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Q.
Let $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$. If $\int\limits_{\cos x}^{1} t ^{2} f ( t ) dt =\sin ^{3} x +\cos x$ then $\frac{1}{\sqrt{3}} f ^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to :
At right hand vicinity of $x =0$ given equation does not satisfy
$\because \text { LHS }=\int\limits_{1^{-}}^{1} t ^{2} f ( t ) dt =0, \text { RHS }=\displaystyle\lim _{ x \rightarrow 0^{+}}\left(\sin ^{3} x +\cos x \right)=1$
LHS $\neq$ RHS hence data given in question is wrong hence BONUS
Correct data should have been
$\int\limits_{\cos x}^{1} t^{2} f(t) d t=\sin ^{3} x+\cos x-1$
Calculation for option
differentiating both sides
$-\cos ^{2} x f(\cos x) \cdot(-\sin x)=3 \sin ^{2} x \cdot \cos x-\sin x$
$\Rightarrow f(\cos x)=3 \tan x-\sec ^{2} x$
$\Rightarrow f^{\prime}(\cos x)(-\sin x)=3 \sec ^{2} x-2 \sec ^{2} x \tan x$
$\Rightarrow f^{\prime}(\cos x) \cos x=\frac{2}{\cos ^{2} x}-\frac{3}{\sin x \cdot \cos x}$
When $\cos x=\frac{1}{\sqrt{3}} ; \sin x=\frac{\sqrt{2}}{\sqrt{3}}$
$\therefore f^{\prime}\left(\frac{1}{\sqrt{3}}\right) \frac{1}{\sqrt{3}}=6-\frac{9}{\sqrt{2}}$