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Other Common Inequalities

CSCA Other Common Inequalities study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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Other Common Inequalities

In the CSCA exam, besides basic inequalities, quadratic inequalities, and rational inequalities, there are other common types of inequalities that need to be mastered. They can usually be solved by **equivalent transformation**, utilizing the known properties and methods of inequalities. This section focuses on **higher-degree polynomial inequalities** and **inequalities involving absolute value**.

1. Higher-Degree Polynomial Inequalities

Higher-degree inequalities are polynomial inequalities with degree greater than 2, e.g., $(x-1)(x-2)(x-3) > 0$. Their solution method is **very similar** to the sign analysis (number line method) for rational inequalities because polynomials can be factored into products of linear or irreducible quadratic factors.

**Core Solution Steps**:

1. **Factor Completely**: Rearrange so one side is 0 and the other side is factored into linear factors $(x-a)$ or irreducible quadratic factors (with $\Delta < 0$).

2. **Plot Critical Points**: Find all roots (zeros) from the linear factors and mark them on a number line in increasing order. Irreducible quadratic factors (with $\Delta < 0$) have **no real roots** and their sign is **always positive** (if the leading coefficient is positive) or always negative; they do not affect sign changes across intervals and can be ignored or their constant sign considered separately.

3. **Partition and Determine Sign**: The zeros divide the number line into intervals. Choose a test value from each interval, substitute it into the factored polynomial, and determine the overall sign based on the sign of each factor.

4. **Write Solution Set**: According to the sign requirement of the original inequality ($>0$, $\ge 0$, $<0$, $\le 0$), select the intervals with the correct sign, paying attention to whether endpoints (zeros) are included.

**Key Principle**: Each linear factor $(x-a)$ changes sign at its zero $a$ (from negative to positive or vice versa). The overall sign of the polynomial is the product of the signs of its linear factors.

**Example 1: Solve $(x+1)(x-2)(x-4) < 0$.**

**Solution**:

1. Zeros: $x=-1, 2, 4$.

2. Partition: Intervals: $(-\infty, -1)$, $(-1, 2)$, $(2, 4)$, $(4, +\infty)$.

3. Test signs:

* $x=-2$: sign $(-) \times (-) \times (-) = (-)$, negative.

* $x=0$: sign $(+) \times (-) \times (-) = (+)$, positive.

* $x=3$: sign $(+) \times (+) \times (-) = (-)$, negative.

* $x=5$: sign $(+) \times (+) \times (+) = (+)$, positive.

4. The inequality requires $<0$ (negative). Solution set: $(-\infty, -1) \cup (2, 4)$.

**Example 2: Solve $x^3 - 4x^2 + x + 6 \le 0$.**

**Solution**:

1. Factor: $x^3 - 4x^2 + x + 6 = (x+1)(x-2)(x-3)$.

2. Zeros: $x=-1, 2, 3$. Since inequality includes equality, zeros are included.

3. Partition: $(-\infty, -1]$, $[-1, 2]$, $[2, 3]$, $[3, +\infty)$.

4. Test signs (same as Ex.1): $(-\infty, -1)$ negative, $(-1, 2)$ positive, $(2, 3)$ negative, $(3, +\infty)$ positive.

5. The inequality requires $\le 0$ (non-positive). Solution set: $(-\infty, -1] \cup [2, 3]$.

2. Inequalities Involving Absolute Value

Basic forms include $|x| < a$, $|x| > a$, $|x| \le a$, $|x| \ge a$, and more general forms like $|f(x)| < g(x)$. The core of solving them is understanding the algebraic definition and geometric meaning of absolute value, and using **case analysis** or formula transformation.

**Definition of Absolute Value**: $|a| = \begin{cases} a, & \text{if } a \ge 0 \\ -a, & \text{if } a < 0 \end{cases}$

**Basic Formulas (for $a > 0$)**:

* $|x| < a \quad \Longleftrightarrow \quad -a < x < a$.

* $|x| \le a \quad \Longleftrightarrow \quad -a \le x \le a$.

* $|x| > a \quad \Longleftrightarrow \quad x < -a \ \text{ or } \ x > a$.

* $|x| \ge a \quad \Longleftrightarrow \quad x \le -a \ \text{ or } \ x \ge a$.

**Geometric Meaning**: $|x|$ represents the distance from point $x$ to the origin $0$ on the number line. Therefore:

* $|x| < a$: All points whose distance to the origin is less than $a$.

* $|x| > a$: All points whose distance to the origin is greater than $a$.

**General Form Solution**: For $|f(x)| < g(x)$ or $|f(x)| > g(x)$, two common methods are:

1. **Formula Method (Squaring)**: Use $|f(x)| < g(x) \Longleftrightarrow -g(x) < f(x) < g(x)$ or $|f(x)| > g(x) \Longleftrightarrow f(x) < -g(x) \ \text{or} \ f(x) > g(x)$. **Prerequisite: $g(x)$ must be non-negative**, otherwise discussion is needed. A safer method is squaring both sides: $|f(x)| < g(x) \Longleftrightarrow [f(x)]^2 < [g(x)]^2$, provided both sides are non-negative (usually assumed).

2. **Case Analysis Method**: Based on the sign of the expression inside the absolute value, $f(x)$ ($\ge 0$ or $< 0$), remove the absolute value in each case, solve the resulting inequalities separately, and then combine the solution sets.

**Example 3: Solve $|2x - 1| < 3$.**

**Solution** (Formula Method):

From $|2x-1| < 3$, we have $-3 < 2x - 1 < 3$.

Solve this compound inequality:

$-3 < 2x - 1$ gives $2x > -2$, so $x > -1$.

$2x - 1 < 3$ gives $2x < 4$, so $x < 2$.

Taking the intersection: $-1 < x < 2$.

Solution set: $(-1, 2)$.

**Example 4: Solve $|x^2 - 5x| > 6$.**

**Solution** (Formula Method):

From $|x^2 - 5x| > 6$, we have $x^2 - 5x < -6$ **or** $x^2 - 5x > 6$.

Solve the two quadratic inequalities:

1. $x^2 - 5x + 6 < 0$, i.e., $(x-2)(x-3) < 0$, solution: $2 < x < 3$.

2. $x^2 - 5x - 6 > 0$, i.e., $(x-6)(x+1) > 0$, solution: $x < -1$ or $x > 6$.

Taking the union, solution set: $(-\infty, -1) \cup (2, 3) \cup (6, +\infty)$.

**Example 5: Solve $|x-1| > 2x$.**

**Solution** (Case Analysis):

1. **Case 1: $x-1 \ge 0$, i.e., $x \ge 1$.** Then $|x-1| = x-1$. The inequality becomes $x-1 > 2x$, giving $x < -1$. This solution $x < -1$ contradicts the condition $x \ge 1$. So **no solution in this case**.

2. **Case 2: $x-1 < 0$, i.e., $x < 1$.** Then $|x-1| = -(x-1) = 1-x$. The inequality becomes $1-x > 2x$, giving $1 > 3x$, i.e., $x < \frac{1}{3}$.

This solution $x < \frac{1}{3}$ intersected with the condition $x < 1$ gives $x < \frac{1}{3}$.

Combining both cases, the solution set is: $(-\infty, \frac{1}{3})$.

**Summary**: The core idea for handling complex inequalities is **transformation**—using factoring, substitution, case analysis, etc., to convert them into inequalities we already know how to solve, such as linear, quadratic, or basic inequalities.