Mathematics · 2024
JEE Main · 27 January 2024, Shift 1 · Q9
If a = lim_x → 0 (√(1+√(1+x^4))-√ 2)/x^4 and b = lim_x → 0 (sin^2 x)/(√ 2 -√(1+ cos x)), then the value of ab^3 is:
If $\displaystyle \mathrm{a}=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $\displaystyle \mathrm{b}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $\displaystyle \mathrm{ab}^3$ is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 32$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.