Vinod Singh and Shweta Pawar Solutions for Chapter: Continuity, Exercise 3: Competitive Thinking

If $f\left(x\right)=\frac{{20}^x+3^x-6^x-{10}^x}{1-\cos 8x}$, for $x\ne 0=\left(\frac{k}{16}\right)\log \left(\frac{10}{3}\right)·\log 2$, for$x=0$ is continuous at $x=0,$ then the value of $k$ is

Note: Assume base of logarithm as $e$ everywhere.

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{x-2}{\left|x-2\right|}+a,& x<2\\ a+b ,& x=2\\ \frac{x-2}{\left|x-2\right|}+b,& x>2\end{array}\right.$ is continuous at $x=2$, then $a+b=$

If $f\left(x\right)=\left|x\right|+\left|x-1\right|$, then

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2-4x+3}{x^2-1},& \mathrm{for} x\ne 1\\ 2,& \mathrm{for} x=1\end{array}\right.\begin{array}{r} \\ \end{array}$, then

If $f\left(x\right)=\left\{\begin{array}{c}\frac{\left|x-a\right|}{x-a}, \mathrm{when} x\ne a\\ 1, \mathrm{when} x=a\end{array}\right.$, then 

The function $f\left(x\right)=\left|x\right|+\frac{\left|x\right|}{x}$ is

The function $f\left(x\right)=\frac{2x^2+7}{x^3+3x^2-x-3}$ is discontinuous for

If $f:R\to R$ is defined by $f\left(x\right)=\left\{\begin{array}{lll}x-1& ,& \mathrm{for} x\le 1\\ 2-x^2& ,& \mathrm{for} 1<x\le 3\\ x-10& ,& \mathrm{for} 3<x<5\\ 2x& ,& \mathrm{for} x\ge 5\end{array}\right.$ then the set of points of discontinuity of $f$ is