Question

If the function \( f(x) \) defined below is continuous at \( x = 2 \), find the value of the constant \( k \): \[ f(x) = \begin{cases} kx^2 & \text{if } x \le 2 3x - 2 & \text{if } x > 2 \end{cases} \]

• A. \( 1 \)
• B. \( 2 \)
• C. \( 4 \)
• D. \( 0 \)

Hint:

For continuity problems involving piecewise functions, locate the boundary point first. Simply plug that boundary value into both expressions and solve the resulting basic algebraic equation to find the missing coefficients quickly.

Answer 1

The Correct Option is A

Concept: For a piecewise function to be continuous at a transition boundary point \( x = c \), the Left-Hand Limit (LHL), the Right-Hand Limit (RHL), and the functional value at that point \( f(c) \) must all be equal: \[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \] If these conditions match, the graph connects smoothly without any breaks or jumps at the boundary.

Step 1: Evaluate the Left-Hand Limit and the functional value at \( x = 2 \).
The left-hand slice of the function handles the region where \( x \le 2 \). Substituting the boundary value directly into the expression: \[ \lim_{x \to 2^-} f(x) = f(2) = k(2)^2 = 4k \]

Step 2: Evaluate the Right-Hand Limit at \( x = 2 \).
The right-hand slice of the function handles the region where \( x > 2 \). Evaluating the limit for this polynomial expression: \[ \lim_{x \to 2^+} f(x) = 3(2) - 2 = 6 - 2 = 4 \]

Step 3: Equate the limits to isolate the value of the constant \( k \).
Since the problem states that the function is continuous at \( x = 2 \), we equate our LHL and RHL results: \[ 4k = 4 \quad \Rightarrow \quad k = 1 \]