The function $f: [-1,1]→R$ is given by $f(x)=\frac{x}{x+2}$

one-one only

The correct answer is Option (3) → one-one only

Given function: $f(x) = \frac{x}{x + 2}, \ x \in [-1, 1]$

Check one-one:

Compute derivative: $f'(x) = \frac{(x + 2)(1) - x(1)}{(x + 2)^2} = \frac{2}{(x + 2)^2}$

Since $(x + 2)^2 > 0$ for all $x \in [-1, 1]$, we have $f'(x) > 0$

⇒ $f(x)$ is strictly increasing ⇒ one-one ✅

Check onto:

At $x = -1$, $f(-1) = \frac{-1}{1} = -1$

At $x = 1$, $f(1) = \frac{1}{3}$

Hence range of $f(x)$ = $[-1, \frac{1}{3}]$

Since codomain is $\mathbb{R}$ but range ⊂ $\mathbb{R}$, $f$ is not onto ❌

Final Answer:

one-one only