Question

In each of the following cases, state whether the function is one-one,onto or bijective. Justify your answer.
(i) $f:R\to R$ defined by $f\left(x\right)=3-4x$
(ii) $f:R\to R$ defined by $f\left(x\right)=1+x^2$

Answer 1

Solve for one-one:
$f\left(x\right)=3-4x$
$f\left(x_1\right)=3-4x_1$
$f\left(x_2\right)=3-4x_2$
For one-one.
$f\left(x_1\right)=f\left(x_2\right)$
$3-4x_1=3-4x_2$
$-4x_1=-4x_2$
$x_1=x_2$
So clearly,if $f\left(x_1\right)=f\left(x_2\right)$
Then, $x_1=x_2$
$\therefore f$ is one-one.

Solve for onto:
$f\left(x\right)=3-4x$
Let $f\left(x\right)=y,$ such that $y\in R$
$3-4x=y$
$-4x=y-3$
$x=\frac{y-3}{-4}$
Now, for $y=f\left(x\right)$
Substituting value of $x$ in $f\left(x\right)$
$f\left(x\right)=f\left(\frac{y-3}{-4}\right)$
$\Rightarrow f\left(x\right)=3-4\left(\frac{y-3}{-4}\right)$
$\Rightarrow f\left(x\right)=3+(y-3)$
$\Rightarrow f\left(x\right)=y$
Thus, for every $y\in R$, there exists $x\in R$ such that
$f\left(x\right)=y$
$\therefore f$ is onto.
$\because f\left(x\right)$ is one-one and onto both.
$\therefore f\left(x\right)$ is bijective.


Solve for one-one.
$f\left(x\right)=1+x^2$
For one-one,
$f\left(x_1\right)=f\left(x_2\right)$
$1+(x_1{)}^2=1+(x_2{)}^2$
$(x_1{)}^2=(x_2{)}^2$
$\therefore x_1=x_2$ or $x_1=-x_2$
Thus, $f\left(x_1\right)=f\left(x_2\right)$ does not only imply that $x_1=x_2$ (additionally, $x_1=-x_2$ also)
So, $f$ is not one-one.

Solve for onto.
$f\left(x\right)=1+x^2$
Let $f\left(x\right)=y$, such that $y\in R$
$1+x^2=y$
$x=\pm \sqrt{y-1}$
Now $y$ is a real number, it canbe negative also,
Which is not possible as root of negative number is not real.
Hence, $x$ is not real or not equal to$y$.
So, $f$ is not onto.
Hence, $f$ is neither one-one nor onto or not bijective.