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:
We are given a piecewise continuous function and need to check its continuity at the boundary points $x = -3$ and $x = 3$.

Step 2: Key Formula or Approach:
A function is continuous at a point $x = a$ if and only if the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at that point are all equal: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^2} f(x) = f(a)$$

Step 3: Detailed Explanation:
Case 1: Checking continuity at $x = -3$
For $x \le -3$, $|x| = -x$. Thus, $f(x) = -x + 3$. $$\text{LHL} = \lim_{x \to -3^-} (-x + 3) = -(-3) + 3 = 6$$ $$\text{RHL} = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$$ $$f(-3) = |-3| + 3 = 3 + 3 = 6$$ Since $\text{LHL} = \text{RHL} = f(-3) = 6$, $f(x)$ is continuous at $x = -3$.
Case 2: Checking continuity at $x = 3$
$$\text{LHL} = \lim_{x \to 3^-} (-2x) = -2(3) = -6$$ $$\text{RHL} = \lim_{x \to 3^+} (6x - 2) = 6(3) - 2 = 16$$ Since $\text{LHL} \neq \text{RHL}$, the limit does not exist at $x = 3$, rendering $f(x)$ discontinuous at $x = 3$.

Step 4: Final Answer:
Therefore, $f(x)$ is continuous at $x = -3$ but discontinuous at $x = 3$, which corresponds to option (B).