G1.1 Assume that the curve $x^{2}+3 y^{2}=12$ and the straight line $m x+y=16$ intersect at only one point. If $a=m^{2}$, find the value of $a$.

G1.2 It is given that $x+y=1$ and $x^{2}+y^{2}=2$. If $x^{3}+y^{3}=b$, find the value of $b$.
G1.3 In the following figure, $A C=A D=A E=E D=D B$
and $\angle B E C=c^{\circ}$. Given that $\angle B D C=26^{\circ}$ and $\angle A D B=46^{\circ}$, find the value of $c$.

G1.4 It is given that $4 \cos ^{4} \theta+5 \sin ^{2} \theta-4=0$, where $0^{\circ}<\theta<360^{\circ}$. If the maximum value of $\theta$ is $d$, find the value of $d$.
Sub. $y=16-m x$ into $x^{2}+3 y^{2}=12$
$$
\begin{array}{l}
\Rightarrow x^{2}+3(16-m x)^{2}=12 \\
x^{2}+3\left(256-32 m x+m^{2} x^{2}\right)=12 \\
\Rightarrow\left(1+3 m^{2}\right) x^{2}-96 m x+756=0
\end{array}
$$

The straight line is a tangent $\Rightarrow \Delta=(-96 m)^{2}-4\left(1+3 m^{2}\right) 756=0$
$$
\begin{array}{l}
576 m^{2}-189\left(1+3 m^{2}\right)=0 \\
\Rightarrow 64 m^{2}-21\left(1+3 m^{2}\right)=0 \\
\Rightarrow m^{2}=21 \\
\Rightarrow a=21
\end{array}
$$

Reference: 2011 FI2.2
$$
\begin{array}{l}
(x+y)^{2}=1 \Rightarrow x^{2}+y^{2}+2 x y=1 \\
\Rightarrow 2+2 x y=1 \\
\Rightarrow x y=-\frac{1}{2} \\
b=x^{3}+y^{3} \\
\quad=(x+y)\left(x^{2}+y^{2}-x y\right) \\
=1\left(2+\frac{1}{2}\right)=\frac{5}{2}
\end{array}
$$
$\triangle A D E$ is an equilateral triangle.
$$
\begin{array}{l}
\angle D A E=\angle A D E=\angle A E D=60^{\circ} \\
\because B D=D E \text { and } \angle B D E=46^{\circ}+60^{\circ}=106^{\circ} \\
\therefore \angle B E D=\left(180^{\circ}-106^{\circ}\right) \div 2=37^{\circ}(\angle \text { s sum of } \triangle) \\
\angle A E B=60^{\circ}-37^{\circ}=23^{\circ} \\
\angle A D C=26^{\circ}+46^{\circ}=72^{\circ}
\end{array}
$$
$$
\begin{array}{l}
\left.\because A C=A D \text { and } \angle A D C=72^{\circ}=\angle A C D \text { (base } \angle, \text { isos. } \triangle\right) \\
\therefore \angle C A D=180^{\circ}-72^{\circ} \times 2=36^{\circ}(\angle \mathrm{s} \text { sum of } \triangle) \\
\because A C=A E \text { and } \angle C A E=36^{\circ}+60^{\circ}=96^{\circ} \\
\therefore \angle A E C=\left(180^{\circ}-96^{\circ}\right) \div 2=42^{\circ}(\angle \mathrm{s} \text { sum of } \triangle) \\
\angle C E D=60^{\circ}-42^{\circ}=18^{\circ} \\
\angle B C E=60^{\circ}-18^{\circ}-23^{\circ}=19^{\circ} \\
c=19 \\
4 \cos ^{4} \theta+5 \sin ^{2} \theta-4=0 \Rightarrow 4 \cos ^{4} \theta+5\left(1-\cos ^{2} \theta\right)-4=0 \Rightarrow 4 \cos ^{4} \theta-5 \cos ^{2} \theta+1=0 \\
\left(4 \cos ^{2} \theta-1\right)\left(\cos ^{2} \theta-1\right)=0 \\
\cos ^{2} \theta=\frac{1}{4} \text { or } 1 \\
\Rightarrow \cos \theta=\frac{1}{2},-\frac{1}{2}, 1 \text { or }-1 . \\
\theta=60^{\circ}, 300^{\circ}, 120^{\circ}, 180^{\circ}, 240^{\circ} .
\end{array}
$$

The maximum value of $\theta=300^{\circ}$
$$
d=300^{\circ}
$$