Given below are two statements:Statement I: T | vedclass

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(B) Statement $I$: $f(x) = \frac{x}{1+|x|}$.We can write this as:$f(x) = \begin{cases} \frac{x}{1+x}, & x \ge 0 \ \frac{x}{1-x}, & x < 0 \end{cases}$

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Both Statement $I$ and Statement $II$ are true.

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(B) Statement $I$: $f(x) = \frac{x}{1+|x|}$.We can write this as:$f(x) = \begin{cases} \frac{x}{1+x}, & x \ge 0 \ \frac{x}{1-x}, & x For $x \ge 0$, $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$.For $x  0$.Since the derivative is always positive, the function is strictly increasing and hence one-one. Thus, Statement $I$ is true.Statement $II$: $f(x) = \frac{x^2+4x-30}{x^2-8x+18}$.To check if it is many-one, we check if there exist $x_1 
eq x_2$ such that $f(x_1) = f(x_2)$.Let $f(x) = k$. Then $k(x^2-8x+18) = x^2+4x-30$.$(k-1)x^2 - (8k+4)x + (18k+30) = 0$.For this to have two distinct roots, the discriminant $D$ must be positive.$D = (8k+4)^2 - 4(k-1)(18k+30) > 0$.$16(2k+1)^2 - 24(k-1)(3k+5) > 0$.$16(4k^2+4k+1) - 24(3k^2+2k-5) > 0$.$64k^2+64k+16 - 72k^2-48k+120 > 0$.$-8k^2+16k+136 > 0 \Rightarrow k^2-2k-17 Since there exists a range of $k$ for which there are two distinct roots, the function is many-one. Thus, Statement $II$ is true.

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