Question

If the function \( f(x) \), defined below, is continuous on the interval \([0,8]\), then: \[ f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases} \]

• A. a = 3, b = −2
• B. a = −3, b = 2
• C. a = −3, b = −2
• D. a = 3, b = 2

Hint:

Continuity at a point requires matching left-hand and right-hand limits.

Answer 1

The Correct Option is A

Since \( f(x) \) is continuous on \([0,8]\), it must be continuous at \( x = 2 \) and \( x = 4 \). At \( x = 2 \), \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] \[ \lim_{x \to 2^-} (x^2 + ax + b) = \lim_{x \to 2^+} (3x + 2) \] \[ 4 + 2a + b = 3(2) + 2 \] \[ 2a + b = 4 \quad {(Equation 1)} \] At \( x = 4 \), \[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) \] \[ \lim_{x \to 4^-} (3x + 2) = \lim_{x \to 4^+} (2ax + 5b) \] \[ 3(4) + 2 = 2a(4) + 5b \] \[ 8a + 5b = 14 \quad {(Equation 2)} \] Solving Equations 1 and 2, we get: \[ a = 3, \quad b = -2 \]