Question

\( Let f(x) = | x^2 - x | + |x|, \text{ where } x \in \mathbb{R} \text{ and } | t | \text{ denotes the greatest integer less than or equal to } t. \text{ Then, } f \text{ is:} \)

• A. Not continuous at \(x = 0\) and at \(x = 1\)
• B. Continuous at \(x = 0\) and at \(x = 1\)
• C. Continuous at \(x = 1\), but not continuous at \(x = 0\)
• D. Continuous at \(x = 0\), but not continuous at \(x = 1\)

Hint:

Remember, a function is continuous at a point if the left-hand limit, right-hand limit, and function value are all equal at that point.

Answer 1

The Correct Option is D

We are given that \( f(x) = \left\lfloor x^2 - x \right\rfloor + \left\lfloor x \right\rfloor \). We need to determine the continuity of this function at specific points.
Step 1: Check continuity at \(x = 0\) At \(x = 0\), \[ f(0) = \left\lfloor 0^2 - 0 \right\rfloor + \left\lfloor 0 \right\rfloor = \left\lfloor 0 \right\rfloor + \left\lfloor 0 \right\rfloor = 0. \] As \(x \to 0^-\) and \(x \to 0^+\), we observe that the function is approaching the same value. Hence, the function is continuous at \(x = 0\).
Step 2: Check continuity at \(x = 1\) At \(x = 1\), \[ f(1) = \left\lfloor 1^2 - 1 \right\rfloor + \left\lfloor 1 \right\rfloor = \left\lfloor 0 \right\rfloor + \left\lfloor 1 \right\rfloor = 0 + 1 = 1. \] Now, check the limit from both sides: \[ \lim_{x \to 1^-} f(x) = \left\lfloor 1^2 - 1 \right\rfloor + \left\lfloor 1 \right\rfloor = 0 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = \left\lfloor 1^2 - 1 \right\rfloor + \left\lfloor 1 \right\rfloor = 0. \] The left-hand limit and right-hand limit are equal, but the function at \(x = 1\) gives a value of 1.
Therefore, the function is not continuous at \(x = 1\).