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

To solve this problem, we begin by evaluating the given integral:

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

Notice that the integral involves a square root and exponential function, which suggests a possible substitution to simplify it. Assume the substitution:

\(e^x - 1 = t^2\).

Differentiating both sides with respect to \(x\) gives:

\[e^x \, dx = 2t \, dt.\]

This means \(dx = \frac{2t}{e^x} \, dt\). Rewrite the integral in terms of \(t\):

\[\int \frac{dx}{\sqrt{e^x - 1}} = \int \frac{1}{t} \cdot \frac{2t}{e^x} \, dt = 2 \int \frac{dt}{e^x}.\]

Back-substitute \(e^x = t^2 + 1\):

\[= 2 \int \frac{dt}{t^2 + 1}.\]

This integral is the standard inverse tangent function:

\[= 2 \tan^{-1}(t).\]

Now, let's use the limits of the integral. When \(x = \log_e a\), \(e^x - 1 = a - 1\), so \(t = \sqrt{a - 1}\). When \(x = 4\), \(t = \sqrt{e^4 - 1}\). Therefore, the definite integral becomes:

\[2 \left( \tan^{-1}(\sqrt{e^4 - 1}) - \tan^{-1}(\sqrt{a - 1}) \right) = \frac{\pi}{6}.\]

Thus,

\[\tan^{-1}(\sqrt{e^4 - 1}) - \tan^{-1}(\sqrt{a - 1}) = \frac{\pi}{12}.\]

This indicates that \(\alpha\) satisfies this equation:

\(e^\alpha = e^4 - 1\) and \(e^{-\alpha} = a - 1\).

We are told \(e^\alpha\) and \(e^{-\alpha}\) are roots of a quadratic equation. Using the property of roots:

1. Sum of roots = \(e^\alpha + e^{-\alpha} = 5/2\).
2. Product of roots = \(e^\alpha \times e^{-\alpha} = 1\).

This matches the quadratic:

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

Therefore, the correct equation is:

\(2x^2 - 5x + 2 = 0.\)

Answer 2

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