Inequalities | DSE - Wyatt's Notes

An inequality states that one expression is greater than or less than another. Inequalities arise when finding the domain and range of functions, and are closely related to quadratic functions through their graphical interpretation.

Let $a$, $b$ And $c$ be real numbers. The following properties hold for inequalities:

Addition property:

Adding the same quantity to both sides preserves the inequality.

Multiplication by a positive number:

Multiplication by a negative number (reversal):

This is the most important rule to remember: multiplying or dividing both sides by a negative Number reverses the inequality sign.

• If $a > b > 0$ Then $a^2 > b^2$ and $\dfrac{1}{a} < \dfrac{1}{b}$.
• If $a > b$ and $c > d$ Then $a + c > b + d$.
• If $a > b > 0$ and $c > d > 0$ Then $ac > bd$.

The solution set is $(4, \infty)$.

• Solve $2x + 3 \leq x - 4$:

The solution set is $(-\infty, -7]$.

• Solve $-2x + 6 < 10$:

Note the reversal of the inequality sign when dividing by $-2$. The solution set is $(-2, \infty)$.

• Solve $5(x - 2) \geq 3x + 4$:

The solution set is $[7, \infty)$.

A quadratic inequality has the form $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$Or their non-strict Variants, where $a \neq 0$. Solving quadratic inequalities relies on understanding the graph of the Corresponding quadratic function $f(x) = ax^2 + bx + c$.

The graph of $f(x) = ax^2 + bx + c$ is a parabola. The solution of $f(x) > 0$ corresponds to the $x$-values where the parabola lies above the $x$-axis, and $f(x) < 0$ corresponds to where the Parabola lies below the $x$-axis.

The discriminant $\Delta = b^2 - 4ac$ determines the number of Intersections with the $x$-axis:

Using factorization:

1. Bring all terms to one side so the inequality is in the form $ax^2 + bx + c \gtrless 0$.
2. Factorize (or use the quadratic formula) to find the roots.
3. Draw a sign diagram to determine the sign of the expression in each interval.

Sign diagram method:

1. Find the roots of $ax^2 + bx + c = 0$.
2. Mark the roots on a number line, dividing the real line into intervals.
3. Test the sign of the expression in each interval.
4. Select the intervals satisfying the inequality.

Solution: $(-2, 8)$.

• Solve $|2x + 1| \geq 7$:

Solution: $(-\infty, -4] \cup [3, \infty)$.

• Solve $|3x - 6| > 0$:

Solution: $\mathbb{R} \setminus \{2\}$.

• Solve $|x^2 - 4| < 5$:

Since $x^2 \geq 0$ for all real $x$The left inequality $-1 < x^2$ is always satisfied.

From $x^2 < 9$: $-3 < x < 3$.

Solution: $(-3, 3)$.

• Solve $|x + 2| \leq -1$:

Since $|x + 2| \geq 0$ for all real $x$It can never be $\leq -1$.

Solution: $\varnothing$.

A system of inequalities requires finding the set of values that satisfy all inequalities Simultaneously. The solution set of the system is the intersection of the solution sets of the Individual inequalities.

1. Solve each inequality separately.
2. Find the intersection of all solution sets.
3. Represent the combined solution on a number line or using interval notation.

The same principle applies: solve each inequality independently, then take the intersection of all Solution sets.

Solution: $\left(-\infty, \dfrac{22}{5}\right]$.

1. Question: Solve $x^2 - 6x + 9 \geq 0$.

From $2x + 1 > 0$: $x > -\dfrac{1}{2}$.

Intersection: $-\dfrac{1}{2} < x < 5$I.e., $\left(-\dfrac{1}{2}, 5\right)$.

1. Question: Solve $|3x - 5| < 7$.

Case 1: $2x + 3 \geq 0$ (i.e., $x \geq -\dfrac{3}{2}$), so $|2x + 3| = 2x + 3$:

Roots: $x = \dfrac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.

So $1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}$.

Combined with $x \geq -\dfrac{3}{2}$: $\max\!\left(1 - \sqrt{2},\; -\dfrac{3}{2}\right) \leq x \leq 1 + \sqrt{2}$.

Since $1 - \sqrt{2} \approx -0.414 > -\dfrac{3}{2} = -1.5$The constraint is $1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}$.

Case 2: $2x + 3 < 0$ (i.e., $x < -\dfrac{3}{2}$), so $|2x + 3| = -(2x + 3)$:

Discriminant: $\Delta = 4 - 20 = -16 < 0$. Since $a = 1 > 0$The expression is always positive. No Solution in this case.

Solution: $[1 - \sqrt{2},\; 1 + \sqrt{2}]$.

1. Question: For what values of $k$ does the quadratic equation $x^2 + 2kx + k + 6 = 0$ have two Distinct real roots?

From $|x - 2| \leq 3$: $-3 \leq x - 2 \leq 3 \implies -1 \leq x \leq 5$.

From $x > 0$: $x \in (0, \infty)$.

Intersection of all three:

• From the first: $[1, 4]$.
• From the second: $[-1, 5]$.
• From the third: $(0, \infty)$.

Combined: $[1, 4]$.

Solution: $[1, 4]$.

