Let (2 1- a +2 1+ a ), f( a ), (3 a +3 - a ) be a A. P. and α be the minimum value of f( a ) . Then the value of the integral int log e (α-1) log e (α) frac dx ( e 2 x - e -2 x ) is:

Mathematics · Sequence and Series

JEE Main 2026 — 5 April, Evening Shift — Question 42

Let $\left(2^{1-\mathrm{a}}+2^{1+\mathrm{a}}\right), f(\mathrm{a}),\left(3^{\mathrm{a}}+3^{-\mathrm{a}}\right)$ be a A. P. and $\alpha$ be the minimum value of $f(\mathrm{a})$ . Then the value of the integral $\int_{\log _{\mathrm{e}}(\alpha-1)}^{\log _{\mathrm{e}}(\alpha)} \frac{\mathrm{dx}}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)}$ is:

Option A:
$\frac{1}{2} \log _{\mathrm{e}}\left(\frac{4}{3}\right)$
Option B:
$\frac{1}{4} \log _{\mathrm{e}}\left(\frac{4}{3}\right)$
Correct
Option C:
$\frac{1}{2} \log _{\mathrm{e}}\left(\frac{8}{5}\right)$
Option D:
$\frac{1}{4} \log _{\mathrm{e}}\left(\frac{8}{5}\right)$

Answer: B

Step-by-step solution

$f(a)=\frac{1}{2}\left(2\left(2^{a}+2^{-a}\right)+\left(3^{a}+3^{-a}\right)\right)$ $\mathrm{a}=\frac{1}{2}(2 \times 2+2)=3$ $I=\int_{\ln (3-1)}^{\ln 3} \frac{e^{2 x}}{e^{4 x}-1} d x$ put $\mathrm{e}^{2 \mathrm{x}}=\mathrm{t}$ $I=\frac{1}{2} \int_{4}^{9} \frac{d t}{t^{2}-1}$ $I=\frac{1}{4}\left(\ln \left(\frac{t-1}{t+1}\right)\right)_{4}^{9}$ $I=\frac{1}{4} \ln \left(\frac{4}{3}\right)$

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

Answer: B

Step-by-step solution

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