Question

If \[ f(x)= \begin{cases} x^2, & \text{if } x\leq 2,\\[4pt] 4x-\alpha, & \text{if } x>2, \end{cases} \] is continuous at \(x=2\), then the value of \(\alpha\) is equal to: 

• A. 2
• B. 5
• C. 3
• D. 4
• E. 0

Hint:

For piecewise linear or quadratic functions, continuity simply means "the two parts must meet at the boundary". Just plug the boundary value into both pieces and set them equal.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a function to be continuous at a point \(x = c\), the left-hand limit (LHL), the right-hand limit (RHL), and the function value at that point must all be equal.
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \]

Step 2: Key Formula or Approach:
Set the limits from both sides of \(x = 2\) equal to each other.

Step 3: Detailed Explanation:
1. Left-hand limit (LHL) at \(x = 2\):
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4 \]
2. Right-hand limit (RHL) at \(x = 2\):
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (4x - \alpha) = 4(2) - \alpha = 8 - \alpha \]
3. Since the function is continuous at \(x = 2\):
\[ LHL = RHL \implies 4 = 8 - \alpha \]
\[ \alpha = 8 - 4 = 4 \]

Step 4: Final Answer:
The value of \(\alpha\) is 4.