🛠️ JEE➗ Maths

The value of \(k \in N\) for which the integral \(I_n=\int_0^1\left(1-x^k\right)^n d x, n \in N\), satisfies \(147 I_{20…

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The value of \(k \in N\) for which the integral \(I_n=\int_0^1\left(1-x^k\right)^n d x, n \in N\), satisfies \(147 I_{20}=148 I_{21}\) is

[JEE Main 2024, 8 Apr (Shift 1)]

a

8

b

7

c

14

d

10

✓ Correct answer: b)

7

Explanation

\({I}_{n}={\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n}\cdot 1dx\)

\({I}_{n}={\left[{\left(1-{x}^{k}\right)}^{n}\cdot x\right]}_{0}^{1}-nk{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(-{x}^{k-1}\right)\cdot xdx\)

\(=-kn{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(1-{x}^{k}-1\right)dx\)

\({I}_{n}=-nk{\int }_{0}^{1}\left[{\left(1-{x}^{k}\right)}^{n}-{\left(1-{x}^{k}\right)}^{n-1}\right]dx\)

\({I}_{n}=nk{I}_{n-1}-nk{I}_{n}\)

\(\frac{{I}_{n}}{{I}_{n-1}}=\frac{nk}{nk+1}\)

\(\frac{{I}_{21}}{{I}_{20}}=\frac{21k}{1+21k}\)

\(=\frac{147}{148}\Rightarrow k=7\)

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