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Question:

If $\mathrm{I}_{1}=\int_{0}^{1}\left(1-x^{50}\right)^{100} d x$ and $\mathrm{I}_{2}=\int_{0}^{1}\left(1-x^{50}\right)^{101} d x$ such that $\mathrm{I}_{2}=\alpha \mathrm{I}_{1}$ then $\alpha$ equals to :

1. (1) $\frac{5049}{5050}$

2. (2) $\frac{5050}{5049}$

3. (3) $\frac{5050}{5051}$

4. (4) $\frac{5051}{5050}$

Correct Option: , 3

Solution:

$I_{2}=\int_{0}^{1}\left(1-x^{50}\right)^{101} d x=\int_{0}^{1}\left(1-x^{50}\right)\left(1-x^{50}\right)^{100} d x$

$I_{2}=\int_{0}^{1}\left(1-x^{50}\right)^{100} d x-\int_{0}^{1} x \underbrace{x^{49}\left(1-x^{50}\right)^{100}}_{\text {II }} d x$

$I_{2}=I_{1}+\left[\frac{x}{5050}\left(1-x^{50}\right)^{101}\right]_{0}^{1}-\int_{0}^{1} \frac{\left(1-x^{50}\right)^{101}}{5050} d x$

$I_{2}=I_{1}+0-\frac{I_{2}}{5050}$

$\Rightarrow \frac{5051}{5050} I_{2}=I_{1} \Rightarrow I_{2}=\frac{5050}{5051} I_{1}$

$\Rightarrow \alpha=\frac{5050}{5051}$