For a positive number $a$, there is a function $$f ( x ) = \tan \frac { \pi x } { a }$$ defined on the set $\left\{ x \left\lvert \, - \frac { a } { 2 } < x \leq a \right. , x \neq \frac { a } { 2 } \right\}$. As shown in the figure, there is a line passing through three points $\mathrm { O } , \mathrm { A } , \mathrm { B }$ on the graph of $y = f ( x )$. Let $\mathrm { C }$ be the point other than $\mathrm { A }$ where the line parallel to the $x$-axis passing through point $\mathrm { A }$ meets the graph of $y = f ( x )$. When triangle $\mathrm { ABC }$ is equilateral, what is the area of triangle $\mathrm { ABC }$? (Here, $\mathrm { O }$ is the origin.) [4 points]
(1) $\frac { 3 \sqrt { 3 } } { 2 }$
(2) $\frac { 17 \sqrt { 3 } } { 12 }$
(3) $\frac { 4 \sqrt { 3 } } { 3 }$
(4) $\frac { 5 \sqrt { 3 } } { 4 }$
(5) $\frac { 7 \sqrt { 3 } } { 6 }$

Q2 Suppose that the curve $y=2\cos 2x$ and the curve $y=4\cos x+k$ have a common tangent at $x=a\left(0(1) We set $f(x)=2\cos 2x$ and $g(x)=4\cos x+k$. Since we have assumed that $y=f(x)$ and $y=g(x)$ have a common tangent at $x=a$, we see that
$$f'(a)=g'(a), \quad f(a)=g(a).$$
Since $f'(a)=g'(a)$ and $0Hence the coordinates of the tangent point are $\left(\frac{\pi}{\mathbf{O}},-\mathbf{Q}\right)$, and the equation of the common tangent line is
$$y=-\mathbf{R}\sqrt{\mathbf{S}}\left(x-\frac{\pi}{\mathbf{T}}\right)-\mathbf{U}.$$
(2) We are to find the area $S$ of the region bounded by these two curves over the range $-\frac{\pi}{2}\leqq x\leqq\frac{\pi}{2}$.
Since both of these curves are symmetric with respect to the $y$-axis, by putting $b=\mathbf{V}$ and $c=\frac{\pi}{\mathbf{O}}$ we have
$$S=\mathbf{W}\int_{b}^{c}(2\cos 2x-4\cos x-k)\,dx$$
By calculating this, we obtain
$$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$