Definite IntegrationmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Definite Integration: Let Minimum Value Value (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

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

\displaystyle\int_{\ln(\alpha-1)}^{\ln\alpha}\dfrac{dx}{e^{2x}-e^{-2x}}

is:

\dfrac{1}{2}\log_e\!\left(\dfrac{4}{3}\right)

correct

\dfrac{1}{4}\log_e\!\left(\dfrac{4}{3}\right)

\dfrac{1}{2}\log_e\!\left(\dfrac{8}{5}\right)

\dfrac{1}{4}\log_e\!\left(\dfrac{8}{5}\right)

Solution

Step 1: Determine

f(a)

from the A.P. condition

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

Step 2: Find the minimum

\alpha

via AM-GM By AM-GM,

2^a+2^{-a}\geq2

and

3^a+3^{-a}\geq2

, with equality at

a=0

f(a)_{\min} = f(0) = \frac{1}{2}(2\cdot2+2) = 3 \implies \alpha=3

Step 3: Set up the integral with the known limits

I = \int_{\ln2}^{\ln3}\frac{dx}{e^{2x}-e^{-2x}}

Step 4: Evaluate using the substitution

t=e^{2x}

dx=\dfrac{dt}{2t}

I = \int_4^9\frac{1}{t-t^{-1}}\cdot\frac{dt}{2t} = \frac{1}{2}\int_4^9\frac{dt}{t^2-1} = \frac{1}{4}\!\left[\ln\frac{t-1}{t+1}\right]_4^9

= \frac{1}{4}\!\left(\ln\frac{8}{10}-\ln\frac{3}{5}\right) = \frac{1}{4}\ln\!\left(\frac{8}{10}\cdot\frac{5}{3}\right) = \frac{1}{4}\ln\frac{4}{3}

Answer: (2)

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