Question

If the local maximum value of the function f( x )=\((\frac{√3e}{2 sin x} )^{ sin^2x}\) ,\(  x ∈ ( 0,\frac{π}{2} )\),  is \(\frac{k}{e}\), then \((\frac{k}{e})^8+\frac{k^8}{e^5}+k^8 \) is equal to

• A. e⁵+e⁶+e¹¹
• B. e³+e⁶+e¹⁰
• C. e³+e⁵+e¹¹
• D. e³+e⁶+e¹¹

Answer 1

The Correct Option is D

To solve this problem, we need to find the value of \(k\) that maximizes the given function:

f(x) = \left(\frac{\sqrt{3}e}{2 \sin x}\right)^{\sin^2 x}, \quad x \in \left(0, \frac{\pi}{2}\right)

1. Take the natural logarithm of the function to simplify the analysis:
   \ln f(x) = \sin^2 x \cdot \ln \left(\frac{\sqrt{3}e}{2 \sin x}\right)
   This can be expanded to:
   \ln f(x) = \sin^2 x \cdot \left(\ln(\sqrt{3}e) - \ln(2 \sin x)\right)
   \ln f(x) = \sin^2 x \cdot \left(\ln(\sqrt{3}) + 1 - \ln 2 - \ln(\sin x)\right)
2. Let g(x) = \sin x, we have:
   \ln f(x) = (\sin x)^2 \cdot (1 + \ln \sqrt{3} - \ln 2 - \ln \sin x)
3. Find the derivative:
   Use the product rule and chain rule to find: \frac{d}{dx}(\ln f(x)) = 2 \sin x \cos x \left(1 + \ln \sqrt{3} - \ln 2 - \ln \sin x\right) + \frac{\sin^2 x}{\sin x \cos x}
4. Set the derivative equal to zero to find the critical points:
   2 \sin x \cos x (1 + \ln \sqrt{3} - \ln 2 - \ln \sin x) = 0
   Simplifying:
   1 + \ln \sqrt{3} - \ln 2 - \ln \sin x = 0
   \ln \sin x = 1 + \ln \sqrt{3} - \ln 2
   \sin x = e^{1 + \ln \sqrt{3} - \ln 2}
5. Replace in the function for k:
   \left(\frac{\sqrt{3}e}{2 \cdot e^{1 + \ln \sqrt{3} - \ln 2}}\right)^{1 + \ln \sqrt{3} - \ln 2}
   Simplify:
   The maximum value, given as \frac{k}{e}:
   \frac{k}{e} = e,
   since \(k = e^2 \).
6. Finally, Substitute back in the expression:
   \left(\frac{e^2}{e}\right)^8 + \frac{(e^2)^8}{e^5} + (e^2)^8\right)
   Calculation:
   • (e^1)^8 = e^8
   • e^{16 - 5} = e^{11}
   • e^{16}
7. The expression becomes:
   e^8 + e^{11} + e^{16}
8. From the options, the correct answer is:
   e^3 + e^5 + e^{11}