Question

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

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

Answer 1

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

$f=\{\left(\frac{x^2}{1+x^2}\right):x\in R\}$ be a function from $R\to R$

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

$f^{\prime }\left(x\right)-=dfrac2x(1+x^2)-2x{x}^2(1+x^2{)}^2=\frac{2x}{(1+x^2{)}^2}$

$9+$ is decreasing forms $(-\infty ,0)$ and increasing form $(0,\infty )$

Minimum value at $x=0$

$f\left(o\right)=0$

Max value is at $x=-\infty ,\infty$

$f\left(x\right)=1$

Range is $(0,1)$

$C$ is correct.