Question

Let
$S = {f : ℝ \to ℝ : f is polynomial and f (f (x)) = {(f (x))}^{2024} for x \in ℝ} .$
Then, the number of elements in S is _____________

Answer: $\underset{_____________}{3}$

Let $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2024} x^{2024} + \dots$ , where all $a_{i} \in ℝ, i \in ℕ \cup {0}$ is a general polynomial.

\begin{array}{rcl} Now, f (f (x)) & = & f (a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2024} x^{2024} + \dots) \\ ∴ f (f (x)) & = & a_{0} + a_{1} (a_{0} + a_{1} x + a_{2} x^{2} + \dots) + a_{2} {(a_{0} + a_{1} x + a_{2} x^{2} + \dots)}^{2} + \dots + \\ a_{2024} {(a_{0} + a_{1} x + a_{2} x^{2} + \dots)}^{2024} + \dots \end{array}

Also, {(f (x))}^{2024} = {(a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2024} x^{2024} + \dots)}^{2024}

Now, $f (f (x)) = {(f (x))}^{2024}$ forall $x \in ℝ$ ony if

Case(I) : all ${a_{i}}^{'} s$ are zero

i . e ., a_{0} = a_{1} = a_{2} = \dots = a_{2024} = \dots = 0

i . e ., f (x) = 0

Case(II) : $a_{0} = 1$ and all ${a_{i}}^{'} s$ are 0, where $i \in ℕ$

i . e ., a_{0} = 1 and a_{1} = a_{2} = \dots = a_{2024} = \dots = 0

i . e ., f (x) = 1

Case(III) : $a_{2024} = 1$ and all ${a_{i}}^{'} s$ are 0, where $i \in ℕ \cup {0} - {2024}$

i . e ., a_{2024} = 1 and a_{0} = a_{1} = a_{2} = \dots = a_{2023} = a_{2025} = \dots = 0

i . e ., f (x) = x^{2024}

So, S contains only three polynomials : $f (x) = 0, f (x) = 1$ , and $f (x) = x^{2024} .$
Hence, the number of elements in S is 3.

So, the correct answer is $\to$ $\underset{_____________}{3}$