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:

(B) $\frac{1}{4} \log _{\mathrm{e}}\left(\frac{4}{3}\right)$

(A) $\frac{1}{2} \log _{\mathrm{e}}\left(\frac{4}{3}\right)$

(C) $\frac{1}{2} \log _{\mathrm{e}}\left(\frac{8}{5}\right)$

(D) $\frac{1}{4} \log _{\mathrm{e}}\left(\frac{8}{5}\right)$

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Sequences and Series

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_{l o g_{e} (α - 1)}^{l o g_{e} (α)} \frac{dx}{( e ^{2 x} - e ^{- 2 x} )}

is:

Answer: (B)

Since the three quantities are in A.P.,

2 f (a) = (2^{1 - a} + 2^{1 + a}) + (3^{a} + 3^{- a})

, so

f (a) = (2^{a} + 2^{- a}) + \frac{1}{2} (3^{a} + 3^{- a})

. Each bracket is least at

a = 0

, so

α = f (0) = 2 + 1 = 3

. The integral runs from

ln (α - 1) = ln 2

ln α = ln 3

, and

\frac{1}{e ^{2 x} - e ^{- 2 x}} = \frac{e ^{2 x}}{e ^{4 x} - 1}

. With

t = e^{2 x}

d t = 2 e^{2 x} d x

, the limits become

t = 4

and

t = 9

, so

I = \frac{1}{2} \int_{4}^{9} \frac{d t}{t ^{2} - 1} = \frac{1}{4} [ln \frac{t - 1}{t + 1}]_{4}^{9} = \frac{1}{4} (ln \frac{8}{10} - ln \frac{3}{5}) = \frac{1}{4} ln \frac{4}{3}

. The value of the integral is

\frac{1}{4} lo g_{e} \frac{4}{3}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option