If n(2 n+1) \int_0^1(1-x^n)^2 n d x=1177 \int_0^1(1-x^n)^2 n+1 ~d x , then n \in N is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 26 Jul · Shift 1 · Q42 with Solution

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Correct answer: 24

Let $I_k=\int_0^1 (1-x^n)^k\,dx.$ The given equation is $n(2n+1)I_{2n}=1177\,I_{2n+1}.$ We need to find $n\in\mathbb N$ .
Transform the integral using substitution. Let $t=x^n \implies x=t^{1/n},\qquad dx=\frac1n t^{\frac1n-1}dt.$ Hence,

=\frac1n B\left(\frac1n,k+1\right),$$ where $B$ is the Beta function. So, $$I_{2n}=\frac1n B\left(\frac1n,2n+1\right), \qquad I_{2n+1}=\frac1n B\left(\frac1n,2n+2\right).$$ 3. Take the ratio using the Beta function property $$B(x,y+1)=\frac{y}{x+y}B(x,y).$$ With $x=\frac1n$ and $y=2n+1$, we get $$B\left(\frac1n,2n+2\right)=\frac{2n+1}{\frac1n+2n+1}B\left(\frac1n,2n+1\right).$$ Therefore, $$I_{2n+1}=\frac{2n+1}{2n+1+\frac1n}I_{2n}.$$ 4. Substitute into the given equation: $$n(2n+1)I_{2n}=1177\,I_{2n+1} =1177\cdot \frac{2n+1}{2n+1+\frac1n}I_{2n}.$$ Since $I_{2n}>0$ and $2n+1>0$, cancel them: $$n=\frac{1177}{2n+1+\frac1n}.$$ Multiply through: $$n\left(2n+1+\frac1n\right)=1177.$$ So, $$2n^2+n+1=1177.$$ Hence, $$2n^2+n-1176=0.$$ 5. Solve the quadratic: $$2n^2+n-1176=0.$$ Discriminant: $$\Delta=1+4\cdot 2\cdot 1176=1+9408=9409=97^2.$$ Thus, $$n=\frac{-1\pm 97}{4}.$$ So, $$n=24 \quad \text{or} \quad n=-\frac{49}{2}.$$ Since $n\in\mathbb N$, we get $$\boxed{24}.$$ 6. Comparison with stored answer: Stored correct answer = $24$, which matches our derived answer.

View written solutionFree

Correct answer: 24

Let $I_k=\int_0^1 (1-x^n)^k\,dx.$ The given equation is $n(2n+1)I_{2n}=1177\,I_{2n+1}.$ We need to find $n\in\mathbb N$ .
Transform the integral using substitution. Let $t=x^n \implies x=t^{1/n},\qquad dx=\frac1n t^{\frac1n-1}dt.$ Hence,

=\frac1n B\left(\frac1n,k+1\right),$$ where $B$ is the Beta function. So, $$I_{2n}=\frac1n B\left(\frac1n,2n+1\right), \qquad I_{2n+1}=\frac1n B\left(\frac1n,2n+2\right).$$ 3. Take the ratio using the Beta function property $$B(x,y+1)=\frac{y}{x+y}B(x,y).$$ With $x=\frac1n$ and $y=2n+1$, we get $$B\left(\frac1n,2n+2\right)=\frac{2n+1}{\frac1n+2n+1}B\left(\frac1n,2n+1\right).$$ Therefore, $$I_{2n+1}=\frac{2n+1}{2n+1+\frac1n}I_{2n}.$$ 4. Substitute into the given equation: $$n(2n+1)I_{2n}=1177\,I_{2n+1} =1177\cdot \frac{2n+1}{2n+1+\frac1n}I_{2n}.$$ Since $I_{2n}>0$ and $2n+1>0$, cancel them: $$n=\frac{1177}{2n+1+\frac1n}.$$ Multiply through: $$n\left(2n+1+\frac1n\right)=1177.$$ So, $$2n^2+n+1=1177.$$ Hence, $$2n^2+n-1176=0.$$ 5. Solve the quadratic: $$2n^2+n-1176=0.$$ Discriminant: $$\Delta=1+4\cdot 2\cdot 1176=1+9408=9409=97^2.$$ Thus, $$n=\frac{-1\pm 97}{4}.$$ So, $$n=24 \quad \text{or} \quad n=-\frac{49}{2}.$$ Since $n\in\mathbb N$, we get $$\boxed{24}.$$ 6. Comparison with stored answer: Stored correct answer = $24$, which matches our derived answer.