$\because \quad f(x) \text{ is continuous}$

$\therefore \quad (\text{LHL})_{x=0} = (\text{RHL})_{x=0} = (\text{VF})_{x=0} \quad \implies \quad \lim_{h \to 0} \left( \frac{a - b \cos 2h}{h^2} \right) = 23$

$\text{For this } a = b$

$\implies \lim_{h \to 0} a \left( \frac{1 - \cos 2h}{h^2} \right) = 2 \quad \implies \quad \lim_{h \to 0} a \left( \frac{2 \sin^2 h}{h^2} \right) = 2 \quad \implies \quad 2a = 2$

$\implies a = 1 \implies b = 1$

$\text{Now } (\text{LHL})_{x=1} = (\text{RHL})_{x=1} = (\text{VF})_{x=1} \quad \implies \quad c + 3 = 3 = 3 \quad \implies \quad c = 0$

$\text{Now } (\text{LHD})_{x=1} = (2x)_{x=1} = 2 \quad \implies \quad f(x) \text{ is differentiable at } x = 1 \qquad (\text{RHD})_{x=1} = 2$

$\text{Now } (\text{RHD})_{x=0} = (2x)_{x=0} = 0$

$(\text{LHD})_{x=0} = \lim_{h \to 0} \left( \frac{\frac{2 \sin^2 h}{h^2} - 2}{-h} \right) = 2 \lim_{h \to 0} \frac{(\sin h + h)(\sin h - h)}{-h^3} = 0 \quad \implies \quad f(x) \text{ is differentiable at } x = 0$

$m = 0, \; a = b = 1, \; c = 0 \quad \implies \quad m + a + b + c = 2$

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