The value of x such that x > 1, satisfying the equation 1x t ln t dt = (1)/(4) is A) √(e) B) e C) e2 D) e - 1

Official answer: A) √(e). Explanation: Evaluate the definite integral using integration by parts: t ln t dt Let u = ln t du = (1)/(t) dt and dv = t dt v = (t2)/(2). t ln t dt = (t2)/(2) ln t - (t2)/(2) · (1)/(t) dt = (t2)/(2) ln t - (t2)/(4) Applying the limits from 1 to x: 1x t ln t dt = [ (t2)/(2) ln t - (t2)/(4) ]1x = ( (x2)/(2) ln x - (x2)/(4) ) - ( 0 - (1)/(4) ) = (x2)/(2) ln x - (x2)/(4) + (1)/(4) Equating this to (1)/(4): (x2)/(2) ln x - (x2)/(4) + (1)/(4) = (1)/(4) (x2)/(2) ln x - (x2)/(4) = 0 (x2)/(4)(2ln x - 1) = 0 Since x > 1, x2 ≠ 0. Therefore: 2ln x - 1 = 0 ln x = (1)/(2) x = e1/2 = √(e) Thus, option (A) is correct.

Calculus

GATE CSE 2025 Set 2 · Q12

MCQ1 markEasy

Time00:00

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

Select one answerSingle correct

Pick an option to check

Solution

Step-by-step verified solution

Official answer: A) √(e)

Calculus

Evaluate the definite integral using integration by parts:

\int t \ln t \, dt

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

\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}

Applying the limits from $1$ to $x$ :

\int_1^x t \ln t \, dt = \left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_1^x = \left( \frac{x^2}{2} \ln x - \frac{x^2}{4} \right) - \left( 0 - \frac{1}{4} \right) = \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4}

Equating this to $\frac{1}{4}$ :

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

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

Since $x > 1$ , $x^2 \neq 0$ . Therefore:

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

Thus, option (A) is correct.

Try the question first, then check your work.

Think the official key or this write-up is wrong?

Report error or challenge

Examiner trap

The misconception this question tests

This Calculus previous year question is Q12 of the GATE CSE 2025 Set 2 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

Free diagnostic

Where do you stand for GATE?

Take the 20-minute diagnostic: the 3 topics that cost you the most marks, and a rank range. Or sit 5 related Calculus PYQs now with no account.

One GATE CSE PYQ every morning

An official question by email at 7 AM IST, with a link to its verified solution. Unsubscribe in one click.

Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 14 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2025 Set 2.

How GATE tests Calculus

More from GATE CSE 2025 Set 2

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.