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Solving inequalities

Practice section 5 of 7

Questions 25–30 in this collection

Practice questions

Review practice · 1 of 6

A rational identity and a scaled input

Show that 2x2+2x−8x2+x−6−1=x2+x−2x2+x−6\dfrac{2x^2+2x-8}{x^2+x-6}-1=\dfrac{x^2+x-2}{x^2+x-6}. [1]

Hence, without using a calculator, solve the inequality 2x2+2x−8x2+x−6≤1\dfrac{2x^2+2x-8}{x^2+x-6}\leq1. [4]

Deduce the solution to the inequality 4x2+2x−42x2+x−3≤1\dfrac{4x^2+2x-4}{2x^2+x-3}\leq1. [2]

[2016/PJC/Promo/3]

CJC · Review Problems · Question 2

Solution

Review practice · 2 of 6

Transfer a rational inequality to logarithmic inputs

(i) Without using a calculator, solve the inequality 5x−1x2+x−6≤1\dfrac{5x-1}{x^2+x-6}\leq1. [3]

(ii) Hence solve the inequality 1+1−5ln⁡x(ln⁡x)2+ln⁡x−6≥01+\dfrac{1-5\ln x}{(\ln x)^2+\ln x-6}\geq0. [3]

[2013/YJC/I/4]

CJC · Review Problems · Question 3

Solution

Review practice · 5 of 6

Prove positivity and retain an isolated equality

Given that xx is real, prove algebraically that x2−2x+3x^2-2x+3 is always positive. [1]

Hence, without using a graphing calculator, find the range of values of xx that satisfy x+4x2−2x+3≥32−x\dfrac{x+4}{x^2-2x+3}\geq\dfrac3{2-x}. [5]

[2016/MJC/Promo/1]

CJC · Review Problems · Question 6

Solution

Review practice · 6 of 6

A positive quadratic with an absolute-value input

Without using a calculator, solve the inequality 3x2x2−3x+1>−1\dfrac{3x}{2x^2-3x+1}>-1. [4]

Hence solve 3∣x∣2x2−3∣x∣+1>−1\dfrac{3|x|}{2x^2-3|x|+1}>-1. [3]

[2016/YJC/Promo/10(b)]

CJC · Review Problems · Question 7

Solution

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