JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Definite Integral of Absolute Logarithm Function

Q: Why do we split the integral at x = 1?

Because ln x < 0 when 0 < x < 1 and ln x ≥ 0 when x ≥ 1. The absolute value definition requires |u| = -u when u < 0.

If integral of _(1/3)^3 |log_e x|dx = m/nlog_e((n^2)/e), where m and n are coprime natural numbers, then m^2 + n^2 - 5 is equal to

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Correct answer

The value of $m^{2} + n^{2} - 5$ is $20$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Integral $I = \int_{1 / 3}^{3} ∣ \ln x ∣ d x = \frac{m}{n} \ln (\frac{n^{2}}{e})$ with $m, n \in N$ and $\gcd (m, n) = 1$ .

02Goal

Evaluate the integral, match it to $\frac{m}{n} \ln (\frac{n^{2}}{e})$ to determine $m$ and $n$ , and compute $m^{2} + n^{2} - 5$ .

03Approach

Split the domain at $x = 1$ , where $\ln x$ changes sign. For $x \in [1 / 3, 1]$ , $∣ \ln x ∣ = - \ln x$ , and for $x \in [1, 3]$ , $∣ \ln x ∣ = \ln x$ . Then use $\int \ln x d x = x \ln x - x$ .

04Execute

Split the integral: $I = - \int_{1 / 3}^{1} \ln x d x + \int_{1}^{3} \ln x d x$ Antiderivative of $\ln x$ is $F (x) = x \ln x - x$ . Evaluating: $- [x \ln x - x]_{1 / 3}^{1} = - [(0 - 1) - (\frac{1}{3} \ln (1 / 3) - \frac{1}{3})] = - [- 1 + \frac{1}{3} \ln 3 + \frac{1}{3}] = - [- \frac{2}{3} + \frac{1}{3} \ln 3] = \frac{2}{3} - \frac{1}{3} \ln 3$ $[x \ln x - x]_{1}^{3} = (3 \ln 3 - 3) - (0 - 1) = 3 \ln 3 - 2$ Adding both parts: $I = (\frac{2}{3} - \frac{1}{3} \ln 3) + (3 \ln 3 - 2) = \frac{8}{3} \ln 3 - \frac{4}{3} = \frac{4}{3} (2 \ln 3 - 1) = \frac{4}{3} \ln (\frac{3^{2}}{e})$

05Execute

Equating $\frac{4}{3} \ln (\frac{3^{2}}{e})$ with $\frac{m}{n} \ln (\frac{n^{2}}{e})$ gives $m = 4$ and $n = 3$ , which are coprime natural numbers. Compute the required expression: $m^{2} + n^{2} - 5 = 4^{2} + 3^{2} - 5 = 16 + 9 - 5 = 20$

✓Verify

Check values: $m = 4, n = 3 ⟹ \gcd (4, 3) = 1$ . The formula inside the logarithm has $n^{2} / e = 9 / e$ , exactly matching $\ln (9) - 1 = 2 \ln 3 - 1$ .

Hints that build this answer step by step

How should the integral $\int_{1 / 3}^{3} ∣ \ln x ∣ d x$ be split to remove the absolute value?
$- \int_{1 / 3}^{1} \ln x d x + \int_{1}^{3} \ln x d x$
Using the antiderivative $\int \ln x d x = x \ln x - x$ , what is the evaluated value of the integral?
$\frac{8}{3} \ln 3 - \frac{4}{3}$
Rewriting $\frac{8}{3} \ln 3 - \frac{4}{3}$ in the form $\frac{m}{n} \ln (\frac{n^{2}}{e})$ , what are the coprime values of $m$ and $n$ ?
$m = 4$ and $n = 3$
With $m = 4$ and $n = 3$ , what is the final value of $m^{2} + n^{2} - 5$ ?
$20$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we split the integral at x = 1?

Because $\ln x < 0$ when $0 < x < 1$ and $\ln x \geq 0$ when $x \geq 1$ . The absolute value definition requires $∣ u ∣ = - u$ when $u < 0$ .

Answer

The value of $m^{2} + n^{2} - 5$ is $20$ .

Why each option works or fails

Step-by-step solution

Given: Integral $I = \int_{1 / 3}^{3} ∣ \ln x ∣ d x = \frac{m}{n} \ln (\frac{n^{2}}{e})$ with $m, n \in N$ and $\gcd (m, n) = 1$ .
Goal: Evaluate the integral, match it to $\frac{m}{n} \ln (\frac{n^{2}}{e})$ to determine $m$ and $n$ , and compute $m^{2} + n^{2} - 5$ .
Approach: Split the domain at $x = 1$ , where $\ln x$ changes sign. For $x \in [1 / 3, 1]$ , $∣ \ln x ∣ = - \ln x$ , and for $x \in [1, 3]$ , $∣ \ln x ∣ = \ln x$ . Then use $\int \ln x d x = x \ln x - x$ .
Execute: Split the integral: $I = - \int_{1 / 3}^{1} \ln x d x + \int_{1}^{3} \ln x d x$ Antiderivative of $\ln x$ is $F (x) = x \ln x - x$ . Evaluating: $- [x \ln x - x]_{1 / 3}^{1} = - [(0 - 1) - (\frac{1}{3} \ln (1 / 3) - \frac{1}{3})] = - [- 1 + \frac{1}{3} \ln 3 + \frac{1}{3}] = - [- \frac{2}{3} + \frac{1}{3} \ln 3] = \frac{2}{3} - \frac{1}{3} \ln 3$ $[x \ln x - x]_{1}^{3} = (3 \ln 3 - 3) - (0 - 1) = 3 \ln 3 - 2$ Adding both parts: $I = (\frac{2}{3} - \frac{1}{3} \ln 3) + (3 \ln 3 - 2) = \frac{8}{3} \ln 3 - \frac{4}{3} = \frac{4}{3} (2 \ln 3 - 1) = \frac{4}{3} \ln (\frac{3^{2}}{e})$
Execute: Equating $\frac{4}{3} \ln (\frac{3^{2}}{e})$ with $\frac{m}{n} \ln (\frac{n^{2}}{e})$ gives $m = 4$ and $n = 3$ , which are coprime natural numbers. Compute the required expression: $m^{2} + n^{2} - 5 = 4^{2} + 3^{2} - 5 = 16 + 9 - 5 = 20$
Verify: Check values: $m = 4, n = 3 ⟹ \gcd (4, 3) = 1$ . The formula inside the logarithm has $n^{2} / e = 9 / e$ , exactly matching $\ln (9) - 1 = 2 \ln 3 - 1$ .

Definite Integral of Absolute Logarithm Function

What feels right?

The value of $m^{2} + n^{2} - 5$ is $20$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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