JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 20)
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 ___________.
$$\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R$$, then $$\alpha$$ equals ___________.
Answer
1
Explanation
$${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}} .{(1 - x)^{n - 1}}dx$$
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
$$dy = - {1 \over {{z^2}}}dz$$
$$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)} $$
$$ \Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1$$
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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