JEE Main 2026 Apr 5 Shift 2, Mathematics Q18: Definite Integration And Fundamental Theorem

JEE Main2026Apr 5, Shift 2Mathematics

Let $(2^{1 - a} + 2^{1 + a})$ , $f (a)$ , $(3^{a} + 3^{- a})$ be in A.P. and $α$ be the minimum value of $f (a)$ . Then the value of the integral

$\int_{l o g_{e} (α - 1)}^{l o g_{e} (α)} \frac{d x}{e ^{2 x} - e ^{- 2 x}}$ is:

A
$\frac{1}{2} lo g_{e} (\frac{4}{3})$
B
$\frac{1}{4} lo g_{e} (\frac{4}{3})$
C
$\frac{1}{2} lo g_{e} (\frac{8}{5})$
D
$\frac{1}{4} lo g_{e} (\frac{8}{5})$

Solution

For an AP, the middle term equals the average: $f (a) = \frac{1}{2} (2 (2^{a} + 2^{- a}) + (3^{a} + 3^{- a}))$ .

Each of $2^{a} + 2^{- a}$ and $3^{a} + 3^{- a}$ is minimised at $a = 0$ , giving values $2$ and $2$ . So $f_{m i n} = \frac{1}{2} (2 \cdot 2 + 2) = 3 \Rightarrow α = 3$ .

Limits become $ln 2$ to $ln 3$ . Substitute $t = e^{2 x}$ , $d t = 2 e^{2 x} d x$ :

$I = \int_{l n 2}^{l n 3} \frac{e ^{2 x} d x}{e ^{4 x} - 1} = \frac{1}{2} \int_{4}^{9} \frac{d t}{t ^{2} - 1}$ .

$= \frac{1}{2} \cdot \frac{1}{2} [ln \frac{t - 1}{t + 1}]_{4}^{9} = \frac{1}{4} [ln \frac{8}{10} - ln \frac{3}{5}] = \frac{1}{4} ln \frac{4}{3}$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q18: Definite Integration And Fundamental Theorem

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