Question

Show that the function $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1},\forall xϵR$ is neither one-one nor onto. Also, if $g:R\to R$ is defined as $g\left(x\right)=2x-1$, find $fog\left(x\right)$.

Answer 1

We have, the function $f:R\to R$ defined by

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

For one-one:
Let $x_1,x_2\in R$
Now,
$f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1}{x_1^2+1}=\frac{x_2}{x_2^2+1}$

$\Rightarrow x_1{x}_2^2+x_1=x_2{x}_1^2+x_2$

$\Rightarrow x_1{x}_2^2-x_2{x}_1^2+x_1-x_2=0$

$\Rightarrow -x_1{x}_2[x_1-x_2]+(x_1-x_2)=0$

$\Rightarrow (x_1-x_2)(1-x_1{x}_2)=0$

$\Rightarrow x_1=x_{2}or{x}_1{x}_2=1$

But, there exists some values of $x_1$ and $x_2$ so that $x_1\ne x_2$ but $f\left(x_1\right)=f\left(x_2\right)$

Like $x_1=2$ and $x_2=\frac{1}{2}$ then,

$f\left(x_1\right)=\frac{2}{5}$ and $f\left(x_2\right)=\frac{2}{5}$ but $x_1\ne x_2$

Hence, $f\left(x\right)$ is not one-one.

For onto:
Again, consider a value '1' as element in co-domain R.
$\Rightarrow \frac{x}{x^2+1}=1$

$\Rightarrow x^2+1=x$

$x^2-x+1=0$

i.e., quadratic equation in $x$

Here, discriminant D < 0.

Hnece, there is no real value of $x\in R$ for which $f\left(x\right)=1$.

$\therefore f$ is not an onto function.

Thus, $f$ is neither one-one nor onto.

Also, $f\left(x\right)=\frac{x}{x^2+1},\forall x\in R$ and $g\left(x\right)=2x-1$

$\therefore fog\left(x\right)=f\left(g\right(x\left)\right)$$=\frac{2x-1}{(2x-1{)}^2+1}$$=\frac{2x-1}{4x^2+1-4x+1}=\frac{2x-1}{4x^2-4x+2}$