QJEEM21Ma0e6c: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0e6c Solution: If , for m, , and , then equals ___________.

Question

Verified Solution • 2025–26 Session moderate

If

{I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx}

, for m,

n \ge 1

, and

\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R

, then

\alpha

equals ___________.

Accepted Answer

{I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}} .{(1 - x)^{n - 1}}dx

Put

x = {1 \over {y + 1}} \Rightarrow dx = {{ - 1} \over {{{(y + 1)}^2}}}dy

1 - x = {y \over {y + 1}}

\therefore

{I_{m,n}} = \int\limits_\infty ^0 {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}( - 1)dy = } \int\limits_0^\infty {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy}

.... (i)

Similarly,

{I_{m,n}} = \int\limits_0^1 {{x^{n - 1}}.{{(1 - x)}^{m - 1}}dx}

\Rightarrow {I_{m,n}} = \int\limits_0^\infty {{{{y^{m - 1}}} \over {{{(y + 1)}^{m + n}}}}dy}

.... (ii)

From (i) & (ii)

2{I_{m,n}} = \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy}

\Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} = {I_1} + {I_2}

Put

y = {1 \over z}

in I₂

dy = - {1 \over {{z^2}}}dz

\Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)}

\Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1

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