The value of _ x arrow 0 _ e ( (e x) · (e^ 2 x) · · (e^ 10 x)) e^ 2 -e^ 2 x is equal to

Correct answer: A. (e^ 20 -1) 2 (e^ 2 -1) Numerator: _e( (ex) (e^2x) (e^ 10 x)) = ∑_ k=1 ^ 10 _e( (e^kx)) . Using (u) 1 + (u^2)/(2) for small u , we get _e( (e^kx)) e^ 2k x^2 2 . Thus the numerator is (x^2)/(2)∑_ k=1 ^ 10 e^ 2k = x^2 e^2(e^ 20 -1) 2(e^2-1) . Denominator: Using x 1 - (x^2)/(2) , we have e^ 2 x e^2(1-x^2) , so e^2 - e^ 2 x e^2 x^2 . Therefore: _ x → 0 = x^2 e^2(e^ 20 -1) 2(e^2-1) e^2 x^2 = e^ 20 -1 2(e^2-1) .

JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Limits

The value of _ x arrow 0 _ e ( (e x) · (e^ 2 x) · · (e^ 10 x)) e^ 2...

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

Options

A $\frac{\left(e^{20}-1\right)}$ {2 $\left(e^{2}-1\right)}$
B $\frac{\left(e^{10}-1\right)}$ {2 $\left(e^{2}-1\right)}$
C $\frac{\left(e^{10}-1\right)}$ {2 e^{2} $\left(e^{2}-1\right)}$
D $\frac{\left(e^{20}-1\right)}$ {2 e^{2} $\left(e^{2}-1\right)}$

Correct answer

A. $\frac{\left(e^{20}-1\right)}$ {2 $\left(e^{2}-1\right)}$

Step-by-step solution

Numerator:

\log_e(\sec(ex)\sec(e^2x)\cdots\sec(e^{10}x)) = \sum_{k=1}^{10} \log_e(\sec(e^kx))

.
Using

\sec(u) \approx 1 + \frac{u^2}{2}

for small

u

, we get

\log_e(\sec(e^kx)) \approx \frac{e^{2k}x^2}{2}

.
Thus the numerator is

\approx \frac{x^2}{2}\sum_{k=1}^{10} e^{2k} = \frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}

.
Denominator: Using

\cos x \approx 1 - \frac{x^2}{2}

, we have

e^{2\cos x} \approx e^2(1-x^2)

, so

e^2 - e^{2\cos x} \approx e^2 x^2

.
Therefore:

\lim_{x \to 0} = \frac{\frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}}{e^2 x^2} = \frac{e^{20}-1}{2(e^2-1)}

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