The function $f: [-1,1]→ R$ (set of real numbers) given by $f(x)=\frac{x}{x+3}$ is

one-one only

The correct answer is Option (1) → one-one only

Given function:

$f(x) = \frac{x}{x + 3}$, with domain $[-1, 1]$.

Step 1: Check if $f$ is one-one

Compute derivative:

$f'(x) = \frac{3}{(x + 3)^2}$

Since $(x + 3)^2 > 0$ for all $x \in [-1, 1]$, we have $f'(x) > 0$ everywhere.

⇒ $f(x)$ is strictly increasing.

⇒ Therefore, $f(x)$ is one-one.

Step 2: Check if $f$ is onto $\mathbb{R}$

Range of $f$ for $x \in [-1, 1]$:

$f(-1) = \frac{-1}{2} = -0.5$, $f(1) = \frac{1}{4} = 0.25$

Since $f$ is increasing, the range is:

$[-\frac{1}{2}, \frac{1}{4}]$

But the codomain is $\mathbb{R}$, and $f(x)$ does not take values outside this interval.

⇒ $f(x)$ is not onto $\mathbb{R}$.

Final conclusion:

$f(x) = \frac{x}{x+3}$ on $[-1, 1]$ is one-one but not onto $\mathbb{R}$.