Question:
The number of points where the function
\(f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \le -1 \\ [4x^2 - 1] & \text{if } -1 \lt x \lt 1 \\ |x + 1| + |x - 2| & \text{if } x \ge 1 \end{cases}\)
\([t]\) denotes the greatest integer \(\le t\), is discontinuous is ____________.
The number of points where the function
\(f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \le -1 \\ [4x^2 - 1] & \text{if } -1 \lt x \lt 1 \\ |x + 1| + |x - 2| & \text{if } x \ge 1 \end{cases}\)
\([t]\) denotes the greatest integer \(\le t\), is discontinuous is ____________.
\(f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \le -1 \\ [4x^2 - 1] & \text{if } -1 \lt x \lt 1 \\ |x + 1| + |x - 2| & \text{if } x \ge 1 \end{cases}\)
\([t]\) denotes the greatest integer \(\le t\), is discontinuous is ____________.
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The absolute-value pieces are continuous; count the jumps of the greatest integer function on (-1, 1) and check the junctions x = -1 and x = 1.
Updated On: Nov 4, 2026
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Correct Answer: 7
Solution and Explanation
Piece 1 (\(x \le -1\)): |2x^2 - 3x - 7| is continuous. Piece 3 (\(x \ge 1\)): |x+1| + |x-2| is continuous.
Middle piece, \(-1 \lt x \lt 1\): \(4x^2 - 1\) takes values in \([-1, 3)\). The greatest integer function is discontinuous where \(4x^2 - 1\) crosses an integer value:
\(4x^2 - 1 = 0 \Rightarrow x = \pm\frac12\); \(4x^2 - 1 = 1 \Rightarrow x = \pm\frac{1}{\sqrt2}\); \(4x^2 - 1 = 2 \Rightarrow x = \pm\frac{\sqrt3}{2}\).
That gives 6 points. At x = 0, \(4x^2 - 1 = -1\) is the minimum and \([4x^2 - 1] = -1\) on both sides, so it is continuous there. (The value 3 is reached only at x = +/-1, outside the open interval.)
Junction x = -1: f(-1) = |2 + 3 - 7| = 2, left limit = 2, right limit \(= \lim [4x^2 - 1]\) with \(4x^2 - 1 \to 3^-\), so = 2. Continuous.
Junction x = 1: f(1) = |2| + |-1| = 3, right limit = 3, left limit = \([3^-] = 2\). Discontinuous.
Total points of discontinuity = 6 + 1 = 7.
Answer: 7
Middle piece, \(-1 \lt x \lt 1\): \(4x^2 - 1\) takes values in \([-1, 3)\). The greatest integer function is discontinuous where \(4x^2 - 1\) crosses an integer value:
\(4x^2 - 1 = 0 \Rightarrow x = \pm\frac12\); \(4x^2 - 1 = 1 \Rightarrow x = \pm\frac{1}{\sqrt2}\); \(4x^2 - 1 = 2 \Rightarrow x = \pm\frac{\sqrt3}{2}\).
That gives 6 points. At x = 0, \(4x^2 - 1 = -1\) is the minimum and \([4x^2 - 1] = -1\) on both sides, so it is continuous there. (The value 3 is reached only at x = +/-1, outside the open interval.)
Junction x = -1: f(-1) = |2 + 3 - 7| = 2, left limit = 2, right limit \(= \lim [4x^2 - 1]\) with \(4x^2 - 1 \to 3^-\), so = 2. Continuous.
Junction x = 1: f(1) = |2| + |-1| = 3, right limit = 3, left limit = \([3^-] = 2\). Discontinuous.
Total points of discontinuity = 6 + 1 = 7.
Answer: 7
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