(a) 通过画出一组合适的两个函数图像的草图，证明方程

只有一个根。

(b) 通过计算证明此根位于 $1$ 与 $2$ 之间。

(c) 证明：如果由迭代公式

给出的数列收敛，那么它收敛于 (a) 问中方程的根。

(d) 使用此迭代公式计算此根，结果保留 $2$ 位小数。每次迭代的结果保留 $4$ 位小数。

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。

画 $y=\sqrt x$ 与 $y=e^x-3$，注意前者定义域为 $x\ge0$。指数曲线从较低处出发并最终超过根式曲线；交点右侧指数曲线的斜率始终更大，因此不会再相交。

Sketch $y=\sqrt x$ for $x\ge0$, starting at $(0,0)$ and increasing with decreasing gradient.

On the same axes sketch $y=e^x-3$, passing through $(0,-2)$ and $(\ln3,0)$, and increasing with increasing gradient. Mark their single intersection.

For $0\le x\le\ln3$, $e^x-3\le0\le\sqrt x$, so no intersection occurs there. For $x>\ln3$,

Thus after $x=\ln3$ the exponential curve rises faster throughout and eventually lies above $\sqrt x$. The two graphs therefore cross exactly once, so the equation has $\boxed{\text{only one root}}$.

把方程移项成连续函数为零，在 $1$、$2$ 处计算函数值并比较符号。

Let $f(x)=e^x-3-\sqrt x$. Then

Since $f$ is continuous on $[1,2]$, there is a root between $1$ and $2$. By part (a), it is the only root.

若迭代收敛，连续函数作用后可取极限，极限在相邻两项中相同。把极限方程指数化，就得到(a)中的原方程。

Suppose $x_n\to L$. The real iterates have $x_n\ge0$, so $L\ge0$. By continuity of $\ln(3+\sqrt x)$ on $x\ge0$,

Exponentiating,

Thus the limit satisfies the equation in part (a), and so it is its unique root.

从区间内的值开始迭代，计算时保留精度，显示到四位小数，直到相邻结果稳定。

Starting with $x_0=1$, use

Keeping unrounded values internally:

Hence the root is $\boxed{1.43}$ to 2 decimal places.