Question

If \([\cdot]\) denotes the greatest integer function, then \(f(x) = \left[\frac{1}{2} - x\right] + [x]\)

• A. is continuous at \(x = \frac{1}{2}\)
• B. is discontinuous at \(x = \frac{1}{2}\)
• C. \(\lim_{x \to (1/2)^+} f(x) = 2\)
• D. \(\lim_{x \to (1/2)^-} f(x) = 1\)

Hint:

Greatest integer function has jump discontinuities at integer points.

Answer 1

The Correct Option is B

To determine the continuity or discontinuity of the function \( f(x) \) at \( x = \frac{1}{2} \), we need to analyze the limits from both the left and right directions and compare them with the function value at \( x = \frac{1}{2} \).

The given function is:

\(f(x) = \left[\frac{1}{2} - x\right] + [x]\)

where \([\cdot]\) denotes the greatest integer function (or floor function).

Let's calculate the left-hand limit as \( x \to \left(\frac{1}{2}\right)^- \):

• For \( x \to \left(\frac{1}{2}\right)^- \), \( x \) is slightly less than \( \frac{1}{2} \).
• \(\left[\frac{1}{2} - x\right] = \left[\text{a small positive number}\right] = 0\)
• \([x] = [\text{slightly less than }\frac{1}{2}] = 0\)
• Thus, the left-hand limit is: \(f(x) = 0 + 0 = 0\)

Let's calculate the right-hand limit as \( x \to \left(\frac{1}{2}\right)^+ \):

• For \( x \to \left(\frac{1}{2}\right)^+ \), \( x \) is slightly more than \( \frac{1}{2} \).
• \(\left[\frac{1}{2} - x\right] = [-\text{a small positive number}] = -1\)
• \([x] = [\text{slightly more than }\frac{1}{2}] = 0\)
• Thus, the right-hand limit is: \(f(x) = -1 + 0 = -1\)

Value of the function at \( x = \frac{1}{2} \):

• \(\left[\frac{1}{2} - \frac{1}{2}\right] = [0] = 0\)
• f\left(\frac{1}{2}\right) = 0 + 0 = 0

Since the left-hand limit (\(0\)) is not equal to the right-hand limit (\(-1\)), the function \( f(x) \) is discontinuous at \( x = \frac{1}{2} \).

Therefore, the correct answer is: is discontinuous at \( x = \frac{1}{2} \).