the common difference is d, then the nth term a_n is given by:a_n = a_1 + (n - 1)dRight. So, for example, a_3 would be a_1 + 2d, and a_8 would be a_1 + 7d. The problem states that a_8 = 2a_3. Let me write that down:a_8 = 2a_3Substituting the expressions for those terms:a_1 + 7d = 2(a_1 + 2d)Let me solve this equation. Expanding the right side:a_1 + 7d = 2a_1 + 4dNow, subtract a_1 and 4d from both sides to get:7d - 4d = 2a_1 - a_1Which simplifies to:3d = a_1So, a_1 = 3d. Alright, that relates the first term and the common difference. So the first term is three times the common difference. Interesting.Now, moving on to the sum of the first n terms. The formula for S_n is:S_n = n/2 [2a_1 + (n - 1)d]Alternatively, sometimes written as S_n = n(a_1 + a_n)/2. Either formula is fine. Let me stick with the first one.So, S_n = (n/2)[2a_1 + (n - 1)d]We need to compute S_{15}/S_5. Let's compute each of these separately.First, S_{15}:S_{15} = (15/2)[2a_1 + 14d]Similarly, S_5:S_5 = (5/2)[2a_1 + 4d]So, the ratio S_{15}/S_5 is:[ (15/2)(2a_1 + 14d) ] / [ (5/2)(2a_1 + 4d) ) ]Simplify this fraction. The 15/2 divided by 5/2 is 15/5 = 3. So, the ratio simplifies to:3*(2a_1 + 14d)/(2a_1 + 4d)Now, we can substitute a_1 = 3d from earlier. Let's do that.First, substitute a_1 = 3d into the numerator:2*(3d) + 14d = 6d + 14d = 20dAnd the denominator:2*(3d) + 4d = 6d + 4d = 10dTherefore, the ratio becomes:3*(20d)/(10d) = 3*(20/10) = 3*2 = 6Wait, that seems straightforward. Let me verify the steps again to make sure I didn't make a mistake.Starting with a_8 = 2a_3:a_1 + 7d = 2(a_1 + 2d)Expanding: a1 +7d = 2a1 +4dSubtract a1 and 4d: 3d = a1. Correct.Then, sum formulas:S_n = n/2 [2a1 + (n-1)d]Thus S15 = 15/2 [2a1 +14d], S5=5/2[2a1 +4d]Ratio: (15/2)/(5/2) = 3. Then numerator is 2a1 +14d, denominator is 2a1 +4d.Substitute a1=3d: numerator is 6d +14d=20d, denominator is 6d +4d=10d. So 20d/10d=2. Multiply by 3 gives 6. So yes, answer is 6.Alternatively, maybe use the other formula for S_n: n/2*(a1 + an). Let me try that.For S_15, it's 15/2*(a1 + a15). Similarly, S5 is 5/2*(a1 + a5).Compute a15 = a1 +14d, and a5 = a1 +4d.So S15/S5 = [15/2*(a1 + a1 +14d)] / [5/2*(a1 + a1 +4d)] = [15*(2a1 +14d)]/[5*(2a1 +4d)] = 3*(2a1 +14d)/(2a1 +4d) which is the same as before. Then substituting a1=3d gives the same result.So, seems like 6 is the answer.Wait, let me check if maybe I made a mistake in the arithmetic. Let's do it step by step.Given a1=3d.Compute S5:First term a1=3d, common difference d.Sum of first 5 terms: 5/2 [2*3d +4d] =5/2 [6d +4d] =5/2 [10d] =5*5d=25dSum of first 15 terms:15/2 [2*3d +14d] =15/2 [6d +14d] =15/2 [20d] =15*10d=150dTherefore, S15/S5 =150d /25d=6. Yep, same result. So definitely 6.Therefore, the answer is 6. So, in boxed form, boxed{6}Final Answerboxed{6}</think>Given an arithmetic sequence ({a_n}) with the sum of the first (n) terms denoted by (S_n), and the condition (a_8 = 2a_3), we need to find the value of (frac{S_{15}}{S_5}).First, recall the general term of an arithmetic sequence:[ a_n = a_1 + (n - 1)d ]Given (a_8 = 2a_3), we substitute the expressions for these terms:[ a_1 + 7d = 2(a_1 + 2d) ]Expanding and simplifying:[ a_1 + 7d = 2a_1 + 4d ][ 3d = a_1 ]Thus, the first term (a_1) is (3d).Next, we use the formula for the sum of the first (n) terms of an arithmetic sequence:[ S_n = frac{n}{2} [2a_1 + (n - 1)d] ]Calculate (S_{15}) and (S_5):[ S_{15} = frac{15}{2} [2a_1 + 14d] ][ S_5 = frac{5}{2} [2a_1 + 4d] ]Substitute (a_1 = 3d) into these expressions:[ S_{15} = frac{15}{2} [2(3d) + 14d] = frac{15}{2} [6d + 14d] = frac{15}{2} cdot 20d = 15 cdot 10d = 150d ][ S_5 = frac{5}{2} [2(3d) + 4d] = frac{5}{2} [6d + 4d] = frac{5}{2} cdot 10d = 5 cdot 5d = 25d ]Finally, compute the ratio:[ frac{S_{15}}{S_5} = frac{150d}{25d} = 6 ]Thus, the value of (frac{S_{15}}{S_5}) is (boxed{6}).]

🔑:Since {a_n} is an arithmetic sequence and a_8 = 2a_3, we have:a_1 + 7d = 2(a_1 + 2d),Therefore, a_1 = 3d.Thus, frac{S_{15}}{S_{5}} = frac{15a_1 + 105d}{5a_1 + 10d} = 6.Hence, the answer is boxed{6}.From a_8 = 2a_3, we deduce a_1 = 3d. Then, by using the formula for the sum of the first n terms of an arithmetic sequence, we can reach the conclusion.This question tests the student's mastery of the general term formula and the sum formula of an arithmetic sequence, and it is considered a medium-level problem.

🧠:[<think>Okay, let's tackle this problem step by step. The function given is y equals the absolute value of sine(2x minus π/6). We need to check which of the statements A, B, C, or D is correct. Let's go through each option one by one.First, let's recall some basics about trigonometric functions and their transformations. The standard sine function, y = sin(x), has a period of 2π, is symmetric about the origin (odd function), and its graph oscillates between -1 and 1. When we introduce transformations like horizontal compression/stretch, phase shifts, and absolute values, these properties change.The given function is y = |sin(2x - π/6)|. Let's break down the transformations here. The argument of the sine function is 2x - π/6, which can be rewritten as 2(x - π/12). This indicates a horizontal compression by a factor of 2 (which affects the period) and a phase shift (horizontal shift) of π/12 to the right. Additionally, taking the absolute value of the sine function will reflect any negative parts of the wave above the x-axis, effectively creating a "wave" that repeats every half period of the original sine function.Starting with option A: The period is π/4. To find the period of the function, let's first consider the period of the inner sine function without the absolute value. The general period of sin(Bx + C) is 2π / |B|. Here, B is 2, so the period of sin(2x - π/6) would be 2π / 2 = π. However, since we take the absolute value of the sine function, the period halves. This is because the absolute value causes the graph to repeat every half period. Normally, sin(x) has a period of 2π, but |sin(x)| has a period of π. Similarly, sin(2x - π/6) has a period of π, so |sin(2x - π/6)| should have a period of π/2. Wait, but hold on, is that correct?Wait, let's double-check. The standard |sin(x)| has a period of π because the absolute value reflects the negative half-wave to positive, making each period half the original. So, if the original function sin(Bx + C) has period T = 2π / B, then the period of |sin(Bx + C)| would be T/2 = π / B. So in our case, B is 2, so the period should be π / 2. Therefore, the period of y = |sin(2x - π/6)| is π/2. But option A says the period is π/4, which is half of π/2. So that's incorrect. So A is wrong.Moving on to option B: One of the axes of symmetry of the function graph is the line x = π/3.To check if x = π/3 is an axis of symmetry, we need to verify if the function is symmetric about that vertical line. A function is symmetric about the line x = a if f(a + h) = f(a - h) for any h. Let's test this with a = π/3.Let’s compute f(π/3 + h) and f(π/3 - h):First, f(x) = |sin(2x - π/6)|Compute f(π/3 + h):= |sin(2*(π/3 + h) - π/6)|= |sin(2π/3 + 2h - π/6)|= |sin(π/2 + 2h)|Similarly, f(π/3 - h):= |sin(2*(π/3 - h) - π/6)|= |sin(2π/3 - 2h - π/6)|= |sin(π/2 - 2h)|Now, sin(π/2 + 2h) = cos(2h) because sin(π/2 + θ) = cosθ. Similarly, sin(π/2 - 2h) = cos(2h) because sin(π/2 - θ) = cosθ. So both expressions inside the absolute value become cos(2h), which is already non-negative because cosine is between -1 and 1, but wait, cos(2h) can be positive or negative depending on h. Wait, but we have absolute value of sin(π/2 + 2h) and sin(π/2 - 2h). Wait, sin(π/2 + 2h) is cos(2h), and sin(π/2 - 2h) is cos(2h). So both are equal to |cos(2h)|. Therefore, f(π/3 + h) = |cos(2h)| and f(π/3 - h) = |cos(2h)|, so they are equal. Therefore, the function is symmetric about x = π/3. Therefore, option B is correct.But let me confirm once again. Alternatively, maybe there's another way. Let's check for h = 0, both sides should be equal. For h = π/6:f(π/3 + π/6) = f(π/2) = |sin(2*(π/2) - π/6)| = |sin(π - π/6)| = |sin(5π/6)| = |1/2| = 1/2f(π/3 - π/6) = f(π/6) = |sin(2*(π/6) - π/6)| = |sin(π/3 - π/6)| = |sin(π/6)| = |1/2| = 1/2. So they are equal here.Another h, say h = π/12:f(π/3 + π/12) = f(5π/12) = |sin(2*(5π/12) - π/6)| = |sin(5π/6 - π/6)| = |sin(4π/6)| = |sin(2π/3)| = |√3/2| = √3/2f(π/3 - π/12) = f(π/4) = |sin(2*(π/4) - π/6)| = |sin(π/2 - π/6)| = |sin(π/3)| = |√3/2| = √3/2. Again equal.So it seems symmetric. Therefore, B is correct.Option C: The function is decreasing on [2π/3, 5π/6].To check if the function is decreasing on that interval, we need to analyze the derivative or look at the behavior of the function. Let's first consider the original function without the absolute value: sin(2x - π/6). The absolute value will affect the increasing/decreasing nature by reflecting parts of the graph.But perhaps it's easier to break down the absolute value. The function y = |sin(2x - π/6)| can be considered as a wave where all negative parts are flipped to positive. The original sine function sin(2x - π/6) has a period of π, so the absolute version will have a period of π/2. Let's find the critical points where sin(2x - π/6) = 0, which occur when 2x - π/6 = kπ for integers k. Solving for x: 2x = kπ + π/6 => x = (kπ)/2 + π/12. These are the points where the absolute value function has corners (i.e., points where it changes direction).So, between these critical points, the function |sin(2x - π/6)| will either be increasing or decreasing depending on the original sine wave's behavior. Let's consider the interval [2π/3, 5π/6]. Let's convert these to decimals for easier understanding: 2π/3 ≈ 2.094, 5π/6 ≈ 2.618. Let's see where this interval falls in terms of critical points.Find critical points near this interval. Let's solve for x when 2x - π/6 = kπ:x = (kπ + π/6)/2 = (k + 1/6)π/2.Let's compute x for k=1: (1 + 1/6)π/2 = (7/6)π/2 = 7π/12 ≈ 1.832 (which is less than 2π/3 ≈ 2.094)k=2: (2 + 1/6)π/2 = (13/6)π/2 = 13π/12 ≈ 3.400 (which is more than 5π/6 ≈ 2.618). Wait, maybe k=1 gives 7π/12 ≈ 1.832, which is less than 2π/3 ≈ 2.094. Then next critical point after 7π/12 would be when k=2: 13π/12 ≈ 3.400, which is beyond our interval. Wait, maybe I made a mistake here.Wait, 2x - π/6 = kπ => 2x = kπ + π/6 => x = (kπ + π/6)/2. Let's compute for k=1: x = (π + π/6)/2 = (7π/6)/2 = 7π/12 ≈ 1.833. For k=2: x = (2π + π/6)/2 = (13π/6)/2 = 13π/12 ≈ 3.40. But our interval is from 2π/3 ≈ 2.094 to 5π/6 ≈ 2.618. So between 2π/3 and 5π/6, is there any critical point?Let's check when k=1: 7π/12 ≈ 1.833, which is before 2π/3 ≈ 2.094. Next critical point is 13π/12 ≈ 3.40, which is after 5π/6 ≈ 2.618. Therefore, in the interval [2π/3, 5π/6], there are no critical points. Therefore, the function |sin(2x - π/6)| is either entirely increasing or decreasing on this interval.To determine if it's increasing or decreasing, let's analyze the derivative. The derivative of |sin(u)| where u = 2x - π/6 is:dy/dx = (d/dx)|sin(u)| = (cos(u)/|sin(u)|) * du/dx when sin(u) ≠ 0. But since we are considering intervals where sin(u) doesn't cross zero (as there are no critical points in [2π/3, 5π/6]), we can determine the sign.Alternatively, let's consider the original sine function sin(2x - π/6) on the interval [2π/3, 5π/6]. Let's compute 2x - π/6 when x is in [2π/3, 5π/6]:For x = 2π/3: 2*(2π/3) - π/6 = 4π/3 - π/6 = 8π/6 - π/6 = 7π/6For x = 5π/6: 2*(5π/6) - π/6 = 10π/6 - π/6 = 9π/6 = 3π/2Therefore, the argument of the sine function goes from 7π/6 to 3π/2. Let's analyze sin(θ) where θ is from 7π/6 to 3π/2. In this interval, sin(θ) is negative because 7π/6 is in the third quadrant and 3π/2 is the negative y-axis. Therefore, the original sine function is negative here, so |sin(θ)| = -sin(θ).Therefore, y = -sin(2x - π/6) on this interval. Let's find the derivative of this:dy/dx = -cos(2x - π/6) * 2So dy/dx = -2cos(2x - π/6)Now, we need to determine the sign of dy/dx on [2π/3, 5π/6]. Let's compute cos(2x - π/6) when θ = 2x - π/6 is between 7π/6 and 3π/2.In the interval θ ∈ [7π/6, 3π/2], cosine is:- From 7π/6 (third quadrant) to 3π/2: cosine is negative in the third quadrant (7π/6 to 4π/3) and becomes zero at 3π/2.Therefore, cos(θ) is negative in [7π/6, 3π/2). Therefore, dy/dx = -2*(negative) = positive. Therefore, the derivative is positive on this interval, meaning the function y is increasing. But the question states that the function is decreasing on [2π/3, 5π/6]. Therefore, option C is incorrect.Wait, but hold on. Wait, the original sine function is negative here, so y = |sin(θ)| = -sin(θ). Therefore, derivative is -cos(theta)*d(theta)/dx. Wait, but theta = 2x - π/6, so d(theta)/dx = 2. Therefore, dy/dx = -cos(theta)*2. Since theta is between 7π/6 and 3π/2, cos(theta) is negative (third and fourth quadrants, but from 7π/6 to 3π/2, it's third and fourth? Wait, 7π/6 is 210 degrees, 3π/2 is 270 degrees. So theta moves from 210 to 270 degrees. In this interval, cosine is negative (third quadrant: 180-270, but 210 to 270 is from third to