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 ∫_ _e(α-1) ^ _e(α) dx (e^ 2x - e^ -2x ) is :

Correct answer: B. (1)/(4) _e((4)/(3)) Since (2^ 1-a + 2^ 1+a ) , f(a) , and (3^a + 3^ -a ) are in A.P., we have: 2f(a) = 2^ 1-a + 2^ 1+a + 3^a + 3^ -a f(a) = (2^ -a + 2^a) + (1)/(2)(3^a + 3^ -a ) Using the AM-GM inequality, x + (1)/(x) ≥ 2 for x > 0 . 2^a + 2^ -a ≥ 2 and 3^a + 3^ -a ≥ 2 . The minimum value of f(a) occurs when a = 0 . α = f(0) = (2^0 + 2^0) + (1)/(2)(3^0 + 3^0) = 2 + 1 = 3 . Now, we evaluate the integral: I = ∫_ _e 2 ^ _e 3 dx e^ 2x - e^ -2x = ∫_ _e 2 ^ _e 3 e^ 2x e^ 4x - 1 dx Let e^ 2x = t 2e^ 2x dx = dt . When x = _e 2 , t = e^ 2 _e 2 = 4 . When x = _e 3 , t = e^ 2 _e 3 = 9 . I = (1)/(2) ∫_ 4 ^ 9 (dt)/(t^2 - 1) I = (1)/(2) [ (1)/(2) _e | (t-1)/(t+1) | ]_ 4 ^ 9 I = (1)/(4) ( _e (8)/(10) - _e (3)/(5) ) I = (1)/(4) ( _e (4)/(5) - _e (3)/(5) ) = (1)/(4) _e ( (4/5)/(3/5) ) = (1)/(4) _e ( (4)/(3) ) Answer: (1)/(4) _e((4)/(3))

JEE Main 20265 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Definite Integration

Let (2^ 1-a + 2^ 1+a ) , f(a) , (3^a + 3^ -a ) be in A.P. and α be...

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)} \dfrac{dx}{(e^{2x} - e^{-2x})}

is :

Options

A $\dfrac{1}{2} \log_e\left(\dfrac{4}{3}\right)$
B $\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$
C $\dfrac{1}{2} \log_e\left(\dfrac{8}{5}\right)$
D $\dfrac{1}{4} \log_e\left(\dfrac{8}{5}\right)$

Correct answer

B. $\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$

Step-by-step solution

Since

(2^{1-a} + 2^{1+a})

f(a)

, and

(3^a + 3^{-a})

are in A.P., we have:

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

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

Using the AM-GM inequality,

x + \dfrac{1}{x} \ge 2

for

x > 0

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

and

3^a + 3^{-a} \ge 2

.

The minimum value of

f(a)

occurs when

a = 0

\alpha = f(0) = (2^0 + 2^0) + \dfrac{1}{2}(3^0 + 3^0) = 2 + 1 = 3

.

Now, we evaluate the integral:

I = \int_{\log_e 2}^{\log_e 3} \dfrac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e 2}^{\log_e 3} \dfrac{e^{2x}}{e^{4x} - 1} dx

Let

e^{2x} = t \Rightarrow 2e^{2x} dx = dt

.

When

x = \log_e 2

t = e^{2\log_e 2} = 4

.

When

x = \log_e 3

t = e^{2\log_e 3} = 9

I = \dfrac{1}{2} \int_{4}^{9} \dfrac{dt}{t^2 - 1}

I = \dfrac{1}{2} \left[ \dfrac{1}{2} \log_e \left| \dfrac{t-1}{t+1} \right| \right]_{4}^{9}

I = \dfrac{1}{4} \left( \log_e \dfrac{8}{10} - \log_e \dfrac{3}{5} \right)

I = \dfrac{1}{4} \left( \log_e \dfrac{4}{5} - \log_e \dfrac{3}{5} \right) = \dfrac{1}{4} \log_e \left( \dfrac{4/5}{3/5} \right) = \dfrac{1}{4} \log_e \left( \dfrac{4}{3} \right)

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

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