JEE Main 2024 — Mathematics Calculus

Given: $f(x)=\frac{({2}^{x}+{2}^{-x})\mathrm{tan}x\sqrt{{\mathrm{tan}}^{-1}({x}^{2}-x+1)}}{{(7{x}^{2}+3x+1)}^{3}}$f(x)=(7x2+3x+1)3(2x+2−x)tanxtan−1(x2−x+1)​​

We know that, ${f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(h)-f(0)}{h}$f′(0)=h→0lim​hf(h)−f(0)​

$$\Rightarrow {f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\frac{({2}^{h}+{2}^{-h})\mathrm{tan}h\sqrt{{\mathrm{tan}}^{-1}({h}^{2}-h+1)}}{{(7{h}^{2}+3h+1)}^{3}}-\frac{({2}^{0}+{2}^{0})\mathrm{tan}0\sqrt{{\mathrm{tan}}^{-1}(0-0+1)}}{{(0+0+1)}^{3}}}{h}$$⇒f′(0)=h→0lim​h(7h2+3h+1)3(2h+2−h)tanhtan−1(h2−h+1)​​−(0+0+1)3(20+20)tan0tan−1(0−0+1)​​​

$$\Rightarrow {f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{({2}^{h}+{2}^{-h})\mathrm{tan}h\sqrt{{\mathrm{tan}}^{-1}({h}^{2}-h+1)}}{h{(7{h}^{2}+3h+1)}^{3}}$$⇒f′(0)=h→0lim​h(7h2+3h+1)3(2h+2−h)tanhtan−1(h2−h+1)​​

$$\Rightarrow {f}^{'}(0)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{({2}^{h}+{2}^{-h})\sqrt{{\mathrm{tan}}^{-1}({h}^{2}-h+1)}}{{(7{h}^{2}+3h+1)}^{3}}$$⇒f′(0)=h→0lim​(7h2+3h+1)3(2h+2−h)tan−1(h2−h+1)​​

$$\Rightarrow {f}^{'}(0)=\frac{({2}^{0}+{2}^{0})\sqrt{{\mathrm{tan}}^{-1}(1)}}{{(0+1)}^{3}}$$⇒f′(0)=(0+1)3(20+20)tan−1(1)​​

$$\Rightarrow {f}^{'}(0)=(2)\sqrt{\frac{\pi }{4}}$$⇒f′(0)=(2)4π​​

$$\Rightarrow {f}^{'}(0)=\sqrt{\pi }$$⇒f′(0)=π​