(a) 通过画出合适的两个函数图像的草图，证明方程

在区间 $0<x<\frac12\pi$ 内只有一个根。

(b) 通过计算证明这个根位于 $0.9$ 与 $1$ 之间。

(c) 证明：若由迭代公式

给出的数列收敛，则它收敛于第 (a) 问中方程的根。

(d) 使用第 (c) 问中给出的迭代公式计算 $x$，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。

正割在区间内有渐近线，必须分两段讨论。左段为正，不可能等于负指数；右段递增而负指数递减，只相交一次。

Sketch $y=\sec2x$ and $y=-e^x$ on $0<x<\pi/2$, marking the asymptote $x=\pi/4$ and their single intersection.

For $0<x<\pi/4$, $\sec2x>0$ whereas $-e^x<0$, so there is no intersection.

For $\pi/4<x<\pi/2$, $\sec2x$ increases from $-\infty$ towards $-1$, whereas $-e^x$ decreases. Just to the right of $\pi/4$, the secant curve is below the exponential curve. Near $\pi/2$, it is above it since $-1>-e^{\pi/2}$.

By continuity and strict opposite monotonicity, there is exactly one intersection on this second interval. Hence there is $\boxed{\text{exactly one root in }(0,\pi/2)}$.

把方程等价地变成余弦与负指数的交点问题，可避开正割的渐近线。两个函数在整个区间单调方向相反。

The equation is equivalent to

Sketch $y=\cos2x$ and $y=-e^{-x}$ for $0<x<\pi/2$.

The cosine curve strictly decreases from $1$ towards $-1$, crossing the horizontal axis at $x=\pi/4$. The other curve strictly increases from $-1$ towards $-e^{-\pi/2}$.

Near $x=0$, $\cos2x>-e^{-x}$; near $x=\pi/2$, $\cos2x<-e^{-x}$. Thus the continuous curves intersect, and their strict opposite monotonicity gives exactly one intersection.

Therefore the original equation has $\boxed{\text{exactly one root}}$ in the stated interval.

计算两边之差并检查变号。这里只在不跨渐近线的闭区间上使用连续性。

Let $f(x)=\sec2x+e^x$. Using radians,

Since $[0.9,1]$ lies entirely to the right of $\pi/4$, $f$ is continuous on this interval. The sign change therefore proves that the root lies in $\boxed{0.9<x<1}$.

若迭代收敛，连续的迭代函数在极限处满足不动点方程。依次取余弦和倒数可恢复原方程。

If the sequence converges to $L$, continuity of the iteration function gives

Hence

Taking reciprocals,

The fixed-point equation gives $L\in[\pi/4,\pi/2]$. The value $L=\pi/4$ would require $e^{-L}=0$, and $L=\pi/2$ would require $e^{-L}=1$; both are impossible. Thus $L\in(\pi/4,\pi/2)$, so by (a) it is the unique required root.

取初值一，使用弧度制迭代。每步计算保留完整精度，表格输出五位小数。

Using $x_0=1$ and

the iterations are

Thus the root is $\boxed{x=0.978}$ to 3 decimal places.