Inequalities — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for inequalities.

UT-1: Sign Flip When Dividing by Negative

Question:

Solve the inequality $\dfrac{3x - 1}{2 - x} \geq 0$.

Solution:

Critical values: $x = \dfrac{1}{3}$ (numerator zero) and $x = 2$ (denominator zero).

Do NOT simply cross-multiply, because the sign of $(2 - x)$ is unknown.

Sign chart:

Interval	$3x - 1$	$2 - x$	Quotient
$x < \dfrac{1}{3}$	$-$	$+$	$-$
$\dfrac{1}{3} < x < 2$	$+$	$+$	$+$
$x > 2$	$+$	$-$	$-$

At $x = \dfrac{1}{3}$: quotient is $0$ (included since $\geq 0$).

At $x = 2$: undefined (excluded).

Solution: $x \in \left[\dfrac{1}{3},\; 2\right)$.

A common mistake is cross-multiplying by $(2 - x)$ without considering the sign, which would give the wrong inequality direction for $x > 2$.

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UT-2: Absolute Value Inequality

Question:

Solve $|2x - 5| \lt 3x + 1$.

Solution:

Since the RHS involves $x$, we cannot simply split into two cases without considering the sign of the RHS.

Case 1: $3x + 1 \leq 0$, i.e. $x \leq -\dfrac{1}{3}$.

$|2x - 5| \geq 0$ and $3x + 1 \leq 0$, so $|2x - 5| \geq 0 > 3x + 1$ is possible only if $|2x - 5| < 3x + 1$. But $3x + 1 \leq 0$ while $|2x - 5| \geq 0$, so $|2x - 5| < 3x + 1$ is impossible when $3x + 1 \leq 0$ (since LHS $\geq 0$ and RHS $\leq 0$, equality requires both zero, but $|2x-5|=0 \implies x=5/2 \not\leq -1/3$).

No solution in this case.

Case 2: $3x + 1 > 0$, i.e. $x > -\dfrac{1}{3}$.

$-(3x + 1) < 2x - 5 < 3x + 1$

Left inequality: $-3x - 1 < 2x - 5 \implies -5x < -4 \implies x > \dfrac{4}{5}$.

Right inequality: $2x - 5 < 3x + 1 \implies -x < 6 \implies x > -6$ (always true when $x > -\dfrac{1}{3}$).

Combining with $x > -\dfrac{1}{3}$: $x > \dfrac{4}{5}$.

Solution: $x \in \left(\dfrac{4}{5},\; \infty\right)$.

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UT-3: Quadratic Inequality with Non-Standard Leading Coefficient

Question:

Solve $-2x^2 + 3x + 5 > 0$.

Solution:

Factorise: $-2x^2 + 3x + 5 = -(2x^2 - 3x - 5) = -(2x - 5)(x + 1)$.

So $(2x - 5)(x + 1) < 0$.

Critical values: $x = -1$ and $x = \dfrac{5}{2}$.

Since the parabola $2x^2 - 3x - 5$ opens upward, it is negative between the roots.

Solution: $-1 < x < \dfrac{5}{2}$, i.e. $x \in (-1,\; \tfrac{5}{2})$.

A common mistake is forgetting to reverse the inequality when factoring out the negative sign.

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UT-4: System of Linear Inequalities — Feasible Region

Question:

Find the region satisfying all of the following:

$x + y \leq 6, \quad 2x - y \geq 1, \quad x \geq 0, \quad y \geq 0$

Find the maximum value of $P = 3x + 2y$ in this region.

Solution:

Corner points of the feasible region:

1. Intersection of $x + y = 6$ and $2x - y = 1$: adding gives $3x = 7$, so $x = \dfrac{7}{3}$, $y = 6 - \dfrac{7}{3} = \dfrac{11}{3}$. Point: $\left(\dfrac{7}{3},\; \dfrac{11}{3}\right)$.

