First slide

Functions (XII)

Question

Let f : R → R be a function defined by, $f (x) = x + \sqrt{x^{2}}, then f is$

Moderate

A

injective
B

surjective
C

bijective
D

None of these

Solution

We have, $f (x) = x + \sqrt{x^{2}} = x + | x |$
Clearly, f is not one-one as f (– 1) = f (– 2) = 0 but – 1 ≠ –2.
Also, f is not onto as f (x) ≥ 0 ∀ x ∈ R,
$∴$ range of f = (0, ∞) ⊂ R.

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