Let $\mathrm{S}=(0,1)\cup (1,2)\cup (3,4)$ and $\mathrm{T}=\{0,1,2,3\}$. Then which of the following statements is(are) true?

$\begin{array}{l}\mathrm{S}=(0,1)\cup (1,2)\cup (3,4)\\ \mathrm{T}=\{0,1,2,3\}\end{array}$
Domain of function $\mathrm{f}:\mathrm{S}\to \mathrm{T}$ has infinitely many elements in set S
$\Rightarrow$Strictly increasing function is not possible as number of points in domain is infinite but co–domain has only 4 elements.
If number of points in domain are n, then number of functions possible from S to T are 4ⁿ as $\mathrm{n}\to \mathrm{\infty }\Rightarrow 4^{\mathrm{n}}\to \mathrm{\infty }$
Number of continuous functions possible are $4{\mathrm{C}}_1\times 4{\mathrm{C}}_1\times 4{\mathrm{C}}_1=64$
Because domain has 3 regions, so continuous functions are 64 
As the range of continuous function is a single element in the set T which is a constant, so every function is differentiable in all intervals as the derivative is 0 always.