JEE Main 2024 — 27 January, Shift 1 — Question 16

$\mathrm{a}=\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\sqrt{1+\mathrm{x}^{4}}}-\sqrt{2}}{\mathrm{x}^{4}}$ $\begin{aligned}&=\lim_{x\rightarrow0}\frac{\sqrt{1+x^{4}}1}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)}&\end{aligned}$

$=\lim_{x\rightarrow0}\frac{x^{4}}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)\left(\sqrt{1+x^{4}}+1\right)}$

Applying limit $a=\frac{1}{4 \sqrt{2}}$

$b=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$

$=\lim _{x \rightarrow 0} \frac{\left(1-\cos ^{2} x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$

$b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$

Applying limits $b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}$

Now, $\mathrm{ab}^{3}=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^{3}=32$