fourth? Wait, 270 is the negative y-axis. So from 7π/6 (210 degrees) to 3π/2 (270 degrees), cosine is negative in 7π/6 to 4π/3 (240 degrees) and zero at 3π/2. Wait, actually, 7π/6 is 210 degrees, 4π/3 is 240 degrees, 3π/2 is 270 degrees. So in 210 to 270 degrees, cosine is negative from 210 to 270 except at 270 where it's zero. Therefore, cos(theta) is negative throughout [7π/6, 3π/2). Therefore, dy/dx = -2 * (negative) = positive. So the derivative is positive, meaning the function is increasing. Therefore, option C says the function is decreasing, which is incorrect. So C is wrong.Option D: The function is an even function.An even function satisfies f(x) = f(-x) for all x. Let's check if |sin(2x - π/6)| = |sin(-2x - π/6)|.Compute f(-x):|sin(2*(-x) - π/6)| = |sin(-2x - π/6)| = |sin(- (2x + π/6))| = | - sin(2x + π/6) | = |sin(2x + π/6)|Compare this to f(x) = |sin(2x - π/6)|For f(x) to be even, we need |sin(2x - π/6)| = |sin(2x + π/6)| for all x. Let's test x = 0:f(0) = |sin(-π/6)| = | -1/2 | = 1/2f(-0) = f(0) = 1/2. Wait, but x=0 is symmetric. Let's try another x, say x = π/12:f(π/12) = |sin(2*(π/12) - π/6)| = |sin(π/6 - π/6)| = |sin(0)| = 0f(-π/12) = |sin(2*(-π/12) - π/6)| = |sin(-π/6 - π/6)| = |sin(-π/3)| = | -√3/2 | = √3/2. So f(-π/12) = √3/2 ≠ f(π/12) = 0. Therefore, f(-x) ≠ f(x) here. Hence, the function is not even. Therefore, option D is incorrect.So summarizing:A: Incorrect (period is π/2)B: CorrectC: IncorrectD: IncorrectTherefore, the correct answer is B.Final Answerboxed{B}</think>Given the function ( y = |sin(2x - frac{pi}{6})| ), we need to determine which of the statements A, B, C, or D is correct.Option A: The period is (frac{pi}{4}).The period of ( sin(2x - frac{pi}{6}) ) is ( frac{2pi}{2} = pi ). Taking the absolute value of a sine function halves its period, so the period becomes ( frac{pi}{2} ). Therefore, statement A is incorrect.Option B: One of the axes of symmetry of the function graph is the line ( x = frac{pi}{3} ).To check symmetry about ( x = frac{pi}{3} ), we verify if ( f(frac{pi}{3} + h) = f(frac{pi}{3} - h) ). Calculating both sides:- ( f(frac{pi}{3} + h) = |sin(2(frac{pi}{3} + h) - frac{pi}{6})| = |sin(frac{pi}{2} + 2h)| = |cos(2h)| )- ( f(frac{pi}{3} - h) = |sin(2(frac{pi}{3} - h) - frac{pi}{6})| = |sin(frac{pi}{2} - 2h)| = |cos(2h)| )Both sides are equal, confirming symmetry. Therefore, statement B is correct.Option C: The function is decreasing on ([frac{2pi}{3}, frac{5pi}{6}]).Analyzing the interval ([frac{2pi}{3}, frac{5pi}{6}]):- The argument ( 2x - frac{pi}{6} ) ranges from ( frac{7pi}{6} ) to ( frac{3pi}{2} ), where ( sin(theta) ) is negative. Thus, ( y = -sin(2x - frac{pi}{6}) ).- The derivative ( frac{dy}{dx} = -2cos(2x - frac{pi}{6}) ). Since ( cos(theta) ) is negative in this interval, the derivative is positive, indicating the function is increasing. Therefore, statement C is incorrect.Option D: The function is an even function.An even function satisfies ( f(x) = f(-x) ). Testing ( x = frac{pi}{12} ):- ( f(frac{pi}{12}) = |sin(0)| = 0 )- ( f(-frac{pi}{12}) = |sin(-frac{pi}{3})| = frac{sqrt{3}}{2} )Since ( f(x) neq f(-x) ), the function is not even. Therefore, statement D is incorrect.The correct answer is (boxed{B}).,<think>Okay, let me try to figure out this problem. The function given is y equals the absolute value of sine(2x minus π/6). We need to check which of the statements A, B, C, or D is correct. Let me go through each option one by one, but first, maybe I should recall some properties of trigonometric functions, especially when they're inside an absolute value.Starting with option A: The period is π/4. Hmm, normally the period of sine function is 2π. If there's a coefficient inside the sine function, like sin(Bx), the period becomes 2π/B. Here, the coefficient of x is 2, so without the absolute value, the period would be 2π/2 = π. But since there's an absolute value, the period might be halved because the absolute value makes the negative parts of the sine wave positive. For example, the regular sine function has a period of 2π, but |sin x| has a period of π. So applying that here, the original period was π, so with absolute value, would the period become π/2? Wait, but maybe not. Wait, let me think.The absolute value reflects all the negative parts of the sine wave to positive, effectively making the graph repeat every half period. So if the original period was π, then the period of |sin(2x - π/6)| would be π/2. Because after half the original period, the pattern repeats due to the absolute value. So if the original function had period π, then |sin(2x - π/6)| would have period π/2. But option A says the period is π/4. That would mean the period is half of π/2, which isn't correct. So maybe A is incorrect. Let me verify that.Wait, let's take an example. Let's consider the standard |sin(x)|. The regular sin(x) has period 2π, and |sin(x)| has period π. So the period is halved. So applying that logic here: the original function sin(2x - π/6) has period 2π/2 = π. So its absolute value should have period π/2. Therefore, the period should be π/2, not π/4. So option A is incorrect. So A is wrong.Moving on to option B: One of the axes of symmetry of the function graph is the line x = π/3. Hmm, symmetry axes. Since this is an absolute value of a sine function, which is a periodic function with certain symmetries. The original sine function sin(2x - π/6) has phase shift. Let's write it as sin(2(x - π/12)). So the phase shift is π/12 to the right. But when taking the absolute value, the graph will be reflected above the x-axis, so the symmetry axes might be related to the points where the original sine function crosses the x-axis or reaches maxima/minima.An axis of symmetry would be a vertical line x = a such that the graph is symmetric around that line. For the absolute value of a sine function, the graph will have symmetry axes at the midpoint between two consecutive points where the original sine function is zero. Let's find the zeros of sin(2x - π/6). The function inside the absolute value, sin(2x - π/6), is zero when 2x - π/6 = kπ, where k is an integer. Solving for x: 2x = kπ + π/6 ⇒ x = (kπ + π/6)/2 = kπ/2 + π/12. So zeros occur at x = π/12, π/2 + π/12 = 7π/12, π + π/12 = 13π/12, etc. The midpoints between these zeros would be the points halfway between π/12 and 7π/12, which is (π/12 + 7π/12)/2 = 8π/24 = π/3. So the midpoint is π/3. Therefore, the line x = π/3 is indeed an axis of symmetry for the graph of |sin(2x - π/6)|. So option B is correct.Wait, but let me check again. Let's take two zeros at x = π/12 and 7π/12. The midpoint is (π/12 + 7π/12)/2 = 8π/12 / 2 = (2π/3)/2 = π/3. Yes, so x = π/3 is the midpoint. Therefore, the graph should be symmetric around x = π/3. Let me visualize it. If you take the original sine wave between π/12 and 7π/12, the absolute value would make it a "hill" from π/12 to 7π/12, peaking in the middle. The middle is at π/3. So the axis of symmetry is at π/3. Hence, B is correct. So option B is correct. So maybe B is a correct statement.But let's hold on and check the other options as well.Option C: The function is decreasing on [2π/3, 5π/6]. Let's see. To check if the function is decreasing on that interval, we can look at the derivative. However, since the function is absolute value, we have to be careful with the derivative. Alternatively, we can analyze the behavior without calculus.First, let's consider the expression y = |sin(2x - π/6)|. Let's analyze the interval [2π/3, 5π/6]. Let's convert these x-values into the argument of the sine function. Let's compute 2x - π/6 when x is in [2π/3, 5π/6].First, when x = 2π/3: 2*(2π/3) - π/6 = 4π/3 - π/6 = 8π/6 - π/6 = 7π/6.When x = 5π/6: 2*(5π/6) - π/6 = 10π/6 - π/6 = 9π/6 = 3π/2.So the argument of the sine function goes from 7π/6 to 3π/2. Let's recall that sin(θ) is negative in this interval (from 7π/6 to 3π/2), so |sin(θ)| = -sin(θ) in this interval. Therefore, y = -sin(2x - π/6) in this interval. Therefore, the function y = -sin(2x - π/6) on [2π/3, 5π/6].To check if it's decreasing, we can look at the derivative. The derivative of y with respect to x is -cos(2x - π/6)*2. So dy/dx = -2cos(2x - π/6). We need to check the sign of this derivative on [7π/6, 3π/2] for θ = 2x - π/6.Wait, θ goes from 7π/6 to 3π/2. Let's see what cos(θ) is in this interval. The interval θ from 7π/6 to 3π/2 is from 210 degrees to 270 degrees. Cosine is negative in the third quadrant (180 to 270 degrees) but becomes zero at 270 degrees. So cos(θ) is negative in [7π/6, 3π/2). Therefore, dy/dx = -2cos(θ) = -2*(negative) = positive. So the derivative is positive in this interval, meaning the function is increasing. Therefore, the function is increasing on [2π/3, 5π/6], which contradicts option C which says it's decreasing. Therefore, option C is incorrect.Alternatively, without calculus, let's think about the behavior of |sin(θ)| where θ is from 7π/6 to 3π/2. In this interval, sin(θ) goes from sin(7π/6) = -1/2 to sin(3π/2) = -1. So |sin(θ)| goes from 1/2 to 1. But wait, sin(7π/6) is -1/2, so |sin(7π/6)| = 1/2. Then as θ increases from 7π/6 to 3π/2, sin(θ) goes from -1/2 to -1, so |sin(θ)| goes from 1/2 up to 1. So the absolute value is increasing. Therefore, the function y is increasing on that interval. So option C is wrong.Therefore, C is incorrect.Option D: The function is an even function. An even function satisfies y(x) = y(-x) for all x. Let's check if |sin(2x - π/6)| = |sin(-2x - π/6)|.Compute y(-x): |sin(2*(-x) - π/6)| = |sin(-2x - π/6)| = |sin(- (2x + π/6))| = | - sin(2x + π/6) | = |sin(2x + π/6)|.Compare this with y(x) = |sin(2x - π/6)|. For the function to be even, we need |sin(2x - π/6)| = |sin(2x + π/6)| for all x. Let's check for a specific x value. Let's take x = 0. Then y(0) = |sin(-π/6)| = | -1/2 | = 1/2. y(-0) is the same as y(0), but let's check y(-x) when x is not zero. Wait, even function requires y(x) = y(-x). Let me take x = π/6. Then y(π/6) = |sin(2*(π/6) - π/6)| = |sin(π/3 - π/6)| = |sin(π/6)| = 1/2. Now y(-π/6) = |sin(2*(-π/6) - π/6)| = |sin(-π/3 - π/6)| = |sin(-π/2)| = | -1 | = 1. So y(π/6) = 1/2 and y(-π/6) = 1. These are not equal, so the function is not even. Therefore, option D is incorrect.Therefore, the correct statement is B.Final Answerboxed{B}</think>Given the function ( y = |sin(2x - frac{pi}{6})| ), we need to determine which of the statements A, B, C, or D is correct.Option A: The period is (frac{pi}{4}).The original function (sin(2x - frac{pi}{6})) has a period of (pi) because the coefficient of (x) is 2. When taking the absolute value, the period is halved, resulting in a period of (frac{pi}{2}). Therefore, the period is not (frac{pi}{4}), so option A is incorrect.Option B: One of the axes of symmetry of the function graph is the line (x = frac{pi}{3}).The zeros of (sin(2x - frac{pi}{6})) occur at (x = frac{pi}{12} + frac{kpi}{2}) for integer (k). The midpoint between consecutive zeros (x = frac{pi}{12}) and (x = frac{7pi}{12}) is (frac{pi}{3}). This midpoint is an axis of symmetry for the graph of (|sin(2x - frac{pi}{6})|). Therefore, option B is correct.Option C: The function is decreasing on ([frac{2pi}{3}, frac{5pi}{6}]).Analyzing the interval ([frac{2pi}{3}, frac{5pi}{6}]), the argument (2x - frac{pi}{6}) ranges from (frac{7pi}{6}) to (frac{3pi}{2}). In this interval, (sin(theta)) is negative, so (|sin(theta)| = -sin(theta)). The function (-sin(theta)) is increasing in this interval because (sin(theta)) becomes more negative. Therefore, the function is increasing, not decreasing, so option C is incorrect.Option D: The function is an even function.An even function satisfies (y(x) = y(-x)). Checking (y(0)) and (y(frac{pi}{6})) versus (y(-frac{pi}{6})) shows discrepancies, indicating the function is not even. Therefore, option D is incorrect.The correct answer is (boxed{B}).]

🔑:The function y=|sin (2x- frac {pi}{6})| has a period of frac {pi}{2} since the initial phase is not 0, so option A is incorrect.Substituting x= frac {pi}{3} into the function's expression, the function reaches its maximum value of 1, so B is correct.The function increases and decreases on the interval [frac {2pi}{3}, frac {5pi}{6}], so C is incorrect.When x=0, the function does not reach its extremum, which clearly indicates that it is not an even function, so D is incorrect.Therefore, the correct answer is boxed{B}.To determine the correctness of option A, calculate the period of the function. To judge option B, substitute x= frac {pi}{3} into the function's expression and check if it is an extremum. To assess option C, evaluate the monotonicity of the function on the interval [frac {2pi}{3}, frac {5pi}{6}]. To verify option D, directly judge the evenness or oddness of the function. This question is a basic one, focusing on the application of the fundamental properties of trigonometric functions. Mastering the basic knowledge of trigonometric functions is key to solving the problem, which tests computational ability.

