(3)

Here, $f\left(x\right)$$=\{\begin{array}{cc}x,& x<–1\\ x^{21},& –1\le x<0\\ x,& 0\le x<1\\ x^{21},& x\ge 1\end{array}$ is continuous everywhere.

Then, n = 0

$\Rightarrow f\text{'}\left(x\right)$$=\{\begin{array}{cc}1,& x<–1\\ 21x^{20},& –1\le x<0\\ 1,& 0\le x<1\\ 21x^{20},& x\ge 1\end{array}$ is not differentiable at

$x=–1,0,1 \Rightarrow m=3$

So, m + n = 3.

(8)

Values of x, where $\left[\frac{x^2}{2}\right]$ may be discontinuous on $x\in [0,4]$ are $\sqrt{2},2,\sqrt{6},2\sqrt{2},\sqrt{10},2\sqrt{3},\sqrt{14},4$

And for $\left[\sqrt{x}\right]$, values of x are = 1, 4

On checking for continuity at these points, we get f(x) is discontinuous at $x=1,\sqrt{2},2,\sqrt{6},2\sqrt{2},\sqrt{10},2\sqrt{3},\sqrt{14}$ and continuous at x = 4.

Hence, f(x) is discontinuous for 8 values of $x\in [0,4]$.

(2)

$f\left(0\right)=\underset{x\to 0}{\lim }\frac{\tan \left(\tan x\right)–\sin \left(\sin x\right)–\tan x+\tan x–\sin x+\sin x}{\tan x–\sin x}$

$=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\tan \left(\tan x\right)–\tan x}{{\tan }^{3}x}}\times {\displaystyle \frac{{\tan }^{3}x}{x^3}}+{\displaystyle \frac{\tan x–\sin x}{x^3}}+{\displaystyle \frac{\sin x–\sin \left(\sin x\right)}{{\sin }^{3}x}}\times {\displaystyle \frac{{\sin }^{3}x}{x^3}}}{{\displaystyle \frac{\tan x–\sin x}{x^3}}}$

$=1+\frac{({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}})}{({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}})}=2$.          [From L'Hospital's Rule]

(5)

$f\left(x\right)$$=\{\begin{array}{cc}3x,& x<0\\ \min \{1+x+[x], x+2[x\left]\right\},& 0\le x\le 2\\ 5,& x>2\end{array}$

$\Rightarrow$$f\left(x\right)$$=\{\begin{array}{cc}3x,& x<0\\ x,& 0\le x<1\\ x+2,& 1\le x<2\\ 5,& x>2\end{array}$

$\therefore$   By graph, we have f(x) is not continuous at $x\in \{1,2\} \Rightarrow \alpha =2$

f(x) is not differentiable at $x\in \{0,1,2\} \Rightarrow \beta =3$

$\therefore \alpha +\beta =5$.