(a) 画出 $y=|x+3a|$ 的图像的草图，其中 $a$ 为正常数。

(b) 由此或使用其他方法，解不等式 $|x+3a|>a-2x$。

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。

绝对值图像是向上开的折线，先标出顶点和纵轴截距，两条射线的斜率分别为正负一。

The graph is V-shaped, with vertex $(-3a,0)$ and $y$-intercept $(0,3a)$.

Its two straight branches are

Draw both branches as solid lines, entirely on or above the horizontal axis.

按绝对值内部的正负分段。每段得到的不等式必须与该段范围取交集。

For $x\geqslant-3a$, the inequality becomes

Thus $x>-2a/3$, which already implies $x\geqslant-3a$ because $a>0$.

For $x<-3a$, it becomes

This is incompatible with $x<-3a$, so there are no solutions on this branch. Combining the cases,

平方比较前先检查右边正负。右边为负时所有值都成立，右边非负时才能等价平方。

If $x>a/2$, then $a-2x<0$, so $|x+3a|>a-2x$ holds automatically.

If $x\leqslant a/2$, both sides are nonnegative, so squaring is equivalent:

Taking the difference of the squares gives

Since $x\leqslant a/2$ and $a>0$, the factor $4a-x$ is positive. Hence $x>-2a/3$ in this case.

The two cases combine to give