Let $f(x)=2+|x|-|x-1|+|x+1|, x \in R$. Consider

$(S1)$: $f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$

$( S 2): \int_{-2}^{2} f ( x ) dx =12$Then,

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in[0,1]$ and $\int_0^1 f(x) d x=\frac{\pi}{4}$ Then, $\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} d x$ equals

The true solution set of the inequality,$\sqrt{5x-6-x^2}+\left( \frac{\pi }{2}\int\limits_{0}^{x}{dz} \right)>x\int\limits_{0}^{\pi }{{{\sin }^{2}}xdx}$ is:

The true solution set of the inequality,

$\sqrt {5\,x\,\, - \,\,6\,\, - \,\,{x^2}} \,\, + \,\,\frac{\pi }{2}\,\,\int\limits_0^x {} $$dz > x \int\limits_0^\pi  {} sin^2 x \,dx$ is :

If $b _{ n }=\int \limits_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} nx }{\sin x } dx , n \in N$, then

Let $I_n=\int_0^{\pi / 2} x^n \cos x d x$, where $n$ is a non-negative integer. Then, $\sum \limits_{n=2}^{\infty}\left(\frac{I_n}{n !}+\frac{I_n-2}{(n-2) !}\right)$ equals