If f(x) = begin{cases} frac{a|x|+x^2-2(sin |x | vedclass

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(B) For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = b$.First, consider the right-hand limit

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$2$

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(B) For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = b$.First, consider the right-hand limit $(x > 0)$:$\lim_{x 	o 0^+} \frac{ax + x^2 - 2\sin x \cos x}{x} = \lim_{x 	o 0^+} \left( a + x - \frac{\sin(2x)}{x} ight) = a + 0 - 2 = a - 2$.Next, consider the left-hand limit $(x Let $x = -h$ where $h > 0$. As $x 	o 0^-$, $h 	o 0^+$.$\lim_{h 	o 0^+} \frac{a|-h| + (-h)^2 - 2\sin|-h|\cos|-h|}{-h} = \lim_{h 	o 0^+} \frac{ah + h^2 - 2\sin h \cos h}{-h} = \lim_{h 	o 0^+} \left( -a - h + \frac{\sin(2h)}{h} ight) = -a - 0 + 2 = -a + 2$.For continuity, $a - 2 = -a + 2 = b$.From $a - 2 = -a + 2$, we get $2a = 4$, so $a = 2$.Substituting $a = 2$ into $b = a - 2$, we get $b = 2 - 2 = 0$.Therefore, $a + b = 2 + 0 = 2$.

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:

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