Given below are two statements :

Statement I : The function $$f:R\rightarrow R $$ defined by $$f(x)=\frac{x}{1+\mid x\mid}$$ is one-one.

Statement II : The function $$f:R\rightarrow R $$ defined by $$f(x)=\frac{x^{2}+4x-30}{x^{2}-8x+18}$$ is many-one.

In the light of the above statements, choose the correct answer from the options given below :

The given function is:
$$f(x) = \frac{x}{1+|x|}$$

To check if it is one-one (injective), we can analyze its behavior in two cases based on the definition of the absolute value |x|:

• Case 1: For $$x \geq 0$$

  $$f(x) = \frac{x}{1+x}$$
  Differentiating with respect to x:

  $$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$$
• Case 2: For x < 0

  $$f(x) = \frac{x}{1-x}$$
  Differentiating with respect to x:

  $$f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$$

Since f'(x) > 0 for all $$x \neq 0$$ and the function is continuous at x = 0, the function is strictly increasing across its entire domain $\mathbb{R}$. Any strictly monotonic function is always one-one.

$$f(x) = \frac{x^2+4x-30}{x^2-8x+18}$$

First, let's look at the denominator: $$x^2-8x+18 = (x-4)^2 + 2$$, which is always positive for all real values of x. Thus, the domain is $$\mathbb{R}$$.

Now, let's evaluate the limits of the function at infinity:

• As $$x \to \infty$$:

  $$\lim_{x \to \infty} \frac{x^2+4x-30}{x^2-8x+18} = 1$$

• As $$x \to -\infty$$:

  $$\lim_{x \to -\infty} \frac{x^2+4x-30}{x^2-8x+18} = 1$$

Since the function is continuous on $$\mathbb{R}$$ and approaches the same value (1) at both $$+\infty$$ and $$-\infty$$, it must rise/fall and then return toward 1. By the Intermediate Value Theorem (or Rolle's Theorem), the function will output the same value for multiple distinct inputs, which means it is many-one.

Both Statement I and Statement II are true.

Correct Option: D