(a) 通过画出合适的两个函数图像的草图，证明方程 $\sec2x=-2x-\frac12$ 在区间 $0\le x\le\frac12\pi$ 内恰有一个根。

(b) 通过计算证明这个根位于 $0.8$ 与 $1.2$ 之间。

(c) 证明：如果由迭代公式

给出的实数序列收敛，则它收敛于 (a) 问中方程的根。

(d) 使用此迭代公式计算这个根，答案精确到小数点后 $3$ 位。每次迭代的结果保留小数点后 $5$ 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。

正割图像有渐近线，要把区间分成两段。左段正负号相反，右段两图单调方向相反，因此只有一次相交。

Sketch $y=\sec2x$ and $y=-2x-\frac12$ on $[0,\pi/2]$, marking the asymptote $x=\pi/4$ and the single intersection.

For $0\leqslant x<\pi/4$, the secant curve is positive and the line is negative, so there is no intersection.

For $\pi/4<x\leqslant\pi/2$, $\sec2x$ strictly increases from $-\infty$ to $-1$, while the line decreases. Near $\pi/4$ the secant curve is below the line; at $\pi/2$ it is above it since

Continuity gives an intersection on this second interval, and opposite strict monotonicity makes it unique. Thus the graphs have exactly one point of intersection, giving $\boxed{\text{exactly one root}}$.

用连续函数检查端点变号。给定小区间在渐近线右边，不跨过不连续点。

Let $f(x)=\sec2x+2x+\frac12$. Using radians,

Since $0.8>\pi/4$, $f$ is continuous on $[0.8,1.2]$. The sign change proves that the root lies in $\boxed{0.8<x<1.2}$.

取极限得到不动点方程，再依次取余弦和倒数，恢复原方程。

Every iterated value lies in $[0,\pi/2]$, so a limit $L$ also lies there. In particular, $4L+1\geqslant1$, so the denominator is nonzero. Continuity at the limit gives

Thus

Taking reciprocals,

Since the inverse cosine gives $0\leqslant L\leqslant\pi/2$, this is the unique root found in (a).

用区间内初值开始迭代，内部保留完整精度，输出五位小数。

Starting with $x_0=1$, apply the iteration in radians:

The results are

Hence the root is $\boxed{x=0.992}$ to 3 decimal places.