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

$\begin{array}{r}Differentiating with respect to x\\ -{\cos }^{2}xf(\cos x)\cdot (-\sin x)=3{\sin }^{2}x\cdot \cos x-\sin x\\ \Rightarrow f(\cos x)=3\tan x-{\sec }^{2}x\\ \Rightarrow f^{\mathrm{\prime }}(\cos x)(-\sin x)=3{\sec }^{2}x-2{\sec }^{2}x\tan x\\ \Rightarrow f^{\mathrm{\prime }}(\cos x)\cos x=\frac{2}{{\cos }^{2}x}-\frac{3}{\sin x\cdot \cos x}\\ \therefore f^{\mathrm{\prime }}\left(\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}}=6-\frac{9}{\sqrt{2}}.\end{array}$

$$\begin{gathered} \text{ At right hand vicinity of }\mathrm{x}=0\text{ given equation does not satisfy } \\ \because \text{ LHS }={\int }_{1^{-}}^1 {\mathrm{t}}^2\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=0,\mathrm{RHS}=\underset{\mathrm{x}\to 0^{+}}{lim} \left({\sin }^3\mathrm{x}+\cos \mathrm{x}\right)=1 \\ \mathrm{LHS}\ne \mathrm{RHS}\text{ hence data given in question is wrong } \end{gathered}$$

$$\begin{gathered} Correct question must be \\ {\int }_{\cos x}^1 t^{2}f\left(t\right)dt={\sin }^{3}x+\cos x-1 \end{gathered}$$