Functions Ques 2
- If $f:[0, \infty) \rightarrow[0, \infty)$ and $f(x)=\frac{x}{1+x}$, then $f$ is
(2003, 2M)
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) neither one-one nor onto
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Answer:
Correct Answer: 2.(b)
Solution: (b) Given, $f:[0, \infty) \rightarrow[0, \infty)$
Here, domain is $[0, \infty)$ and codomain is $[0, \infty)$.
Thus, to check one-one
Since, $f(x)=\frac{x}{1+x} \Rightarrow f^{\prime}(x)=\frac{1}{(1+x)^2}>0, \forall x \in[0, \infty)$
$\therefore f(x)$ is increasing in its domain. Thus, $f(x)$ is one-one in its domain. To check onto (we find range)
Again, $ y=f(x)=\frac{x}{1+x} $
$\Rightarrow \quad y+y x=x$
$\Rightarrow \quad x=\frac{y}{1-y} \Rightarrow \frac{y}{1-y} \geq 0$
Since, $x \geq 0$, therefore $0 \leq y<1$
i.e. Range $\neq$ Codomain
$\therefore f(x)$ is one-one but not onto.
BETA
