Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Level 1

If $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-ax+3,& x \text{is rational} \\ 2-x,& x \text{is irrational }\end{array}\right.$ is continuous at exactly two points, then the possible values of $\text{'}a\text{'}$ are

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

A function is defined as follows: $f\left(x\right)=\left\{\begin{array}{cc}-\frac{x}{\sqrt{x^2}},& x\ne 0\\ 0,& x=0\end{array}\right.$ . Which one of the following is correct in respect of the above function $?$

If $f\left(x\right)=\left|x\right|+|x-1|,$ then which one of the following is correct?

Consider the function: $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin 2x}{5x},& \text{ if }x\ne 0\\ \frac{2}{15},& \text{ if }x=0\end{array}\right.$ 

Which one of the following is correct in respect to the function?

If $f\left(x\right)=\left\{\begin{array}{ll}{e}^{x^2+x},& x>0\\ ax+b,& x\le 0\end{array}\right.$ is differentiable at $x=0,$ then

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x\mathrm{logcos}x}{\log \left(1+x^2\right)};& x\ne 0\\ 0;& x=0\end{array}\right.$, then

The Rolle's theorem is not applicable to$f\left(x\right)=\left\{\begin{array}{cl}x,& 0\le x\le 1\\ 2-x,& 1\le x\le 2\end{array}\right.$ on $[0, 2]$ because