(a) 画出 $y=|3x-2a|$ 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 $|3x-2a|<x+5a$。

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。

绝对值图像的顶点是内部一次式等于零的位置，纵截距由代入零得到。

The graph is V-shaped, with vertex $(2a/3,0)$ and $y$-intercept $(0,2a)$.

Draw the solid branches $y=2a-3x$ for $x<2a/3$ and $y=3x-2a$ for $x\geqslant2a/3$, extending upwards on both sides.

把绝对值小于一个量改写成双边不等式，分别求解两边再取交集。

The inequality is equivalent to

The left inequality gives

and the right gives

Throughout this interval, $x+5a>0$, as required for the modulus comparison. Hence

先限制右边为正，再平方并用平方差分解。最后把二次不等式的区间与正值条件取交集。

The inequality requires $x+5a>0$, so $x>-5a$. On this domain, both sides are nonnegative, and squaring is equivalent:

The two roots are $-3a/4$ and $7a/2$. Since $a>0$, the product is negative between them:

Every point of this interval also satisfies $x>-5a$. Therefore