The value of $\mathrm{k} \in \mathbb{N}$ for which the integral $I_{n}=\int_{0}^{1}\left(1-x^{k}\right)^{n} d x, n \in \mathbb{N}$, satisfies $147 I_{20}=148 I_{21}$ is :

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Definite Integrals

The value of

k \in N

for which the integral

I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x, n \in N

, satisfies

147 I_{20} = 148 I_{21}

is :

Answer: (D)

Integrating

I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x

by parts gives the reduction

I_{n} = nk (I_{n - 1} - I_{n})

, i.e.

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

. Then

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

. Setting

147 I_{20} = 148 I_{21}

gives

\frac{I _{21}}{I _{20}} = \frac{147}{148}

, so

\frac{21 k}{21 k + 1} = \frac{147}{148} \Rightarrow 21 k = 147 \Rightarrow k = 7

. Option (4).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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