Question:
Let \(f : [-1, 3] \to \mathbb{R}\) be defined as
\(f(x) = \begin{cases}|x| + [x], & -1 \le x \lt 1\\ x + |x|, & 1 \le x \lt 2\\ x + [x], & 2 \le x \le 3\end{cases}\)
where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at:
Let \(f : [-1, 3] \to \mathbb{R}\) be defined as
\(f(x) = \begin{cases}|x| + [x], & -1 \le x \lt 1\\ x + |x|, & 1 \le x \lt 2\\ x + [x], & 2 \le x \le 3\end{cases}\)
where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at:
\(f(x) = \begin{cases}|x| + [x], & -1 \le x \lt 1\\ x + |x|, & 1 \le x \lt 2\\ x + [x], & 2 \le x \le 3\end{cases}\)
where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at:
Show Hint
Rewrite \(f\) on each unit interval and check the jump at the integers \(0, 1, 2, 3\) (and the end point \(-1\)).
Updated On: Nov 5, 2026
- only three points
- four or more points
- only two points
- only one point
Show Solution
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The Correct Option is A
Solution and Explanation
Simplify each piece:
\(-1 \le x \lt 0\): \(|x| + [x] = -x - 1\); \(0 \le x \lt 1\): \(x\); \(1 \le x \lt 2\): \(2x\); \(2 \le x \lt 3\): \(x + 2\); \(f(3) = 6\).
At \(x = 0\): left limit \(= -1\), \(f(0) = 0\) : discontinuous.
At \(x = 1\): left limit \(= 1\), \(f(1) = 2\) : discontinuous.
At \(x = 2\): left limit \(= 4\), \(f(2) = 4\), right limit \(= 4\) : continuous.
At \(x = 3\): left limit \(= 5\), \(f(3) = 6\) : discontinuous.
At \(x = -1\): \(f(-1) = 0\) equals the right limit \(-(-1) - 1 = 0\) : continuous. Within each open piece \(f\) is continuous.
So \(f\) is discontinuous at exactly three points: \(0, 1, 3\).
Answer: (1) only three points
\(-1 \le x \lt 0\): \(|x| + [x] = -x - 1\); \(0 \le x \lt 1\): \(x\); \(1 \le x \lt 2\): \(2x\); \(2 \le x \lt 3\): \(x + 2\); \(f(3) = 6\).
At \(x = 0\): left limit \(= -1\), \(f(0) = 0\) : discontinuous.
At \(x = 1\): left limit \(= 1\), \(f(1) = 2\) : discontinuous.
At \(x = 2\): left limit \(= 4\), \(f(2) = 4\), right limit \(= 4\) : continuous.
At \(x = 3\): left limit \(= 5\), \(f(3) = 6\) : discontinuous.
At \(x = -1\): \(f(-1) = 0\) equals the right limit \(-(-1) - 1 = 0\) : continuous. Within each open piece \(f\) is continuous.
So \(f\) is discontinuous at exactly three points: \(0, 1, 3\).
Answer: (1) only three points
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