Let $f : [0,1] \to [0,1]$ be a continuous function,then the equation $f(x) = x$

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} a^2 \cos^2 x + b^2 \sin^2 x, & x \leq 0 \\ e^{ax+b}, & x > 0 \end{cases}$ and is a continuous function,then:

If $f(x)$ is continuous at point $x=0$ where $f(x) = \begin{cases} \frac{3 \sin x + 5 \tan x}{a^x - 1} & , x < 0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x \cos x}{b^x - 1} & , x > 0 \end{cases}$ then the values of $a$ and $b$,respectively,are

Consider $f(x) = [x]|x^3 - 2x^2 - x + 2|$ in $[-\frac{3}{2}, \frac{9}{2}]$. The number of points where $f(x)$ is discontinuous is (where $[.]$ denotes the greatest integer function).

Let $f : [a, b] \rightarrow [1, \infty)$ be a continuous function and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) dt & \text{if } a \leq x \leq b \\ \int_a^b f(t) dt & \text{if } x > b \end{cases}$. Then:

If $f(x) = \begin{cases} \frac{(e^{3x}-1) \sin x^{\circ}}{x^2} & x \neq 0 \\ \frac{\pi}{60} & x = 0 \end{cases}$,then: