Continuity Condition

For the function $f(x)$ to be continuous at $x = 0$, the limit of $f(x)$ as $x$ approaches 0 must exist and be equal to $f(0)$.

$ \lim_{x \to 0} f(x) = f(0) $

Given $f(0) = b$, we require:

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

Simplify the trigonometric term: $2(\sin|x|)(\cos|x|) = \sin(2|x|)$.

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

Limit Calculation

We evaluate the limit using the left-hand limit (LHL) and the right-hand limit (RHL).

Right-Hand Limit (RHL)

For $x \to 0^+$, $|x| = x$. The limit becomes:

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

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

Using the standard limit $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$, we get:

$ \text{RHL} = a + 0 - 2 = a - 2 $

Left-Hand Limit (LHL)

For $x \to 0^-$, $|x| = -x$. The limit becomes:

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

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

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

Using the standard limit, we get:

$ \text{LHL} = -a + 0 + 2 = -a + 2 $

Determining Constants a and b

For continuity, LHL = RHL:

$ a - 2 = -a + 2 $

Solving for $a$:

$ 2a = 4 \implies a = 2 $

The value of the limit is found by substituting $a=2$ into either LHL or RHL:

$ \text{Limit} = a - 2 = 2 - 2 = 0 $

Since the limit must equal $f(0)$, we have $b = 0$.

Final Calculation

We found $a = 2$ and $b = 0$. The question asks for $a + b$.

$ a + b = 2 + 0 = 2 $