Amit M Agarwal Solutions for Chapter: Continuity and Differentiability, Exercise 1: Work Book Exercise 5.1

Let $f\left(x\right)=\frac{\sin x}{x},x\ne 0.$ Then $f\left(x\right)$ can be continuous at $x=0$ if

If $f\left(x\right)=\left\{\begin{array}{ll}a{x}^2+b,& 0\le x<1\\ 4,& x=1\\ x+3,& 1<x\le 2\end{array}\right.$, then the value of $a \& b$ for which $f\left(x\right)$ cannot be continuous at $x=1,$ is

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\sqrt{2}\sin x}{\pi -4x},& x\ne \frac{\pi }{4}\\ a,& x=\frac{\pi }{4}\end{array}\right.$ is continuous at $x=\frac{\pi }{4},$ then $a$ is equal to

If $f\left(x\right)=\frac{x^2-\left(a+2\right)x+a}{x-2}, \mathrm{for} x\ne 2$ and $f\left(2\right)=2$ is continuous at $x=2$, then $a$ is equal to

If $f\left(x\right)=\frac{a\cos x-\cos bx}{x^2}, x\ne 0$ and $f\left(0\right)=4$ is continuous at $x=0,$ then the ordered pair $\left(a,b\right)$ is

The set of points of discontinuity of the function $f\left(x\right)=\left|\sin x\right|$ is

The set of points of discontinuity of the function $f\left(x\right)=\frac{\left|\sin x\right|}{\sin x}$ is

If the function $f\left(x\right)={\left(\sin 2x\right)}^{{\tan }^{2}2x}$ is not defined at $x=\frac{\pi }{4}$. Then, the value of $f\left(\frac{\pi }{4}\right),$ so that $f\left(x\right)$ is continuous at $x=\frac{\pi }{4}$, is