Question

Find the range of $f\left(x\right)=\frac{x^2}{(1+x^2)}$ ($x$ real).

• A. $0\le y<1$
• B. $0\le y\le 1$
• C. $y<1$
• D. $y<-1$

Answer 1

Answer: (A)

$0\le y<1$

Since for every real x, $1+x^2\ne 0$, therefore $\frac{x^2}{(1+x^2)}$is a real number for all real $x$.

Hence the domain of f is the set R of all real numbers. The range of f consists of all real numbers y such that $f\left(x\right)=y$ for real $x.$

Now $f\left(x\right)=y=\frac{x^2}{(1+x^2)}=y=x^2=y+y{x}^2=x=\sqrt{\frac{y}{(1-y)}}...$

$\left(1\right)$ Since x is real, we must have $\frac{y}{(1-y)}\ge 0(y\ne 1)$which is satisfied if $0\le y\le 1$

Hence range of $f=y:yisrealand0\le y<1$