2020 EJC Promo Q2

2020 EJC Promo Q2

5 marks

Promo

Solve exactly the inequality \(\frac{{{x}^{2}}-x-1}{x+1}\le 1\).[3]
Hence, solve exactly the inequality \(\frac{{{x}^{2}}+x-1}{1-x}\le 1\).[2]

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Answer:(i) \(x<-1\) or \(1-\sqrt{3}\le x\le 1+\sqrt{3}\) (ii) \({\therefore x < 1 \text{or} -\sqrt{3} - 1 \le x \le \sqrt{3} - 1}\)

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