JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Definite Integrals

Differentiating both sides with respect to $x$, the left side (by the fundamental theorem of calculus, with the variable lower limit $\cos x$) becomes ${\cos }^{2}x\sin xf(\cos x)$, while the right side becomes $3{\sin }^{2}x\cos x-\sin x$, the constant contributing nothing. Cancelling $\sin x$ and dividing by ${\cos }^{2}x$ gives $f(\cos x)=3\tan x-{\sec }^{2}x$. Differentiating this relation once more, $-\sin x{f}^{\prime }(\cos x)=3{\sec }^{2}x-2{\sec }^{2}x\tan x$, and multiplying through by $\cos x$ (then using $\tan x\cos x/\sin x=1$) leads to $\cos x{f}^{\prime }(\cos x)=2{\sec }^{2}x-\frac{3}{\sin x\cos x}$. The quantity asked for is exactly $\cos x{f}^{\prime }(\cos x)$ evaluated where $\cos x=\frac{1}{\sqrt{3}}$; there ${\sec }^{2}x=3$ and $\sin x=\sqrt{\frac{2}{3}}$, so $\sin x\cos x=\frac{\sqrt{2}}{3}$ and the value is $2(3)-\frac{3}{\sqrt{2}/3}=6-\frac{9}{\sqrt{2}}$.