Let $$f : R \to R$$ be defined by $$f(x) = \frac{|x|-1}{|x|+1}$$, then f is:

We are given a function $$ f : \mathbb{R} \to \mathbb{R} $$ defined by $$ f(x) = \frac{|x| - 1}{|x| + 1} $$. We need to determine if this function is one-one (injective), onto (surjective), both, or neither. A function is one-one if different inputs always give different outputs, meaning if $$ f(a) = f(b) $$, then $$ a = b $$. A function is onto if every element in the codomain is mapped to by some element in the domain, meaning for every real number $$ y $$, there exists some $$ x $$ such that $$ f(x) = y $$. The codomain here is all real numbers, $$ \mathbb{R} $$.

First, we check if the function is one-one. Notice that the expression depends on $$ |x| $$, which is always non-negative. For any $$ x $$, $$ |x| = |-x| $$, so let us compute $$ f(-x) $$:

$$ f(-x) = \frac{|-x| - 1}{|-x| + 1} = \frac{|x| - 1}{|x| + 1} = f(x). $$

This shows that $$ f(-x) = f(x) $$ for all $$ x $$. Therefore, for any $$ x \neq 0 $$, we have $$ f(x) = f(-x) $$, but $$ x $$ and $$ -x $$ are different if $$ x \neq 0 $$. For example, let $$ x = 1 $$:

$$ f(1) = \frac{|1| - 1}{|1| + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0. $$

Now, $$ f(-1) = \frac{|-1| - 1}{|-1| + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 $$. So $$ f(1) = f(-1) = 0 $$, but $$ 1 \neq -1 $$. Since two different inputs ($$ 1 $$ and $$ -1 $$) give the same output ($$ 0 $$), the function is not one-one.

Next, we check if the function is onto. To be onto, for every real number $$ y $$, there must be some $$ x $$ such that $$ f(x) = y $$. Since $$ f(x) $$ depends only on $$ |x| $$, we set $$ t = |x| \geq 0 $$. Then the function becomes:

$$ g(t) = \frac{t - 1}{t + 1}, \quad t \geq 0. $$

We need to find the range of $$ g(t) $$ as $$ t $$ varies from 0 to infinity. This range will be the same as the range of $$ f(x) $$, because for each $$ t \geq 0 $$, there is at least one $$ x $$ (for example, $$ x = t $$ or $$ x = -t $$) such that $$ |x| = t $$.

Evaluate $$ g(t) $$ at key points:

• When $$ t = 0 $$: $$ g(0) = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1 $$.
• When $$ t = 1 $$: $$ g(1) = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 $$.
• As $$ t \to \infty $$: $$ g(t) = \frac{t - 1}{t + 1} = \frac{1 - \frac{1}{t}}{1 + \frac{1}{t}} \to \frac{1 - 0}{1 + 0} = 1 $$.

Now, check if $$ g(t) $$ can equal 1: set $$ \frac{t - 1}{t + 1} = 1 $$. Then $$ t - 1 = t + 1 $$, which simplifies to $$ -1 = 1 $$, a contradiction. So $$ g(t) \neq 1 $$ for any $$ t \geq 0 $$.

To see how $$ g(t) $$ behaves, compute its derivative. Let $$ g(t) = \frac{t - 1}{t + 1} $$. Using the quotient rule:

$$ g'(t) = \frac{(1)(t + 1) - (t - 1)(1)}{(t + 1)^2} = \frac{t + 1 - (t - 1)}{(t + 1)^2} = \frac{t + 1 - t + 1}{(t + 1)^2} = \frac{2}{(t + 1)^2}. $$

Since $$ (t + 1)^2 > 0 $$ for all $$ t \geq 0 $$, we have $$ g'(t) > 0 $$, so $$ g(t) $$ is strictly increasing for $$ t \geq 0 $$.

As $$ t $$ increases from 0 to infinity, $$ g(t) $$ increases continuously from $$ g(0) = -1 $$ to the limit 1 (but never reaches 1). Therefore, the range of $$ g(t) $$ is all real numbers from -1 inclusive to 1 exclusive, denoted as $$ [-1, 1) $$.

Since $$ f(x) $$ has the same range $$ [-1, 1) $$, we can see that:

• $$ y = -1 $$ is achieved when $$ x = 0 $$ (since $$ f(0) = \frac{|0| - 1}{|0| + 1} = \frac{-1}{1} = -1 $$).
• Values between -1 and 1 are achieved; for example, for $$ y = 0.5 $$, solve $$ \frac{t - 1}{t + 1} = 0.5 $$: $$ t - 1 = 0.5(t + 1) \implies t - 1 = 0.5t + 0.5 \implies t - 0.5t = 0.5 + 1 \implies 0.5t = 1.5 \implies t = 3. $$ So $$ |x| = 3 $$, meaning $$ x = 3 $$ or $$ x = -3 $$, and $$ f(3) = \frac{|3| - 1}{|3| + 1} = \frac{3 - 1}{3 + 1} = \frac{2}{4} = 0.5 $$.
• But $$ y = 1 $$ is not achieved, as shown earlier.

Now, consider values outside $$ [-1, 1) $$. For example, $$ y = 2 $$:

Set $$ f(x) = 2 $$: $$ \frac{|x| - 1}{|x| + 1} = 2 \implies |x| - 1 = 2(|x| + 1) \implies |x| - 1 = 2|x| + 2 \implies -1 - 2 = 2|x| - |x| \implies -3 = |x| $$. But $$ |x| \geq 0 $$, so $$ |x| = -3 $$ is impossible. Similarly, for $$ y = -2 $$: $$ \frac{|x| - 1}{|x| + 1} = -2 \implies |x| - 1 = -2(|x| + 1) \implies |x| - 1 = -2|x| - 2 \implies |x| + 2|x| = -2 + 1 \implies 3|x| = -1 $$, again impossible.

Thus, the range of $$ f $$ is $$ [-1, 1) $$, which is a proper subset of the codomain $$ \mathbb{R} $$. For instance, $$ y = 1 $$ and $$ y = 2 $$ are in $$ \mathbb{R} $$ but not in the range. Therefore, $$ f $$ is not onto.

Since $$ f $$ is not one-one and not onto, it is neither injective nor surjective.

Hence, the correct answer is Option B.