1. Question: A ball is thrown upward from a height of $2$ m with an initial velocity of $20$ M/s. The height $h$ (in metres) after $t$ seconds is given by $h(t) = -5t^2 + 20t + 2$. During what Time interval is the ball at a height greater than $17$ m?

Factorize the numerator: $\dfrac{(x-2)(x+2)}{x-1} \geq 0$.

Critical points: $x = -2$, $x = 1$, $x = 2$.

Sign diagram:

The expression is $\geq 0$ when $-2 \leq x < 1$ or $x \geq 2$.

(Note: $x = -2$ is included because the numerator is zero there; $x = 1$ is excluded; $x = 2$ is Included.)

Solution: $ contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Inequalities with other DSE mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

To solve $\dfrac{f(x)}{g(x)} > 0$ (or $\geq, <, \leq$):

1. Find the zeros of $f(x)$ and $g(x)$ (the critical points).
2. Note that $g(x) \neq 0$ — these points are always excluded.
3. Construct a sign diagram across all intervals defined by the critical points.
4. Select intervals satisfying the inequality.
5. Include critical points from the numerator (where $f(x) = 0$) only for $\geq$ or $\leq$.

Solve $\dfrac{x^2 - 3x + 2}{x + 1} \leq 0$.

For inequality problems in DSE Paper 1:

1. When solving quadratic inequalities, always find the roots and sketch the parabola or draw a sign chart.
2. When multiplying or dividing by a negative, explicitly state that the inequality sign reverses.
3. For rational inequalities, identify points where the denominator is zero and exclude them.
4. For system of inequalities, draw each solution on a number line and identify the intersection.

Exact answers are preferred. If an approximate numerical answer is required, use 3 significant figures.

1. Quadratic inequalities with parameters (find the range of a parameter).
2. Absolute value inequalities (split into cases).
3. Rational inequalities (sign diagram method).
4. Systems of inequalities (intersection of solution sets).
5. Inequalities involving the discriminant (condition for real roots).

Worked Example: Inequality with quadratic and absolute value

Solve $|x^2 - 4x + 3| < 3x - 5$.

Worked Example: Quadratic inequality with parameter

Find the range of $m$ such that $mx^2 + (m - 1)x + m > 0$ for all real $x$.

Worked Example: System with three inequalities

Solve the system: x^2 - 2x - 15 \leq 0$$|x - 1| \leq 4$$x > 0.

DSE Practice 1. Solve $\dfrac{2x - 1}{x + 3} \geq 1$.

DSE Practice 2. Find the range of $k$ for which $kx^2 - 2kx + 3 > 0$ for all real $x$.

DSE Practice 3. Solve $|2x - 3| > |x + 1|$.

DSE Practice 4. Find all real values of $x$ satisfying $x^2 - 2|x| - 8 < 0$.

DSE Practice 5. Given that $x^2 + 2(k + 1)x + 9 > 0$ for all real $x$Find the range of $k$.

DSE Practice 6. Solve the inequality $\dfrac{x^2 - x - 6}{x^2 - 4} \leq 0$.

• Forgetting to flip the inequality when multiplying/dividing by a negative. This is the single most common error.

• Including excluded values from the domain. For $\frac{f(x)}{g(x)} \geq 0$, values where $g(x) = 0$ are excluded even though the inequality is non-strict.

• Wrong quadratic inequality solution. For $ax^2 + bx + c > 0$ with $a > 0$ and $\Delta < 0$, the solution is all real numbers, not “no solution.”

• Always identify the domain before solving inequalities involving fractions or square roots.

• Use sign charts for rational and polynomial inequalities — plot critical values and test intervals.

• When multiplying by a variable, split into cases based on sign, or use the fact that $\frac{f}{g} > 0$ is equivalent to $fg > 0$ (with $g \neq 0$).

• A quadratic $ax^2 + bx + c$ is always positive iff $a > 0$ and $\Delta < 0$.

Problem. Find all values of $k$ for which $x^2 + 2kx + k + 8 > 0$ for all real $x$.

Solution. For the quadratic to be always positive with leading coefficient $1 > 0$: $\Delta < 0$. $(2k)^2 - 4(1)(k+8) < 0 \implies 4k^2 - 4k - 32 < 0 \implies k^2 - k - 8 < 0$

Roots: $k = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2}$

Solution: $\frac{1 - \sqrt{33}}{2} < k < \frac{1 + \sqrt{33}}{2}$.

$\blacksquare$

Problem. Solve $\dfrac{x - 1}{x^2 - 4} \leq 0$.

Solution. Domain: $x \neq \pm 2$.

Critical values: $x = -2,\; 1,\; 2$.

Sign chart:

The quotient $\leq 0$ on $(-\infty, -2) \cup [1, 2)$. Note $x = 1$ is included (numerator $= 0$), but $x = \pm 2$ are excluded.

$\blacksquare$

======= 3. Misreading the question, particularly with “hence’ vs ‘hence or otherwise’ — the former requires using previous work.

1. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

   Stashed changes:docs/docs_dse/Maths/compulsory/inequalities.md

flowchart TD
    A[5_Inequalities] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Behind every scientific discovery and technological innovation lies mathematics. Functions model relationships between variables, statistics reveals patterns in data, and logic ensures rigorous reasoning. Mathematics teaches us to think precisely, solve systematically, and communicate evidently - skills that are valuable far beyond the classroom.