Question

If \(f(x) = \begin{cases} 2x + 3, & \text{if } x \leq 1 \\ 2ax + bx, & \text{if } x > 1 \end{cases}\) is differentiable \(\forall x \in \mathbb{R}\), then \(f(2) =\)

• A. \(5\)
• B. \(4\)
• C. \(-4\)
• D. \(-10\)

Hint:

To ensure the function is differentiable at \( x = 1 \), you need to check both continuity and differentiability at that point. For continuity, the left-hand and right-hand limits at \( x = 1 \) should be equal. For differentiability, the derivatives from both sides must match at \( x = 1 \). Solve these conditions to find the values of \( a \) and \( b \), and then calculate \( f(2) \) using the second piece of the function for \( x>1 \).

Answer 1

The Correct Option is B

Since \( f(x) \) is differentiable everywhere, it is particularly continuous and differentiable at \( x = 1 \). This implies both the function and its derivative must match at \( x = 1 \). Thus, we must ensure that: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x), \quad \text{and} \quad f'(1^-) = f'(1^+). \] Step 1: Ensure continuity at \( x = 1 \). For continuity at \( x = 1 \), we equate the values of the function from both sides: \[ \lim_{x \to 1^-} (2x + 3) = \lim_{x \to 1^+} (2ax + bx). \] Substitute \(x = 1\) into both expressions: \[ 2(1) + 3 = 2a(1) + b(1), \quad 5 = 2a + b. \] Thus, we obtain the first equation: \[ 2a + b = 5. \tag{1} \] Step 2: Ensure differentiability at \( x = 1 \). For differentiability, the derivatives from both sides must also be equal: \[ f'(x) = \frac{d}{dx} \left( 2x + 3 \right) = 2 \quad \text{for } x \leq 1, \] \[ f'(x) = \frac{d}{dx} \left( 2ax + bx \right) = 2a + b \quad \text{for } x > 1. \] At \( x = 1 \), we set the derivatives equal to each other: \[ 2 = 2a + b. \tag{2} \] Step 3: Solve the system of equations. We have the system of equations: \[ 2a + b = 5 \quad \text{(from continuity)}, \] \[ 2a + b = 2 \quad \text{(from differentiability)}. \] The system yields: \[ 2a + b = 5 \quad \text{and} \quad 2a + b = 2, \] which is contradictory. Thus, there seems to be a misstep in this approach, rechecking.