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