GATE CS 2025 Set 2 Q2 — Maxima and Minima, Mean Value Theorem, Integration

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

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Answer: A

To find the value of $x > 1$ that satisfies the given equation, we first evaluate the definite integral using integration by parts:

\int_1^x t \ln t \, dt = \frac{1}{4}

Let $u = \ln t$ and $dv = t \, dt$ . Then, $du = \frac{1}{t} \, dt$ and $v = \frac{t^2}{2}$ .

Applying the 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 - \frac{t^2}{4}

Now, evaluate this expression from the limits $1$ to $x$ :

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

\left( \frac{x^2}{2} \ln x - \frac{x^2}{4} \right) - \left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right) = \frac{1}{4}

Since $\ln 1 = 0$ , this simplifies to:

\frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} = \frac{1}{4}

Subtracting $\frac{1}{4}$ from both sides gives:

\frac{x^2}{2} \ln x - \frac{x^2}{4} = 0

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

Since we are given $x > 1$ , we have $x^2 \neq 0$ . Therefore, we must have:

2 \ln x - 1 = 0 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}

Thus, the correct option is A.

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Published 10 Oct 2026

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