I2.1 It is given that $m, n>0$ and $m+n=1$. If the minimum value of $\left(1+\frac{1}{m}\right)\left(1+\frac{1}{n}\right)$ is $a$, find the value of $a$.

I2.2 If the roots of the equation $x^{2}-(10+a) x+25=0$ are the square of the roots of the equation $x^{2}+b x=5$, find the positive value of $b$. (Reference: 2001 FI3.4)

I2.3 If $(x y-2)^{b-1}+(x-2 y)^{b-1}=0$ and $c=x^{2}+y^{2}-1$, find the value of $c$.
I2.4 If $f(x)$ is a polynomial of degree two, $f(f(x))=x^{4}-2 x^{2}$ and $d=f(c)$, find the value of $d$.
$$
\begin{array}{l}
(m+n)^{2}=1 \\
\Rightarrow m^{2}+n^{2}+2 m n=1 \\
\Rightarrow m^{2}+n^{2}=1-2 m n \\
(m-n)^{2} \geq 0 \\
\Rightarrow m^{2}-2 m n+n^{2} \geq 0 \\
\Rightarrow 1-2 m n-2 m n \geq 0 \\
\Rightarrow m n \leq \frac{1}{4} \Rightarrow \frac{1}{m n} \geq 4 \\
\left(1+\frac{1}{m}\right)\left(1+\frac{1}{n}\right)=\frac{(1+m)(1+n)}{m n}=\frac{1+m+n+m n}{m n}=\frac{1+1+m n}{m n}=\frac{2}{m n}+1 \geq 2 \times 4+1=9 \\
x^{2}-19 x+25=0, \text { roots }=\alpha, \beta ; \alpha+\beta=19, \alpha \beta=25 \\
x^{2}+b x=5, \text { roots } r, s ; r+s=-b, r s=-5
\end{array}
$$

Now $r^{2}=\alpha, s^{2}=\beta$
$$
\begin{array}{l}
19=\alpha+\beta=r^{2}+s^{2}=(r+s)^{2}-2 r s=b^{2}-2(-5) \\
b^{2}=9 \\
\Rightarrow b=3 \text { (positive root) }
\end{array}
$$

Method 2 Replace $x$ by $\sqrt{x}$ in $x^{2}+b x=5$
$$
\begin{array}{l}
x+b \sqrt{x}=5 \\
b \sqrt{x}=5-x \\
b^{2} x=25-10 x+x^{2} \\
x^{2}-\left(10+b^{2}\right) x+25=0, \text { which } \\
\Rightarrow b=3 \text { (positive root) } \\
(x y-2)^{2}+(x-2 y)^{2}=0 \\
\Rightarrow x y=2 \text { and } x=2 y \\
\Rightarrow 2 y^{2}=2 \\
\Rightarrow y= \pm 1, x= \pm 2 \\
c=x^{2}+y^{2}-1=4+1-1=4
\end{array}
$$
$$
x^{2}-\left(10+b^{2}\right) x+25=0 \text {, which is identical to } x^{2}-19 x+25=0
$$

Let $f(x)=p x^{2}+q x+r$
$$
f(f(x))=p\left(p x^{2}+q x+r\right)^{2}+q\left(p x^{2}+q x+r\right)+r \equiv x^{4}-2 x^{2}
$$

Compare coefficient of $x^{4}: p=1$
Compare coefficient of $x^{3}: 2 q=0 \Rightarrow q=0$
Compare the constant: $r^{2}+q r+r=0 \Rightarrow r=0$ or $r+q+1=0$
Compare coefficient of $x: 2 q r+q^{2}=0 \Rightarrow q=0$ or $2 r+q=0 \ldots \ldots$ (2)
Sub. $q=0$ into (1): $r=0$ or -1
But $q=0, r=-1$ does not satisfy $2 r+q=0$ in (2)
$$
\therefore(p, q, r)=(1,0,0) \text { or }(1,0,-1)
$$

Sub. $(p, q, r)=(1,0,0)$ into $f(f(x))=x^{4}-2 x^{2} \equiv x^{4}$, which is a contradiction $\therefore$ rejected
Sub. $(p, q, r)=(1,0,-1)$ into $f(f(x))=x^{4}-2 x^{2} \equiv\left(x^{2}-1\right)^{2}-1$
RHS $=x^{4}-2 x^{2}+1-1=$ LHS
$$
\therefore f(x)=x^{2}-1 ; d=f(4)=4^{2}-1=15
$$