Let f: [0, ∞) → [0, 2] be defined by f (x)=(2x/1+x), then f is

Question

Let f: [0, ∞) → [0, 2] be defined by f (x)=(2x/1+x), then f is (A) one-one but not onto (B) onto but not one-one (C) both one-one and onto (D) neither one-one nor onto

Tardigrade Admin · Accepted Answer

We have, y=(2x/x+1) ⇒ 2x = yx + y ⇒ x=(y/2-y) Now, x ∈ [0, ∞) ⇒ 0 le y < 2 ⇒ Range of f(x) = [0, 2) Since range ⊂ co-domain ⇒ f is into. Clearly, f is one-one also.

Q. Let $f$ : $[0$ , $\infty) \to [0$ , $2]$ be defined by $f (x) = \frac{2 x}{1 + x}$ , then $f$ is

one-one but not onto

onto but not one-one

both one-one and onto

neither one-one nor onto

We have, $y = \frac{2 x}{x + 1}$
$\Rightarrow 2 x = y x + y$
$\Rightarrow x = \frac{y}{2 - y}$
Now, $x \in [0$ , $\infty)$
$\Rightarrow 0 \leq y < 2$
$\Rightarrow$ Range of $f (x) = [0$ , $2)$
Since range $\subset$ co-domain
$\Rightarrow f$ is into. Clearly, $f$ is one-one also.

Q. Let f : [0, ∞)→[0, 2] be defined by f(x)=1+x2x​, then f is