Vinod Singh and Shweta Pawar Solutions for Chapter: Continuity, Exercise 2: Critical Thinking

The function $f\left(x\right)=\frac{x^2-4}{\sin x-2}$ is

If the function $f\left(x\right)$ is continuous in $\left[0,8\right]$, where $f\left(x\right)=\left\{\begin{array}{l}{x}^2+ax+6, 0\le x<2\\ 3x+2, 2\le x\le 4\\ 2ax+5b, 4<x\le 8\end{array}\right.$, then 

If $f\left(x\right)$ is continuous in $\left[-2,2\right]$, where $f\left(x\right)=\left\{\begin{array}{c}\frac{\sin ax}{x}-2, \mathrm{for}-2\le x<0\\ 2x+1, \mathrm{for} 0\le x\le 1\\ 2b\sqrt{x^2+3}-1, \mathrm{for} 1<x\le 2\end{array}\right.$, then the value of $\left(a+b\right)$ is

The values of $a$ and $b$ such that the function defined by $f\left(x\right)=\left\{\begin{array}{cll}7& ,& \mathrm{if} x\le 2\\ ax+b& ,& \mathrm{if} 2<x<9\\ 21& ,& \mathrm{if} x\ge 9\end{array}\right.$ is continuous in its domain are

A function $f\left(x\right)$ is defined by $f\left(x\right)=\frac{e^x+e^{-x}-2}{x\sin x}$ for $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\left\{0\right\}$

The value of $f\left(0\right)$ so that $f$ will be continuous in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is

Let $f\left(x\right)=\left\{\begin{array}{l}\frac{x^4-5x^2+4}{\left|\left(x-1\right)\left(x-2\right)\right|}; x\ne 1, 2\\ \begin{array}{r} 6; x=1\\ 12; x=2\end{array}\end{array}\right.$, then $f\left(x\right)$ is continuous on the set

If $f\left(x\right)$ is continuous in $\left[0,\pi \right]$, where $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}x+a\sqrt{2}\sin x, 0\le x<\frac{\pi }{4}\\ 2x\cot x+b, \frac{\pi }{4}\le x\le \frac{\pi }{2}\end{array}\\ a\cos 2x-b\sin x, \frac{\pi }{2}<x\le \pi \end{array}\right.$, then 

If the function $f\left(x\right)=\left\{\begin{array}{c}1+\sin \frac{\pi x}{2}, \text{for}-\infty <x\le 1\\ ax+b, \text{for} 1<x<3\\ 6 \tan \frac{\pi x}{12}, \text{for} 3\le x<6\end{array}\right.$ is continuous in the interval $\left(-\infty ,6\right)$, then the values of $a$ and $b$ are respectively