Question

Let \[ \int_{\log_e a}^{4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}. \] Then \(e^\alpha\) and \(e^{-\alpha}\) are the roots of the equation:

• A. \(2x^2 - 5x + 2 = 0\)
• B. \(x^2 - 2x - 8 = 0\)
• C. \(2x^2 - 5x - 2 = 0\)
• D. \(x^2 + 2x - 8 = 0\)

Answer 1

The Correct Option is A

Given: \[ \int_{\log_e \alpha}^{\log_e 4} \frac{dx}{\sqrt{e^x - 1}} = \frac{\pi}{6}. \]

Let: \[ e^x - 1 = t^2 \implies e^x dx = 2t dt \quad \text{and} \quad dx = \frac{2t dt}{t^2 + 1}. \] 

Substituting into the integral: \[ \int \frac{dx}{\sqrt{e^x - 1}} = \int \frac{2t dt}{(t^2 + 1) \cdot t} = \int \frac{2 dt}{t^2 + 1} = 2 \tan^{-1} t. \] 

Reverting the substitution: \[ 2 \tan^{-1} (\sqrt{e^x - 1}) \Big|_{\log_e \alpha}^{\log_e 4}. \] 

Evaluating at the limits: \[ 2 \left[ \tan^{-1} \left( \sqrt{e^{\log_e 4} - 1} \right) - \tan^{-1} \left( \sqrt{e^\alpha - 1} \right) \right] = \frac{\pi}{6}. \] Simplifying: \[ 2 \left[ \tan^{-1} (\sqrt{3}) - \tan^{-1} (\sqrt{e^\alpha - 1}) \right] = \frac{\pi}{6}. \] 

Dividing by 2: \[ \tan^{-1} (\sqrt{3}) - \tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{12}. \] 

Since \(\tan^{-1} (\sqrt{3}) = \frac{\pi}{3}\), we have: \[ \frac{\pi}{3} - \tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{12}. \] 

Rearranging: \[ \tan^{-1} (\sqrt{e^\alpha - 1}) = \frac{\pi}{4}. \] 

Thus: \[ \sqrt{e^\alpha - 1} = 1 \implies e^\alpha - 1 = 1 \implies e^\alpha = 2. \] 

Therefore: \[ e^{-\alpha} = \frac{1}{2}. \] 

The roots \( e^\alpha = 2 \) and \( e^{-\alpha} = \frac{1}{2} \) satisfy the equation: \[ x^2 - \left( 2 + \frac{1}{2} \right) x + 1 = 0 \implies 2x^2 - 5x + 2 = 0. \] 

Therefore: \[ 2x^2 - 5x + 2 = 0. \]