Mathematics · 2026
JEE Main · 28 January 2026, Shift 1 · Q18
The value of lim_x → 0 (log_e( sec (e x) · sec (e^2 x) · … · sec (e^10 x)))/(e^2-e^2 cos x) is equal to
The value of
$$\lim _{x \rightarrow 0} \frac{\log _e\left(\sec (e x) \cdot \sec \left(e^2 x\right) \cdot \ldots \cdot \sec \left(e^{10} x\right)\right)}{e^2-e^{2 \cos x}}
$$
is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{\left(e^{20}-1\right)}{2\left(e^2-1\right)}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.