A-Level H2 Math | 5 Essential Questions
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.
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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
$\left( 10-x \right)\left( 10-\left| x \right| \right)>11$,
giving your answers in exact form.
[5]
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