Question

If $f(x) = \begin{cases} 3x - 2, & 0 \leq x \leq 1\\ 2x^2 + ax, & 1<x<2 \end{cases}$ is continuous for $x \in (0, 2)$, then $a$ is equal to :

• A. -4
• B. -7
• C. -2
• D. -1

Hint:

For continuity at a point, equate the left-hand and right-hand limits at that point.

Answer 1

The Correct Option is B

For the function to be continuous at $x = 1$, the left-hand limit and right-hand limit must be equal at $x = 1$. The left-hand limit is: \[ f(1) = 3(1) - 2 = 1 \] The right-hand limit is: \[ f(1) = 2(1)^2 + a(1) = 2 + a \] Setting these equal, we get: \[ 1 = 2 + a \quad \Rightarrow \quad a = -1 \] Thus, the correct answer is $a = -1$.