Let \(f(x)= \begin{cases}|x|+3 & \text{if } x \le -3\\ -2x & \text{if } -3\lt x\lt 3\\ 6x+2 & \text{if } x\ge 3\end{cases}\), then which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(f(x)\) is discontinuous at \(x = 3\)
f is continuous at a point when the left-hand limit, the right-hand limit and the value there are all equal. Only the joining points x = −3 and x = 3 need checking.
At \(x = -3\):
\( \displaystyle \lim_{x\to -3^-} f(x) = |-3| + 3 = 6 \)
\( \displaystyle \lim_{x\to -3^+} f(x) = -2(-3) = 6 \)
\(f(-3) = |-3| + 3 = 6\)
All three are equal, so f is continuous at \(x = -3\).
At \(x = 3\):
\( \displaystyle \lim_{x\to 3^-} f(x) = -2(3) = -6 \)
\( \displaystyle \lim_{x\to 3^+} f(x) = 6(3) + 2 = 20 \)
The one-sided limits differ, so f is discontinuous at \(x = 3\).
NCERT (2026-27 edition):
Continuity
