MCQ
$f(x) = x + \sqrt {{x^2}} $ is a function from $R \to R$ , then $f(x)$ is
- AInjective
- BSurjective
- CBijective
- ✓None of these
✓
Answer
Correct option: D.
None of these
d
(d) We have $f(x) = x + \sqrt {{x^2}} = x + |x|$
(d) We have $f(x) = x + \sqrt {{x^2}} = x + |x|$
Clearly f is not one-one as $f( - 1) = f( - 2) = 0$ but $ - 1 \ne 2$.
Also $f$ is not onto as $f(x) \ge 0,\,\forall x \in R,$
Also, range of $f = (0,\infty ) \subset R$.
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