Question

Let \[ f(x)= \begin{cases} ax+3, & x \leq 2 \\ a^2x-1, & x > 2 \end{cases} \] Then the values of \(a\) for which \(f\) is continuous for all \(x\) are:

• A. \( 1 \) and \( -2 \)
• B. \( 1 \) and \( 2 \)
• C. \( -1 \) and \( 2 \)
• D. \( -1 \) and \( -2 \)
• E. \( 0 \) and \( 3 \)

Hint:

When solving continuity problems for piecewise functions, focus exclusively on the "break point." If the individual pieces are polynomials, they are already continuous everywhere else.

Answer 1

The Correct Option is C

Understanding the Concept: For a piecewise function to be continuous at a point \( x=c \), the left-hand limit, the right-hand limit, and the function's value must all be equal: \[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \]

Step 1: Evaluate the limits at the point of transition (\( x = 2 \)).
Left-hand limit (LHL): \[ \lim_{x \to 2^-} (ax + 3) = 2a + 3 \] Right-hand limit (RHL): \[ \lim_{x \to 2^+} (a^2x - 1) = 2a^2 - 1 \]

Step 2: Set the limits equal to each other for continuity.
\[ 2a^2 - 1 = 2a + 3 \] \[ 2a^2 - 2a - 4 = 0 \]

Step 3: Solve the quadratic equation for \( a \).
Divide the entire equation by 2: \[ a^2 - a - 2 = 0 \] Factor the quadratic: \[ (a - 2)(a + 1) = 0 \] Thus, \( a = 2 \) or \( a = -1 \).