Q36 / 90 · KVPY 2011 (SB/SX) · Mathematics
For each positive integer $\mathrm{n}$, define $\mathrm{f}_{\mathrm{n}}(\mathrm{x})=$ minimum $\left(\frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n} !}, \frac{(1-\mathrm{x})^{\mathrm{n}}}{\mathrm{n} !}\right)$, for $0 \leq \mathrm{x} \leq 1$. Let $\mathrm{I}_{\mathrm{n}}=\int_{0}^{1} \mathrm{f}_{\mathrm{n}}(\mathrm{x}) \mathrm{dx}, \mathrm{n} \geq 1$. Then $I_{n}=\sum_{n=1}^{\infty} I_{n}$ is equal to-
- Correct answerA$2 \sqrt{\mathrm{e}}-3$
- B$2 \sqrt{\mathrm{e}}-2$
- C$2 \sqrt{\mathrm{e}}-1$
- D$2 \sqrt{\mathrm{e}}$
Solution
$I_{n}=\int_{0}^{1 / 2} \frac{x^{n}}{n !} d x+\int_{1 / 2}^{1} \frac{(1-x)^{n}}{n !} d x=\frac{1}{(n+1) !}\left(\left(\frac{1}{2}\right)^{n+1}+\left(\frac{1}{2}\right)^{n+1}\right)=\frac{\left(\frac{1}{2}\right)^{n}}{(n+1) !}$
$\sum_{n=1}^{\infty} I_{n}=\left(\frac{1 / 2}{2 !}+\frac{(1 / 2)^{2}}{3 !}+\ldots \ldots\right)=2 \sqrt{e}-3$
Hence, the correct answer is option: 'A'.