2. Intersection of $2x - y = 1$ with $y = 0$: $2x = 1$, $x = \dfrac{1}{2}$. Point: $\left(\dfrac{1}{2},\; 0\right)$.

3. Intersection of $x + y = 6$ with $x = 0$: $y = 6$. Point: $(0,\; 6)$.

4. Origin: $(0,\; 0)$.

Evaluate $P = 3x + 2y$:

• $\left(\dfrac{7}{3},\; \dfrac{11}{3}\right)$: $P = 7 + \dfrac{22}{3} = \dfrac{43}{3} \approx 14.33$
• $\left(\dfrac{1}{2},\; 0\right)$: $P = \dfrac{3}{2} = 1.5$
• $(0,\; 6)$: $P = 12$
• $(0,\; 0)$: $P = 0$

Maximum value: $\dfrac{43}{3}$ at $\left(\dfrac{7}{3},\; \dfrac{11}{3}\right)$.

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UT-5: Inequality Involving Reciprocals

Question:

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

Solution:

Bring to one side:

$\frac{1}{x - 1} - \frac{2}{x + 1} \leq 0$

$\frac{(x + 1) - 2(x - 1)}{(x-1)(x+1)} \leq 0$

$\frac{x + 1 - 2x + 2}{(x-1)(x+1)} \leq 0$

$\frac{3 - x}{(x-1)(x+1)} \leq 0$

$\frac{x - 3}{(x-1)(x+1)} \geq 0$

Critical values: $x = -1$, $x = 1$, $x = 3$.

Sign chart:

Interval	Test	Sign
$x < -1$	$x = -2$	$-$
$-1 < x < 1$	$x = 0$	$+$
$1 < x < 3$	$x = 2$	$-$
$x > 3$	$x = 4$	$+$

Including zeros ($x = 3$), excluding poles ($x = -1, 1$):

Solution: $(-1,\; 1) \cup [3,\; \infty)$.

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Integration Tests

Tests synthesis of inequalities with other topics.

IT-1: Inequalities and Quadratics (with Quadratics)

Question:

Find the range of values of $k$ such that the quadratic expression $x^2 + 2kx + k^2 - 2k + 5$ is always positive for all real $x$.

Solution:

The expression is always positive if the discriminant is negative (since the leading coefficient $1 > 0$).

$\Delta = (2k)^2 - 4(k^2 - 2k + 5) = 4k^2 - 4k^2 + 8k - 20 = 8k - 20$

$\Delta < 0 \implies 8k - 20 < 0 \implies k < \dfrac{5}{2}$.

Solution: $k \in (-\infty,\; \tfrac{5}{2})$.

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IT-2: Inequalities and Functions (with Functions)

Question:

Let $f(x) = x^2 - 6x + 5$. Find the set of values of $x$ for which $f(x) \leq f(2x)$.

Solution:

$f(x) = x^2 - 6x + 5$ and $f(2x) = 4x^2 - 12x + 5$.

$x^2 - 6x + 5 \leq 4x^2 - 12x + 5$

$0 \leq 3x^2 - 6x$

$3x^2 - 6x \geq 0$

$3x(x - 2) \geq 0$

Critical values: $x = 0$, $x = 2$.

Solution: $x \leq 0$ or $x \geq 2$, i.e. $x \in (-\infty,\; 0] \cup [2,\; \infty)$.

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IT-3: Inequalities and Logarithms (with Logarithms)

Question:

Solve $\log_2(x + 3) \lt \log_2(5 - x)$.

Solution:

Since the logarithm function is strictly increasing, we can compare arguments directly:

$x + 3 < 5 - x$

$2x < 2 \implies x < 1$

But we also need the domain: $x + 3 > 0$ and $5 - x > 0$, giving $-3 < x < 5$.

Combining: $-3 < x < 1$.

Solution: $x \in (-3,\; 1)$.

A common mistake is forgetting the domain restriction. If the base were between 0 and 1, the inequality would reverse.

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Worked Examples

WE-1: Solving a System of Inequalities

Question:

Solve the simultaneous inequalities:

$2x + 3 > x + 7 \quad \text{and} \quad 3x - 1 \leq 2x + 5$

Solution:

First inequality: $2x + 3 > x + 7 \implies x > 4$.

Second inequality: $3x - 1 \leq 2x + 5 \implies x \leq 6$.

Both must hold: $4 < x \leq 6$, i.e. $x \in (4,\; 6]$.

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WE-2: Quadratic Inequality with Equal Roots

Question:

Find the range of values of $k$ for which $x^2 - 6x + k > 0$ for all real $x$.

Solution:

For the quadratic to be always positive (since the leading coefficient $1 > 0$), we need $\Delta < 0$.

