JEE Advanced 2023 — Paper 1 — Question 1

Set $S$ is the union of three disjoint open intervals: $(0,1)$, $(1,2)$, and $(3,4)$. It has infinitely many elements.

Set $T$ has 4 elements. A function from $S$ to $T$ assigns to each element of $S$ an element of $T$.

Since $S$ is infinite, there are infinitely many such assignments (each of the infinitely many points can be mapped to any of the 4 values). Hence option A is true. A strictly increasing function from $S$ to $T$ would require that if $x<\ y$ then $f(x)<\ f(y)$.

Since $S$ has infinitely many points but $T$ has only 4 distinct values, it is impossible to have a strictly increasing function (the function would need infinitely many distinct values). Hence option B is false. A continuous function from $S$ to $T$ must be constant on each connected component of $S$.

The connected components are the three intervals $(0,1)$, $(1,2)$, and $(3,4)$.

On each interval, the function can take any of the 4 values in $T$, giving $4 \times 4 \times 4 = 64$ possible continuous functions. Since $64 \le 120$, option C is true. Every continuous function from $S$ to $T$ is constant on each open interval, hence its derivative exists and is 0 on each interval. Therefore it is differentiable on $S$. Option D is true. Thus the correct options are A, C, D.