Question

Let $$\begin{array}{ll} f(x) = |x| + 3, & \text{if } x \le -3 \\ f(x) = -2x, & \text{if } -3 < x < 3 \\ f(x) = 6x - 2, & \text{if } x \ge 3 \end{array}$$ then

• A. $f(x)$ is discontinuous at both $x = -3$ as well as $x = 3$
• B. $f(x)$ is continuous at $x = -3$ but discontinuous at $x = 3$
• C. $f(x)$ is continuous at $x = -3$ as well as $x = 3$
• D. $f(x)$ is discontinuous at $x = -3$ but $f(x)$ is continuous at $x = 3$

Hint:

To quickly test continuity in piecewise functions, directly substitute the boundary values into the adjacent equations. For $x = -3$, both paths yield $6$. For $x = 3$, one yields $-6$ and the other yields $16$. The mismatch immediately signals a discontinuity.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Examine the continuity of a piecewise function at the transition points x = -3 and x = 3.

Step 2: Key Formula or Approach:
Continuity at x = a demands that the left-hand limit, right-hand limit, and function value f(a) are all identical: lim_{x→a⁻} f(x) = lim_{x→a⁺} f(x) = f(a).

Step 3: Detailed Explanation:
At x = -3: For x ≤ -3, |x| = -x, so f(x) = -x + 3. LHL = -(-3) + 3 = 6. RHL using -2x: -2(-3) = 6. f(-3) = |-3| + 3 = 6. All three equal 6, so continuous at x = -3. At x = 3: LHL = -2(3) = -6. RHL using 6x - 2: 6(3) - 2 = 16. Since -6 ≠ 16, the limit does not exist; the function is discontinuous at x = 3.

Step 4: Final Answer:
Continuous at x = -3 but discontinuous at x = 3, option (B).