Question

If the function
\[ f(x) = \begin{cases} \sin(2x), & \text{for } x > 0, \\ a + bx, & \text{for } x \leq 0, \end{cases} \]
where \( a \) and \( b \) are constants, is differentiable at \( x = 0 \), then \( a + b \) is equal to:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For a function to be differentiable at a point, it must be continuous at that point, and the derivative must be the same from both sides.

Answer 1

The Correct Option is C

We are given the function \( f(x) \), and we are asked to find the value of \( a + b \) such that the function is differentiable at \( x = 0 \).

Step 1: Continuity at \( x = 0 \)
For \( f(x) \) to be differentiable at \( x = 0 \), it must first be continuous at \( x = 0 \). Thus, we must have:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0). \]
For \( x > 0 \), \( f(x) = \sin(2x) \), and for \( x \leq 0 \), \( f(x) = a + bx \).

From the definition of \( f(x) \), we have:
\[ f(0) = a + b \cdot 0 = a. \]
Also,
\[ \lim_{x \to 0^+} \sin(2x) = \sin(0) = 0. \]
Thus, for continuity at \( x = 0 \), we must have:
\[ a = 0. \]

Step 2: Differentiability at \( x = 0 \)
Next, we check the differentiability condition. The derivative of \( f(x) \) at \( x = 0 \) must be the same from both sides:
\[ \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}. \]
For \( x > 0 \), \( f(x) = \sin(2x) \), so:
\[ \lim_{x \to 0^+} \frac{\sin(2x)}{x} = \lim_{x \to 0^+} 2 \cos(2x) = 2. \]
For \( x \leq 0 \), \( f(x) = a + bx \), so:
\[ \frac{f(x) - f(0)}{x} = \frac{bx}{x} = b. \]
For differentiability at \( x = 0 \), we need:
\[ b = 2. \]

Thus, \( a = 0 \) and \( b = 2 \), so:
\[ a + b = 0 + 2 = 2. \]

Final Answer:
The value of \( a + b \) is \( \boxed{2} \).