Verified Solution • 2025–26 Session moderate

QJEEM22Maf540: JEE Main 2022 Mathematics - Class 12 JEEM22Maf540 Solution: If , then is equal to ______________.

If n(2n+1)∫01(1−xn)2ndx=1177∫01(1−xn)2n+1 dx\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x , then n∈N\mathrm{n} \in \mathbf{N} is equal to ______________.

Step-by-Step Verified Solution

∫01(1−xn)2n+1dx=∫011 . (1−xn)2n+1dx\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} }

=[(1−xn)2n+1 . x]01−∫01x . (2n+1)(1−xn)2n . −nxn−1dx= \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx}

=n(2n+1)∫01(1−(1−xn))(1−xn)2ndx= n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx}

=n(2n+1)∫01(1−xn)2ndx−n(2n+1)∫01(1−xn)2n+1dx= n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx - n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} }

(1+n(2n+1))∫01(1−xn)2n+1dx=n(2n+1)∫01(1−xn)2ndx(1 + n(2n + 1))\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx} }

(2n2+n+1)∫01(1−xn)2n+1dx=1177∫01(1−xn)2n+1dx(2{n^2} + n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = 1177\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} }

∴\therefore 2n2+n+1=11772{n^2} + n + 1 = 1177

2n2+n−1176=02{n^2} + n - 1176 = 0

∴\therefore n=24n = 24 or −492- {{49} \over 2}

∴\therefore n=24n = 24

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