🧠:[<think>Okay, so I need to find the fourth derivative of the function y = (5x - 8) * 2^(-x). Hmm, derivatives can get complicated, especially higher-order ones. Let me start by recalling some differentiation rules. Since this function is a product of two functions, I'll probably need to use the product rule multiple times. The product rule states that the derivative of u*v is u’v + uv’. First, let me identify the two functions involved here. Let’s let u = 5x - 8 and v = 2^(-x). So, y = u * v. To find the first derivative, y', I need to find u’ and v’ and apply the product rule. Let’s compute u’ first. The derivative of 5x - 8 with respect to x is just 5. That’s straightforward. Now, for v = 2^(-x), I need to find its derivative. The derivative of a^t with respect to t is a^t * ln(a), right? So here, the base is 2, and the exponent is -x. So, the derivative of 2^(-x) with respect to x should be 2^(-x) * ln(2) * (-1), because of the chain rule. The derivative of -x is -1. So, v’ = -ln(2) * 2^(-x). Therefore, the first derivative y’ is u’v + uv’:y’ = 5 * 2^(-x) + (5x - 8) * (-ln(2) * 2^(-x))Simplify that a bit: y’ = 5*2^(-x) - (5x - 8)*ln(2)*2^(-x). Maybe factor out 2^(-x):y’ = 2^(-x) [5 - (5x - 8)ln(2)]Alright, that’s the first derivative. Now, I need to find the second, third, and fourth derivatives. Since each derivative will involve a product of functions (probably similar terms each time), this might get repetitive, but let me take it step by step.Let’s denote the first derivative as y’ = 2^(-x) [5 - (5x - 8)ln(2)]. To find y’’, I need to differentiate y’. Let me rewrite y’ as:y’ = 5*2^(-x) - (5x - 8)*ln(2)*2^(-x)But maybe it's easier to treat the entire expression as a single product. Let’s consider the general form: y’ = A*2^(-x) + B*(5x - 8)*2^(-x). Wait, actually, no. Wait, in the first derivative, after factoring, it's 2^(-x) multiplied by [5 - (5x -8)ln(2)]. So, perhaps write it as:y’ = [5 - (5x -8)ln(2)] * 2^(-x)So, this is again a product of two functions: let’s call the first part, [5 - (5x -8)ln(2)], as a new function, say, f(x) = 5 - (5x -8)ln(2), and the second part is still 2^(-x). Therefore, to differentiate y’, we can use the product rule again.First, find f’(x). The derivative of 5 is 0. The derivative of -(5x -8)ln(2) is -5*ln(2). So, f’(x) = -5 ln(2). And the derivative of 2^(-x) is still -ln(2)*2^(-x). Therefore, applying the product rule to f(x)*2^(-x):y'' = f’(x)*2^(-x) + f(x)*(-ln(2))*2^(-x)Plugging in f’(x) and f(x):y'' = [-5 ln(2)] * 2^(-x) + [5 - (5x -8)ln(2)] * (-ln(2)) * 2^(-x)Factor out 2^(-x) and -ln(2):Wait, let me compute term by term:First term: -5 ln(2) * 2^(-x)Second term: -ln(2) * [5 - (5x -8)ln(2)] * 2^(-x)So combining both terms:y'' = -5 ln(2) * 2^(-x) -5 ln(2) * 2^(-x) + (5x -8)(ln(2))^2 * 2^(-x)Wait, hold on, let me expand the second term:Second term is -ln(2) * 5 * 2^(-x) + ln(2) * (5x -8)(ln(2)) * 2^(-x)So, total:First term: -5 ln(2) * 2^(-x)Second term: -5 ln(2) * 2^(-x) + (5x -8)(ln(2))^2 * 2^(-x)Therefore, combining like terms:-5 ln(2) -5 ln(2) = -10 ln(2), so:y'' = [-10 ln(2) + (5x -8)(ln(2))^2] * 2^(-x)Hmm, that seems a bit complex. Let me check my steps again to make sure I didn't make a mistake.Original y’ = [5 - (5x -8)ln(2)] * 2^(-x)Differentiating:First, derivative of [5 - (5x -8)ln(2)] is derivative of 5 is 0, derivative of -(5x -8)ln(2) is -5 ln(2). So f’(x) = -5 ln(2). Then, derivative of 2^(-x) is -ln(2) 2^(-x). Then product rule gives:y'' = (-5 ln(2)) * 2^(-x) + [5 - (5x -8)ln(2)] * (-ln(2)) * 2^(-x)So, combining terms:= (-5 ln(2) -5 ln(2) + (5x -8)(ln(2))^2) * 2^(-x)Wait, how?Wait, the second term is [5 - (5x -8)ln(2)] multiplied by -ln(2). Let me expand that:= -5 ln(2)*2^(-x) + (5x -8)(ln(2))^2 * 2^(-x) -5 ln(2)*2^(-x)Wait, no, actually:Wait, let me re-express:y'' = (-5 ln(2)) * 2^(-x) + [5 - (5x -8)ln(2)] * (-ln(2)) * 2^(-x)= [ -5 ln(2) -5 ln(2) + (5x -8)(ln(2))^2 ] * 2^(-x)Wait, no. Let me factor out -ln(2) from the second term:= -5 ln(2) * 2^(-x) -ln(2) * [5 - (5x -8)ln(2)] * 2^(-x)= -5 ln(2) * 2^(-x) -5 ln(2) * 2^(-x) + (5x -8)(ln(2))^2 * 2^(-x)Ah, yes, that's correct. So combining the first two terms:-5 ln(2) -5 ln(2) = -10 ln(2). Therefore:y'' = [ -10 ln(2) + (5x -8)(ln(2))^2 ] * 2^(-x)Okay, so the second derivative is that. Now, moving on to the third derivative, y''':We need to differentiate y''. Let me write y'' as:y'' = [ (5x -8)(ln(2))^2 -10 ln(2) ] * 2^(-x)So again, this is a product of two functions: the polynomial part (let's call it g(x) = (5x -8)(ln(2))^2 -10 ln(2)) and 2^(-x). Applying the product rule again.First, find g’(x). The derivative of (5x -8)(ln(2))^2 is 5*(ln(2))^2. The derivative of -10 ln(2) is 0. So, g’(x) = 5*(ln(2))^2.Then, derivative of 2^(-x) is -ln(2)*2^(-x). So, applying product rule:y''' = g’(x)*2^(-x) + g(x)*(-ln(2))*2^(-x)Plugging in:y''' = 5*(ln(2))^2 *2^(-x) + [ (5x -8)(ln(2))^2 -10 ln(2) ] * (-ln(2)) *2^(-x)Let me expand that:First term: 5*(ln(2))^2 *2^(-x)Second term: - (5x -8)(ln(2))^3 *2^(-x) +10 (ln(2))^2 *2^(-x)Combine terms:5*(ln(2))^2 *2^(-x) +10 (ln(2))^2 *2^(-x) - (5x -8)(ln(2))^3 *2^(-x)Adding the first two terms: 15*(ln(2))^2 *2^(-x)So,y''' = [15*(ln(2))^2 - (5x -8)(ln(2))^3 ] *2^(-x)Alright, third derivative done. Now, the fourth derivative y'''':Let me write y''' as:y''' = [ - (5x -8)(ln(2))^3 +15*(ln(2))^2 ] *2^(-x)Again, product of two functions: h(x) = - (5x -8)(ln(2))^3 +15*(ln(2))^2 and 2^(-x). Differentiate using product rule.First, find h’(x):Derivative of - (5x -8)(ln(2))^3 is -5*(ln(2))^3.Derivative of 15*(ln(2))^2 is 0. So, h’(x) = -5*(ln(2))^3.Derivative of 2^(-x) is -ln(2)*2^(-x). Therefore,y'''' = h’(x)*2^(-x) + h(x)*(-ln(2))*2^(-x)Substitute h’(x) and h(x):y'''' = -5*(ln(2))^3 *2^(-x) + [ - (5x -8)(ln(2))^3 +15*(ln(2))^2 ]*(-ln(2)) *2^(-x)Let me expand the second term:= -5*(ln(2))^3 *2^(-x) + [ (5x -8)(ln(2))^4 -15*(ln(2))^3 ] *2^(-x)Combine like terms:The first term is -5*(ln(2))^3 *2^(-x), the second term has -15*(ln(2))^3 *2^(-x) and (5x -8)(ln(2))^4 *2^(-x).So combining the constants:-5*(ln(2))^3 -15*(ln(2))^3 = -20*(ln(2))^3Therefore:y'''' = [ -20*(ln(2))^3 + (5x -8)(ln(2))^4 ] *2^(-x)We can factor out (ln(2))^3 *2^(-x):y'''' = (ln(2))^3 *2^(-x) [ -20 + (5x -8)ln(2) ]Alternatively, writing it in another form:y'''' = [ (5x -8)ln(2) -20 ] * (ln(2))^3 *2^(-x )Or, to make it look cleaner:y^{(IV)} = (5x -8)(ln 2)^4 cdot 2^{-x} - 20(ln 2)^3 cdot 2^{-x}We can factor 2^{-x} and (ln 2)^3:= 2^{-x} (ln 2)^3 [ (5x -8)(ln 2) -20 ]Alternatively, expand the terms inside:= 2^{-x} (ln 2)^3 [5x ln 2 -8 ln2 -20]So, depending on how the answer is expected, maybe either form is acceptable, but perhaps the factored form is preferable.Let me check my steps again to make sure I didn't make any errors.Starting from y'''':After differentiating y''', which was [15*(ln(2))^2 - (5x -8)(ln(2))^3 ] *2^(-x)Differentiate h(x) = 15*(ln(2))^2 - (5x -8)(ln(2))^3h’(x) = 0 -5*(ln(2))^3 = -5*(ln(2))^3Multiply by 2^(-x): -5*(ln(2))^3 *2^(-x)Then h(x)*(-ln(2))*2^(-x) = [15*(ln(2))^2 - (5x -8)(ln(2))^3 ]*(-ln(2))*2^(-x)= -15*(ln(2))^3 *2^(-x) + (5x -8)*(ln(2))^4 *2^(-x)So, total y'''' = (-5*(ln(2))^3 -15*(ln(2))^3 + (5x -8)*(ln(2))^4 ) *2^(-x)Which gives -20*(ln(2))^3 + (5x -8)*(ln(2))^4 multiplied by 2^(-x). Yes, that's correct.Therefore, the fourth derivative is:y^{(IV)} = [ (5x -8)(ln 2)^4 -20(ln 2)^3 ] 2^{-x}Alternatively, factoring out (ln 2)^3:= (ln 2)^3 [ (5x -8) ln 2 -20 ] 2^{-x}Which is the same as:= 2^{-x} (ln 2)^3 (5x ln 2 -8 ln2 -20 )So, to present this neatly, perhaps writing all the constants in terms of ln2. Let me check if the problem expects the answer factored or expanded. The problem says "Find the derivative of the specified order," and usually, for such problems, both forms are acceptable unless specified. But maybe they prefer it factored with 2^{-x} and the polynomial in x.Alternatively, maybe we can write it as:y^{(IV)} = 2^{-x} ( ln 2 )^3 [5 x ln 2 - 8 ln 2 - 20 ]Alternatively, factor 5x ln2 -8 ln2 -20 can be written as 5x ln2 - (8 ln2 +20). But I don't think it can be factored further. So probably the most compact form is:y^{(IV)} = 2^{-x} ( ln 2 )^3 [5 x ln 2 - 8 ln 2 - 20 ]Alternatively, factor out a common term? Let's see:5x ln2 -8 ln2 -20 = ln2(5x -8) -20. So that's how it's written in the previous step. So perhaps:y^{(IV)} = 2^{-x} ( ln 2 )^3 [ (5x -8) ln 2 -20 ]That seems concise. So I think that's the answer.Let me verify by computing each derivative step by step once again quickly:Original function: y = (5x -8)2^{-x}First derivative:y’ = 5*2^{-x} + (5x -8)(-ln2)2^{-x} = 2^{-x}[5 - (5x -8)ln2]Second derivative:y'' = [-5 ln2 - (5x -8)(ln2)^2 -5 ln2 ]2^{-x}? Wait, no. Wait, no. Wait, first derivative is [5 - (5x -8)ln2]2^{-x}Differentiate:Derivative of [5 - (5x -8)ln2] is -5 ln2Derivative of 2^{-x} is -ln2 2^{-x}So y'' = (-5 ln2)2^{-x} + [5 - (5x -8)ln2](-ln2)2^{-x}= (-5 ln2 -5 ln2 + (5x -8)(ln2)^2 )2^{-x}Wait, expanding the second term:[5*(-ln2) + (5x -8)(ln2)^2]2^{-x}Wait, no, the second term is [5 - (5x -8)ln2] * (-ln2) = -5 ln2 + (5x -8)(ln2)^2So total:y'' = (-5 ln2)2^{-x} -5 ln2 2^{-x} + (5x -8)(ln2)^2 2^{-x}Which combines to:(-10 ln2 + (5x -8)(ln2)^2)2^{-x}Yes, that's correct.Third derivative:Differentiate y'':Terms: -10 ln2 *2^{-x} and (5x -8)(ln2)^2 *2^{-x}Derivative of first term: -10 ln2*(-ln2)2^{-x} = 10 (ln2)^2 2^{-x}Derivative of second term: 5*(ln2)^2 *2^{-x} + (5x -8)(ln2)^2*(-ln2)2^{-x}So total:10 (ln2)^2 2^{-x} +5 (ln2)^2 2^{-x} - (5x -8)(ln2)^3 2^{-x}= 15 (ln2)^2 2^{-x} - (5x -8)(ln2)^3 2^{-x}Thus, y''' = [15 (ln2)^2 - (5x -8)(ln2)^3 ]2^{-x}Fourth derivative:Differentiate y''':Terms: 15 (ln2)^2 *2^{-x} and - (5x -8)(ln2)^3 *2^{-x}Derivative of first term: 15 (ln2)^2*(-ln2)2^{-x} = -15 (ln2)^3 2^{-x}Derivative of second term: -5 (ln2)^3 *2^{-x} + (5x -8)(ln2)^4 *2^{-x}So total:-15 (ln2)^3 2^{-x} -5 (ln2)^3 2^{-x} + (5x -8)(ln2)^4 2^{-x}= -20 (ln2)^3 2^{-x} + (5x -8)(ln2)^4 2^{-x}Which factors to:[ (5x -8)(ln2)^4 -20 (ln2)^3 ]2^{-x} = (ln2)^3 [ (5x -8)ln2 -20 ]2^{-x}Yes, that's consistent with what I had before. So, I think that's correct.Therefore, the fourth derivative is y^{(IV)} = [ (5x -8)(ln2)^4 -20(ln2)^3 ]2^{-x}, which can be written as 2^{-x}(ln2)^3[5x ln2 -8 ln2 -20]. Alternatively, if we want to write it as a combination of terms:= 5x (ln2)^4 2^{-x} -8 (ln2)^4 2^{-x} -20 (ln2)^3 2^{-x}But unless the problem specifies a particular form, the factored version is probably better.So, the final answer is:boxed{2^{-x} (ln 2)^3 left(5 x (ln 2) - 8 ln 2 - 20right)}Or, alternatively:boxed{(5 x (ln 2)^4 - 8 (ln 2)^4 - 20 (ln 2)^3) cdot 2^{-x}}But the first boxed expression seems cleaner.Final Answerboxed{2^{-x} (ln 2)^3 left(5 x ln 2 - 8 ln 2 - 20right)}</think>To find the fourth derivative of the function ( y = (5x - 8) cdot 2^{-x} ), we proceed step by step using the product rule and chain rule.1. First Derivative: [ y' = frac{d}{dx}[(5x - 8) cdot 2^{-x}] = 5 cdot 2^{-x} + (5x - 8) cdot (-ln 2) cdot 2^{-x} ] Simplifying: [ y' = 2^{-x} [5 - (5x - 8)ln 2] ]2. Second Derivative: [ y'' = frac{d}{dx}[2^{-x} (5 - (5x - 8)ln 2)] ] Applying the product rule: [ y'' = -5 ln 2 cdot 2^{-x} + (5 - (5x - 8)ln 2) cdot (-ln 2) cdot 2^{-x} ] Simplifying: [ y'' = 2^{-x} [-10 ln 2 + (5x - 8)(ln 2)^2] ]3. Third Derivative: [ y''' = frac{d}{dx}[2^{-x} (-10 ln 2 + (5x - 8)(ln 2)^2)] ] Applying the product rule: [ y''' = 5 (ln 2)^2 cdot 2^{-x} + (-10 ln 2 + (5x - 8)(ln 2)^2) cdot (-ln 2) cdot 2^{-x} ] Simplifying: [ y''' = 2^{-x} [15 (ln 2)^2 - (5x - 8)(ln 2)^3] ]4. Fourth Derivative: [ y'''' = frac{d}{dx}[2^{-x} (15 (ln 2)^2 - (5x - 8)(ln 2)^3)] ] Applying the product rule: [ y'''' = -5 (ln 2)^3 cdot 2^{-x} + (15 (ln 2)^2 - (5x - 8)(ln 2)^3) cdot (-ln 2) cdot 2^{-x} ] Simplifying: [ y'''' = 2^{-x} [ -20 (ln 2)^3 + (5x - 8)(ln 2)^4 ] ] Factoring out ((ln 2)^3): [ y'''' = 2^{-x} (ln 2)^3 [5x ln 2 - 8 ln 2 - 20] ]Thus, the fourth derivative is:[boxed{2^{-x} (ln 2)^3 left(5 x ln 2 - 8 ln 2 - 20right)}],<think>Okay, so I need to find the fourth derivative of the function y = (5x - 8) * 2^(-x). Hmm, derivatives of products can be tricky. I remember that the product rule is necessary here, right? The product rule says that the derivative of u*v is u’v + uv’. But since we need the fourth derivative, this might get complicated because we'll have to apply the product rule multiple times. Maybe there's a pattern that emerges after each differentiation? Let me try to work it out step by step.First, let's define u = 5x - 8 and v = 2^(-x). The first derivative y' would be u’v + uv’. Let's compute u’ first. The derivative of 5x -8 is just 5. So u’ = 5. Now, what's the derivative of v = 2^(-x)? The derivative of a^t with respect to t is a^t ln(a), but here we have a negative exponent. So dv/dx = -2^(-x) ln(2). Therefore, applying the product rule:y' = 5 * 2^(-x) + (5x -8)*(-2^(-x) ln(2))Hmm, maybe factor out 2^(-x) to simplify?y' = 2^(-x) [5 - (5x -8) ln(2)]That's a bit cleaner. Now, let's move on to the second derivative y''. Let's differentiate y' again. So y' is 2^(-x) times [5 - (5x -8) ln(2)]. Let