$\Delta = 36 - 4k < 0 \implies k > 9$

When $k = 9$: $\Delta = 0$, and $x^2 - 6x + 9 = (x-3)^2 \geq 0$. The inequality is strict ($>$), so $x = 3$ gives $0 \not> 0$.

Therefore $k > 9$ (strictly).

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WE-3: Absolute Value Inequality

Question:

Solve $|3x - 4| \leq 8$.

Solution:

$-8 \leq 3x - 4 \leq 8$

$-4 \leq 3x \leq 12$

$-\frac{4}{3} \leq x \leq 4$

Solution: $x \in \left[-\dfrac{4}{3},\; 4\right]$.

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WE-4: Inequality with Rational Expression

Question:

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

Solution:

The denominator $x^2 + 1 > 0$ for all real $x$ (always positive).

Therefore the sign of the expression is determined by the numerator alone:

$x^2 - 1 > 0 \implies (x-1)(x+1) > 0 \implies x < -1$ or $x > 1$.

Solution: $x \in (-\infty,\; -1) \cup (1,\; \infty)$.

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WE-5: Non-Linear Inequality

Question:

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

Solution:

$x^3 - x^2 - x + 1 = x^2(x - 1) - (x - 1) = (x - 1)(x^2 - 1) = (x - 1)^2(x + 1)$

Critical values: $x = 1$ (double root) and $x = -1$.

Sign chart:

Interval	Test	$(x-1)^2$	$(x+1)$	Product
$x < -1$	$x = -2$	$+$	$-$	$-$
$-1 < x < 1$	$x = 0$	$+$	$+$	$+$
$x > 1$	$x = 2$	$+$	$+$	$+$

The expression is non-positive when $x \leq -1$ or $x = 1$.

Solution: $x \in (-\infty,\; -1] \cup \{1\}$.

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WE-6: Quadratic Inequality with No Real Roots

Question:

Show that $x^2 + 4x + 5 > 0$ for all real $x$.

Solution:

$\Delta = 16 - 20 = -4 < 0$

Since the discriminant is negative and the leading coefficient is positive, the quadratic is always positive.

Alternatively, completing the square:

$x^2 + 4x + 5 = (x + 2)^2 + 1 \geq 1 > 0$

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WE-7: Inequality Involving Square Roots

Question:

Solve $\sqrt{2x + 1} \leq x + 1$.

Solution:

Domain: $2x + 1 \geq 0 \implies x \geq -\dfrac{1}{2}$. Also RHS $= x + 1$.

Since $\sqrt{2x+1} \geq 0$, we need $x + 1 \geq 0$, i.e. $x \geq -1$.

Combined domain: $x \geq -\dfrac{1}{2}$.

Squaring both sides: $2x + 1 \leq x^2 + 2x + 1 \implies 0 \leq x^2$, which is true for all real $x$.

So the solution is the domain: $x \in \left[-\dfrac{1}{2},\; \infty\right)$.

DSE Exam Technique: When squaring both sides of an inequality, always check the domain and the sign of both sides. Squaring is only valid when both sides are non-negative.

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WE-8: Product of Two Linear Inequalities

Question:

Solve $(2x - 3)(x + 4) > 0$.

Solution:

Critical values: $x = \dfrac{3}{2}$ and $x = -4$.

Since the quadratic opens upward (leading coefficient $= 2 > 0$):

The product is positive outside the roots.

Solution: $x < -4$ or $x > \dfrac{3}{2}$, i.e. $x \in (-\infty,\; -4) \cup \left(\dfrac{3}{2},\; \infty\right)$.

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Common Pitfalls

1. Forgetting to flip the inequality when multiplying or dividing by a negative number. If you multiply or divide both sides of an inequality by a negative quantity, you MUST reverse the inequality sign. This is the single most common error in inequality problems.

2. Cross-multiplying without considering the sign of the denominator. When solving $\dfrac{A}{B} > 0$, you cannot simply write $A > 0$ because the sign depends on $B$. Use a sign chart or consider cases.

3. Including values that make the denominator zero. When solving rational inequalities, the values that make the denominator zero must be EXCLUDED from the solution set, even if the numerator is also zero at those points.

4. Incorrectly handling double roots. A double root does not change the sign of the expression (it "bounces off" the axis). So at a double root, the expression equals zero, and the inequality direction determines whether to include or exclude it.

