Mathematics · 2020
JEE Main · 3 September 2020, Shift 1 · Q74
If lim_x → 0{1/x^8(1- cos x^2/2- cos x^2/4....+ cos x^2/2 cos x^2/4)}=2^- k, then the value of k is \ \ \ \.
If $\displaystyle \lim _{x \rightarrow 0}\left\{\frac{1}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}\right.\right.$
$\displaystyle \left.\left.+\cos \frac{x^2}{2} \cos \frac{x^2}{4}\right)\right\}=2^{-\mathrm{k}}$, then the value of k is \_\_\_\_.
Official answer
8
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.