me denote the second part as a new function, maybe call it w = 5 - (5x -8) ln(2). Then y' = 2^(-x) * w. Applying the product rule again, the derivative of y' will be derivative of 2^(-x) * w + 2^(-x) * derivative of w.First, derivative of 2^(-x) is -2^(-x) ln(2), as before. Derivative of w is derivative of 5 which is 0, minus derivative of (5x -8) ln(2). The derivative of (5x -8) is 5, so derivative of w is -5 ln(2). Therefore, putting it all together:y'' = -2^(-x) ln(2) * [5 - (5x -8) ln(2)] + 2^(-x) * (-5 ln(2))Let me factor out -2^(-x) ln(2) from both terms:y'' = -2^(-x) ln(2) [5 - (5x -8) ln(2) + 5]Wait, hold on. Let me check that again. The first term is -2^(-x) ln(2) multiplied by [5 - (5x -8) ln(2)], and the second term is 2^(-x) multiplied by (-5 ln(2)). So if I factor out -2^(-x) ln(2), the first term is just [5 - (5x -8) ln(2)], and the second term is (-5 ln(2)) divided by (- ln(2)) which is 5. Wait, maybe another approach.Alternatively, let's expand the terms first:y'' = -2^(-x) ln(2) * 5 + 2^(-x) ln(2) * (5x -8) ln(2) -5 ln(2) * 2^(-x)Combine like terms:The first term is -5 ln(2) * 2^(-x), the third term is also -5 ln(2) * 2^(-x), so combined they are -10 ln(2) * 2^(-x). The middle term is 2^(-x) * (ln(2))^2 * (5x -8). So:y'' = -10 ln(2) * 2^(-x) + (5x -8)(ln(2))^2 * 2^(-x)Factor out 2^(-x):y'' = 2^(-x) [ -10 ln(2) + (5x -8)(ln(2))^2 ]Alternatively, factor out (ln(2))^2 maybe? Not sure if that helps. Let's see:Wait, the coefficient of (5x -8) is (ln(2))^2, and the constant term is -10 ln(2). Maybe not. Let's leave it as is for now.Moving on to the third derivative y'''. Differentiate y''. So y'' is 2^(-x) times [ -10 ln(2) + (5x -8)(ln(2))^2 ]. Let's denote the bracketed term as A = -10 ln(2) + (5x -8)(ln(2))^2. Then y'' = 2^(-x) * A. The derivative of y'' will be derivative of 2^(-x) * A + 2^(-x) * derivative of A.Derivative of 2^(-x) is -2^(-x) ln(2). Derivative of A is derivative of -10 ln(2) which is 0, plus derivative of (5x -8)(ln(2))^2. The derivative of (5x -8) is 5, so derivative of A is 5 (ln(2))^2. Therefore:y''' = -2^(-x) ln(2) * [ -10 ln(2) + (5x -8)(ln(2))^2 ] + 2^(-x) * 5 (ln(2))^2Let's expand this:First term: -2^(-x) ln(2) * (-10 ln(2)) = 10 (ln(2))^2 * 2^(-x)First term part two: -2^(-x) ln(2) * (5x -8)(ln(2))^2 = - (5x -8)(ln(2))^3 * 2^(-x)Second term: 5 (ln(2))^2 * 2^(-x)Combine the terms:10 (ln(2))^2 * 2^(-x) + 5 (ln(2))^2 * 2^(-x) - (5x -8)(ln(2))^3 * 2^(-x)Combine the first two terms:15 (ln(2))^2 * 2^(-x) - (5x -8)(ln(2))^3 * 2^(-x)Factor out 2^(-x):y''' = 2^(-x) [15 (ln(2))^2 - (5x -8)(ln(2))^3 ]Hmm, I see a pattern here. Let me check the first few derivatives to see if there's a trend.First derivative:y' = 2^(-x) [5 - (5x -8) ln(2)]Second derivative:y'' = 2^(-x) [ -10 ln(2) + (5x -8)(ln(2))^2 ]Third derivative:y''' = 2^(-x) [15 (ln(2))^2 - (5x -8)(ln(2))^3 ]Wait, the coefficients of the constant terms are alternating in sign and multiplying by 5 each time? Let's see:First derivative: 5 (constant term) and -(5x -8) ln(2)Second derivative: -10 ln(2) (constant term coefficient) and (5x -8)(ln(2))^2Third derivative: 15 (ln(2))^2 and -(5x -8)(ln(2))^3So the coefficients of the constant terms are 5, -10 ln(2), 15 (ln(2))^2. Hmm, 5, -10, 15... that's 5*1, 5*(-2), 5*3, but multiplied by ln(2) to the power of 0, 1, 2 respectively. Wait, maybe the pattern is:For the nth derivative, the constant term is 5*(-1)^{n-1} * n * (ln(2))^{n-1}Wait, let's check:First derivative (n=1): 5*(-1)^{0} *1 * (ln(2))^0 = 5*1*1*1 = 5. Correct.Second derivative (n=2): 5*(-1)^{1} *2 * (ln(2))^1 = 5*(-1)*2*ln(2) = -10 ln(2). Correct.Third derivative (n=3): 5*(-1)^{2} *3 * (ln(2))^2 = 5*1*3*(ln(2))^2 = 15 (ln(2))^2. Correct.So the constant term for the nth derivative is 5*(-1)^{n-1} *n*(ln(2))^{n-1}And the other term is (5x -8)*(-1)^n*(ln(2))^n. Let me check:First derivative: [5 - (5x -8) ln(2)] = 5 + (5x -8)*(-ln(2)). So the second term is (5x -8)*(-ln(2)) which is (-1)^1*(5x -8)*(ln(2))^1. So for n=1, yes.Second derivative: [ -10 ln(2) + (5x -8)(ln(2))^2 ] = (5x -8)*(ln(2))^2 -10 ln(2). The first term is (5x -8)*(ln(2))^2 = (5x -8)*(-1)^2*(ln(2))^2. So yes, for n=2, the coefficient is (-1)^2*(ln(2))^2. The constant term is -10 ln(2) which matches 5*(-1)^{2-1}*2*(ln(2))^{2-1} = 5*(-1)^1*2*ln(2) = -10 ln(2). Correct.Third derivative: [15 (ln(2))^2 - (5x -8)(ln(2))^3 ] = - (5x -8)(ln(2))^3 +15 (ln(2))^2. The coefficient for (5x -8) is - (ln(2))^3 = (-1)^3*(ln(2))^3. The constant term is 15 (ln(2))^2 = 5*(-1)^{3-1}*3*(ln(2))^{3-1} = 5*1*3*(ln(2))^2. Correct.Therefore, the pattern seems to hold. So for the nth derivative, y^(n) = 2^(-x) [ 5*(-1)^{n-1} *n*(ln(2))^{n-1} + (5x -8)*(-1)^n*(ln(2))^n ]Alternatively, factoring out (-1)^{n}*(ln(2))^{n-1}, we can write:y^(n) = 2^(-x)*(-1)^n*(ln(2))^{n-1} [ -5n + (5x -8) ln(2) ]Wait, let's see:Start with the expression:5*(-1)^{n-1} *n*(ln(2))^{n-1} + (5x -8)*(-1)^n*(ln(2))^nFactor out (-1)^{n-1}*(ln(2))^{n-1}:= (-1)^{n-1}*(ln(2))^{n-1} [5n + (5x -8)*(-1)*ln(2) ]= (-1)^{n-1}*(ln(2))^{n-1} [5n - (5x -8) ln(2) ]But then we can write this as:(-1)^{n}*(ln(2))^{n-1} [ -5n + (5x -8) ln(2) ]Hmm, perhaps. Alternatively, since (-1)^{n-1} = -(-1)^n, so:= -(-1)^n*(ln(2))^{n-1} [5n - (5x -8) ln(2) ]But maybe this isn't necessary. The main point is that we can express the nth derivative as:y^(n) = 2^(-x)*(-1)^n*(ln(2))^{n}*(5x -8) + 2^(-x)*5*(-1)^{n-1}*n*(ln(2))^{n-1}Wait, perhaps another way. Let me separate the terms:y^(n) = 2^(-x)*(5x -8)*(-1)^n*(ln(2))^n + 2^(-x)*5*n*(-1)^{n-1}*(ln(2))^{n-1}Factor out 2^(-x)*(-1)^{n-1}*(ln(2))^{n-1}:= 2^(-x)*(-1)^{n-1}*(ln(2))^{n-1} [ (5x -8)*(-1)*(ln(2)) +5n ]= 2^(-x)*(-1)^{n-1}*(ln(2))^{n-1} [ - (5x -8) ln(2) +5n ]But this might not be particularly useful. Alternatively, let's write it as:y^(n) = 2^{-x} (ln(2))^{n} (-1)^n (5x -8) + 5n 2^{-x} (-1)^{n -1} (ln(2))^{n -1}But perhaps even better to write it as:y^{(n)} = 2^{-x} (-1)^n (ln 2)^n (5x - 8) + 5n 2^{-x} (-1)^{n -1} (ln 2)^{n -1}But combining these terms:= 2^{-x} (-1)^n (ln 2)^{n -1} [ (ln 2)(5x -8) -5n ]Yes, because:First term: 2^{-x} (-1)^n (ln 2)^n (5x -8) = 2^{-x} (-1)^n (ln 2)^{n -1} * ln 2 (5x -8)Second term: 5n 2^{-x} (-1)^{n -1} (ln 2)^{n -1} = 2^{-x} (-1)^{n -1} (ln 2)^{n -1} *5nFactor out 2^{-x} (-1)^n (ln 2)^{n -1}:= 2^{-x} (-1)^n (ln 2)^{n -1} [ ln 2 (5x -8) + (-1) *5n ]Because (-1)^{n -1} = (-1)^n * (-1)^{-1} = (-1)^n * (-1). So:Second term is 2^{-x} (-1)^{n} (-1)^{-1} (ln 2)^{n -1} *5n = 2^{-x} (-1)^n (ln 2)^{n -1} * (-5n)Therefore, combined:= 2^{-x} (-1)^n (ln 2)^{n -1} [ ln 2 (5x -8) -5n ]So that simplifies to:y^{(n)} = 2^{-x} (-1)^n (ln 2)^{n -1} [5 (x ln 2 - n) -8 ln 2 ]Wait, let's expand the bracket:ln 2 (5x -8) -5n =5x ln 2 -8 ln 2 -5n=5x ln 2 -5n -8 ln 2=5(x ln 2 -n) -8 ln 2But not sure if that's helpful. Alternatively, maybe leave it as:ln 2 (5x -8) -5nSo in any case, the nth derivative is:2^{-x} (-1)^n (ln 2)^{n -1} [5x ln 2 -8 ln 2 -5n]Therefore, for the fourth derivative, n=4:y^{(IV)} = 2^{-x} (-1)^4 (ln 2)^{4 -1} [5x ln 2 -8 ln 2 -5*4]Simplify:(-1)^4 =1, (ln 2)^3 remains, and the bracket becomes:5x ln 2 -8 ln 2 -20Therefore:y^{(IV)} =2^{-x} (ln 2)^3 [5x ln 2 -8 ln 2 -20]We can factor out ln 2 from the first two terms in the bracket:=2^{-x} (ln 2)^3 [ln 2 (5x -8) -20]Alternatively, leave it as is. But let's check if this formula works with our previous derivatives.For n=1:y' =2^{-x} (-1)^1 (ln 2)^{0} [5x ln2 -8 ln2 -5*1] = 2^{-x} (-1)(1)[5x ln2 -8 ln2 -5] = -2^{-x} [5x ln2 -8 ln2 -5]But from before, we had y' =2^{-x} [5 - (5x -8) ln2]. Let's see if these are equal:Expand [5 - (5x -8) ln2] =5 -5x ln2 +8 ln2. Comparing to - [5x ln2 -8 ln2 -5] = -5x ln2 +8 ln2 +5. So yes, same as 5 -5x ln2 +8 ln2. Correct.Similarly, n=2:y''=2^{-x}(-1)^2(ln2)^1[5x ln2 -8 ln2 -10]=2^{-x}(ln2)[5x ln2 -8 ln2 -10]Which matches the second derivative we calculated earlier:y''=2^{-x} [ -10 ln2 + (5x -8)(ln2)^2 ].Wait, let's compute:2^{-x} (ln2)[5x ln2 -8 ln2 -10] =2^{-x} [5x (ln2)^2 -8 (ln2)^2 -10 ln2]But in our previous calculation, y'' was:2^{-x} [ -10 ln(2) + (5x -8)(ln(2))^2 ] =2^{-x} [5x (ln2)^2 -8 (ln2)^2 -10 ln2]Which is the same as above. Correct.Therefore, the formula holds. Therefore, the fourth derivative would be as above:y^{(IV)}=2^{-x}(ln2)^3[5x ln2 -8 ln2 -20]We can factor out ln2 from the first two terms:=2^{-x}(ln2)^3[ln2(5x -8) -20]Alternatively, factor out 5 from the terms inside the bracket:=2^{-x}(ln2)^3[5(x ln2 -4) -8 ln2]But not sure if that's helpful. Alternatively, leave it as is.Alternatively, present the answer as:y^{(IV)} =2^{-x}(ln2)^3(5x ln2 -8 ln2 -20)Or factor out common terms. Let's see:5x ln2 -8 ln2 -20 =5x ln2 - (8 ln2 +20)But maybe that's not necessary. The problem just asks for the fourth derivative, so any of these forms is acceptable, but perhaps we can factor the expression a bit more.Alternatively, factor out ln2 from the entire expression:But 5x ln2 -8 ln2 -20 can't be factored by ln2 entirely. Wait, unless we write:= ln2*(5x -8) -20Yes, exactly. So:5x ln2 -8 ln2 -20 = ln2*(5x -8) -20Therefore, we can write:y^{(IV)} =2^{-x}(ln2)^3[ln2*(5x -8) -20]Which is a more compact form. Alternatively, factor out a negative sign:=2^{-x}(ln2)^3[-20 + ln2*(5x -8)]But probably either form is acceptable.Alternatively, expand the terms:=2^{-x}(ln2)^3*(5x ln2 -8 ln2 -20)So, all these are equivalent. Let me check if this matches with the pattern we saw before.If n=4, then according to our previous formula:y^{(4)} =2^{-x} [5*(-1)^{4-1} *4*(ln2)^{4-1} + (5x -8)*(-1)^4*(ln2)^4 ]Compute each term:First term:5*(-1)^3*4*(ln2)^3=5*(-1)*4*(ln2)^3= -20*(ln2)^3Second term:(5x -8)*(1)*(ln2)^4=(5x -8)(ln2)^4Therefore:y^{(4)}=2^{-x}[ -20(ln2)^3 + (5x -8)(ln2)^4 ]Factor out (ln2)^3:=2^{-x}(ln2)^3[ -20 + (5x -8)ln2 ]Which is the same as before. So yes, this matches.Therefore, the fourth derivative is y^{(IV)}=2^{-x}(ln2)^3[ (5x -8)ln2 -20 ]Alternatively, written as:y^{(IV)}= (ln2)^3 cdot 2^{-x} [5x ln2 -8 ln2 -20]Either form is correct. To present the answer, I can write it in the factored form with the constants calculated if possible, but since ln2 is just a constant multiplier, we can leave it as is.Therefore, the final answer should be in a boxed format. Let me check once again to ensure I didn't make a mistake in the signs or coefficients.Starting from n=4:y^{(IV)}=2^{-x}(ln2)^3[5x ln2 -8 ln2 -20]Yes, that seems correct. The coefficients are 5, -8, -20 multiplied by appropriate powers of ln2 and 2^{-x}. I don't see any errors in the derivation steps, and cross-checked with the nth derivative formula. So I think this is correct.Final AnswerThe fourth derivative is boxed{2^{-x} (ln 2)^3 left(5x (ln 2) - 8 ln 2 - 20right)}.</think>To find the fourth derivative of the function ( y = (5x - 8) cdot 2^{-x} ), we can use the product rule and observe a pattern in the derivatives.1. First Derivative: [ y' = 5 cdot 2^{-x} + (5x - 8) cdot (-2^{-x} ln 2) = 2^{-x} left[ 5 - (5x - 8) ln 2 right] ]2. Second Derivative: [ y'' = -2^{-x} ln 2 left[ 5 - (5x - 8) ln 2 right] + 2^{-x} cdot (-5 ln 2) ] Simplifying, we get: [ y'' = 2^{-x} left[ -10 ln 2 + (5x - 8) (ln 2)^2 right] ]3. Third Derivative: [ y''' = -2^{-x} (ln 2) left[ -10 ln 2 + (5x - 8) (ln 2)^2 right] + 2^{-x} cdot 5 (ln 2)^2 ] Simplifying, we get: [ y''' = 2^{-x} left[ 15 (ln 2)^2 - (5x - 8) (ln 2)^3 right] ]4. Fourth Derivative: Using the pattern observed, the nth derivative can be expressed as: [ y^{(n)} = 2^{-x} (-1)^n (ln 2)^{n-1} left[ 5n - (5x - 8) ln 2 right] ] For ( n = 4 ): [ y^{(IV)} = 2^{-x} (ln 2)^3 left[ 5 cdot 4 - (5x - 8) ln 2 right] ] Simplifying, we get: [ y^{(IV)} = 2^{-x} (ln 2)^3 left[ 20 - (5x - 8) ln 2 right] ] Rearranging the terms, we get: [ y^{(IV)} = 2^{-x} (ln 2)^3 left( 5x ln 2 - 8 ln 2 - 20 right) ]Thus, the fourth derivative of the function is:[boxed{2^{-x} (ln 2)^3 left(5x (ln 2) - 8 ln 2 - 20right)}]]

🔑:## Solutiony^{prime}=left((5 x-8) cdot 2^{-x}right)^{prime}=5 cdot 2^{-x}+(5 x-8) cdot 2^{-x} cdot(-ln 2)=2^{-x} cdot(-5 ln 2 cdot x+8 ln 2+5)y^{prime prime}=left(y^{prime}right)^{prime}=left(2^{-x} cdot(-5 ln 2 cdot x+8 ln 2+5)right)^{prime}==2^{-x} cdot(-ln 2) cdot(-5 ln 2 cdot x+8 ln 2+5)+2^{-x} cdot(-5 ln 2 cdot)==2^{-x} cdot ln 2 cdot(5 ln 2 cdot x-8 ln 2-10)y^{prime prime prime}=left(y^{prime prime}right)^{prime}=left(2^{-x} cdot ln 2 cdot(5 ln 2 cdot x-8 ln 2-10)right)^{prime}==2^{-x} cdot(-ln 2) cdot ln 2 cdot(5 ln 2 cdot x-8 ln 2-10)+2^{-x} cdot ln 2 cdot(5 ln 2)==2^{-x} cdot ln ^{2} 2 cdot(-5 ln 2 cdot x+8 ln 2+15)y^{(I V)}=left(y^{prime prime prime}right)^{prime}=left(2^{-x} cdot ln ^{2} 2 cdot(-5 ln 2 cdot x+8 ln 2+15)right)^{prime}==2^{-x} cdot(-ln 2) cdot ln ^{2} 2 cdot(-5 ln 2 cdot x+8 ln 2+15)+2^{-x} cdot ln ^{2} 2 cdot(-5 ln 2)==2^{-x} cdot ln ^{3} 2 cdot(5 ln 2 cdot x-8 ln 2-20)## Problem Kuznetsov Differentiation 19-26

