Let \(f : [0, \infty) \to [0, \infty)\) be defined by
\( \displaystyle f(x) = \frac{x}{x + 2} \)
Then f is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → one-one but not onto
One-one: let \(f(a) = f(b)\). Cross-multiplying,
\(a(b + 2) = b(a + 2)\)
\(ab + 2a = ab + 2b\)
\(a = b\)
Not onto: 1 is in the co-domain \([0, \infty)\). Suppose \(f(x) = 1\) for some \(x \ge 0\):
\(x = x + 2\)
\(0 = 2\)
This is impossible, so 1 has no pre-image. In fact \(x \lt x + 2\), so every value of f is less than 1.
Hence f is one-one but not onto.