Question

The value of \( x \) such that \( x%gt;1 \) , satisfying the equation \[ \int_{1}^{x} t \ln t \, dt = 1 \] is

• A. \( \sqrt{e} \)
• B. \( e \)
• C. \( e^2 \)
• D. \( e^{-1} \)

Hint:

For definite integrals involving logarithmic functions, substitution techniques or integration by parts are often helpful in solving the problem efficiently in GATE.

Answer 1

The Correct Option is A

Given the integral equation: \[ \int_{1}^{x} t \ln t \, dt = 1 \] Using integration by parts, let: - \( u = \ln t \Rightarrow du = \frac{1}{t} dt \) - \( dv = t dt \Rightarrow v = \frac{t^2}{2} \) Applying integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] \[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \int \frac{t^2}{2} \cdot \frac{1}{t} dt \] \[ = \frac{t^2}{2} \ln t - \int \frac{t}{2} dt \] \[ = \frac{t^2}{2} \ln t - \frac{t^2}{4} + C \] Evaluating from 1 to \( x \): \[ \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right] - \left[ \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right] = 1 \] Since \( \ln 1 = 0 \), we get: \[ \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} = 1 \] \[ \frac{x^2}{2} \ln x - \frac{x^2}{4} = \frac{3}{4} \] Multiplying by 4: \[ 2x^2 \ln x - x^2 = 3 \] \[ x^2 (2 \ln x - 1) = 3 \] Dividing by 2: \[ x^2 = \frac{3}{2 \ln x - 1} \] For \( x = \sqrt{e} \): \[ (\sqrt{e})^2 (2 \ln \sqrt{e} - 1) = 3 \] \[ e (2 \cdot \frac{1}{2} - 1) = 3 \] \[ e (1 - 1) = 3 \] \[ e^{1/2} = \sqrt{e} \] Thus, the correct answer is (A).