Let S = (0, 1) (1, 2) (3, 4) and T = \ 0, 1, 2, 3\ . Then which of...

Given,

$S = \left(0, 1\right) \cup$\left(1, 2\right) \cup \left(3, 4\right)$and$T =\left\{0, 1, 2, 3\right\}$</p><p>Now, let domain and co-domain of a function$y = f\left(x\right) are $S and T$ respectively.

Now solving option,

\left(A\right) There are infinitely many elements$in domain and four elements in co-domain.</p><p>$⇒$There are infinitely many functions from$S$to$T$.</p><p>$⇒$Option \left(A\right)$ is correct$</p><p>\left(B\right) If number$ 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.}$</p><p>$⇒$Option \left(B\right) is incorrect$

\left(C\right)$Maximum number of$continuous functions= 4 \times 4 \times 4 = 64$<br />(Every subset \left(0, 1\right)$, \left(1, 2\right), \left(3, 4\right) has four choices)
$∵ 64 < 120$

$⇒$Option \left(C\right) is correct.

\left(D\right)$For every point at whichf\left(x\right)$ is continuous, ${f}^{'}\left(x\right) = 0$ {as derivative of constant is always zero}$</p><p>$⇒$Every continuous function from$S$to$T$is differentiable.</p><p>$⇒$Option \left(D\right) is correct$