Let $$f: \mathbf{R} \rightarrow \mathbf{R}$$ be a function given by

$$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0\end{cases}$$

where $$\alpha, \beta \in \mathbf{R}$$. If $$f$$ is continuous at $$x=0$$, then $$\alpha^2+\beta^2$$ is equal to :

Let $f(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}$. If $\mathrm{m}$ and $\mathrm{n}$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $\mathrm{m}+\mathrm{n}$ is equal to :

Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$.

If for some $\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21$, then $\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :

$$ f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\\\ x^2+c x+2 ; & 0 \leq x \leq 1 \\\\ 2 x+1 ; & x>1\end{cases} $$

If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :