(a) 画出一对适当函数的图像草图，说明方程 $\ln x=3x-x^2$ 有且只有一个实根。

(b) 通过计算验证该根位于 $2$ 与 $2.8$ 之间。

(c) 利用迭代公式 $x_{n+1}=\sqrt{3x_n-\ln x_n}$ 求该根，精确到小数点后 2 位。写出每次迭代的结果，精确到小数点后 4 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。

画出对数曲线与开口向下的抛物线，标出唯一交点。结合两曲线在不同区间的位置与单调性，说明不会出现其他交点。

Sketch $y=\ln x$ for $x>0$, with vertical asymptote $x=0$ and intercept $(1,0)$.

On the same axes sketch $y=3x-x^2$, a downward-opening parabola with roots $0,3$ and vertex $(3/2,9/4)$. Mark their intersection.

For $0<x\le1$, $\ln x\le0<3x-x^2$, so there is no intersection.

For $1\le x\le3/2$, $3x-x^2\ge2$ while $\ln x\le\ln(3/2)<2$, so there is again no intersection.

From $x=3/2$ to $x=3$, the parabola decreases strictly while $\ln x$ increases strictly. The parabola starts above the logarithm and ends below it, so they cross exactly once. For $x\ge3$, the parabola is non-positive and $\ln x$ is positive.

Therefore the equation has $\boxed{\text{one real root}}$.

对两边之差作符号检验，注意对数的定义域包含整个检查区间。

Let $f(x)=3x-x^2-\ln x$. Then

Since $f$ is continuous on $[2,2.8]$, it has a root there. By part (a), this is the unique root.

按给出的迭代式计算，保留内部精度；由于早期迭代可能尚未稳定，应继续到两位小数结果明确。

Starting with $x_0=2.5$, use

The results are:

Thus the root is $\boxed{2.63}$ to 2 decimal places.