Let S = 0 , 1 ∪ 1 ,   2 ∪ 3 ,   4 and T = 0 ,   1 ,   2 ,   3 . Then…

Given,

$S=\left(0, 1\right)\cup \left(1, 2\right)\cup \left(3, 4\right)$ and $T=\left\{0, 1, 2, 3\right\}$

Now, let domain and co-domain of a function $y=f\left(x\right)$ are $S$ and $T$ respectively.

Now solving option,

$\left(\mathrm{A}\right)$ There are infinitely many elements in domain and four elements in co-domain.

$\Rightarrow$ There are infinitely many functions from $S$ to $T$.

$\Rightarrow$ Option $\left(\mathrm{A}\right)$ is correct

$\left(\mathrm{B}\right)$ If number of elements in domain is greater than number of elements in co-domain, then number of strictly increasing function is zero.

{ Assume $\sin x$ function its elements in domain is greater than its range hence, it is not an strictly increasing function.}

$\Rightarrow$ Option $\left(\mathrm{B}\right)$ is incorrect

$\left(\mathrm{C}\right)$ Maximum number of continuous functions $=4\times 4\times 4=64$
(Every subset $\left(0, 1\right), \left(1, 2\right), \left(3, 4\right)$ has four choices)
$\because 64<120$

$\Rightarrow$ Option $\left(\mathrm{C}\right)$ is correct.

$\left(\mathrm{D}\right)$ For every point at which $f\left(x\right)$ is continuous, $f^{\text{'}}\left(x\right)=0$ {as derivative of constant is always zero}

$\Rightarrow$ Every continuous function from $S$ to $T$ is differentiable.

$\Rightarrow$ Option $\left(\mathrm{D}\right)$ is correct