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# Mark (√) against the correct answer in the following:

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 $(f)=[0,1)$