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 address this problem, we commence by evaluating the provided integral:

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

The integral contains a square root and an exponential function, suggesting a substitution for simplification. Let us assume the substitution:

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

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

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

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

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

Substitute back \(e^x = t^2 + 1\):

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

This integral evaluates to the standard inverse tangent function:

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

Now, we apply the integration limits. When \(x = \log_e a\), \(e^x - 1 = a - 1\), so \(t = \sqrt{a - 1}\). When \(x = 4\), \(t = \sqrt{e^4 - 1}\). Consequently, the definite integral is:

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

This simplifies to:

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

This implies that \(\alpha\) satisfies the following equations:

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

Given that \(e^\alpha\) and \(e^{-\alpha}\) are roots of a quadratic equation, we utilize the properties of roots:

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

This corresponds to the quadratic equation:

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

Therefore, the definitive equation is:

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