Question

Let \( f(x) = \begin{cases} x^2 - \alpha, & \text{if } x < 1 \\ \beta x - 3, & \text{if } x \geq 1 \end{cases} \). If \( f \) is continuous at \( x = 1 \), then the value of \( \alpha + \beta \) is:

• A. -2
• B. 2
• C. 4
• D. -4
• E. 0

Hint:

To ensure continuity at a point for a piecewise function, always set the limits from the left and right equal to the function value at that point, and solve for any unknown constants.

Answer 1

The Correct Option is C

For \( f \) to be continuous at \( x = 1 \), the left-hand limit as \( x \to 1^- \) must equal the right-hand limit as \( x \to 1^+ \), and both must equal \( f(1) \).
Calculate the left-hand limit: \[ \lim_{x \to 1^-} (x^2 - \alpha) = 1^2 - \alpha = 1 - \alpha \] Calculate the right-hand limit and \( f(1) \): \[ \lim_{x \to 1^+} (\beta x - 3) = \beta \cdot 1 - 3 = \beta - 3 \] \[ f(1) = \beta \cdot 1 - 3 = \beta - 3 \] Set the left-hand limit equal to the right-hand limit for continuity: \[ 1 - \alpha = \beta - 3 \] Solve for \( \alpha + \beta \): \[ 1 - \alpha = \beta - 3 \implies \alpha + \beta = 4 \] Thus, the value of \( \alpha + \beta \) that makes \( f \) continuous at \( x = 1 \) is 4.