QJEEM24Ma649a: JEE Main 2024 Mathematics - Class 12 JEEM24Ma649a Solution: If and , then the value of is :

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

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

$\begin{aligned}& b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\& =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\& b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})\end{aligned}$

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

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