Unit 1 · Sets and Inequalities　集合与不等式

By the end of this chapter you can:

1. Read & write sets in roster, builder, and interval notation
2. Perform union, intersection, subset, and complement operations
3. Solve linear, quadratic, rational, and absolute-value inequalities
4. Compare powers/logs without a calculator using monotonicity

Exam weight on past CSCA papers: ~8% (3–4 of 48 MCQs).

1.1　Set Operations　集合的运算

A set is a collection of distinct objects called elements. We write
$$A = \{6,\ 7,\ 8,\ 9,\ 10\}$$
to mean "the set whose elements are 6, 7, 8, 9, 10." Order doesn't matter; duplicates collapse to one.

The four symbols to memorise

⚠️ The #1 trap on CSCA: $6 \in A$ is true, but $\{6\} \in A$ is false — the set $\{6\}$ is not an element of $A$; $A$ contains only the numbers 6,7,8,9,10. Use $\{6\} \subseteq A$ instead.

Empty set　空集

$$\Phi = \{\}$$

The empty set has no elements. It is a subset of every set ($\Phi \subseteq A$ always). It is not an element of $A$ unless $A$ was specifically defined to contain it.

Union and intersection — number-line picture

$$A \cup B = \{x : x \in A \ \text{or}\ x \in B\} \qquad A \cap B = \{x : x \in A \ \text{and}\ x \in B\}$$

🔑 Number-line trick: draw both sets on the same number line. Intersection = overlap. Union = everything covered.

Worked Example 1.1.A

Let $A = \{x : -2 \le x \le 2\}$ and $B = \{x : x > 0\}$. Find $A \cup B$ and $A \cap B$.

Solution.

Picture on the number line:

A:  ●━━━━━━━━━━●            (closed at −2 and 2)
   −2          2

B:           ○━━━━━━━━━━━━━ (open at 0, extends to ∞)
             0

• $A \cap B$ = where both are shaded = $(0,\ 2]$, i.e. $\{x : 0 < x \le 2\}$.
• $A \cup B$ = everything either covers = $[-2,\ +\infty)$, i.e. $\{x : x \ge -2\}$.

✅ This is exactly question 2 on the January CSCA paper.

Subsets — counting them

A set with $n$ elements has $2^{n}$ subsets (each element is either "in" or "out"). E.g. $A = \{a, b, c\}$ has $2^{3}=8$ subsets, including $\Phi$ and $A$ itself.

1.2　Set-Builder and Interval Notation

Three ways to write the same set:

Interval conventions:

• $[a,\ b]$ — closed, includes both endpoints
• $(a,\ b)$ — open, excludes both endpoints
• $[a,\ b)$ — half-open
• $(-\infty,\ 5]$ — unbounded on the left

🔑 Reading rule: square bracket = "I'm in" (endpoint included). Round bracket = "I'm out" (endpoint excluded). Infinity always gets a round bracket — you can't reach $\infty$.

1.3　Linear & Quadratic Inequalities　一元一次/二次不等式

Linear — solve like equations, with ONE warning

When you multiply or divide both sides by a negative number, flip the inequality sign.

$$-2x + 6 > 0 \implies -2x > -6 \implies x < 3$$

(Divided by $-2$, flipped $>$ to $<$.)

Quadratic — factor + sign chart

Goal: solve $x^{2} - x - 2 > 0$ (Q3 on Jan paper).

Step 1 — factor. $\ x^{2} - x - 2 = (x-2)(x+1)$.

Step 2 — find roots. $x = -1$ and $x = 2$.

Step 3 — sign chart.

Step 4 — pick the regions where the product matches the inequality.

We want $> 0$, so take "+" regions: $x < -1$ or $x > 2$.

$$\boxed{\ \{x : x < -1 \ \text{or}\ x > 2\}\ }$$

🔑 Parabola shortcut: For $(x-a)(x-b)$ with $a < b$ (parabola opens up):

• $> 0$ → outside the roots: $x < a$ or $x > b$
• $< 0$ → between the roots: $a < x < b$

Worked Example 1.3.A — A trickier sign

Solve $-x^{2} + 4x - 3 \ge 0$.

Multiply by $-1$ and flip: $x^{2} - 4x + 3 \le 0$. Factor: $(x-1)(x-3) \le 0$. Between the roots: $1 \le x \le 3$, i.e. $[1, 3]$.

1.4　Rational Inequalities　分式不等式

⚠️ Trap: you cannot simply multiply both sides by the denominator — its sign is unknown and could flip the inequality.

Method: move everything to one side, combine into one fraction $\frac{N(x)}{D(x)}$, then build a sign chart from the zeros of $N$ and $D$.

