Question

Let $f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$ be a function from $R$ to $R$ then range of $f$ is .

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

Answer 1

The correct option is B $[0,1]$

$f\left(x\right)=\frac{x^2}{x^2+1}$

is defined for all real $x$

now, say

$y=\frac{x^2}{x^2+1}\Rightarrow y{x}^2+y=x^2\Rightarrow x^2(1-y)=y\Rightarrow x=\sqrt{\frac{y}{1-y}}$

$x$ is defined only when,

$\frac{y}{1-y}\ge 0\Rightarrow \frac{y(y-1)}{{(1-y)}^2}\ge 0\Rightarrow y(1-y)\ge 0\therefore yϵ[0,1]$