Question

Let f be a differentiable function in \((0, \frac{π}{2})\). If \(∫_{cosx} ^1 t^2f(t)dt=sin^3x+cosx\), then \(\frac{1}{\sqrt3}f'(\frac{1}{\sqrt3})\) is equal to

• A. \(6-9\sqrt2\)
• B. \(6-\frac{9}{\sqrt2}\)
• C. \(\frac{9}{2}-6\sqrt2\)
• D. \(\frac{9}{\sqrt 2}-6\)

Answer 1

The Correct Option is B

 To solve this problem, we need to understand the given equation and use differentiation to find \( f'(t) \), then evaluate it at \( t = \frac{1}{\sqrt{3}} \).

We are given that:

\[\int_{cos x}^1 t^2 f(t) \, dt = \sin^3 x + \cos x\]

Our objective is to find \( \frac{1}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) \).

1. Apply the Leibniz Rule for differentiation under the integral sign. The Leibniz Rule states:
2. In this problem, \( u(x) = \cos x \), \( v(x) = 1 \), \( F(t) = t^2 f(t) \). The boundary derivative terms become:
   • The partial derivative of the integrand \( t^2 f(t) \) with respect to \( x \) is zero because there is no explicit \( x \) dependence in \( t^2 f(t) \).
   • The derivative of the upper limit: \( \frac{d}{dx}(1) = 0 \).
   • The derivative of the lower limit: \( \frac{d}{dx}(\cos x) = -\sin x \).
   • Thus the differentiation gives: \( -(t^2 f(t) \mid_{t = \cos x}) (-\sin x) \). This term simplifies to: \( \cos^2 x \sin x f(\cos x) \).
3. The differentiation of the right side \(\sin^3 x + \cos x\) with respect to \( x \) gives:
4. Setting these derivatives equal, we get:
5. Thus, we express \( f(\cos x) \) as:
6. Let \( t = \cos x \). Then, as \( x \to \frac{\pi}{3} \), \( \cos x \to \frac{1}{2} \) and \( \sin^2 x = 1 - \cos^2 x \).
7. The function simplifies to \( f(t) \). Differentiate with respect to \( t \) to find \( f'(t) \).
8. When \( t = \frac{1}{\sqrt{3}} \), the required expression is:
9. Thus, the expression for \( f'(t) \) results from the derivative at that specific \( t \) value:

Therefore, the correct answer is \(6-\frac{9}{\sqrt2}\).