Question

The function f(x) is given by:

For x < 0:
f(x) = e^x + ax

For x ≥ 0:
f(x) = b(x - 1)²

The function is differentiable at x = 0. Then,

• A. \( a = 3, b = 1 \)
• B. \( a = -3, b = 1 \)
• C. \( a = 3, b = -1 \)
• D. \( a = -3, b = -1 \)

Hint:

For differentiability at a point, remember that both the function value and the first derivative must match from both sides at that point. Always check both the left-hand and right-hand limits carefully.

Answer 1

The Correct Option is B

At \( x = 0 \), for the function to be differentiable, we need both the left-hand limit (LHL) and the right-hand limit (RHL) to be equal, as well as the function values.
For \( x<0 \), \( f(x) = e^x + ax \), and for \( x \geq 0 \), \( f(x) = b(x-1)^2 \). 1. Left-hand limit (LHL): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (e^x + ax) = e^0 + a(0) = 1. \] 2. Right-hand limit (RHL): \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} b(x-1)^2 = b(0-1)^2 = b. \] For the function to be continuous, we require that LHL = RHL. Therefore, we have: \[ 1 = b \quad \Rightarrow \quad b = 1. \] 3. Differentiability at \( x = 0 \): The derivative from the left at \( x = 0 \) is: \[ \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{e^x + ax - 1}{x} = \lim_{x \to 0^-} \frac{e^x - 1 + ax}{x}. \] Applying L'Hopital's Rule: \[ = \lim_{x \to 0^-} \frac{e^x + a}{1} = e^0 + a = 1 + a. \] The derivative from the right at \( x = 0 \) is: \[ \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{b(x-1)^2 - 1}{x} = \lim_{x \to 0^+} \frac{b(x^2 - 2x + 1) - 1}{x}. \] \[ = \lim_{x \to 0^+} \frac{-2bx}{x} = -2b. \] For the function to be differentiable, we require that LHL = RHL. Thus, \[ 1 + a = -2b = -2(1) = -2 \quad \Rightarrow \quad a = -3. \] Thus, the correct values are \( a = -3 \) and \( b = 1 \). Thus, the correct answer is option (2).