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 = $

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=1$ and $\int_0^{\frac{\pi}{3}} f(t) dt = 0$. Then which of the following statements is (are) $TRUE$?
$(A)$ The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(B)$ The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$

List $I$	List $II$
$P.$ The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$,satisfying $f(0)=0$ and $\int_0^1 f(x) dx=1$,is	$1.$ $8$
$Q.$ The number of points in the interval $(-\sqrt{13}, \sqrt{13})$ at which $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

Codes: $P \quad Q \quad R \quad S$

Which of the following inequalities is/are $TRUE$?
$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$
$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$
$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$
$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$

If $f(x) = \int_0^x {t(\sin x - \sin t) dt}$,then which of the following is true?

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 $2$.