Q43 / 75 · JEE Main 2026 (05 April Shift 2) · Mathematics
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 :
- A$\dfrac{1}{2} \log_e\left(\dfrac{4}{3}\right)$
- Correct answerB$\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$
- C$\dfrac{1}{2} \log_e\left(\dfrac{8}{5}\right)$
- D$\dfrac{1}{4} \log_e\left(\dfrac{8}{5}\right)$
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: $\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$
$2f(a) = 2^{1-a} + 2^{1+a} + 3^a + 3^{-a}$
$f(a) = (2^{-a} + 2^a) + \dfrac{1}{2}(3^a + 3^{-a})$
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: $\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$
Hence, the correct answer is option: 'B'.