Question

For real constants $a$ and $b$, let \[ f(x) = \begin{cases} \frac{a \sin x - 2x}{x}, & x < 0 \\ bx, & x \ge 0 \end{cases} \] 

If $f$ is a differentiable function, then the value of $a + b$ is 
 

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For a piecewise function to be differentiable at a point, both continuity and equal derivative limits from both sides must hold true.

Answer 1

The Correct Option is C

Step 1: Continuity at $x = 0$. 
For $f$ to be continuous at $x = 0$, \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x). \] For $x < 0$, \[ \lim_{x \to 0^-} \frac{a \sin x - 2x}{x} = a - 2. \] For $x \ge 0$, \[ \lim_{x \to 0^+} bx = 0. \] So, continuity gives $a - 2 = 0 $\Rightarrow$ a = 2$. 
 

Step 2: Differentiability at $x = 0$. 
For differentiability, \[ \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x). \] For $x < 0$, \[ f'(x) = \frac{a(x \cos x - \sin x) - 2x}{x^2}. \] As $x \to 0$, \[ f'(0^-) = -\frac{a}{3} \text{ (by L'Hôpital's rule or Taylor expansion)}. \] For $x > 0$, $f'(x) = b$. So $f'(0^+) = b$. Differentiability gives $f'(0^-) = f'(0^+) $ $\Rightarrow$ $ b = a - 2$. Substitute $a = 2$, we get $b = 0$. 
 

Step 3: Conclusion. 
\[ a + b = 2 + 0 = 2. \]