JEE Main 2022 — Mathematics Calculus

Let $\displaystyle {I}_{1}={\int }_{0}^{1}{(1-{x}^{n})}^{2n}dx$I1​=∫01​(1−xn)2ndx and $\displaystyle {I}_{2}={\int }_{0}^{1}{(1-{x}^{n})}^{2n+1}dx$I2​=∫01​(1−xn)2n+1dx

Now on solving ${I}_{2}$I2​ we get,

$${I}_{2}={\int }_{0}^{1}{(1-{x}^{n})}^{2n+1}\cdot 1dx$$I2​=∫01​(1−xn)2n+1⋅1dx

Now on using by parts integration we get,

$$\Rightarrow {I}_{2}={(1-{x}^{n})}^{2n+1}.{x}_{0}^{1}-{\int }_{0}^{1}(2n+1){(1-{x}^{n})}^{2n}(-n{x}^{n-1})xdx$$⇒I2​=(1−xn)2n+1.x01​−∫01​(2n+1)(1−xn)2n(−nxn−1)xdx

$\Rightarrow {I}_{2}=-n(2n+1){{I}_{2}-{I}_{1}}$⇒I2​=−n(2n+1)I2​−I1​ $\displaystyle {\text{as }{I}_{1}={\int }_{0}^{1}{(1-{x}^{n})}^{2n}dx}$as I1​=∫01​(1−xn)2ndx

$$\Rightarrow (2{n}^{2}+n+1){I}_{2}=n(2n+1){I}_{1}$$⇒(2n2+n+1)I2​=n(2n+1)I1​

$${\text{Given }n(2n+1){\int }_{0}^{1}{(1-{x}^{n})}^{2n}dx=1177{\int }_{0}^{1}{(1-{x}^{n})}^{2n+1}dx}$$Given n(2n+1)∫01​(1−xn)2ndx=1177∫01​(1−xn)2n+1dx

$$\Rightarrow \frac{{I}_{1}}{{I}_{2}}=\frac{2{n}^{2}+n+1}{n(2n+1)}=\frac{1177}{n(2n+1)}$$⇒I2​I1​​=n(2n+1)2n2+n+1​=n(2n+1)1177​

$$\Rightarrow 2{n}^{2}+n-1176=0$$⇒2n2+n−1176=0

$$\Rightarrow n=\frac{-1\pm \sqrt{1+4\times 2\times 1176}}{4}$$⇒n=4−1±1+4×2×1176​​

$$\Rightarrow n=\frac{-1\pm \sqrt{9409}}{4}$$⇒n=4−1±9409​​

$$\Rightarrow n=\frac{-1\pm 97}{4}$$⇒n=4−1±97​

$$\Rightarrow n=24or\frac{-49}{2}$$⇒n=24or2−49​

But given $n\in N$n∈N so $n=24$n=24