5. Not checking the domain before squaring. When solving $\sqrt{f(x)} > g(x)$, you must first establish that $f(x) \geq 0$ and $g(x) \geq 0$ before squaring both sides. Squaring an inequality where one side is negative gives incorrect results.

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DSE Exam-Style Questions

DSE-1

Find the range of values of $x$ for which:

(a) $x^2 - 5x + 6 < 0$ (2 marks) (b) $2x^2 + 3x - 2 \geq 0$ (3 marks) (c) Both inequalities in (a) and (b) are satisfied simultaneously. (2 marks)

Solution:

(a) $(x - 2)(x - 3) < 0 \implies 2 < x < 3$, i.e. $x \in (2,\; 3)$.

(b) $(2x - 1)(x + 2) \geq 0$.

Critical values: $x = \dfrac{1}{2}$ and $x = -2$.

Opens upward: $x \leq -2$ or $x \geq \dfrac{1}{2}$, i.e. $x \in (-\infty,\; -2] \cup \left[\dfrac{1}{2},\; \infty\right)$.

(c) Intersection of $(2,\; 3)$ and $(-\infty,\; -2] \cup \left[\dfrac{1}{2},\; \infty\right)$:

$(2,\; 3) \cap \left[\dfrac{1}{2},\; \infty\right) = (2,\; 3)$.

Solution: $x \in (2,\; 3)$.

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DSE-2

Find the range of values of $k$ for which the equation $x^2 + 2kx + k^2 + 3 = 0$ has:

(a) Two distinct real roots. (2 marks) (b) No real roots. (1 mark) (c) Real roots that are both positive. (4 marks)

Solution:

(a) $\Delta = (2k)^2 - 4(k^2 + 3) = 4k^2 - 4k^2 - 12 = -12 < 0$ for all $k$.

There are NEVER two distinct real roots. The answer is: no such value of $k$ exists.

(b) $\Delta < 0$ for all $k$, so there are no real roots for all values of $k$.

(c) Since the equation never has real roots, there is no value of $k$ for which both roots are positive.

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DSE-3

Solve the inequality $\dfrac{x^2 - 4x + 3}{x^2 - 9} \leq 0$. (5 marks)

Solution:

$\frac{(x-1)(x-3)}{(x-3)(x+3)} = \frac{x - 1}{x + 3}$

provided $x \neq 3$ (makes denominator zero in original) and $x \neq -3$.

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

Critical values: $x = -3$ (excluded) and $x = 1$ (included).

Sign chart:

Interval	Test	Sign
$x < -3$	$x = -4$	$+$
$-3 < x < 1$	$x = 0$	$-$
$x > 1$	$x = 2$	$+$

Including $x = 1$, excluding $x = -3$ and $x = 3$.

Solution: $x \in (-3,\; 1] \cup (3,\; \infty)$? No -- checking: for $x > 3$, $\dfrac{x-1}{x+3} > 0$, which does not satisfy $\leq 0$.

Correct solution: $x \in (-3,\; 1]$.

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DSE-4

Solve $|x - 3| > |2x + 1|$. (4 marks)

Solution:

Square both sides (both sides are non-negative after taking absolute value):

$(x - 3)^2 > (2x + 1)^2$

$x^2 - 6x + 9 > 4x^2 + 4x + 1$

$0 > 3x^2 + 10x - 8$

$3x^2 + 10x - 8 < 0$

$(3x - 2)(x + 4) < 0$

$-4 < x < \frac{2}{3}$

Solution: $x \in \left(-4,\; \dfrac{2}{3}\right)$.

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DSE-5

Find the range of values of $x$ for which both $\dfrac{x}{x - 2} > 0$ and $x^2 - 4x + 3 < 0$ are satisfied. (5 marks)

Solution:

First inequality: $\dfrac{x}{x - 2} > 0$.

Critical values: $x = 0$ and $x = 2$.

Sign chart: positive for $x < 0$ and $x > 2$.

Solution: $x \in (-\infty,\; 0) \cup (2,\; \infty)$.

Second inequality: $(x - 1)(x - 3) < 0 \implies 1 < x < 3$.

Solution: $x \in (1,\; 3)$.

Intersection: $((-\infty,\; 0) \cup (2,\; \infty)) \cap (1,\; 3) = (2,\; 3)$.

Solution: $x \in (2,\; 3)$.