(a) 画出 $y=|4x-2|$ 的图像草图。

(b) 解不等式 $1+3x<|4x-2|$。

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。

绝对值把原直线在横轴下方的部分向上反射。先标出顶点与纵轴截距，再画两条射线。

The graph is V-shaped:

Its vertex is $(1/2,0)$, its $y$-intercept is $(0,2)$, and the two branches have gradients $-4$ and $4$.

按绝对值内表达式的符号分两段解一次不等式，每段结果都须与该段范围相交。

For $x<1/2$, $|4x-2|=2-4x$. Hence

This satisfies $x<1/2$.

For $x\ge1/2$, $|4x-2|=4x-2$. Hence

Combining the two cases,

平方前先检查左侧 $1+3x$ 是否为负。负数一定小于绝对值；非负时才可等价平方，最后合并。

If $x<-1/3$, then $1+3x<0\le|4x-2|$, so the inequality holds.

If $x\ge-1/3$, both sides are non-negative, and squaring is equivalent:

Thus $x<1/7$ or $x>3$. Within this case, this gives $-1/3\le x<1/7$ or $x>3$.

Together with $x<-1/3$, the complete answer is