The value of \lim_{x \rightarrow 0} {\log _e ( \sec (e x) \cdot \sec (e^2 x) \cdot . . . \cdot \sec (e^{10} x) ) }/{e^2-e^{2 \cos x}} is equal to [28 Jan 2026 Shift 1]

Correct Answer is : (e20−1)2(e2−1)

\frac{ (e^{10}-1) }{2 e^2 (e^2-1) }

\frac{ (e^{20}-1) }{2 e^2 (e^2-1) }

\frac{ (e^{10}-1) }{2 (e^2-1) }

\frac{ (e^{20}-1) }{2 (e^2-1) }

JEE Main 28 Jan 2026 Shift 1 Paper Solved Paper

Section: Mathematics

Question : 15 of 75

Marks: +1, -0

The value of

lim

x⟶0

‌

loge( sec(ex)⋅ sec(e2x)⋅...⋅ sec(e10x))

e2−e2‌cos‌x

is equal to

[28 Jan 2026 Shift 1]

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