Given below are two statements :

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

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

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

Statement I: We start by considering the function $$f: \mathbb{R}\rightarrow \mathbb{R}$$ defined by $$f(x)=\f\frac{x}{1+\mid x\mid}$$ and aim to show that $$f$$ is one-one.

Next, we use the derivative test for monotonicity. For $$x \gt 0$$, we have $$\mid x\mid = x$$ so $$f(x)=\f\frac{x}{1+x}$$. Here $$u(x)=x,\;v(x)=1+x$$. Then $$u'(x)=1,\;v'(x)=1$$. By the quotient rule $$f'(x)=\f\frac{u'v - uv'}{v^2}=\f\frac{(1)(1+x)-(x)(1)}{(1+x)^2}=\f\frac{1}{(1+x)^2}$$ which is positive for all $$x \gt 0$$.

Similarly, for $$x \lt 0$$, we have $$\mid x\mid = -x$$ so $$f(x)=\f\frac{x}{1-x}$$. Here $$u(x)=x,\;v(x)=1-x$$. Then $$u'(x)=1,\;v'(x)=-1$$. By the quotient rule $$f'(x)=\f\frac{(1)(1-x)-(x)(-1)}{(1-x)^2}=\f\frac{1}{(1-x)^2}$$ which is positive for all $$x \lt 0$$.

This gives us $$f'(x) \gt 0$$ on both $$(-\infty,0)$$ and $$(0,\infty)$$, and since $$f$$ is continuous at $$x=0$$, we conclude that $$f$$ is strictly increasing on $$\mathbb{R}$$. Because a strictly increasing function is one-one, Statement I is true.

Statement II: Next, we consider the function $$f: \mathbb{R}\rightarrow \mathbb{R}$$ defined by $$f(x)=\f\frac{x^{2}+4x-30}{x^{2}-8x+18}$$ and show that $$f$$ is many-one.

First, the denominator $$x^{2}-8x+18$$ has discriminant $$(-8)^{2}-4\cdot1\cdot18=64-72=-8$$ which is negative. Since the leading coefficient is positive, the denominator is always positive for all real $$x$$, and thus the domain is $$\mathbb{R}$$.

We compute the derivative using the quotient rule with $$u(x)=x^{2}+4x-30$$ and $$v(x)=x^{2}-8x+18$$:

$$ f'(x)=\f\frac{u'(x)\,v(x)-u(x)\,v'(x)}{[v(x)]^{2}}=\f\frac{(2x+4)(x^{2}-8x+18)-(2x-8)(x^{2}+4x-30)}{(x^{2}-8x+18)^{2}} $$

Expanding and simplifying the numerator gives

$$ (2x+4)(x^{2}-8x+18)-(2x-8)(x^{2}+4x-30)=-12\bigl(x^{2}-8x+14\bigr)=12\bigl(2-(x-4)^{2}\bigr). $$

Thus $$f'(x)=\f\frac{12\bigl(2-(x-4)^{2}\bigr)}{(x^{2}-8x+18)^{2}}\,. $$

Since the denominator is always positive, the sign of $$f'(x)$$ is determined by $$2-(x-4)^{2}$$. We have $$f'(x)\gt 0$$ when $$(x-4)^{2}\lt 2$$ and $$f'(x)\lt 0$$ when $$(x-4)^{2}\gt 2$$. Hence $$f$$ first increases on $$(4-\sqrt{2},4+\sqrt{2})$$ and then decreases outside this interval.

Therefore, $$f$$ is not strictly monotonic on $$\mathbb{R}$$, so it cannot be one-one and is thus many-one. Hence Statement II is true.

Both Statement I and Statement II are true. The correct answer is Option D.