JEE Main 2024 — 29 January, Shift 1 — Question 18

$$f(x)=\frac{\left(2^{x}+2^{-x}\right)\tan x \sqrt{\tan^{-1}(x^{2}-x+1)}}{(7x^{2}+3x+1)^{3}}$$

First evaluate at $x=0$.

$$2^{0}+2^{0}=2$$ $$\tan 0=0$$ $$\tan^{-1}(0^2-0+1)=\tan^{-1}(1)=\frac{\pi}{4}$$ $$\sqrt{\frac{\pi}{4}}=\frac{\sqrt{\pi}}{2}$$ $$(7\cdot0+0+1)^3=1$$

Hence,

$$f(0)=0$$

Now write $f(x)=\frac{N(x)}{D(x)}$.

Since $D(0)=1$ and $N(0)=0$,

$$f'(0)=N'(0)$$

Let

$$N(x)=A(x)\tan x\,B(x)$$

where

$$A(x)=2^x+2^{-x}, \qquad B(x)=\sqrt{\tan^{-1}(x^2-x+1)}$$

Compute values at $x=0$:

$$A(0)=2$$ $$A'(x)=2^x\ln2-2^{-x}\ln2$$ $$A'(0)=\ln2-\ln2=0$$ $$B(0)=\frac{\sqrt{\pi}}{2}$$

Also,

$$\frac{d}{dx}(\tan x)\Big|_{x=0}=1$$

Since $\tan 0=0$, only derivative of $\tan x$ contributes:

$$N'(0)=A(0)\cdot 1 \cdot B(0)$$ $$=2\times\frac{\sqrt{\pi}}{2}$$ $$=\sqrt{\pi}$$

Therefore,

$$\boxed{f'(0)=\sqrt{\pi}}$$