JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 2)
The value of $$k \in \mathbb{N}$$ for which the integral $$I_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N}$$, satisfies $$147 I_{20}=148 I_{21}$$ is
8
14
7
10
Explanation
$$\begin{aligned} & I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\ & =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\ & =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\ & I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x \end{aligned}$$
_8th_April_Morning_Shift_en_2_1.png)
$$I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int_\limits0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor$$
$$\begin{aligned} & I(21)=I(20)-\frac{1}{21 k} I(20) \\ & \Rightarrow[I(21)](21 k+1)=21 K I(20) \\ & \Rightarrow 21 K=147 \Rightarrow K=7 \end{aligned}$$
Comments (0)
