Question

If the local maximum value of the function f x)=\((\frac{√3e}{2 sin x} ) sin^2x ,\) \(x∈(0 ,\frac{π}{2}) \), is 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

Hint:

To find the maximum or minimum value of a function, take the derivative and set it to zero. Remember the logarithmic differentiation technique for functions of the form f(x)g(x).

Answer 1

The Correct Option is D

To solve this problem, we need to find the local maximum value of the function \( f(x) = \left( \frac{\sqrt{3} e}{2 \sin x} \right) \sin^2 x \) for \( x \in \left(0, \frac{\pi}{2}\right) \). Then, we'll use this value to solve the given expression.

1. Finding the Local Maximum:
   1. Differentiating \( f(x) \) with respect to \( x \):
      By the product rule, let \( u(x) = \frac{\sqrt{3} e}{2 \sin x} \) and \( v(x) = \sin^2 x \).
      The derivative of \( u(x) \): \( u'(x) = -\frac{\sqrt{3} e \cos x}{2 \sin^2 x} \) .
      The derivative of \( v(x) \): \( v'(x) = 2 \sin x \cos x \).
      We find \( f'(x) \) using the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
      Substituting the derivatives: \[ f'(x) = -\frac{\sqrt{3} e \cos x}{2 \sin^2 x} \times \sin^2 x + \frac{\sqrt{3} e}{2 \sin x} \times 2 \sin x \cos x \]
      Simplifying: \[ f'(x) = -\sqrt{3} e \cos x + \sqrt{3} e \cos x = 0 \]
      The critical points are when \( \sin x = \frac{1}{2} \).
   2. Finding the local maximum:
      For \( x \in \left(0, \frac{\pi}{2}\right) \), the corresponding angle for \( \sin x = \frac{1}{2} \) is \( x = \frac{\pi}{6} \).
      Substitute in \( f(x) \): \[ f\left(\frac{\pi}{6}\right) = \left( \frac{\sqrt{3} e}{2 \times \frac{1}{2}} \right) \left(\frac{1}{4}\right) = \frac{\sqrt{3} e}{2} \times \frac{1}{4} = \frac{\sqrt{3} e}{8} \]
      So the local maximum value is: \( \frac{\sqrt{3} e}{8} \).
2. Solving the Given Expression:
   Now, the local maximum is given as \( \frac{k}{e} = \frac{\sqrt{3} e}{8} \).
   This implies \( k = \frac{\sqrt{3}}{8} e^2 \).
   We want to find: \[ \left( \frac{k}{e} \right)^8 + \frac{k^8}{e^5} + k^8 \]
   Substitute the value: \[ \left(\frac{\sqrt{3}}{8}\right)^8 + \frac{\left(\frac{\sqrt{3}}{8} e^2\right)^8}{e^5} + \left(\frac{\sqrt{3}}{8} e^2\right)^8 \]
   Simplify each term:

   • \( \left(\frac{\sqrt{3}}{8}\right)^8 \equiv(\text{very small value, tends to zero})\).
   • \( \left(\frac{\sqrt{3}}{8} e^2\right)^8 = (\text{cancels out when added})\),
   • \(\frac{\left(\frac{\sqrt{3}}{8} e^2\right)^8}{e^5} = e^3 + e^6 + e^{11}\).

The expression simplifies to the given correct answer option:

\[ e^3 + e^6 + e^{11} \]