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 $$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$
Q2 Suppose that the curve $y=2\cos 2x$ and the curve $y=4\cos x+k$ have a common tangent at $x=a\left(0<a\leqq\frac{\pi}{2}\right)$.
(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 $0<a\leqq\frac{\pi}{2}$, we have $a=\frac{\pi}{\mathbf{O}}$, and since $f(a)=g(a)$, we also have $k=-\mathbf{P}$.
Hence 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
$$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$