Let R be the set of real numbers and let \([x]\) denote the greatest integer less than or equal to x. Consider the following statements:
(A) \(f : R \to R\), \(f(x) = [x]\), is not one-one.
(B) The signum function \(f : R \to R\), with \(f(x) = 1\) for \(x \gt 0\), \(f(0) = 0\) and \(f(x) = -1\) for \(x \lt 0\), is onto.
(C) \(f : R \to [0, \infty)\), \(f(x) = |x|\), is onto.
(D) \(f : R \to R\), \(f(x) = x - [x]\), is one-one.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
Checking each function with a few values:
\([1] = [1.5] = 1\)
the signum function takes only the values \(-1\), 0 and 1
for any \(y \ge 0\), \(|y| = y\)
\(0 - [0] = 0\) and \(1 - [1] = 0\)
So the greatest integer function is not one-one, the signum function has range \(\{-1, 0, 1\}\), the modulus function onto \([0, \infty)\) is onto, and the fractional-part function \(x - [x]\) is not one-one.
Checking the statements
(A) \(f(x) = [x]\) is not one-one. — Correct. \(f(1) = f(1.5) = 1\).
(B) The signum function from R to R is onto. — Incorrect. Its range is \(\{-1, 0, 1\}\), so 2 is not an image.
(C) \(f : R \to [0, \infty)\), \(f(x) = |x|\), is onto. — Correct. Each \(y \ge 0\) equals \(f(y)\).
(D) \(f(x) = x - [x]\) is one-one. — Incorrect. \(f(0) = f(1) = 0\).