MCQ
For each positive integer $n$, defined $f_n(x)=$ minimum $\left(\frac{x^n}{n !}, \frac{(1-x)^n}{n !}\right)$, for $0 \leq x \leq 1 .$ Let $I_n=\int \limits_0^1 f_n(x) d x, n \geq 1$. Then, $\sum \limits_{n=1}^{\infty} I_n$ is equal to
  • $2 \sqrt{e}-3$
  • B
    $2 \sqrt{e}-2$
  • C
    $2 \sqrt{e}-1$
  • D
    $2 \sqrt{e}$

Answer

Correct option: A.
$2 \sqrt{e}-3$
a
(a)

We have,

$f_n(x) =\operatorname{minimum}\left(\frac{x^n}{n !}, \frac{(1-x)^n}{n !}\right), x \in[0,1]$

$I_n =\int \limits_0^1 f_n(x) d x$

$I_n =\left|\int \limits_0^{1 / 2} \frac{x^n}{n !} d x\right|+\left|\int_{1 / 2}^1 \frac{(1-x)^n}{n !} d x\right|$

$I_n =\left|\left[\frac{x^{n+1}}{(n+1) n !}\right]_0^{1 / 2}\right|+\mid\left[\left.\frac{(1-x)^{n+1}}{(n+1) n !}\right|_{1 / 2} ^1 \mid\right.$

$I_n =\frac{2}{(n+1) !}\left[\left(\frac{1}{2}\right)^{n+1}\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[\frac{1}{2 !}\left(\frac{1}{2}\right)^2+\frac{1}{3 !}\left(\frac{1}{2}\right)^3\right]$

$\sum \limits_{n=1}^{\infty} I_n =2\left[e^{1 / 2}-\left(1+\frac{1}{2}\right)\right]$

$=2 \sqrt{e}-3$

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