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Mathematics · Definite Integration

JEE Main 2024 — 8 April, Shift 1 — Question 1

The value of k∈N\mathrm{k} \in \mathbb{N} for which the integral In=∫01(1−xk)ndx,n∈NI_{n}=\int_{0}^{1}\left(1-x^{k}\right)^{n} d x, n \in \mathbb{N}, satisfies 147I20=148I21147 I_{20}=148 I_{21} is:

  1. Option A:

    10

  2. Option B:

    8

  3. Option C:

    14

  4. Option D:

    7

    Correct

Answer: D

Step-by-step solution

In=∫01(1−xk)n.1dx \mathrm{I}_{\mathrm{n}}=\int_{0}^{1}\left(1-\mathrm{x}^{\mathrm{k}}\right)^{\mathrm{n}} .1 \mathrm{dx}

In=(1−xk)n⋅x−nk∫01(1−xk)n−1⋅xk−1⋅dxI_{n}=\left(1-x^{k}\right)^{n} \cdot x-n k \int_{0}^{1}\left(1-x^{k}\right)^{n-1} \cdot x^{k-1} \cdot d x

In=nk∫01[(1−xk)n−(1−xk)n−1]dxI_{n}=n k \int_{0}^{1}\left[\left(1-x^{k}\right)^{n}-\left(1-x^{k}\right)^{n-1}\right] d x

In=nkIn−nkIn\mathrm{I}_{\mathrm{n}}=\mathrm{nkI}_{\mathrm{n}}-\mathrm{nkI}_{\mathrm{n}}

InIn−1=nknk+1\frac{I_{n}}{I_{n-1}}=\frac{\mathrm{nk}}{\mathrm{nk}+1}

I21I20=21k1+21k\frac{\mathrm{I}_{21}}{\mathrm{I}_{20}}=\frac{21 \mathrm{k}}{1+21 \mathrm{k}}

=147148⇒k=7=\frac{147}{148} \Rightarrow \mathrm{k}=7

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Mathematics
Chapter
Definite Integration
Topic
Reduction Formulae in Definite Integrals