Let $I_n = \int_0^1 e^{-y} y^n \, dy$,where $n$ is a non-negative integer. Then,$\sum_{n=1}^{\infty} \frac{I_n}{n!}$ is

If $A_n = \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^n x \, dx$,then $\frac{A_4 - A_6}{A_4} = $

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

Given that for each $a \in (0,1)$,the limit $g(a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$ exists. In addition,it is given that the function $g(a)$ is differentiable on $(0,1)$.
$1.$ The value of $g\left(\frac{1}{2}\right)$ is
$(A) \pi$ $(B) 2\pi$ $(C) \frac{\pi}{2}$ $(D) \frac{\pi}{4}$
$2.$ The value of $g'\left(\frac{1}{2}\right)$ is
$(A) \frac{\pi}{2}$ $(B) \pi$ $(C) -\frac{\pi}{2}$ $(D) 0$
Select the correct pair of answers for $1$ and $2$.

If $f(x) = \text{Max}\{\sin x, \cos x\}$ and $g(x) = \text{Min}\{\sin x, \cos x\}$,then $\int_{0}^{\pi} f(x) dx + \int_{0}^{\pi} g(x) dx = $