Worked Example 1.4.A — Q12 on Jan paper

Solve $\dfrac{2x+1}{x-2} \le 0$.

Critical points (where numerator or denominator is $0$): $\ x = -\tfrac{1}{2}$ and $x = 2$.

We want $\le 0$, so take the "−" region: $-\tfrac{1}{2} < x < 2$.

Boundary check:

• $x = -\tfrac{1}{2}$ makes the numerator $0$, so the quotient $= 0$. Since we allow $\le 0$, include it.
• $x = 2$ makes the denominator $0$, so the expression is undefined — exclude.

$$\boxed{\ \left[-\tfrac{1}{2},\ 2\right)\ }$$

Worked Example 1.4.B — Absolute value

Solve $|2x - 1| < 5$.

Standard split: $-5 < 2x - 1 < 5 \implies -4 < 2x < 6 \implies -2 < x < 3$. Answer: $(-2, 3)$.

For $|ax + b| > c$ ($c > 0$), the rule reverses: $ax + b > c$ or $ax + b < -c$.

1.5　Comparing Powers and Logs　不等式比较大小

You'll get a "which is bigger" MCQ with no calculator. The trick is always monotonicity.

Three monotonicity facts to memorise

1. Power function $f(x) = x^{p}$ on $(0, \infty)$:

   • $p > 0$ → increasing (bigger base → bigger value)
   • $p < 0$ → decreasing

2. Exponential function $f(x) = a^{x}$:

   • $a > 1$ → increasing in $x$
   • $0 < a < 1$ → decreasing in $x$

3. Logarithmic function $f(x) = \log_{a} x$:

   • $a > 1$ → increasing
   • $0 < a < 1$ → decreasing

Worked Example 1.5.A — Q27 on Jan paper

Which is correct?

(a) $2.1^{-2} > 1.2^{-2}$　(b) $0.75^{-0.2} > 0.75^{-0.4}$　(c) $3^{2.25} > 3^{3}$　(d) $2.1^{2/3} > 1.2^{2/3}$

Solution. Test each one with the right monotonicity rule.

• (a) $f(x) = x^{-2}$: exponent $-2 < 0$ → decreasing on $(0,\infty)$. Since $2.1 > 1.2$, we get $2.1^{-2} < 1.2^{-2}$. ❌
• (b) $f(x) = 0.75^{x}$: base $0.75 < 1$ → decreasing in $x$. Since $-0.2 > -0.4$, we get $0.75^{-0.2} < 0.75^{-0.4}$. ❌
• (c) $f(x) = 3^{x}$: base $3 > 1$ → increasing. Since $2.25 < 3$, $3^{2.25} < 3^{3}$. ❌
• (d) $f(x) = x^{2/3}$: exponent $\tfrac{2}{3} > 0$ → increasing on $(0,\infty)$. Since $2.1 > 1.2$, $2.1^{2/3} > 1.2^{2/3}$. ✅

Answer (d).

🔑 Decision tree:

• Same base, different exponents → look at the base. Base $> 1$: bigger exponent wins. Base $< 1$: smaller exponent wins.
• Same exponent, different bases → look at the exponent. Positive: bigger base wins. Negative: smaller base wins.

Logs at a glance

For $a > 1$:

• $\log_{a} x > 0$ when $x > 1$;　$\log_{a} x < 0$ when $0 < x < 1$
• $\log_{a} x$ has domain $(0, \infty)$ — never take a log of zero or negative.

Try it!　自测练习

Q1. Let $A = \{1,2,3,4\}$ and $B = \{3,4,5,6\}$. Find $A \cup B$ and $A \cap B$.

Q2. Solve $x^{2} - 5x + 6 \le 0$.

Q3. Solve $\dfrac{x-3}{x+1} \ge 0$.

Q4. Without a calculator, order from smallest to largest: $0.8^{3},\ 0.8^{2.5},\ 0.8^{-1}$.

Q5. True or false? $\Phi \in \{\Phi, \{1\}\}$.

📌 Chapter summary

Topic	Key skill
Sets	$\in$ vs $\subseteq$; $\cup$ ∩ Φ; number-line picture
Notation	roster, builder, interval — when to use each
Linear inequality	flip sign on negative multiply
Quadratic	factor → sign chart → outside/between rule
Rational	sign chart on numerator & denominator separately, exclude where denom $= 0$
Comparing	monotonicity of $x^{p}$, $a^{x}$, $\log_{a} x$ — that's the whole game

What's next → Unit 2 builds on this with functions: domain, range, odd/even, monotonicity (the same word, now applied to functions in general).