If $I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ and $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$,then $I_1 : I_2$ is

Let $\operatorname{Max} \limits _{0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\operatorname{Min} \limits _ {0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta$. If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{e}\left(\frac{8}{15}\right)$,then $\alpha_{1}+\alpha_{2}$ is equal to

The number of continuous functions $f : [0, \frac{3}{2}] \rightarrow (0, \infty)$ satisfying the equation $4 \int_0^{3/2} f(x) dx + 125 \int_0^{3/2} \frac{dx}{\sqrt{f(x)+x^2}} = 108$ is

Let $f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x$ for $x \in R$ be a function which satisfies $f(x)=x+\int \limits_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to $............$

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 $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$

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