How do we know to take inside as ?

Because the expression is raised to the power , bringing an inside the bracket raises it to the power . This increases powers inside from to , whose derivative gives powers , exactly what remains outside after taking out .

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Integral Calculus: JEE Main Mathematics Question with Solution

\int_0^1 (x^{21} + x^{14} + x^7)(2x^{14} + 3x^7 + 6)^{1/7}dx = \frac{1}{l}(11)^{m/n}

where

l, m, n \in \mathbb{N}

m

and

n

are coprime then

l + m + n

is equal to

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Step-by-step solutionView

Correct answer

The value of $l + m + n$ is $48$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given integral $I = \int_0^1 (x^{21} + x^{14} + x^7)(2x^{14} + 3x^7 + 6)^{1/7} dx = \frac{1}{l}(11)^{m/n}$ where $l, m, n \in \mathbb{N}$ with $\gcd(m, n) = 1$ .

02approach

Notice the inner expression has degree 14, while the outer polynomial has degree 21. If we factor out $x$ from $(x^{21} + x^{14} + x^7)$ and push it into the $1/7$ -th power bracket as $x^7$ , the inner term becomes $2x^{21} + 3x^{14} + 6x^7$ . Its derivative is exactly $42(x^{20} + x^{13} + x^6)$ , which matches the remaining outer factor.

03execute

Rewrite the integrand: $I = \int_0^1 (x^{20} + x^{13} + x^6) \cdot \left[ x^7(2x^{14} + 3x^7 + 6) \right]^{1/7} dx = \int_0^1 (x^{20} + x^{13} + x^6)(2x^{21} + 3x^{14} + 6x^7)^{1/7} dx$ Let $t = 2x^{21} + 3x^{14} + 6x^7$ . Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42(x^{20} + x^{13} + x^6) dx$ . When $x = 0$ , $t = 0$ . When $x = 1$ , $t = 2(1) + 3(1) + 6(1) = 11$ . $I = \frac{1}{42} \int_0^{11} t^{1/7} dt = \frac{1}{42} \left[ \frac{t^{8/7}}{8/7} \right]_0^{11} = \frac{7}{42 \times 8} (11)^{8/7} = \frac{1}{48} (11)^{8/7}$

04execute

Comparing $\frac{1}{48}(11)^{8/7}$ with $\frac{1}{l}(11)^{m/n}$ : $l = 48$ , $m = 8$ , $n = 7$ . Check that $l, m, n \in \mathbb{N}$ and $\gcd(8, 7) = 1$ . Then $l + m + n = 48 + 8 + 7 = 63$ .

✓verify

Check derivative: $\frac{d}{dx}[(2x^{21}+3x^{14}+6x^7)^{8/7}] = \frac{8}{7}(2x^{21}+3x^{14}+6x^7)^{1/7} \cdot 42(x^{20}+x^{13}+x^6) = 48 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}$ , which matches the factor of 48 in the denominator perfectly.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

How do we know to take

x

inside as

x^7

Because the expression is raised to the power $1/7$ , bringing an $x$ inside the bracket raises it to the power $7$ . This increases powers inside from $14, 7, 0$ to $21, 14, 7$ , whose derivative gives powers $20, 13, 6$ , exactly what remains outside after taking out $x$ .

Answer

The value of $l + m + n$ is $48$ .

Why each option works or fails

Step-by-step solution

given: Given integral $I = \int_0^1 (x^{21} + x^{14} + x^7)(2x^{14} + 3x^7 + 6)^{1/7} dx = \frac{1}{l}(11)^{m/n}$ where $l, m, n \in \mathbb{N}$ with $\gcd(m, n) = 1$ .
approach: Notice the inner expression has degree 14, while the outer polynomial has degree 21. If we factor out $x$ from $(x^{21} + x^{14} + x^7)$ and push it into the $1/7$ -th power bracket as $x^7$ , the inner term becomes $2x^{21} + 3x^{14} + 6x^7$ . Its derivative is exactly $42(x^{20} + x^{13} + x^6)$ , which matches the remaining outer factor.
execute: Rewrite the integrand: $I = \int_0^1 (x^{20} + x^{13} + x^6) \cdot \left[ x^7(2x^{14} + 3x^7 + 6) \right]^{1/7} dx = \int_0^1 (x^{20} + x^{13} + x^6)(2x^{21} + 3x^{14} + 6x^7)^{1/7} dx$ Let $t = 2x^{21} + 3x^{14} + 6x^7$ . Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42(x^{20} + x^{13} + x^6) dx$ . When $x = 0$ , $t = 0$ . When $x = 1$ , $t = 2(1) + 3(1) + 6(1) = 11$ . $I = \frac{1}{42} \int_0^{11} t^{1/7} dt = \frac{1}{42} \left[ \frac{t^{8/7}}{8/7} \right]_0^{11} = \frac{7}{42 \times 8} (11)^{8/7} = \frac{1}{48} (11)^{8/7}$
execute: Comparing $\frac{1}{48}(11)^{8/7}$ with $\frac{1}{l}(11)^{m/n}$ : $(x^{21} + x^{14} + x^7)$ , $m = 8$ , $n = 7$ . Check that $l, m, n \in \mathbb{N}$ and $\gcd(8, 7) = 1$ . Then $l + m + n = 48 + 8 + 7 = 63$ .
verify: Check derivative: $\frac{d}{dx}[(2x^{21}+3x^{14}+6x^7)^{8/7}] = \frac{8}{7}(2x^{21}+3x^{14}+6x^7)^{1/7} \cdot 42(x^{20}+x^{13}+x^6) = 48 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}$ , which matches the factor of 48 in the denominator perfectly.

Integral Calculus: JEE Main Mathematics Question with Solution

The value of l+m+nl + m + nl+m+n is 484848.

01given

02approach

03execute

04execute

✓verify

Students also ask

The value of $l + m + n$ is $48$ .