Question

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:

• A. 6
• B. 3
• C. 4
• D. 5

Answer 1

The Correct Option is C

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.