The value of k \in \mathbbN for which the integral I_n=\int_0^1(1-x^k)^n d x, n \in \mathbbN , satisfies 147 I_20=148 I_21 is

Definite Integration PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q32 with Solution

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Correct answer: C

\int_0^1 (1-x^k)^n\,dx.$$ We must find $k\in\mathbb N$. 2. First, transform the integral into Beta-function form. Let $$t=x^k \implies x=t^{1/k}, \quad dx=\frac{1}{k}t^{\frac1k-1}dt.$$ Then $$I_n=\int_0^1 (1-x^k)^n dx =\frac{1}{k}\int_0^1 t^{\frac1k-1}(1-t)^n dt.$$ This is $$I_n=\frac{1}{k}B\left(\frac1k,n+1\right),$$ where $B$ is the Beta function. Using $$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$ we get $$I_n=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(n+1)}{\Gamma(n+1+1/k)}.$$ 3. Now find the ratio $\dfrac{I_{21}}{I_{20}}$. $$I_{20}=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(21)}{\Gamma(21+1/k)}$$ $$I_{21}=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(22)}{\Gamma(22+1/k)}.$$ Therefore, $$\frac{I_{21}}{I_{20}}=\frac{\Gamma(22)}{\Gamma(21)}\cdot \frac{\Gamma(21+1/k)}{\Gamma(22+1/k)}.$$ Using $\Gamma(m+1)=m\Gamma(m)$, $$\frac{\Gamma(22)}{\Gamma(21)}=21,$$ and $$\Gamma(22+1/k)=(21+1/k)\Gamma(21+1/k).$$ So $$\frac{I_{21}}{I_{20}}=\frac{21}{21+1/k}.$$ 4. Use the given condition: $$147I_{20}=148I_{21}.$$ Divide by $148I_{20}$: $$\frac{I_{21}}{I_{20}}=\frac{147}{148}.$$ Hence, $$\frac{21}{21+1/k}=\frac{147}{148}.$$ 5. Solve for $k$. Cross-multiplying, $$21\cdot 148=147\left(21+\frac1k\right).$$ Since $147=21\cdot 7$, $$148=7\left(21+\frac1k\right).$$ Thus, $$148=147+\frac{7}{k}$$ $$1=\frac{7}{k}$$ $$k=7.$$ 6. Check options: - A: $8$ ❌ - B: $14$ ❌ - C: $7$ ✅ - D: $10$ ❌ Therefore, the correct answer is **Option C**.

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Correct answer: C

\int_0^1 (1-x^k)^n\,dx.$$ We must find $k\in\mathbb N$. 2. First, transform the integral into Beta-function form. Let $$t=x^k \implies x=t^{1/k}, \quad dx=\frac{1}{k}t^{\frac1k-1}dt.$$ Then $$I_n=\int_0^1 (1-x^k)^n dx =\frac{1}{k}\int_0^1 t^{\frac1k-1}(1-t)^n dt.$$ This is $$I_n=\frac{1}{k}B\left(\frac1k,n+1\right),$$ where $B$ is the Beta function. Using $$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$ we get $$I_n=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(n+1)}{\Gamma(n+1+1/k)}.$$ 3. Now find the ratio $\dfrac{I_{21}}{I_{20}}$. $$I_{20}=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(21)}{\Gamma(21+1/k)}$$ $$I_{21}=\frac{1}{k}\cdot \frac{\Gamma(1/k)\Gamma(22)}{\Gamma(22+1/k)}.$$ Therefore, $$\frac{I_{21}}{I_{20}}=\frac{\Gamma(22)}{\Gamma(21)}\cdot \frac{\Gamma(21+1/k)}{\Gamma(22+1/k)}.$$ Using $\Gamma(m+1)=m\Gamma(m)$, $$\frac{\Gamma(22)}{\Gamma(21)}=21,$$ and $$\Gamma(22+1/k)=(21+1/k)\Gamma(21+1/k).$$ So $$\frac{I_{21}}{I_{20}}=\frac{21}{21+1/k}.$$ 4. Use the given condition: $$147I_{20}=148I_{21}.$$ Divide by $148I_{20}$: $$\frac{I_{21}}{I_{20}}=\frac{147}{148}.$$ Hence, $$\frac{21}{21+1/k}=\frac{147}{148}.$$ 5. Solve for $k$. Cross-multiplying, $$21\cdot 148=147\left(21+\frac1k\right).$$ Since $147=21\cdot 7$, $$148=7\left(21+\frac1k\right).$$ Thus, $$148=147+\frac{7}{k}$$ $$1=\frac{7}{k}$$ $$k=7.$$ 6. Check options: - A: $8$ ❌ - B: $14$ ❌ - C: $7$ ✅ - D: $10$ ❌ Therefore, the correct answer is **Option C**.