JEE 2024 — Mathematics PYQ

$a={\lim }_{x\to 0}\frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$
$={\lim }_{x\to 0}\frac{1+\sqrt{1+x^4}-2}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}\text{ [on rationalising]}$
$={\lim }_{x\to 0}\frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}$
$={\lim }_{x\to 0}\frac{1+x^4-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}\text{ [on rationalising]}$
$={\lim }_{x\to 0}\frac{1}{(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}=\frac{1}{4\sqrt{2}}$

$b={\lim }_{x\to 0}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$
$={\lim }_{x\to 0}\frac{(1-{\cos }^{2}x)(\sqrt{2}+\sqrt{1+\cos x})}{(\sqrt{2}-\sqrt{1+\cos x})(\sqrt{2}+\sqrt{1+\cos x})}$
$={\lim }_{x\to 0}\frac{(1-{\cos }^{2}x)(\sqrt{2}+\sqrt{1+\cos x})}{2-1-\cos x}$
$={\lim }_{x\to 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$
$=2(2\sqrt{2})=4\sqrt{2}$

So, $a{b}^3=\frac{1}{4\sqrt{2}}\times 64\times 2\sqrt{2}=32$