🧠:[<think>Okay, let's tackle this problem step by step. There are two parts: first finding the standard equation of the ellipse given some conditions, and then determining if there's a fixed point D on the x-axis such that the dot product of vectors DM and DN is constant. Let's start with part 1.Problem 1: Find the standard equation of the ellipse.We know the ellipse has the standard form ( dfrac{x^2}{a^2} + dfrac{y^2}{b^2} = 1 ) with ( a > b > 0 ). The vertices A and B are the left and right vertices, so their coordinates are (-a, 0) and (a, 0) respectively. The eccentricity e is given as ( dfrac{1}{2} ), and the maximum area of triangle PAB is ( 2sqrt{3} ).First, recall that the eccentricity of an ellipse is ( e = dfrac{c}{a} ), where c is the distance from the center to each focus. Given that ( e = 1/2 ), we have ( c = a/2 ).Also, the relationship between a, b, and c in an ellipse is ( c^2 = a^2 - b^2 ). Plugging c = a/2 into this equation:( (a/2)^2 = a^2 - b^2 )Simplify:( a^2/4 = a^2 - b^2 )Multiply both sides by 4:( a^2 = 4a^2 - 4b^2 )Bring terms to one side:( 4b^2 = 3a^2 )Thus,( b^2 = (3/4)a^2 )So, the relationship between a and b is established. Now, we need to use the maximum area of triangle PAB to find the values of a and b.Point P is a variable on the ellipse. Let's parameterize point P using the parametric equations of the ellipse. For an ellipse, the coordinates of a point P can be written as ( (a cos theta, b sin theta) ), where ( theta ) is the parameter.The vertices A and B are at (-a, 0) and (a, 0), so the base of triangle PAB is the distance between A and B, which is 2a. The height of the triangle is the y-coordinate of point P, since the base is along the x-axis. Therefore, the area of triangle PAB is:( text{Area} = dfrac{1}{2} times text{base} times text{height} = dfrac{1}{2} times 2a times |y_P| = a times |y_P| )Since point P is ( (a cos theta, b sin theta) ), the y-coordinate is ( b sin theta ). Therefore, the area becomes:( text{Area} = a times |b sin theta| = ab |sin theta| )To find the maximum area, we note that the maximum value of ( |sin theta| ) is 1. Therefore, the maximum area is ( ab times 1 = ab ). According to the problem, this maximum area is ( 2sqrt{3} ). Therefore:( ab = 2sqrt{3} )We already have ( b^2 = (3/4)a^2 ), so let's solve for b in terms of a. Taking square roots:( b = dfrac{sqrt{3}}{2}a )Now substitute this into the equation ( ab = 2sqrt{3} ):( a times dfrac{sqrt{3}}{2}a = 2sqrt{3} )Simplify:( dfrac{sqrt{3}}{2}a^2 = 2sqrt{3} )Multiply both sides by 2:( sqrt{3} a^2 = 4sqrt{3} )Divide both sides by ( sqrt{3} ):( a^2 = 4 )Thus, ( a = 2 ), since a > 0.Then, ( b = dfrac{sqrt{3}}{2} times 2 = sqrt{3} )Therefore, the standard equation of the ellipse is:( dfrac{x^2}{4} + dfrac{y^2}{3} = 1 )Let me double-check this. If a = 2, then c = a/2 = 1, so foci are at (-1, 0) and (1, 0). Then, ( b^2 = a^2 - c^2 = 4 - 1 = 3 ), which matches. The maximum area would be ab = 2 * sqrt(3) = 2√3, which matches the given condition. So that's correct.Problem 2: Determine if there's a fixed point D on the x-axis such that ( overrightarrow{DM} cdot overrightarrow{DN} ) is constant.This seems a bit more involved. Let me parse the problem.We have the ellipse C from part 1, so standard equation is ( dfrac{x^2}{4} + dfrac{y^2}{3} = 1 ). The left focus F1 is at (-c, 0) = (-1, 0). A variable line l passes through F1 (-1, 0) and intersects the ellipse at points M and N. The question is: Does there exist a fixed point D on the x-axis such that the dot product of vectors DM and DN is a constant, regardless of the line l? If yes, find D and the constant; if not, explain why.First, I need to recall that the dot product of two vectors ( overrightarrow{DM} ) and ( overrightarrow{DN} ) is given by:( overrightarrow{DM} cdot overrightarrow{DN} = |DM||DN|cos theta ), where θ is the angle between them. But since we need this to be constant for any line l passing through F1, we have to find D such that this product is independent of the choice of line l.Alternatively, in coordinate terms, if D is (d, 0), then vectors DM and DN would be (x_M - d, y_M - 0) and (x_N - d, y_N - 0), so the dot product is (x_M - d)(x_N - d) + y_M y_N.Therefore, we need to find d such that (x_M - d)(x_N - d) + y_M y_N is constant for any line passing through F1 (-1, 0).To approach this, we can parametrize the line l passing through F1 (-1, 0). Let's consider the general equation of a line passing through (-1, 0). The line can be written in parametric form, or we can use a slope parameter. Let's use slope parameter m. However, vertical lines would be a problem with slope, so maybe parametrize using an angle θ or parameter t.Alternatively, let's consider the parametric equation of the line. Let’s denote the line passes through (-1, 0) and has a direction vector. Let’s parameterize the line with parameter t. Let's suppose that the line has direction vector (k, l), so parametric equations are x = -1 + kt, y = 0 + lt. But since the line intersects the ellipse at two points M and N, we can substitute into the ellipse equation to find points M and N.Alternatively, perhaps it's better to use a parameter for the line. Let me consider that the line passes through F1 (-1, 0) and has a slope m. Then, the equation of the line is y = m(x + 1). Substituting this into the ellipse equation:( dfrac{x^2}{4} + dfrac{[m(x + 1)]^2}{3} = 1 )Multiply through by 12 to eliminate denominators:3x² + 4m²(x + 1)^2 = 12Expand the terms:3x² + 4m²(x² + 2x + 1) = 123x² + 4m²x² + 8m²x + 4m² - 12 = 0Combine like terms:(3 + 4m²)x² + 8m²x + (4m² - 12) = 0This is a quadratic in x. Let’s denote coefficients as A, B, C:A = 3 + 4m²B = 8m²C = 4m² - 12The solutions x correspond to the x-coordinates of M and N. Let’s denote them as x1, x2.By Vieta's formula, sum x1 + x2 = -B/A = -8m² / (3 + 4m²)Product x1 x2 = C/A = (4m² - 12)/(3 + 4m²)Similarly, the y-coordinates of M and N are y1 = m(x1 + 1) and y2 = m(x2 + 1)Now, we need to compute (x1 - d)(x2 - d) + y1 y2First, expand (x1 - d)(x2 - d):x1 x2 - d(x1 + x2) + d²Then, y1 y2 = m²(x1 + 1)(x2 + 1) = m²(x1 x2 + x1 + x2 + 1)Therefore, the total expression is:[x1 x2 - d(x1 + x2) + d²] + m²[x1 x2 + x1 + x2 + 1]Let’s combine these terms:= x1 x2 (1 + m²) + (-d(x1 + x2) + m²(x1 + x2)) + (d² + m²)Factor out terms with (x1 + x2):= x1 x2 (1 + m²) + (x1 + x2)(-d + m²) + (d² + m²)Now, substitute the Vieta's expressions:x1 + x2 = -8m² / (3 + 4m²)x1 x2 = (4m² - 12)/(3 + 4m²)Therefore, substitute these into the expression:= [(4m² - 12)/(3 + 4m²)](1 + m²) + [(-8m²)/(3 + 4m²)](-d + m²) + (d² + m²)Let’s compute each term step by step.First term: [(4m² - 12)(1 + m²)] / (3 + 4m²)Second term: [(-8m²)(-d + m²)] / (3 + 4m²) = [8m²(d - m²)] / (3 + 4m²)Third term: (d² + m²)Therefore, the entire expression is:[ (4m² - 12)(1 + m²) + 8m²(d - m²) ] / (3 + 4m²) + (d² + m²)Let’s compute the numerator of the first fraction:(4m² - 12)(1 + m²) + 8m²(d - m²)First, expand (4m² - 12)(1 + m²):= 4m²(1 + m²) - 12(1 + m²)= 4m² + 4m^4 - 12 - 12m²= 4m^4 - 8m² - 12Then, add 8m²(d - m²):= 4m^4 - 8m² - 12 + 8m²d - 8m^4Combine like terms:(4m^4 - 8m^4) + (-8m² + 8m²d) + (-12)= -4m^4 -8m² +8m²d -12Therefore, the numerator is -4m^4 +8m²d -8m² -12So the entire expression is:[ -4m^4 +8m²d -8m² -12 ] / (3 + 4m²) + (d² + m²)Let’s split the fraction into terms divided by (3 + 4m²):= [ -4m^4/(3 + 4m²) ) + (8m²d)/(3 + 4m²) -8m²/(3 + 4m²) -12/(3 + 4m²) ] + d² + m²This looks complicated. Maybe we can combine terms differently.Alternatively, let's write the entire expression as:[ -4m^4 +8m²d -8m² -12 ] / (3 + 4m²) + d² + m²Let’s combine the terms by writing everything over the common denominator (3 + 4m²):= [ -4m^4 +8m²d -8m² -12 + (d² + m²)(3 + 4m²) ] / (3 + 4m²)Expand (d² + m²)(3 + 4m²):= d²(3 + 4m²) + m²(3 + 4m²)= 3d² + 4d² m² + 3m² +4m^4Therefore, the numerator becomes:-4m^4 +8m²d -8m² -12 +3d² +4d² m² +3m² +4m^4Simplify term by term:-4m^4 +4m^4 = 08m²d +4d² m² = m²(8d +4d²)-8m² +3m² = -5m²-12 +3d² = 3d² -12So the numerator is:m²(8d +4d²) -5m² +3d² -12Factor m² terms:m²(8d +4d² -5) +3d² -12Thus, the entire expression is:[ (4d² +8d -5)m² +3d² -12 ] / (3 + 4m²)For this expression to be a constant, independent of m², the coefficient of m² in the numerator must be proportional to the denominator (3 +4m²). That is, there must exist constants k1 and k2 such that:(4d² +8d -5)m² +3d² -12 = k1(3 +4m²) + k2(3 +4m²)Wait, no. Actually, for the entire expression [numerator]/[denominator] to be a constant, the numerator must be a constant multiple of the denominator. So:(4d² +8d -5)m² +3d² -12 = K*(3 +4m²)where K is a constant. Then, equating coefficients:Coefficient of m²: 4d² +8d -5 = 4KConstant term: 3d² -12 = 3KTherefore, we have two equations:1. 4d² +8d -5 = 4K2. 3d² -12 = 3KFrom equation 2, we can solve for K:K = d² -4Substitute into equation 1:4d² +8d -5 = 4(d² -4)Expand the right-hand side:4d² -16So:4d² +8d -5 =4d² -16Subtract 4d² from both sides:8d -5 = -16Add 5 to both sides:8d = -11Therefore:d = -11/8So, if d = -11/8, then K = (-11/8)^2 -4 = (121/64) -4 = (121 -256)/64 = (-135)/64Thus, substituting back, we have:[ (4d² +8d -5)m² +3d² -12 ] / (3 +4m²) = K = -135/64, which is a constant.Therefore, such a point D exists at d = -11/8 on the x-axis, and the constant value is -135/64.Let me verify this step by step.First, set up the equation that (4d² +8d -5)m² +3d² -12 = K*(3 +4m²). Then equate coefficients:Coefficients of m²:4d² +8d -5 =4KConstants:3d² -12 =3K ⇒ K = d² -4Substitute into first equation:4d² +8d -5 =4(d² -4)⇒4d² +8d -5 =4d² -16Subtract 4d²:8d -5 = -16 ⇒8d= -11 ⇒d= -11/8Then K= (-11/8)^2 -4= 121/64 -256/64= -135/64Therefore, the expression simplifies to -135/64, a constant. Thus, the answer is D is (-11/8,0) and the constant is -135/64.But wait, the problem statement says "is there a fixed point D on the x-axis such that ( overrightarrow{DM} cdot overrightarrow{DN} ) is constant". So such a point exists, with coordinates (-11/8,0), and the constant is -135/64.But let me check with specific cases. Let's take a horizontal line through F1 (-1,0). The horizontal line is y=0, which intersects the ellipse at x=±2. So points M and N are (-2,0) and (2,0). Then DM vectors would be (-2 - d,0) and (2 - d,0). The dot product is (-2 - d)(2 - d) +0*0 = (-2)(2 - d) -d(2 - d) = (-4 +2d -2d +d²) = d² -4. Wait, according to our previous result, the dot product should be constant -135/64 when d=-11/8. Let's compute d² -4 when d=-11/8:(-11/8)^2 -4 = 121/64 -256/64= -135/64. So that matches. So in this case, for the horizontal line, the dot product is indeed -135/64. Let's check another line. Let's take the vertical line through F1 (-1,0). But a vertical line x=-1. Substitute x=-1 into the ellipse equation:(-1)^2 /4 + y²/3 =1 ⇒1/4 + y²/3=1⇒y²= 3*(3/4)=9/4⇒y=±3/2. So points M and N are (-1, 3/2) and (-1, -3/2). Then vectors DM and DN would be (-1 - d, 3/2) and (-1 -d, -3/2). The dot product is (-1 -d)^2 + (3/2)(-3/2) = (1 +2d +d²) -9/4. So compute this when d=-11/8:(-1 - (-11/8))^2 -9/4 = ( ( -8/8 +11/8 ) )^2 -9/4= (3/8)^2 -9/4=9/64 -144/64= (-135)/64. So that also matches. Therefore, the computation seems correct.Another test: take a line with slope 1 through F1 (-1,0). Equation y = x +1. Substitute into ellipse:x²/4 + (x +1)^2 /3 =1Multiply by 12:3x² +4(x² +2x +1)=123x² +4x² +8x +4=127x² +8x -8=0Solve:x = [-8 ±sqrt(64 + 224)]/(14) = [-8 ±sqrt(288)]/14 = [-8 ±12√2]/14 = [ -4 ±6√2 ]/7Compute coordinates M and N:x1 = (-4 +6√2)/7, y1 = x1 +1 = (3 +6√2)/7x2 = (-4 -6√2)/7, y2 = x2 +1 = (3 -6√2)/7Compute DM · DN where D is (-11/8,0):DM vector: (x1 - (-11/8), y1 -0) = (x1 +11/8, y1)DN vector: (x2 +11/8, y2)Dot product = (x1 +11/8)(x2 +11/8) + y1 y2First compute x1 x2:x1 x2 = [(-4 +6√2)/7][(-4 -6√2)/7] = [16 +24√2 -24√2 -72]/49 = (16 -72)/49 = (-56)/49 = -8/7Then x1 + x2 = [(-4 +6√2) + (-4 -6√2)] /7 = (-8)/7So (x1 +11/8)(x2 +11/8) = x1 x2 +11/8(x1 + x2) + (11/8)^2= (-8/7) +11/8*(-8/7) + (121/64)= (-8/7) -88/56 +121/64Simplify:Convert all to 64 denominator:-8/7 = - (8*64)/(7*64) = -512/448-88/56 = - (88*8)/(56*8) = -704/448121/64 = 121*7 / 448 = 847/448So total:(-512 -704 +847)/448 = (-1216 +847)/448 = (-369)/448Then y1 y2 = [(3 +6√2)/7][(3 -6√2)/7] = [9 - (6√2)^2]/49 = [9 -72]/49 = (-63)/49 = -9/7Convert to 448 denominator: -9/7 = -576/448Thus total dot product:(-369)/448 + (-576)/448 = (-945)/448But -945/448 simplifies. Let me check if this equals -135/64:Multiply numerator and denominator by 64/448: 945/448 = (945/7)/64 = 135/64. So -945/448 = -135/64. Correct.Therefore, even for this arbitrary line, the dot product is -135/64, confirming the result. Therefore, the answer for part 2 is that such a point D exists at (-11/8, 0) and the constant is -135/64.Final Answer1. The standard equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.2. The fixed point ( D ) is boxed{left( -dfrac{11}{8}, 0 right)} and the constant value is boxed{-dfrac{135}{64}}.