Question

Suppose $f:R\to R$ is defined by $f\left(x\right)=\frac{x^2}{1+x^2}$, the range of the function is

• A. $[0,1)$
• B. $[0,1]$
• C. $(0,1]$
• D. $(0,1)$

Answer 1

The correct option is A $[0,1)$

Let $\frac{x^2}{1+x^2}=y$

Multiply both sides by $(1+x^2)$

$\Rightarrow x^2=y(1+x^2)$

$\Rightarrow x^2=y+x^{2}y$

$\Rightarrow x^2-x^2=y+x^{2}y-x^2$

$\Rightarrow 0=x^2(y-1)+y$

$\Rightarrow (y-1)x^2+y=0$

For a quadratic equation $a{x}^2+bx+c=0$ the discriminant is $b^2-4ac$

Here $a=y-1,b=0,c=y$

$\therefore$ the discriminant is $0-4(y-1)\left(y\right)=-4y(y-1)=-4y^2+4y$

We know that, the range is the set of y for which the discriminant is greater or equal to zero.

$\Rightarrow -4y^2+4y\ge 0$

$\Rightarrow y-y^2\ge 0$

$\Rightarrow -y+y^2\le 0$

$\Rightarrow (y-1)y\le 0$

$\Rightarrow 0\le y\le 1$

Check if the range interval endpoints are included.

Put $y=0$ in $\frac{x^2}{1+x^2}=y$, we get $x=0$. Solution exists, therefore $y=0$ is included in the range.

Put $y=1$ in $\frac{x^2}{1+x^2}=y$, we get no solution for $x\in R$. Therefore $y=1$ is excluded in the range.

Therefore the range is $0\le f\left(x\right)<1$. That is, $[0,1)$