Question

Let $f$ be a differentiable function in $\left(0,\frac{\pi }{2}\right)$. If $\underset{\cos x}{\overset{1}{\int }}{t}^{2}f(t)dt={\sin }^{3}x+\cos x$ then $\frac{1}{\sqrt{3}}{f}^{\prime }\left(\frac{1}{\sqrt{3}}\right)$ is equal to :

• A. $6-9\sqrt{2}$
• B. $6-\frac{9}{\sqrt{2}}$
• C. $\frac{9}{2}-6\sqrt{2}$
• D. $\frac{9}{\sqrt{2}}-6$

Answer 1

Answer: (B)

At right hand vicinity of $x=0$ given equation does not satisfy
$\because \text{ LHS }=\underset{1^{-}}{\overset{1}{\int }}{t}^{2}f(t)dt=0,\text{ RHS }=\underset{x\to 0^{+}}{\lim }\left({\sin }^{3}x+\cos x\right)=1$
LHS $\ne$ RHS hence data given in question is wrong hence BONUS
Correct data should have been
$\underset{\cos x}{\overset{1}{\int }}{t}^{2}f(t)dt={\sin }^{3}x+\cos x-1$
Calculation for option
differentiating both sides
$-{\cos }^{2}xf(\cos x)\cdot (-\sin x)=3{\sin }^{2}x\cdot \cos x-\sin x$
$\Rightarrow f(\cos x)=3\tan x-{\sec }^{2}x$
$\Rightarrow f^{\prime }(\cos x)(-\sin x)=3{\sec }^{2}x-2{\sec }^{2}x\tan x$
$\Rightarrow f^{\prime }(\cos x)\cos x=\frac{2}{{\cos }^{2}x}-\frac{3}{\sin x\cdot \cos x}$
When $\cos x=\frac{1}{\sqrt{3}};\sin x=\frac{\sqrt{2}}{\sqrt{3}}$
$\therefore f^{\prime }\left(\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}}=6-\frac{9}{\sqrt{2}}$