Question

Let \( \alpha \) and \( \beta \) be real numbers such that \( f(x) \) is defined as: \[ f(x) = \begin{cases} 2x^2 + 4x + \alpha, & \text{if } x < 1 \\ \beta x^2 + 5, & \text{if } x \geq 1 \end{cases} \] and is differentiable at \( x = 1 \). Then \( \alpha + \beta \) is equal to:

• A. \( 5 \)
• B. \( 6 \)
• C. \( 7 \)
• D. \( 8 \)
• E. \( 9 \)

Hint:

For piecewise differentiable functions, ensure both continuity and derivative matching at the given point.

Answer 1

The Correct Option is C

Step 1: Continuity at \( x = 1 \) Since \( f(x) \) is differentiable, it must also be continuous at \( x = 1 \), meaning: \[ \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x). \] Substituting \( x = 1 \): \[ 2(1)^2 + 4(1) + \alpha = \beta(1)^2 + 5. \] \[ 2 + 4 + \alpha = \beta + 5. \] \[ \alpha - \beta = -1. \quad {(Equation 1)} \] Step 2: Differentiability at \( x = 1 \) Taking derivatives: \[ f'(x) = \frac{d}{dx} (2x^2 + 4x + \alpha) = 4x + 4, \quad {for } x<1. \] \[ f'(x) = \frac{d}{dx} (\beta x^2 + 5) = 2\beta x, \quad {for } x \geq 1. \] Setting \( f'(1^-) = f'(1^+) \): \[ 4(1) + 4 = 2\beta(1). \] \[ 8 = 2\beta. \] \[ \beta = 4. \quad {(Equation 2)} \] Step 3: Solve for \( \alpha \) Using Equation 1: \[ \alpha - 4 = -1. \] \[ \alpha = 3. \] \[ \alpha + \beta = 3 + 4 = 7. \] Thus, the correct answer is (C) 7.