GATE CS 2025 Set 2 Q.12: Calculus

The value of $x$ such that $x > 1$ , satisfying the equation $\int_1^x t \ln t \, dt = \frac{1}{4}$ is

$\sqrt{e}$

$e$

$e^2$

$e - 1$

Official answer and solution

Solution

Evaluate $\int_1^x t \ln t \, dt$ using integration by parts and solve for $x > 1$ .

Using integration by parts with $u = \ln t$ ( $du = \frac{1}{t}\,dt$ ) and $dv = t\,dt$ ( $v = \frac{t^2}{2}$ ):

\int_1^x t \ln t \, dt = \left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_1^x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4}

Setting this equal to $\frac{1}{4}$ gives:

\frac{x^2}{2} \ln x - \frac{x^2}{4} = 0 \implies \frac{x^2}{4}(2 \ln x - 1) = 0

Since $x > 1$ , we have $2 \ln x = 1 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}$ .

Final answer

Option (A)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 12 of 65
Section: Computer Science and Information Technology

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