Mathematics · 2022
JEE Main · 29 July 2022, Shift 2 · Q12
The number of elements in the set S={x ∈ R: 2 cos ((x^2+x)/6)=4^x+4^-x} is:
The number of elements in the set $\displaystyle S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\right\}$ is :
Official answer
(1)
$\displaystyle 1$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.