Let $$f: \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = \begin{cases} \frac{a - b\cos 2x}{x^2}, & x < 0 \\ x^2 + cx + 2, & 0 \leq x \leq 1 \\ 2x + 1, & x > 1 \end{cases}\\$$ If $$f$$ is continuous everywhere in $$\mathbb{R}$$ and $$m$$ is the number of points where $$f$$ is NOT differentiable then $$m + a + b + c$$ equals:

Continuity at $$x=0$$

$$\lim_{x \to 0^-} \frac{a - b \cos 2x}{x^2} = f(0) = 2$$.

For the limit to exist, the numerator must be $$0$$ at $$x=0$$: $$a - b \cos(0) = 0 \implies a = b$$.

Using L'Hôpital's Rule: $$\lim_{x \to 0} \frac{2b \sin 2x}{2x} = \lim_{x \to 0} \frac{4b \cos 2x}{2} = 2b$$.

Given the limit is $$2$$, then $$2b = 2 \implies \mathbf{b = 1, a = 1}$$.

Continuity at $$x=1$$

$$f(1) = 1^2 + c(1) + 2 = 3 + c$$.

$$\lim_{x \to 1^+} (2x + 1) = 3$$.

Setting them equal: $$3 + c = 3 \implies \mathbf{c = 0}$$.

Differentiability ($$m$$)

• At $$x=0$$: $$f'(x)$$ for $$x > 0$$ is $$2x + c = 2x$$. At $$x=0$$, $$f'(0^+) = 0$$. Using the expansion of $$\cos 2x$$, the left-hand derivative is also $$0$$. So $$f$$ is differentiable at $$x=0$$.

• At $$x=1$$: $$f'(1^-) = 2(1) + 0 = 2$$. $$f'(1^+) = 2$$. It is differentiable at $$x=1$$.

Thus, $$m = 0$$.

Final Calculation: $$m + a + b + c = 0 + 1 + 1 + 0 = 2$$.