IfI(m, n)=∫_0^1 x^ m-1 (1-x)^ n-1 d x, m, n 0 , then I(9,14)+I(10,13) is

Correct answer: D. I(9,13) aligned & I(m, m)=∫_0^1 x^ m-1 (1-x)^ n-1 d x & Let x= ^2 θ dx=2 θ θ d θ & I(m, n)=2 ∫_0^ π / 2 ( θ)^ 2 m-1 ( θ)^ 2 n-1 d θ & I(9,14)+I(10,13)=2 ∫_0^ π / 2 ( θ)^ 17 ( θ)^ 27 d θ & +2 ∫_0^ π / 2 ( θ)^ 19 ( θ)^ 25 d θ & =2 ∫_0^ π / 2 ( θ)^ 17 ( θ)^ 25 d θ & =I(9,13) aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Definite Integration

IfI(m, n)=∫_0^1 x^ m-1 (1-x)^ n-1 d x, m, n 0 , then...

\operatorname{IfI}(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n\gt0

, then

I(9,14)+I(10,13)

Options

A $\mathrm{I}(19,27)$
B $\mathrm{I}(9,1)$
C $I(1,13)$
D $\mathrm{I}(9,13)$

Correct answer

D. $\mathrm{I}(9,13)$

Step-by-step solution

\begin{aligned} & \mathrm{I}(\mathrm{m}, \mathrm{m})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} d \mathrm{x} \\ & \text{ Let } \mathrm{x}=\sin ^2 \theta \quad \mathrm{dx}=2 \sin \theta \cos \theta \mathrm{ d} \theta \\ & \mathrm{I}(\mathrm{m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin \theta)^{2 \mathrm{ m}-1}(\cos \theta)^{2 \mathrm{n}-1} d \theta \\ & \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} d \theta \\ & +2 \int_0^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} d \theta \\ & =2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} d \theta \\ & =\mathrm{I}(9,13)\end{aligned}

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