Inequalities

In mathematics, the inequality relation is one of the fundamental building blocks for inequalities. It deals with the mathematical comparison of two elements, and it serves as a key to understanding why numbers behave the way they do. Inequalities allow us to describe sets of numbers and the relationship between those numbers.

2019 NJC P1 Q2

Without using a calculator

(i)

Solve the inequality $\frac{2{{x}^{2}}+5x+9}{x-7}\le x-4$.

[4]

(i) Solve the inequality $\frac{2{{x}^{2}}+5x+9}{x-7}\le x-4$.

[4]

(ii)

Hence, solve the inequality $\frac{2+5{{e}^{x}}+9{{e}^{2x}}}{7{{e}^{x}}-1}\ge 4{{e}^{x}}-1$.

[2]

(ii) Hence, solve the inequality $\frac{2+5{{e}^{x}}+9{{e}^{2x}}}{7{{e}^{x}}-1}\ge 4{{e}^{x}}-1$.

[2]

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2013 CJC Promo Q8

(a)

(i)

Without using a calculator, solve the inequality $\frac{x+6}{{{x}^{2}}-3x-4}\le \frac{1}{4-x}$.

[3]

(i) Without using a calculator, solve the inequality $\frac{x+6}{{{x}^{2}}-3x-4}\le \frac{1}{4-x}$.

[3]

(ii)

Hence, deduce the range of values of $x$ that satisfies

$\frac{\left| x \right|+6}{{{x}^{2}}-3\left| x \right|-4}\le \frac{1}{4-\left| x \right|}$.

[2]

(ii) Hence, deduce the range of values of $x$ that satisfies

$\frac{\left| x \right|+6}{{{x}^{2}}-3\left| x \right|-4}\le \frac{1}{4-\left| x \right|}$.

[2]

(b)

Solve the inequality $\ln \left( x+6 \right)\le -\frac{x}{3}$.

[3]

(b) Solve the inequality $\ln \left( x+6 \right)\le -\frac{x}{3}$.

[3]

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2020 CJC Promo Q1

Without using a calculator, solve

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

[3]

Hence solve

${{\text{e}}^{-x}}<\frac{3}{{{\text{e}}^{-x}}-2}$.

[2]

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2019 TJC Promo Q1

(i)

Sketch the curve with equation $y=\left| \frac{\alpha x}{x+1} \right|$, where $\alpha$ is a positive constant, stating the equations of the asymptotes. On the same diagram, sketch the line with equation $y=\alpha x-2$.

[3]

(i) Sketch the curve with equation $y=\left| \frac{\alpha x}{x+1} \right|$, where $\alpha$ is a positive constant, stating the equations of the asymptotes. On the same diagram, sketch the line with equation $y=\alpha x-2$.

[3]

(ii)

Solve the inequality $\left| \frac{\alpha x}{x+1} \right|\ge \alpha x-2$, giving your answers in term of $\alpha$.

[3]

(ii) Solve the inequality $\left| \frac{\alpha x}{x+1} \right|\ge \alpha x-2$, giving your answers in term of $\alpha$.

[3]

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2013 ACJC P1 Q1

Do not use a graphic calculator in answering this question.
By considering the cases $x<0$ and $x\ge 0$, find the range of values of $x$ that satisfy the inequality

$\left( 10-x \right)\left( 10-\left| x \right| \right)>11$,

[5]

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