Column A		Column B
1	Let $f\left(x\right)=-x^2+10x-20$. The last digit of the sum of all  $2^{2024}$ solutions to the equation $f\left(f\right(\cdots \left(x\right)\cdots \left)\right)=2$, where there are 2024 compositions of  $f$.	A	4
2	The number of irrational roots $x$  satisfying $x^9+\frac{9}{8}{x}^6+\frac{27}{64}{x}^3-x+\frac{219}{512}=0$.	B	3
3	If  $f\left(x\right)$ is a monic quartic polynomial such that  $f(-1)=-1$, $f\left(1\right)=1$ , $f\left(2\right)=3$ , and  $f(-2)=1$, find   $f\left(0\right)$.	C	2
4	For $x$  a real number, let $f\left(x\right)=0$  if $x<1$  and $f\left(x\right)=2x-2$ if  $x\ge 1$. The number of real solutions are there to the equation $f\left(f\right(f\left(f\right(x\left)\right)\left)\right)=x$?	D	0

Note: A monic polynomial is a polynomial whose leading coefficient (the coefficient of the highest power of 𝑥 is 1.

1.Solution:
Define $g\left(x\right)=f\left(f\right(\mathrm{...}\left(x\right)\mathrm{...}\left)\right)$ .
First, we    
calculate:   $f(10-x)=-{(10-x)}^2+10(10-x)-20=-100+20x-x^2+100-10x-20=-x^2+10x-20=f\left(x\right)$
     This implies that $g(10-x)=g\left(x\right).$  Hence, if  $g\left(x\right)=2,$ then  $g(10-x)=2.$
We can also calculate  $f\left(5\right):$ $f\left(5\right)=-5^2+10.5-20=-25+50-20=5$
Therefore,  $g\left(5\right)=5\ne 2.$
Given that$g\left(x\right)=2,$  the possible solutions can be grouped into pairs:
$(x_1,10-x_1),(x_2,10-x_2),\mathrm{...}$
Each pair sums to 10, and there are  $2^{2023}$ pairs. Therefore, the sum of all solutions is: $10\cdot 2^{2023}=10\cdot 2^{2023}.$
The last digit of $10\cdot 2^{2023}$  is determined by the last digit of 10, which is 0.
Thus, the last digit of the sum of all solutions is:0
2.Solution:
Note that we can rewrite the given equation as: $3\sqrt{x-\frac{3}{8}}=x^3+\frac{3}{8}.$
Furthermore, the functions of  $x$  on either side are inverses of each other and increasing. Let  $f\left(x\right)=3\sqrt{x-\frac{3}{8}}.$Suppose that $f\left(x\right)=y=f^{-1}\left(x\right)$ . Then, $f\left(y\right)=x$ .
However, if $x<y$ , we have $f\left(x\right)>f\left(y\right)$, contradicting the fact that $f$  is increasing, and similarly, if $y<x$ , we have $f\left(x\right)<f\left(y\right)$ , again a contradiction. Therefore, if  $f\left(x\right)=f^{-1}\left(x\right)$ and both are increasing functions in $x$ , we require $f\left(x\right)=x$.
This gives the cubic equation:  
                        $x^3-x+\frac{3}{8}=0$                                     
Solving this cubic equation, we have:
                     $(x-\frac{1}{2})(x^2+\frac{1}{2}x-\frac{3}{4})=0$                            
Thus, the solutions are:
                                    $x-\frac{1}{2},x=\frac{-1\pm \sqrt{13}}{4}$               
These are three distinct real roots.
Therefore, the number of distinct real roots is: 3
3.Solution:
The given data tells us that the roots of  $f\left(x\right)+x^2$ are -1, 1, 2, and -2. Since  $f$ is monic and quartic, we can write:
                      $f\left(x\right)+x^2=(x+1)(x-1)(x-2)(x+2)$                        
Therefore:
        $f\left(x\right)=(x+1)(x-1)(x-2)(x+2)-x^2$                                    
We need to find $f\left(0\right)$ :
                          $f\left(0\right)=(0+1)(0-1)(0-2)(0+2)-0^2$              
Calculating this:
             $f\left(0\right)=\left(1\right)(-1)(-2)(20-0=1\cdot (-1)\cdot (-2)\cdot 2=1\cdot 2\cdot 2=4$                
Thus:
$f\left(0\right)=4$                                                    
Therefore, the value of  is $f\left(0\right)$ : 4
4.Solution:
There are 2 solutions.
Certainly, 0 and 0 are fixed points of $f$  and therefore solutions.
1. For  $x<0,$, there can be no solutions since $f$  is non-negative-valued.
2. For $0\le x<1$ , we have $0\le f\left(x\right)<x<1$ . Since  $f\left(0\right)=0$, iteration only produces values below  .
3. For  $1\le x<2$, we have  $0\le f\left(x\right)<2$. Since $f\left(0\right)=0$ , iteration only produces values below  .
4. For $x\ge 2$ ,  $f\left(x\right)>x$, and iteration produces higher values.
Therefore, the only solutions are the fixed points 0 and 2.