If $$f(x)= \begin{cases}\frac{a|x|+x^{2}-2(\sin|x|)(\cos|x|)}{x} & ,x \neq 0\\b & ,x = 0\end{cases}$$
is continuous at x = 0, then a + b is equal to

For continuity at x = 0: $$\lim_{x \to 0^+} \frac{ax + x^2 - 2\sin x\cos x}{x} = \lim_{x \to 0^+} \frac{ax + x^2 - \sin 2x}{x}$$

$$= a + 0 - 2 = a - 2$$ (using $$\sin 2x/x \to 2$$)

$$\lim_{x \to 0^-} \frac{-ax + x^2 - 2\sin(-x)\cos(-x)}{x} = \frac{-ax + x^2 + \sin 2x}{x} = -a + 0 + 2 = 2 - a$$

For continuity: $$a - 2 = 2 - a \implies 2a = 4 \implies a = 2$$. Then $$b = a - 2 = 0$$.

$$a + b = 2 + 0 = 2$$

The answer is Option 3: 2.