</think>Given an ellipse ( C: dfrac{x^2}{a^2} + dfrac{y^2}{b^2} = 1 ) with ( a > b > 0 ), left and right vertices ( A ) and ( B ), eccentricity ( e = dfrac{1}{2} ), and the maximum area of ( triangle PAB ) is ( 2sqrt{3} ):1. Standard Equation of the Ellipse: - The eccentricity ( e = dfrac{1}{2} ) gives ( c = dfrac{a}{2} ). - Using ( c^2 = a^2 - b^2 ), we find ( left( dfrac{a}{2} right)^2 = a^2 - b^2 ), leading to ( b^2 = dfrac{3}{4}a^2 ). - The maximum area of ( triangle PAB ) is ( ab = 2sqrt{3} ). Substituting ( b = dfrac{sqrt{3}}{2}a ), we solve ( a times dfrac{sqrt{3}}{2}a = 2sqrt{3} ), resulting in ( a = 2 ) and ( b = sqrt{3} ). - The standard equation is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).2. Fixed Point ( D ) on the x-axis: - The left focus ( F_1 ) is ((-1, 0)). - For a line passing through ( F_1 ) intersecting the ellipse at ( M ) and ( N ), we need to find if there exists a fixed point ( D ) such that ( overrightarrow{DM} cdot overrightarrow{DN} ) is constant. - Parametrizing the line and using Vieta's formulas, we derive the expression for the dot product in terms of ( d ) (coordinate of ( D )). - Solving for ( d ) such that the expression is constant, we find ( d = -dfrac{11}{8} ) and the constant value (-dfrac{135}{64}).The final answers are:1. (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1})2. The fixed point ( D ) is (boxed{left( -dfrac{11}{8}, 0 right)}) and the constant value is (boxed{-dfrac{135}{64}}).,<think>Okay, let's tackle this problem step by step. There are two parts: first, finding the standard equation of the ellipse given some conditions, and second, determining if there's a fixed point D on the x-axis such that the dot product of vectors DM and DN is constant. Let's start with part 1.Problem 1: Find the standard equation of the ellipse.Given an ellipse with equation x²/a² + y²/b² = 1 (a > b > 0), vertices at A (left) and B (right). The eccentricity e = 1/2, and the maximum area of triangle PAB is 2√3. We need to find a and b.First, recall that the vertices of an ellipse are at (±a, 0), so points A and B are (-a, 0) and (a, 0) respectively. The triangle PAB has these two points as vertices, and point P is any point on the ellipse. We need to find the maximum area of this triangle.The area of triangle PAB can be calculated using the formula for the area of a triangle with base AB and height equal to the y-coordinate of point P. Since AB is the major axis, its length is 2a. The height is the vertical distance from P to the x-axis, which is |y|. Therefore, the area A of triangle PAB is (1/2)*base*height = (1/2)*(2a)*|y| = a|y|.To maximize the area, we need to maximize |y|. The maximum value of |y| on the ellipse occurs at the top and bottom vertices of the ellipse, which are (0, b) and (0, -b). So the maximum area is a*b. According to the problem, this maximum area is 2√3. Therefore, we have:a*b = 2√3.Additionally, the eccentricity e is given as 1/2. For an ellipse, the eccentricity is e = c/a, where c is the distance from the center to each focus, and c² = a² - b².Given e = 1/2, so c = a/2. Therefore:c = a/2c² = a²/4But also c² = a² - b², so:a² - b² = a²/4Subtract a²/4 from both sides:(3/4)a² = b²Thus, b = (√3/2)aNow we can substitute b into the equation a*b = 2√3:a*(√3/2 a) = 2√3Multiply the left side:(√3/2)a² = 2√3Divide both sides by √3:(1/2)a² = 2Multiply both sides by 2:a² = 4Therefore, a = 2 (since a > 0)Then, b² = (3/4)a² = (3/4)*4 = 3So, b = √3Therefore, the standard equation of the ellipse is:x²/4 + y²/3 = 1So that's part 1 done. Now onto part 2.Problem 2: Determine if there's a fixed point D on the x-axis such that vectors DM and DN have a constant dot product.Given that a variable line l passes through the left focus F₁ of the ellipse C and intersects the ellipse at points M and N. We need to check if there's a fixed point D on the x-axis such that the dot product DM · DN is constant, regardless of the line l. If such a D exists, find its coordinates and the constant value.First, let's recall the properties of the ellipse. From part 1, we have a = 2, b = √3, so c = a*e = 2*(1/2) = 1. The foci are at (±c, 0) = (±1, 0). So the left focus F₁ is at (-1, 0).The line l passes through F₁ (-1, 0) and intersects the ellipse at points M and N. We need to see if there's a point D on the x-axis such that DM · DN is constant for any such line l.First, let's parametrize line l. Since it passes through F₁ (-1, 0), we can write its equation as y = m(x + 1), where m is the slope. However, m can vary, making the line variable. Alternatively, we can parametrize using a parameter θ for the angle, but maybe using parametric equations would be better. However, perhaps using coordinates to express M and N.Alternatively, let's use parametric equations of the ellipse. The standard parametrization is x = a cos θ, y = b sin θ. But since the line passes through (-1, 0), maybe we need to find the points of intersection of this line with the ellipse.Alternatively, let's consider the line passing through (-1, 0) with a general slope m. Let's find the points of intersection between the line and the ellipse.Given the ellipse equation x²/4 + y²/3 = 1.The line equation: y = m(x + 1)Substitute y into the ellipse equation:x²/4 + [m²(x + 1)²]/3 = 1Multiply both sides by 12 to eliminate denominators:3x² + 4m²(x + 1)^2 = 12Expand (x + 1)^2:3x² + 4m²(x² + 2x + 1) = 12Expand the terms:3x² + 4m²x² + 8m²x + 4m² = 12Combine like terms:(3 + 4m²)x² + 8m²x + (4m² - 12) = 0This is a quadratic equation in x. Let's denote coefficients:A = 3 + 4m²B = 8m²C = 4m² - 12Then, solving for x using quadratic formula:x = [-B ± √(B² - 4AC)]/(2A)But perhaps instead of solving for x, we can use Vieta's formulas to find the sum and product of roots.Let x₁ and x₂ be the x-coordinates of points M and N. Then:x₁ + x₂ = -B/A = -8m² / (3 + 4m²)x₁x₂ = C/A = (4m² - 12)/(3 + 4m²)Similarly, the corresponding y-coordinates are y₁ = m(x₁ + 1) and y₂ = m(x₂ + 1)Now, we need to find a point D on the x-axis, say D = (d, 0), such that DM · DN is constant.Vectors DM and DN would be (x₁ - d, y₁) and (x₂ - d, y₂) respectively. The dot product is:DM · DN = (x₁ - d)(x₂ - d) + y₁y₂Compute this:= (x₁x₂ - d(x₁ + x₂) + d²) + [m(x₁ + 1)][m(x₂ + 1)]= (x₁x₂ - d(x₁ + x₂) + d²) + m²(x₁x₂ + x₁ + x₂ + 1)Let me compute each term step by step.First term: (x₁x₂ - d(x₁ + x₂) + d²)Second term: m²(x₁x₂ + x₁ + x₂ + 1)Therefore, the total dot product is:x₁x₂ - d(x₁ + x₂) + d² + m²x₁x₂ + m²(x₁ + x₂) + m²Group terms:(1 + m²)x₁x₂ + (-d + m²)(x₁ + x₂) + (d² + m²)Now, from Vieta's formulas, we can substitute x₁ + x₂ and x₁x₂.Given that:x₁ + x₂ = -8m²/(3 + 4m²)x₁x₂ = (4m² - 12)/(3 + 4m²)Let's substitute these into the expression.First, compute (1 + m²)x₁x₂:(1 + m²)*(4m² - 12)/(3 + 4m²)Second, compute (-d + m²)(x₁ + x₂):(-d + m²)*(-8m²)/(3 + 4m²)Third term: (d² + m²)So putting all together:DM · DN = [(1 + m²)(4m² - 12) + (-d + m²)(-8m²)]/(3 + 4m²) + (d² + m²)Let me compute the numerator of the first fraction:(1 + m²)(4m² - 12) + (-d + m²)(-8m²)First term:(4m² - 12 + 4m⁴ - 12m²) = 4m⁴ - 8m² - 12Second term:(-d + m²)(-8m²) = 8dm² - 8m⁴Adding both terms:4m⁴ - 8m² - 12 + 8dm² - 8m⁴ = (-4m⁴) + ( -8m² + 8dm² ) -12Factor terms:= -4m⁴ + 8m²(d - 1) - 12Therefore, DM · DN = [ -4m⁴ + 8m²(d - 1) -12 ] / (3 + 4m²) + d² + m²Now, to combine the terms, let's write the entire expression:DM · DN = [ -4m⁴ + 8m²(d - 1) -12 ] / (3 + 4m²) + d² + m²Let me combine this into a single fraction. Let's write the second part as [d² + m²]*(3 + 4m²)/(3 + 4m²):So,DM · DN = [ -4m⁴ + 8m²(d -1) -12 + (d² + m²)(3 + 4m²) ] / (3 + 4m²)Expand (d² + m²)(3 + 4m²):= 3d² + 4d²m² + 3m² + 4m⁴Therefore, numerator becomes:-4m⁴ + 8m²(d -1) -12 + 3d² + 4d²m² + 3m² + 4m⁴Simplify term by term:-4m⁴ + 4m⁴ = 08m²(d -1) + 3m² = 8dm² -8m² + 3m² = 8dm² -5m²-12 + 3d² + 4d²m²So overall:0 + (8dm² -5m²) + (-12 + 3d² + 4d²m²)Combine like terms:For m⁴: 0For m² terms: (8d -5)m²For constants: -12 +3d²For m² terms multiplied by d²: 4d²m²So total numerator:4d²m² + (8d -5)m² + 3d² -12Factor m² terms:m²(4d² + 8d -5) + (3d² -12)Therefore, DM · DN = [ m²(4d² +8d -5) + 3d² -12 ] / (3 +4m²)We need this expression to be constant for all m. That is, the expression must be independent of m. Therefore, the coefficient of m² in the numerator must be proportional to the denominator's coefficient of m². Let's see:The denominator is 3 +4m². The numerator is m²(4d² +8d -5) +3d² -12.For the entire expression to be constant, the ratio [ numerator ] / [ denominator ] must not depend on m. Therefore, the numerator must be a scalar multiple of the denominator plus a constant. Let's set:Numerator = k*(3 +4m²) for some constant k.Therefore:m²(4d² +8d -5) +3d² -12 = 3k +4k m²Equate coefficients:Coefficient of m²: 4d² +8d -5 =4kConstant term: 3d² -12 =3kSo we have two equations:1. 4d² +8d -5 =4k2. 3d² -12 =3k => k = d² -4Substitute k from equation 2 into equation 1:4d² +8d -5 =4*(d² -4)Expand the right side:4d² +8d -5 =4d² -16Subtract 4d² from both sides:8d -5 = -16Then,8d = -16 +5 = -11d = -11/8Therefore, d = -11/8. Let's check if this satisfies the equations.First, compute k = d² -4 = (121/64) -4 = (121 - 256)/64 = (-135)/64Now check equation 1:4d² +8d -5 =4*(121/64) +8*(-11/8) -5 = (484/64) - (88/8) -5 = (484/64) -11 -5 = (484/64) -16Convert 16 to 1024/64:484/64 - 1024/64 = (-540)/64 = (-135)/164k =4*(-135/64) = (-540)/64 = (-135)/16Yes, they are equal. So d = -11/8 and k = -135/64.Therefore, DM · DN = k = -135/64 for all m. Wait, but according to our setup, we set numerator = k*(denominator). Therefore,DM · DN = [k*(3 +4m²)]/(3 +4m²) = k = -135/64.But wait, actually, if numerator =k*(denominator), then DM · DN = k. So yes, it is constant. Therefore, such a point D exists at d = -11/8 on the x-axis, and the constant value is -135/64.But wait, let me confirm.Wait, the numerator is equal to k*(denominator). Therefore, DM · DN = numerator / denominator = k. Therefore, k is the constant value. So, with d = -11/8, k = -135/64. Therefore, the dot product is -135/64.But the problem states "is there a fixed point D on the x-axis such that DM · DN is constant". So according to this, yes, D is at (-11/8, 0) and the constant is -135/64.But let's verify this with an example. Let's take a specific line l through F₁ (-1, 0) and compute DM · DN for D=(-11/8, 0) and check if it's indeed -135/64.Take line l as the x-axis itself. Wait, but the x-axis passes through F₁ (-1,0) and intersects the ellipse at vertices A (-2,0) and B (2,0). But wait, the x-axis is the major axis. Wait, but in our case, the line is passing through F₁ (-1,0). The x-axis passes through F₁, so it's a horizontal line. So points M and N would be A (-2,0) and another point? Wait, but x-axis intersects the ellipse at (-2,0) and (2,0). But since the line passes through (-1,0), which is the focus. Wait, the x-axis passes through (-1,0), but it's a valid line. Wait, but if we take the x-axis as line l, then points M and N are (-2,0) and (2,0). Then DM · DN would be vectors from D=(-11/8,0) to (-2,0) and (2,0).Compute vectors DM and DN:DM = (-2 - (-11/8), 0) = (-2 +11/8, 0) = (-5/8, 0)DN = (2 - (-11/8), 0) = (2 +11/8, 0) = (27/8, 0)Dot product DM · DN = (-5/8)(27/8) + 0*0 = (-135)/64. Which matches the constant value. So that's correct.Take another line, say vertical line through F₁. Wait, a vertical line through (-1,0) would be x = -1. Substituting into the ellipse equation:(-1)^2/4 + y²/3 =1 => 1/4 + y²/3 =1 => y² = 3*(3/4) = 9/4 => y= ±3/2. So points M and N are (-1, 3/2) and (-1, -3/2).Compute DM and DN:DM = (-1 - (-11/8), 3/2 -0) = (3/8, 3/2)DN = (-1 - (-11/8), -3/2 -0) = (3/8, -3/2)Dot product DM · DN = (3/8)(3/8) + (3/2)(-3/2) = 9/64 - 9/4 = (9 - 144)/64 = (-135)/64. Again, matches. So this seems to hold.Another example: take a line with slope 1 through F₁ (-1,0). Equation: y = x +1. Substitute into the ellipse:x²/4 + (x +1)^2/3 =1Multiply by 12:3x² +4(x² +2x +1)=123x² +4x² +8x +4 =127x² +8x -8=0Solve:x = [-8 ±√(64 +224)]/(14) = [-8 ±√288]/14 = [-8 ±12√2]/14 = [-4 ±6√2]/7So points M and N have x-coordinates (-4 +6√2)/7 and (-4 -6√2)/7, and y-coordinates x +1.Compute vectors DM and DN.First, coordinates of M and N:M: x1 = (-4 +6√2)/7, y1 = x1 +1 = (3 +6√2)/7N: x2 = (-4 -6√2)/7, y2 = x2 +1 = (3 -6√2)/7Coordinates of D: (-11/8,0)Compute DM = (x1 - (-11/8), y1 -0) = (x1 +11/8, y1)Similarly, DN = (x2 +11/8, y2)Compute DM · DN:= (x1 +11/8)(x2 +11/8) + y1 y2First, compute x1 +11/8 and x2 +11/8:x1 +11/8 = [(-4 +6√2)/7 +11/8] = [(-32 +48√2 +77)/56] = (45 +48√2)/56Wait, let's compute numerator:-4*8 +6√2*8 +11*7 = -32 +48√2 +77 =45 +48√2So (45 +48√2)/56Similarly, x2 +11/8 = [(-4 -6√2)/7 +11/8] = (-32 -48√2 +77)/56 = (45 -48√2)/56Multiply these two terms:[(45 +48√2)(45 -48√2)] / (56*56) = [45² - (48√2)²]/3136Compute numerator:2025 - (48² * 2) = 2025 - (2304 * 2) = 2025 -4608 = -2583So first part: -2583/3136Now compute y1 y2:y1 y2 = [(3 +6√2)/7][(3 -6√2)/7] = [9 - (6√2)^2]/49 = [9 -72]/49 = (-63)/49 = -9/7Therefore, total dot product:-2583/3136 + (-9/7) = -2583/3136 - 4032/3136 = (-2583 -4032)/3136 = (-6615)/3136But wait, according to our earlier conclusion, it should be -135/64 ≈ -2.109375But let's compute -6615/3136:Divide numerator and denominator by 7: -945/448 ≈ -2.109375. And -135/64 ≈ -2.109375. Wait, but -945/448 simplifies.Wait, 945 ÷ 7 = 135, 448 ÷7=64. Therefore, -945/448 = -135/64. So indeed, -6615/3136 = -135/64.Therefore, the dot product is indeed -135/64. So this holds. Therefore, point D=(-11/8,0) works.Therefore, the answer to part 2 is yes, such a point D exists at (-11/8, 0), and the constant is -135/64.Final Answer1. The standard equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.2. The fixed point ( D ) exists at coordinates (boxed{left( -dfrac{11}{8}, 0 right)}) with the constant value of the dot product being boxed{-dfrac{135}{64}}.</think>Given an ellipse ( C: dfrac{x^2}{a^2} + dfrac{y^2}{b^2} = 1 ) with ( a > b > 0 ), vertices ( A ) and ( B ), eccentricity ( e = dfrac{1}{2} ), and the maximum area of ( triangle PAB ) being ( 2sqrt{3} ):1. Standard Equation of the Ellipse: - The vertices are ( A(-a, 0) ) and ( B(a, 0) ). - The area of ( triangle PAB ) is maximized when ( P ) is at ( (0, b) ) or ( (0, -b) ), giving the maximum area ( a cdot b = 2sqrt{3} ). - The eccentricity ( e = dfrac{1}{2} ) implies ( c = dfrac{a}{2} ) and ( c^2 = a^2 - b^2 ). - Solving ( a^2 - b^2 = left( dfrac{a}{2} right)^2 ) gives ( b^2 = dfrac{3}{4}a^2 ). - Substituting ( b = dfrac{sqrt{3}}{2}a ) into ( a cdot b = 2sqrt{3} ) yields ( a = 2 ) and ( b = sqrt{3} ). - The standard equation is ( boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1} ).2. Fixed Point ( D ) on the x-axis: - The left focus ( F_1 ) is at ( (-1, 0) ). - A line through ( F_1 ) intersects the ellipse at points ( M ) and ( N ). - Using Vieta's formulas and parametric equations, we derive the coordinates of ( M ) and ( N ). - The dot product ( overrightarrow{DM} cdot overrightarrow{DN} ) is found to be constant when ( D ) is at ( left( -dfrac{11}{8}, 0 right) ). - The constant value of the dot product is ( -dfrac{135}{64} ).Final answers:1. The standard equation of the ellipse is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).2. The fixed point ( D ) exists at coordinates (boxed{left( -dfrac{11}{8}, 0 right)}) with the constant value of the dot product being (boxed{-dfrac{135}{64}}).]

