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(D) For $f(x)$ to be continuous at $x=1$,we must have $f(1^-) = f(1) = f(1^+)$.$f(1) = 1^2 + c(1) + 2 = 3+c$.$f(1^+) = \lim_{h \rightarrow 0} [2(1+h)+

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$2$

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(D) For $f(x)$ to be continuous at $x=1$,we must have $f(1^-) = f(1) = f(1^+)$.$f(1) = 1^2 + c(1) + 2 = 3+c$.$f(1^+) = \lim_{h \rightarrow 0} [2(1+h)+1] = 3$.Thus,$3+c = 3 \implies c = 0$.For $f(x)$ to be continuous at $x=0$,we must have $f(0^-) = f(0) = f(0^+)$.$f(0) = 0^2 + 0(0) + 2 = 2$.$f(0^-) = \lim_{h \rightarrow 0} \frac{a-b \cos 2h}{h^2} = 2$.Using the expansion $\cos 2h = 1 - \frac{(2h)^2}{2!} + \frac{(2h)^4}{4!} - \dots = 1 - 2h^2 + \frac{2}{3}h^4 - \dots$,we get:$\lim_{h \rightarrow 0} \frac{a-b(1 - 2h^2 + \frac{2}{3}h^4)}{h^2} = \lim_{h \rightarrow 0} \frac{(a-b) + 2bh^2 - \frac{2b}{3}h^4}{h^2} = 2$.For the limit to exist,$a-b=0 \implies a=b$. Then $2b = 2 \implies b=1$. Thus $a=1$.Now check differentiability at $x=0$:$LHD = \lim_{h \rightarrow 0} \frac{f(0-h) - f(0)}{-h} = \lim_{h \rightarrow 0} \frac{\frac{1-\cos 2h}{h^2} - 2}{-h} = \lim_{h \rightarrow 0} \frac{1-(1-2h^2+\frac{2}{3}h^4) - 2h^2}{-h^3} = \lim_{h \rightarrow 0} \frac{-\frac{2}{3}h^4}{-h^3} = 0$.$RHD = \lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{h^2+0h+2-2}{h} = \lim_{h \rightarrow 0} h = 0$.Since $LHD = RHD$,$f$ is differentiable at $x=0$.At $x=1$,$LHD = \frac{d}{dx}(x^2+0x+2)|_{x=1} = 2(1) = 2$. $RHD = \frac{d}{dx}(2x+1)|_{x=1} = 2$. Since $LHD=RHD$,$f$ is differentiable at $x=1$.Thus,$m=0$. Therefore,$m+a+b+c = 0+1+1+0 = 2$.

Let $f: R \rightarrow R$ 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 $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

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