Let $f: [-1, 2] \rightarrow \mathbb{R}$ be given by $f(x) = 2x^2 + x + [x^2] - [x]$, where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points, where $f$ is not continuous, is:
- 6
- 3
- 4
- 5
The Correct Option is C
Approach Solution - 1
To determine the number of points where the function \( f(x) = 2x^2 + x + [x^2] - [x] \) is not continuous, we analyze the potential discontinuities caused by the greatest integer function, denoted by \([t]\). Discontinuities for such expressions typically occur at integer points where the floor function \([x]\) or \([x^2]\) transitions.
Let's first identify the key transition points within the domain \([-1, 2]\):
- The points where \([x]\) changes are integer values: \(-1, 0, 1, 2\).
- For \([x^2]\), consider where \(x^2\) is an integer: within the domain \([-1, 2]\), these points are \(x^2 = 0, 1, 4\), corresponding to \(x = 0, \pm1, \pm2\). However, since \([-1, 2]\) is the domain, feasible values are \(-1, 0, 1, 2\).
By evaluating the potential discontinuities at these points:
- \(x = -1\): Check left and right hand values for continuity: \([x] = -1\), \([x^2] = 1\). Moving across \(-1\) within the domain's scope shows a discontinuity due to the piecewise nature of \([x]\) and \([x^2]\).
- \(x = 0\): The function becomes \(2 \times 0^2 + 0 + [0^2] - [0] = 0\) at both points. Hence, potential discontinuity due to a jumping point in \([x]\).
- \(x = 1\): Both \([x]\) and \([x^2]\) change, further implying a discontinuity.
- \(x = 2\): The transition of \([x]\) at an edge of the domain indicates potential discontinuity.
Summing these, the points of discontinuity are \(-1, 0, 1, 2\), leading to a total of 4 points where the function is not continuous.
Approach Solution -2
Given \( f(x) = 2x^2 + x + \lfloor x^2 \rfloor - \lfloor x \rfloor \), we analyze its continuity. The floor function, \( \lfloor x \rfloor \), introduces discontinuities at integer points since it changes its value abruptly.
Step 1: Points of discontinuity from \( \lfloor x \rfloor \) and \( \lfloor x^2 \rfloor \):
1. The term \( \lfloor x \rfloor \) is discontinuous at all integer values of \( x \) within the interval \([-1, 2]\), i.e., at \( x = -1, 0, 1, 2 \). 2. The term \( \lfloor x^2 \rfloor \) introduces discontinuities at points where \( x^2 \) crosses an integer value. On solving: - For \( x^2 = 0, 1, 4 \): - \( x = -2, -1, 0, 1, 2 \) (within \([-1, 2]\), only \( x = -1, 0, 1 \)).
Step 2: Combine discontinuities: Overall, the combined discontinuities occur at the union of these points:
\[ \text{Points: } x = -1, 0, 1, 2. \]
- Since these four points involve jumps in either \( \lfloor x \rfloor \) or \( \lfloor x^2 \rfloor \), \( f(x) \) is discontinuous at exactly 4 points.
Thus, the total number of discontinuities is 4.
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