I4.1 Given that $f(x)=\left(x^{2}+x-2\right)^{2002}+3$ and $f\left(\frac{\sqrt{5}}{2}-\frac{1}{2}\right)=P$, find the value of $P$.
I4.2 In the following figure, $A B C D$ is a rectangle. $E$ and $F$ are points on $A B$ and $B C$ respectively. The areas of triangles $A E D, E B F$ and $F C D$ are $P, 3$ and 5 respectively. If the area of $\triangle E F D$ is $Q$, find the value of $Q$.

I4.3 It is given that $x$ and $y$ are positive integers. If the number of solutions $(x, y)$ of the inequality $x^{2}+y^{2} \leq Q$ is $R$, find the value of $R$. (Reference: 2007 FG1.2)

I4.4 It is given that $\alpha$ and $\beta$ are roots of the equation $x^{2}-a x+a-R=0$, where $a$ is real. If the minimum value of $(\alpha+1)^{2}+(\beta+1)^{2}$ is $S$, find the value of $S$.
$$
\begin{array}{l}
x=\frac{\sqrt{5}}{2}-\frac{1}{2} \\
\Rightarrow 2 x=\sqrt{5}-1 \\
\Rightarrow(2 x+1)=\sqrt{5} \\
\Rightarrow(2 x+1)^{2}=5 \\
\Rightarrow 4 x^{2}+4 x-4=0 \\
\Rightarrow x^{2}+x=1 \\
f(x)=\left(x^{2}+x-2\right)^{2002}+3=(1-2)^{2002}+3=1+3=4
\end{array}
$$

Let $A E=x, C F=y, A D=b, C D=a$.
Then $B E=a-x, B F=b-y$
Given the area of $\triangle A D E=4 \Rightarrow b x=8$..
the area of $\triangle C D F=5 \Rightarrow a y=10$.

The area of $\triangle B E F=3 \Rightarrow(a-x)(b-y)=6$
$$
\Rightarrow a b-b x-a y+x y=6 \text {. }
$$
$\qquad$
Sub. (1), (2) into (3) $a b-8-10+x y=6$
Sub. (1), (2) into the equation again: $a b-18+\frac{80}{a b}=6$
Solving for $a b, a b=20$ or 4 (rejected)
The area of $\triangle D E F=20-3-4-5=8$
$$
\begin{array}{l}
Q=8 \\
x^{2}+y^{2} \leq 8 \\
\Rightarrow(x, y)=(1,1),(1,2),(2,1),(2,2) \\
R=4 \\
x^{2}-a x+a-4=0 ; \alpha+\beta=a, \alpha \beta=a-4 \\
(\alpha+1)^{2}+(\beta+1)^{2}=\alpha^{2}+2 \alpha+1+\beta^{2}+2 \beta+1 \\
=(\alpha+\beta)^{2}-2 \alpha \beta+2(\alpha+\beta)+2 \\
=a^{2}-2(a-4)+2 a+2=a^{2}+10 \geq 10 \\
\end{array}
$$

The minimum value is $S=10$.