Given below are two statements : Statement I : The function f:R R defined by f(x)=(x)/(1+|…

Question

Given below are two statements :

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

Statement II : The function $f:R	o R$ 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 :

Accepted Answer

Statement I: $f(x)=\frac{x}{1+|x|}$ is one-one

To check if a function is one-one (injective), we can check its monotonicity (whether it's always increasing or always decreasing).

For $x\ge 0:f(x)=\frac{x}{1+x}$. The derivative is ${f}^{'}(x)=\frac{1}{(1+x{)}^{2}}$, which is always positive.

For $x

Since the derivative is positive for all $x$ in the domain, the function is strictly increasing. A strictly monotonic function is always one-one.

Statement I is True.

Statement II: $f(x)=\frac{{x}\^{2}+4x-30}{{x}\^{2}-8x+18}$ is many-one

Discriminant of Denominator: For ${x}\^{2}-8x+18,D=(-8{)}\^{2}-4(1)(18)=64-72=-8$. Since $DBehavior at Infinity: As $x	o \infty$ and $x	o -\infty$, the value of $f(x)$ approaches 1 (the ratio of the leading coefficients).

Turning Points: Since the function starts near $y=1$ at $-\infty$, moves through various values, and returns toward $y=1$ at $+\infty$, it must have at least one local maximum or minimum. This means a horizontal line will intersect the graph at more than one point.

Also, let

$y=\frac{{x}^{2}+4x-30}{{x}^{2}-8x+18}$

Cross multiply:

$y({x}^{2}-8x+18)={x}^{2}+4x-30$

Rearrange:

$(y-1){x}^{2}+(-8y-4)x+(18y+30)=0$

This is a quadratic in $x$.

For many values of $y$, this quadratic has two distinct real solutions. Function is MANY-ONE

Solutions · 1