Let the function $f:R \to R$ be defined
by $f(x) = \cos x,\forall x \in R$. Show that
$f$ is neither one-one nor onto.

Given function, $f(x) = \cos x,\forall x \in R$

Now, $f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} = 0$

$\Rightarrow$ $f\left( {\frac{{ - \pi }}{2}} \right) = \cos \frac{\pi }{2} = 0$

$\Rightarrow$ $f\left( {\frac{\pi }{2}} \right) = f\left( {\frac{{ - \pi }}{2}} \right)$

Bi $\frac{\pi }{2} \ne \frac{{ - \pi }}{2}$

So, $f(x)$ is not one-one.

Now, $f(x) = \cos x,\forall x \in R$ is not onto

as there is no pre-image for any real number.

Which does not belonging to the intervals

[-1,1], the range of $\cos x$.