图中显示了曲线 $y=\sqrt x\sin2x$，其中 $0\le x\le\frac12\pi$。曲线在点 $M$ 处达到最大值，该点的 $x$ 坐标为 $a$。

(a) 证明 $\tan2a=-4a$。

(b) 通过计算证明 $0.9<a<0.95$。

(c) 证明：如果由迭代公式

给出的数值序列收敛，则它收敛于 $a$。

(d) 使用 (c) 问中的迭代公式计算 $a$，答案精确到小数点后 $4$ 位。每次迭代的结果保留小数点后 $6$ 位。

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。

用乘积法则求导，乘去根号分母后得到三角方程。除以余弦前说明它不可能在驻点为零。

For $0<x<\pi/2$,

At the maximum $x=a$, the derivative is zero. Multiplying by $2\sqrt a$ gives

Here $\cos2a\ne0$: otherwise $2a=\pi/2$ and the left side would equal $1$. Dividing by $\cos2a$,

用两边之差检查变号，确认区间内没有正切渐近线。

Let $f(x)=\tan2x+4x$. Using radians,

The interval is to the right of $x=\pi/4$, and $f$ is continuous on $[0.9,0.95]$. The sign change gives a root there.

Moreover, for $0<a<\pi/4$, $\tan2a>0$ cannot equal $-4a$. For $\pi/4<a<\pi/2$,

so the root is unique. Hence $\boxed{0.9<a<0.95}$.

对收敛序列取极限，再取正切恢复驻点方程。反正切主值保证极限落在正确的驻点区间。

If the sequence converges to $L$, continuity gives

Thus

Taking tangents,

Since $-\pi/2<\tan^{-1}(4L)<\pi/2$, the fixed-point equation first gives

In particular, $L>0$, so $0<\tan^{-1}(4L)<\pi/2$. Substituting this sharper range into the same equation gives $\pi/4<L<\pi/2$. By the uniqueness established in (b), $\boxed{L=a}$.

用(b)的区间端点作为初值，迭代时保留完整精度，输出六位小数，直到四位小数稳定。

Starting with $x_0=0.9$, use

in radians. The iterations are

Thus $\boxed{a=0.9183}$ to 4 decimal places.