The function f: [0, ∞) → R defined by, f(x) = 2x^2 + 3 , is

The correct answer is Option (2) → one-one but not onto

One-one: let \(f(x_1) = f(x_2)\) with \(x_1, x_2 \in [0, \infty)\):

    \(2x_1^2 + 3 = 2x_2^2 + 3\)
    \(x_1^2 = x_2^2\)

Both numbers are non-negative, so \(x_1 = x_2\). Hence f is one-one. The domain \([0, \infty)\) is what makes this work: on all of R, \(f(-1) = f(1) = 5\).

Not onto: for every x in the domain,

    \(f(x) = 2x^2 + 3 \ge 3\)

So the range is \([3, \infty)\), not the whole codomain R; for example, 0 has no pre-image. Hence f is one-one but not onto.