Question

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

$f:R\to R$ defined by $f\left(x\right)=1+x^2$.

Answer 1

(i) Here, $f:R\to R$ is defined ny f(x) =3 -4x
Let $x_1,x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 3-4x_1=3-4x_2\Rightarrow -4x_1=-4x_2\Rightarrow x_1=x_2$
Therefore, f is one -one.
For any real number y in R, there exists $\frac{3-y}{4}$ in R such that $f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=y$
Therefore, f is onto. Hence, f is bijective.

Here, $f:R\to R$ is defined as $f\left(x\right)=1+x^2$
Let $x_1,x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 1+x_1^2=1+x_2^2\Rightarrow x_1^2\Rightarrow x_1=\pm x_2$
For instance, f(1)=f(-2)=2
Therefore, $f\left(x_1\right)=f\left(x_2\right)$ does not imply that $x_1=x_2$
Therefore, f is not one-one.
Consider an element -2 in co-domain R.
It is seen that $f\left(x\right)=1+x^2$ is positive for all x in R.
Thus, there does not exist any x in domain R such that f(x)=-2.
Therefore, f is not onto. Hence, f is neither one-one nor onto.