Question

f(x) =x2÷1+x2 find domain and range​

Answer 1

We need to find the domain and range of the function,

$\longrightarrow f(x)=\dfrac{x^2}{1+x^2}$

Since the denominator is non - zero,

$\longrightarrow 1+x^2\neq0$

$\longrightarrow x^2\neq-1$

Here $x$ is not a real number, so there are no domain restrictions. Therefore,

$\longrightarrow\underline{\underline{x\in\mathbb{R}}}$

This is the domain of the function.

Let $x=\tan\theta.$ Then,

$\longrightarrow f(\theta)=\dfrac{\tan^2\theta}{1+\tan^2\theta}$

Since $\sec^2\theta-\tan^2\theta=1,$

$\longrightarrow f(\theta)=\dfrac{\tan^2\theta}{\sec^2\theta}$

$\longrightarrow f(\theta)=\dfrac{\left(\dfrac{\sin^2\theta}{\cos^2\theta}\right)}{\left(\dfrac{1}{\cos^2\theta}\right)}$

$\longrightarrow f(\theta)=\sin^2\theta$

We know the range of $\sin\theta$ is,

$\longrightarrow\sin\theta\in[-1,\ 1]$

or,

$\longrightarrow\sin\theta\in[-1,\ 0]\cup[0,\ 1]$

On squaring we get,

$\longrightarrow\sin^2\theta\in[0,\ 1]\cup[0,\ 1]$

$\longrightarrow\sin^2\theta\in[0,\ 1]$

or,

$\longrightarrow f(\theta)\in[0,\ 1]$

Hence the range is,

$\longrightarrow\underline{\underline{f(x)\in[0,\ 1]}}$