如图所示，方程为 $y=2x^{\frac{1}{2}}+1$ 和 $y=\dfrac{1}{2}x^2-x+1$ 的两条曲线相交于 $A(0,1)$ 和 $B(4,5)$。

(a) 求两条曲线之间区域的面积。

两条曲线在 $B$ 点处的切线之间的锐角记为 $\alpha^\circ$，两条坐标轴的刻度比例相同。

(b) 求 $\alpha$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

图中根式曲线在上。由两个交点确定积分端点，对纵坐标之差积分。

The area is

用导数求两条切线的斜率。分别把斜率换成切线与正横轴的夹角，再取差。

At $x=4$, the gradients are $4^{-1/2}=1/2$ and $4-1=3$. Both are positive, so the acute angle is the difference of their inclinations to the positive $x$-axis:

也可以从 $B$ 出发，在两条切线上各取横向位移 $2$。用斜率求纵向位移，再对三角形使用余弦定理。

Moving two units to the right along the two tangents gives displacements $(2,1)$ and $(2,6)$. The resulting triangle has adjacent sides of lengths $\sqrt5$ and $\sqrt{40}$ and opposite side of length $5$. The cosine rule gives

so $\alpha=45$.