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 :

At $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ is continuous therefore, <br/><br/>$\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$ <br/><br/>$f(1)=3+c$ .........(1) <br/><br/>$$ \begin{aligned} & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\\\ & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3 .........(2) \end{aligned} $$ <br/><br/>from (1) and (2) <br/><br/>$\mathrm{c}=0$ <br/><br/>at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$ is continuous therefore, <br/><br/>$$ \begin{aligned} & \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right) ........(3) \\\\ & \mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2 ...........(4) \end{aligned} $$ <br/><br/>$\mathrm{f}\left(0^{-}\right)$has to be equal to 2 <br/><br/>$\lim\limits_{h \rightarrow 0} \frac{a-b \cos (2 h)}{h^2}$ <br/><br/>= $\lim\limits_{h \rightarrow 0} \frac{a-b\left\{1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !}+\ldots\right\}}{h^2}$ <br/><br/>= $\lim\limits_{h \rightarrow 0} \frac{a-b+b\left\{2 h^2-\frac{2}{3} h^4 \ldots\right\}}{h^2}$ <br/><br/>for limit to exist $a-b=0$ and limit is $2 b$ from (3), (4) and (5) <br/><br/>$\mathrm{a}=\mathrm{b}=1$ <br/><br/>checking differentiability at $\mathrm{x}=0$ <br/><br/>$$ \begin{aligned} & \text { LHD : } \lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^2}-2}{-h} \\\\ & \lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !} \ldots\right)-2 h^2}{-h^3}=0 \\\\ & \text { RHD : } \lim _{h \rightarrow 0} \frac{(0+h)^2+2-2}{h}=0 \end{aligned} $$ <br/><br/>Function is differentiable at every point in its domain <br/><br/>$\therefore \mathrm{m}=0$ <br/><br/>m + a + b + c = 0 + 1 + 1 + 0 = 2