Functions Ques 2

  1. 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.



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