Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 2: Exercise 2

The left-hand derivative of $f\left(x\right)=\left[x\right] \sin \left(\mathrm{\pi }x\right)$at$x=k, k$ is an integer and $\left[·\right]$ denotes the greatest integer function, is

If $f\left(x\right)={\left(x+1\right)}^{\cot x}$ be continuous at $x=0$, then $f\left(0\right)$ is equal to

If $f\left(x\right)=a\left|\sin x\right|+b{e}^{\left|x\right|}+c{\left|x\right|}^3$  and if $f\left(x\right)$ is differentiable at $x=0$, then

If $f\left(x\right)=\left\{\begin{array}{cc}\left(\frac{x^2}{a}\right)-a;& \text{when} x<a\\ 0;& \text{when} x=a\\ a-\left(\frac{x^2}{a}\right);& \text{when} x>a\end{array}\right.$, then

If $f\left(x\right)=\left\{\begin{array}{c}\frac{\sin \left[x\right]}{\left[x\right]+1},\text{ for }x>0\\ \frac{\cos \frac{\mathrm{\pi }}{2}\left[x\right]}{\left[x\right]},\text{ for }x<0\\ k,\text{ at }x=0\end{array}\right.$; where $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then in order that $f$ be continuous at $x=0,$ the value of $k$ is

The function defined by $f\left(x\right)=\left\{\begin{array}{cc}{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1}& ,x\ne 2,\\ k& x=2\end{array}\right.$ is continuous from right at the point $x=2$, then $k$ is equal to

If $f\left(x\right)=\left\{\begin{array}{cc}x\sin \frac{1}{x},& x\ne 0\\ k,& x=0\end{array}\right.$ is continuous at  $x=0,$ then the value of $k$ will be

The function $f\left(x\right)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$ is not defined at $x=\mathrm{\pi }$. The value of $f\left(\mathrm{\pi }\right)$, so that $f\left(x\right)$ is continuous at $x=\mathrm{\pi }$, is