Question:medium
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 :
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 :
Updated On: Apr 12, 2026
- $\frac{1}{2} \log_e \left(\frac{4}{3}\right)$
- $\frac{1}{4} \log_e \left(\frac{4}{3}\right)$
- $\frac{1}{2} \log_e \left(\frac{8}{5}\right)$
- $\frac{1}{4} \log_e \left(\frac{8}{5}\right)$
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The Correct Option is B
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