Question:
Mark (√) against the correct answer in the following:
Let $f(x)=\frac{x^{2}}{\left(1+x^{2}\right)} .$ Then, range $(f)=?$
A. $[1, \infty)$
B. $[0,1)$
C. $[-1,1]$
D. $(0,1]$
Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$
$\Rightarrow \mathrm{y}+\mathrm{yx}^{2}=\mathrm{x}^{2}$
$\Rightarrow \mathrm{y}=\mathrm{x}^{2}(1-\mathrm{y})$
$\Rightarrow x=\sqrt{\frac{y}{1-y}}$
$\frac{y}{1-y} \geq 0$
$\Rightarrow y \geq 0$
And
$1-y>0$
$\Rightarrow y<1$
Taking intersection we get
range $(f)=[0,1)$
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