Question

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}

• A. 0
• B. 1
• C. 4
• D. 2

Hint:

Use Maclaurin series expansion or L'Hospital's rule for limits of the form $0/0$.

Answer 1

The Correct Option is D

Simplify $f(x)$ for $x \neq 0$: $f(x) = \frac{a|x| + x^2 - \sin(2|x|)}{x}$.
Calculate Right Hand Limit ($x \to 0^+$): $|x|=x$.
$\lim_{x \to 0^+} \frac{ax + x^2 - \sin(2x)}{x} = a + 0 - 2 = a-2$.
Calculate Left Hand Limit ($x \to 0^-$): $|x|=-x$.
$\lim_{x \to 0^-} \frac{-ax + x^2 - \sin(-2x)}{x} = \lim \frac{-ax + x^2 + \sin(2x)}{x} = -a + 0 + 2 = -a+2$.
For continuity, RHL = LHL. $a-2 = -a+2 \implies 2a=4 \implies a=2$.
The value of the limit is $a-2 = 0$. So $b=0$.
$a+b = 2+0 = 2$.