🔑:1. From the given conditions, we have the eccentricity e = dfrac{c}{a} = dfrac{1}{2}, and the maximum area of the triangle triangle PAB is dfrac{1}{2} times 2ab = ab = 2sqrt{3}. We also know that a^2 = b^2 + c^2.By solving these equations, we get a = 2, b = sqrt{3}, and c = 1.Therefore, the standard equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.2. Let's assume there is a fixed point D(m, 0) such that the scalar product overrightarrow{DM} cdot overrightarrow{DN} is a constant n.i. When the slope of line l is not zero, the left focus of the ellipse C is F_1(-1, 0). Let's suppose the equation of line l is x = ty - 1.Combining this with the ellipse equation, we have: begin{cases} dfrac{x^2}{4} + dfrac{y^2}{3} = 1 x = ty - 1 end{cases}By eliminating x, we get (3t^2 + 4)y^2 - 6ty - 9 = 0.Let M(x_1, y_1) and N(x_2, y_2).Thus, y_1 + y_2 = dfrac{6t}{3t^2 + 4} and y_1 y_2 = dfrac{-9}{3t^2 + 4}.The vectors overrightarrow{DM} = (x_1 - m, y_1) and overrightarrow{DN} = (x_2 - m, y_2), and their dot product is overrightarrow{DM} cdot overrightarrow{DN} = (x_1 - m)(x_2 - m) + y_1 y_2 = x_1 x_2 - m(x_1 + x_2) + m^2 + y_1 y_2.After substituting x_1, x_2, y_1, and y_2 and simplifying, we find overrightarrow{DM} cdot overrightarrow{DN} = dfrac{(-6m - 15)t^2 - 9}{3t^2 + 4} + (m + 1)^2.For overrightarrow{DM} cdot overrightarrow{DN} to be a constant n, we need dfrac{-6m - 15}{3} = dfrac{-9}{4}, which implies m = -dfrac{11}{8}. At this value of m, we get n = -dfrac{135}{64}.ii. When the slope of line l is zero, the dot product overrightarrow{DM} cdot overrightarrow{DN} = -dfrac{5}{8} times dfrac{27}{8} = -dfrac{135}{64}, which is consistent with the above result.Therefore, there exists a point boxed{Dleft(-dfrac{11}{8}, 0right)} such that the scalar product overrightarrow{DM} cdot overrightarrow{DN} is a constant boxed{n = -dfrac{135}{64}}.

🧠:[<think>Okay, so I have this problem here: Let m be an integer, and the two roots of the equation 3x² + mx - 2 = 0 are both greater than -9/5 and less than 3/7. Then m equals what? Hmm, let's see. I need to find the integer m such that both roots of the quadratic equation lie between -9/5 and 3/7. First, let me recall some properties of quadratic equations and their roots. The roots of a quadratic equation ax² + bx + c = 0 can be found using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a). But maybe there's a way to approach this problem without calculating the roots directly? Maybe using inequalities and the relationships between coefficients and roots?Given that both roots are greater than -9/5 and less than 3/7, I need to ensure that both roots satisfy -9/5 < α < 3/7 and -9/5 < β < 3/7. Since the quadratic is 3x² + mx - 2 = 0, the leading coefficient is positive (3), so the parabola opens upwards. Therefore, the quadratic will have a minimum point between its two roots (if they are real and distinct). Wait, but in this case, the roots are both between -9/5 and 3/7. So the entire parabola is "squeezed" between these two values? Hmm, not exactly, because the parabola opens upwards, but the roots are where it crosses the x-axis. So between the two roots, the parabola is below the x-axis, and outside the roots, it's above. But since both roots are between -9/5 and 3/7, that means that outside of this interval, the parabola is positive. But the interval given is from -9/5 to 3/7. Wait, but the roots are both inside that interval? Wait, no, the roots are both greater than -9/5 and less than 3/7. So both roots are within (-9/5, 3/7). So the entire part of the parabola where it crosses the x-axis is within that interval. Therefore, the quadratic must cross the x-axis twice within that interval. So first, I need to ensure that the quadratic has two real roots. The discriminant should be positive. The discriminant D = m² - 4*3*(-2) = m² + 24. Since m² is always non-negative, D = m² + 24 is always positive. So the quadratic will always have two real roots, regardless of the value of m. So that condition is automatically satisfied. Next, both roots must lie within (-9/5, 3/7). Let me denote the roots as α and β. Since the quadratic is 3x² + mx - 2 = 0, by Vieta's formulas, the sum of the roots α + β = -m/3, and the product αβ = -2/3. But how do I use the fact that both roots are within that interval? Maybe I can use inequalities based on the positions of the roots. For a quadratic equation ax² + bx + c = 0, if both roots are less than a certain value, say k, then certain conditions must hold. Similarly, if both roots are greater than a certain value, say h, then other conditions must hold. There's a theorem or method for this, I think.Alternatively, I can consider the quadratic function f(x) = 3x² + mx - 2. If both roots are between -9/5 and 3/7, then f(x) must cross the x-axis twice within that interval. Since the parabola opens upwards, the function must be negative between the two roots and positive outside. Therefore, at the endpoints x = -9/5 and x = 3/7, the function should be positive (since outside the roots, it's positive). Also, the minimum of the parabola, which occurs at x = -b/(2a) = -m/(2*3) = -m/6, should lie within the interval (-9/5, 3/7) because the roots are on both sides of the vertex. So the vertex's x-coordinate should be within the interval where the roots are. Additionally, since both roots are within the interval, the function evaluated at the lower bound (-9/5) and upper bound (3/7) must be positive. Also, the minimum value of the function (at x = -m/6) should be negative because the parabola dips below the x-axis between the two roots. So, let's outline the conditions:1. The function f(-9/5) > 0.2. The function f(3/7) > 0.3. The vertex x-coordinate -m/6 is between -9/5 and 3/7.4. The minimum value of the function (at x = -m/6) is negative.Additionally, since both roots are real and distinct, which we already know from the discriminant being positive.Let me write these conditions down step by step.First, calculate f(-9/5):f(-9/5) = 3*(-9/5)² + m*(-9/5) - 2.Compute that:3*(81/25) - (9m/5) - 2 = (243/25) - (9m/5) - 2.Convert 2 to 50/25:= 243/25 - 9m/5 - 50/25 = (243 - 50)/25 - 9m/5 = 193/25 - 9m/5.This must be greater than 0:193/25 - 9m/5 > 0.Multiply both sides by 25 to eliminate denominators:193 - 45m > 0 → 193 > 45m → m < 193/45 ≈ 4.288... Since m is an integer, m ≤ 4.Second, compute f(3/7):f(3/7) = 3*(3/7)² + m*(3/7) - 2.Calculate:3*(9/49) + (3m/7) - 2 = 27/49 + 3m/7 - 2.Convert 2 to 98/49:= 27/49 + 3m/7 - 98/49 = (27 - 98)/49 + 3m/7 = (-71)/49 + 3m/7.This must be greater than 0:-71/49 + 3m/7 > 0.Multiply both sides by 49:-71 + 21m > 0 → 21m > 71 → m > 71/21 ≈ 3.380... Since m is an integer, m ≥ 4.So from the first two conditions, m must be an integer that's ≤4 and ≥4, so m=4. But wait, maybe we need to check other conditions as well? Because this is just the endpoints. There could be more restrictions from the vertex position and the minimum value.Third condition: The vertex x-coordinate is -m/6, which must lie between -9/5 and 3/7. So:-9/5 < -m/6 < 3/7.Let me solve these inequalities.First, -9/5 < -m/6.Multiply both sides by -6, but reversing the inequality:( -9/5 )*(-6) > m → 54/5 > m → m < 10.8.Second inequality: -m/6 < 3/7.Multiply both sides by -6, reversing inequality:m > -18/7 ≈ -2.571... So m > -2.571. Since m is integer, m ≥ -2.But from previous conditions, m was found to be 4. Wait, but this is conflicting? Wait, when solving for the vertex position, we have:-9/5 < -m/6 < 3/7.Let me write each inequality separately.Left inequality: -9/5 < -m/6Multiply both sides by -6 (inequality flips):( -9/5 )*(-6) > m → 54/5 > m → 10.8 > m → m < 10.8.Right inequality: -m/6 < 3/7Multiply both sides by -6 (inequality flips):m > -18/7 → m > approximately -2.571.So from the vertex position, m must be greater than -2.571 and less than 10.8. Since m is integer, m ≥ -2 and m ≤ 10. But earlier from the first two conditions, m had to be 4. So even though the vertex allows m from -2 to 10, the first two conditions narrow it down to m=4. But maybe m=4 is the only candidate, but we need to check if with m=4, the roots are actually within the interval.Fourth condition: The minimum value of the function is negative. Since the parabola opens upwards, the minimum value occurs at x = -m/6. Let's compute f(-m/6):f(-m/6) = 3*(-m/6)² + m*(-m/6) - 2.Calculate:3*(m²/36) - m²/6 - 2 = (3m²)/36 - (6m²)/36 - 2 = (-3m²)/36 - 2 = -m²/12 - 2.This must be less than 0:- m²/12 - 2 < 0 → -m²/12 < 2 → m²/12 > -2. But since m² is always non-negative, m²/12 is ≥ 0, so -m²/12 - 2 is always less than 0? Wait, if m is any real number, then -m²/12 -2 is always negative because -m²/12 is ≤0, so subtracting 2 makes it ≤ -2 < 0. Therefore, the minimum is always negative. So this condition is automatically satisfied. So we don't need to worry about that.So the key conditions are:1. f(-9/5) > 0 ⇒ m < 193/45 ≈4.288 ⇒ m ≤4.2. f(3/7) > 0 ⇒ m >71/21 ≈3.380 ⇒ m ≥4.3. The vertex x-coordinate must be between -9/5 and 3/7. For m=4, check:x = -m/6 = -4/6 = -2/3 ≈ -0.666...Compare to -9/5 = -1.8 and 3/7 ≈0.428. So -1.8 < -0.666... < 0.428. Yes, it is within the interval. So condition 3 is satisfied.But even with m=4, we need to check if both roots are actually within (-9/5, 3/7). Let's compute the roots when m=4.The equation is 3x² +4x -2=0. Using quadratic formula:x = [-4 ± √(16 - 4*3*(-2))]/(2*3) = [-4 ± √(16 +24)]/6 = [-4 ± √40]/6 = [-4 ± 2√10]/6 = [-2 ± √10]/3.Compute approximate values:√10 ≈3.1623.So,First root: (-2 + 3.1623)/3 ≈1.1623/3≈0.3874.Second root: (-2 -3.1623)/3 ≈-5.1623/3≈-1.7208.Wait, but the second root is approximately -1.7208, which is less than -9/5 = -1.8? Wait, -1.7208 is greater than -1.8. Because -1.7208 is approximately -1.72, which is greater than -1.8. So both roots are:≈0.3874 (which is less than 3/7≈0.428) and ≈-1.7208 (which is greater than -1.8). So both roots are within the interval (-1.8, 0.428). So yes, with m=4, both roots are within the required interval.Wait, but let's check more precisely. Let me compute the exact values.First, compute the roots when m=4:3x² +4x -2=0.Roots:x = [-4 ±√(16 +24)]/6 = [-4 ±√40]/6 = [-4 ±2√10]/6 = (-2 ±√10)/3.So, exact roots: (-2 + √10)/3 and (-2 -√10)/3.Compute numerical values:√10 ≈3.16227766017.First root: (-2 +3.16227766017)/3 ≈1.16227766017/3≈0.3874258867.Second root: (-2 -3.16227766017)/3≈-5.16227766017/3≈-1.72075922005.Compare to -9/5= -1.8 and 3/7≈0.42857142857.So, first root ≈0.3874 < 0.4285, so it's within the upper bound.Second root ≈-1.7208 > -1.8, so within the lower bound.Thus, both roots are indeed within (-9/5, 3/7). Therefore, m=4 satisfies all conditions.But wait, before finalizing, let me check if there are other possible integer values of m. From the first two conditions, m must be ≥4 and ≤4, so only m=4. But just to be thorough, let's check m=3 and m=5 to see if they could possibly satisfy, in case I made an error in reasoning.First, m=3:Check f(-9/5) =193/25 -9*3/5=193/25 -27/5=193/25 -135/25=58/25>0, which is true.f(3/7)= -71/49 +3*3/7= -71/49 +9/7= -71/49 +63/49= -8/49 <0. Which is not greater than 0. Therefore, m=3 doesn't satisfy the second condition.Now