Let \(f : R \to R\) be defined by \(f(x) = x + [x]\), where R is the set of real numbers and \([x]\) denotes the greatest integer less than or equal to x. Then f is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → one-one but not onto
Let n be an integer. For x in \([n, n + 1)\), \([x] = n\), so
\(f(x) = x + n\)
\(2n \le f(x) \lt 2n + 1\)
For example, on \([0, 1)\) the values fill \([0, 1)\), and on \([1, 2)\) they fill \([2, 3)\).
One-one: on each piece \(f(x) = x + n\) is increasing, and the pieces \([2n, 2n + 1)\) for different n do not overlap. So different inputs give different outputs.
Not onto: the values from \(2n + 1\) up to \(2n + 2\) are never reached. For instance 1.5 is not an image:
\(x \lt 1\) gives \(f(x) \lt 1\)
\(x \ge 1\) gives \(f(x) \ge 2\)
So f is one-one but not onto.