JEE Main 2026 (5 Apr, Shift 2) Maths Q18: Integrals

Question

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

Accepted Answer

Option 2: $\frac{1}{4}\log_e\left(\frac{4}{3}\right)$

Answer

$\frac{1}{2}\log_e\left(\frac{4}{3}\right)$

Answer

$\frac{1}{4}\log_e\left(\frac{4}{3}\right)$

Answer

$\frac{1}{2}\log_e\left(\frac{8}{5}\right)$

Answer

$\frac{1}{4}\log_e\left(\frac{8}{5}\right)$

JEE Main 5 April 2026, Shift 2, Maths Q18

Same idea in other shifts