m=5:f(-9/5)=193/25 -45/5=193/25 -9=193/25 -225/25= -32/25 <0. Which doesn't satisfy the first condition.Similarly, m=4 is the only possible integer between 3.38 and 4.288, so m=4.But let me check if the vertex is within the interval for m=4. The vertex x-coordinate is -4/6 = -2/3≈-0.666..., which is between -1.8 and 0.428. Yes, correct.Also, the minimum value is negative, which we already confirmed.Therefore, m=4 is the answer.Final Answerboxed{4}</think>Given the quadratic equation (3x^2 + mx - 2 = 0) with integer (m), we need to find (m) such that both roots are between (-frac{9}{5}) and (frac{3}{7}).1. Discriminant Check: The discriminant (D = m^2 + 24) is always positive, ensuring two real roots.2. Function Values at Endpoints: - For (x = -frac{9}{5}): [ fleft(-frac{9}{5}right) = frac{193}{25} - frac{9m}{5} > 0 implies 193 - 45m > 0 implies m < frac{193}{45} approx 4.288 implies m leq 4 ] - For (x = frac{3}{7}): [ fleft(frac{3}{7}right) = -frac{71}{49} + frac{3m}{7} > 0 implies -71 + 21m > 0 implies m > frac{71}{21} approx 3.380 implies m geq 4 ] Combining these, (m = 4).3. Vertex Position: The vertex (x = -frac{m}{6}) must lie between (-frac{9}{5}) and (frac{3}{7}): [ -frac{9}{5} < -frac{4}{6} < frac{3}{7} implies -1.8 < -0.666ldots < 0.428 ] This is satisfied for (m = 4).4. Roots Check: For (m = 4), solving (3x^2 + 4x - 2 = 0) gives roots: [ x = frac{-2 pm sqrt{10}}{3} ] Approximating roots: - (frac{-2 + sqrt{10}}{3} approx 0.3874) - (frac{-2 - sqrt{10}}{3} approx -1.7208) Both roots lie within (-frac{9}{5} approx -1.8) and (frac{3}{7} approx 0.428).Thus, the integer (m) that satisfies all conditions is (boxed{4}).,<think>Okay, so I need to find the integer m such that the two roots of the quadratic equation 3x² + mx - 2 = 0 are both greater than -9/5 and less than 3/7. Hmm, let's start by recalling some properties of quadratic equations and their roots.First, the general form of a quadratic equation is ax² + bx + c = 0, and the roots can be found using the quadratic formula: [-b ± √(b² - 4ac)]/(2a). In this case, a = 3, b = m, and c = -2. So the roots would be [-m ± √(m² - 4*3*(-2))]/(2*3) = [-m ± √(m² + 24)]/6.But maybe instead of working directly with the roots, I can use the relationships between the coefficients and the roots. Let me denote the roots as α and β. Then by Vieta's formulas, we have:α + β = -m/3andαβ = -2/3.The problem states that both roots are greater than -9/5 and less than 3/7. So, -9/5 < α < 3/7 and -9/5 < β < 3/7. Therefore, both roots lie in the interval (-9/5, 3/7). I need to find m such that this condition is satisfied.Hmm, so how do I translate these inequalities into conditions on m? Let's think. If both roots lie within that interval, then certain conditions on the quadratic function must hold. For example, since the quadratic opens upwards (because the coefficient of x² is positive, 3), the quadratic must be above the x-axis outside the interval (-9/5, 3/7). Wait, no, actually, if both roots are within that interval, then the quadratic must cross the x-axis twice within that interval. But since the leading coefficient is positive, the parabola opens upwards, so it will be negative between the two roots and positive outside. Therefore, if both roots are between -9/5 and 3/7, then the quadratic should be positive when x < -9/5, negative between -9/5 and 3/7 (but since there are two roots in that interval, actually, it would be negative between the two roots and positive outside. Wait, but if both roots are in (-9/5, 3/7), then since the parabola opens upwards, the quadratic will be negative between the two roots and positive outside. Therefore, to ensure that both roots are in (-9/5, 3/7), the quadratic must be positive at x = -9/5 and at x = 3/7, right? Because those points are outside the interval where the roots are, so the quadratic should be positive there.But wait, actually, if both roots are less than 3/7 and greater than -9/5, then the entire interval between -9/5 and 3/7 might contain the two roots. However, depending on the specific roots, they could be both in the left part or both in the right part. Wait, but since the quadratic is positive outside the roots, which are both inside (-9/5, 3/7), then the quadratic should be positive at x = -9/5 and at x = 3/7. Let me verify this. For example, if both roots are between -9/5 and 3/7, then when x < -9/5 (left of the interval), the quadratic is positive (since it opens upwards), and when x > 3/7 (right of the interval), the quadratic is also positive. So yes, substituting x = -9/5 and x = 3/7 into the quadratic equation should give positive values. That's one condition.But also, since both roots are within that interval, the quadratic must have two real roots, so the discriminant must be non-negative. The discriminant here is m² + 24, which is always positive because m² is non-negative and 24 is positive. So the equation always has two distinct real roots, regardless of m. So that's already satisfied.Another condition is that the vertex of the parabola must lie within the interval (-9/5, 3/7). Wait, not necessarily. Because the roots could be on either side of the vertex. Wait, since the parabola opens upwards, the vertex is the minimum point. The roots are on either side of the vertex. But if both roots are in (-9/5, 3/7), then the vertex must also lie between those roots, hence also in (-9/5, 3/7). So the x-coordinate of the vertex, which is -b/(2a) = -m/(2*3) = -m/6, should lie between the two roots. But since the two roots are both in (-9/5, 3/7), then the vertex must also lie in that interval. Therefore, -m/6 must be between α and β. But since α and β are both in (-9/5, 3/7), then -m/6 must be in (-9/5, 3/7). So:-9/5 < -m/6 < 3/7Multiply all parts by -6 (remembering to reverse the inequalities):( -9/5 )*(-6) > m > (3/7)*(-6)Which simplifies to:54/5 > m > -18/754/5 is 10.8, and -18/7 is approximately -2.571. So m must be between -2.571 and 10.8. But m is an integer, so m must be -2, -1, 0, 1, ..., 10. But we need to check which of these integers satisfy the other conditions as well.But perhaps there are more conditions. For example, the quadratic must be positive at x = -9/5 and x = 3/7. Let's compute these values.First, substitute x = -9/5 into 3x² + mx - 2:3*(-9/5)² + m*(-9/5) - 2 = 3*(81/25) - (9m)/5 - 2 = 243/25 - 45m/25 - 50/25 = (243 - 45m - 50)/25 = (193 - 45m)/25.Since this should be positive (as the quadratic is positive outside the roots), we have:(193 - 45m)/25 > 0 => 193 - 45m > 0 => 45m < 193 => m < 193/45 ≈ 4.288. So m < 4.288. Since m is integer, m ≤ 4.Similarly, substitute x = 3/7 into the quadratic:3*(3/7)² + m*(3/7) - 2 = 3*(9/49) + (3m)/7 - 2 = 27/49 + 21m/49 - 98/49 = (27 + 21m - 98)/49 = (21m - 71)/49.This should also be positive, so:(21m - 71)/49 > 0 => 21m - 71 > 0 => 21m > 71 => m > 71/21 ≈ 3.380. So m > 3.380, meaning m ≥ 4 (since m is integer).Wait, but from x = -9/5, we have m < 4.288, so m ≤ 4. From x = 3/7, we have m > 3.380, so m ≥ 4. Therefore, m must be 4. But let's check if m=4 actually satisfies all the conditions.So if m=4, let's check the quadratic equation: 3x² + 4x - 2 = 0.Compute the roots:x = [-4 ± √(16 + 24)]/6 = [-4 ± √40]/6 = [-4 ± 2√10]/6 = [-2 ± √10]/3.Compute approximate values: √10 ≈ 3.162, so:[-2 + 3.162]/3 ≈ 1.162/3 ≈ 0.387[-2 - 3.162]/3 ≈ -5.162/3 ≈ -1.720Wait, but the roots are approximately 0.387 and -1.720. The problem states that both roots must be greater than -9/5 (-1.8) and less than 3/7 (≈0.428). Let's check:First root: 0.387 is less than 3/7 ≈0.428, so that's okay.Second root: -1.720 is greater than -9/5 = -1.8, right? -1.720 is -1.72, which is greater than -1.8. So both roots are within the interval (-1.8, 0.428). Therefore, m=4 seems to satisfy the conditions.But wait, let's verify the values at x = -9/5 and x = 3/7 when m=4.First, at x = -9/5:(193 - 45m)/25 = (193 - 45*4)/25 = (193 - 180)/25 = 13/25 > 0. Good.At x = 3/7:(21m - 71)/49 = (21*4 - 71)/49 = (84 - 71)/49 = 13/49 > 0. Also good.Additionally, the vertex is at x = -m/6 = -4/6 = -2/3 ≈ -0.666, which is within the interval (-9/5, 3/7) since -9/5 = -1.8 and 3/7 ≈0.428, so -0.666 is between them. So the vertex is inside the interval, which is correct.Moreover, we can check the exact roots. Let's compute them precisely.For m=4:The roots are [-4 ± √(16 + 24)]/6 = [-4 ± √40]/6 = [-4 ± 2√10]/6 = (-2 ± √10)/3.Calculate √10 ≈3.16227766.First root: (-2 + 3.16227766)/3 ≈1.16227766/3 ≈0.387425886, which is approximately 0.387, as before. So that's less than 3/7 ≈0.42857.Second root: (-2 - 3.16227766)/3 ≈(-5.16227766)/3 ≈-1.72075922, which is approximately -1.7208. Which is greater than -9/5 = -1.8.Therefore, both roots are indeed within the required interval. So m=4 is a candidate.But let's check if there are other possible integer values for m. From the previous analysis, when we substituted the endpoints x = -9/5 and x = 3/7 into the quadratic, we found that m must satisfy m < 193/45 ≈4.288 and m > 71/21 ≈3.380. Since m is an integer, the only possible integer is m=4.But let's double-check for m=3 and m=5 to see why they don't work.For m=3:Check the quadratic 3x² +3x -2=0.Discriminant: 9 +24=33. Roots: [-3 ±√33]/6.√33≈5.7446.First root: (-3 +5.7446)/6≈2.7446/6≈0.4574, which is greater than 3/7≈0.4286, so that root is outside the upper bound. Therefore, m=3 is invalid.For m=5:The quadratic is 3x² +5x -2=0.Discriminant:25 +24=49. Roots: [-5 ±7]/6.First root: (2)/6=1/3≈0.333, second root: (-12)/6=-2.But the root -2 is less than -9/5=-1.8, so that's outside the lower bound. Therefore, m=5 is invalid.So indeed, only m=4 works. However, let's check m=4 again. Wait, when m=4, the roots are approximately 0.387 and -1.720. Both are within (-1.8, 0.428). So that's correct.But let me just make sure that all conditions are covered. The conditions we considered were:1. Quadratic positive at x = -9/5 and x = 3/7. (Which gave m ≤4 and m ≥4, so m=4.)2. The vertex is within the interval. For m=4, vertex at x=-4/6=-2/3≈-0.666, which is within (-1.8, 0.428).3. The roots are within the interval. As calculated, they are.Additionally, perhaps another condition is that the quadratic has two real roots, but since discriminant is always positive, as we saw, so that's already satisfied.Alternatively, maybe we can use the concept of interval containment for roots. If both roots are greater than -9/5 and less than 3/7, then:Let me recall that for a quadratic equation ax² + bx + c =0 with roots α and β, if both roots are less than a value k, then af(k) >0 and the axis of symmetry is less than k. Similarly, if both roots are greater than h, then af(h) >0 and the axis of symmetry is greater than h.In our case, both roots are greater than -9/5 and less than 3/7. Since a=3>0, the parabola opens upwards, so:1. For both roots to be greater than -9/5: f(-9/5) >0 and the axis of symmetry > -9/5.2. For both roots to be less than 3/7: f(3/7) >0 and the axis of symmetry < 3/7.Which is exactly the conditions we applied earlier. So substituting x=-9/5 and x=3/7 into the quadratic gives positive values, and the axis of symmetry (vertex x-coordinate) is between -9/5 and 3/7.Which we already checked. Therefore, m=4 is the only integer that satisfies all these conditions. So the answer is m=4.Final AnswerThe value of ( m ) is boxed{4}.</think>To find the integer ( m ) such that the two roots of the quadratic equation ( 3x^2 + mx - 2 = 0 ) are both greater than (-frac{9}{5}) and less than (frac{3}{7}), we start by noting the roots must lie within the interval ((- frac{9}{5}, frac{3}{7})). Using Vieta's formulas, the roots (alpha) and (beta) satisfy:[alpha + beta = -frac{m}{3} quad text{and} quad alphabeta = -frac{2}{3}]For the quadratic to be positive at ( x = -frac{9}{5} ) and ( x = frac{3}{7} ) (since the parabola opens upwards), we substitute these values into the quadratic equation:1. Substituting ( x = -frac{9}{5} ):[3left(-frac{9}{5}right)^2 + mleft(-frac{9}{5}right) - 2 > 0 implies frac{193 - 45m}{25} > 0 implies 193 - 45m > 0 implies m < frac{193}{45} approx 4.288]2. Substituting ( x = frac{3}{7} ):[3left(frac{3}{7}right)^2 + mleft(frac{3}{7}right) - 2 > 0 implies frac{21m - 71}{49} > 0 implies 21m - 71 > 0 implies m > frac{71}{21} approx 3.380]Since ( m ) must be an integer, the possible values are ( 4 ).Checking the vertex ( x = -frac{m}{6} ) must lie within ((- frac{9}{5}, frac{3}{7})):[- frac{9}{5} < -frac{m}{6} < frac{3}{7} implies frac{54}{5} > m > -frac{18}{7}]This interval includes ( m = 4 ).Verifying the roots for ( m = 4 ):[3x^2 + 4x - 2 = 0 implies x = frac{-4 pm sqrt{16 + 24}}{6} = frac{-4 pm 2sqrt{10}}{6} = frac{-2 pm sqrt{10}}{3}]The roots are approximately ( 0.387 ) and ( -1.720 ), both within ((-1.8, 0.428)).Thus, the value of ( m ) is (boxed{4}).]

🔑:2.4 .From the problem, we haveleft{begin{array}{l}3 timesleft(-frac{9}{5}right)^{2}+m timesleft(-frac{9}{5}right)-2>0, 3 timesleft(frac{3}{7}right)^{2}+m timesleft(frac{3}{7}right)-2>0 .end{array}right.Solving this, we get (3 frac{8}{21}<m<4 frac{13}{45}). Therefore, (m=4).