Given that $x$ is real, prove that $2{{x}^{2}}-x+2$ is always positive. [1] Solve the

Question

Given that $x$ is real, prove that $2{{x}^{2}}-x+2$ is always positive. [1] Solve the inequality $\frac{3x-1}{2{{x}^{2}}-x+2}\ge \frac{1}{x}$, leaving your answers in exact form. [3] Hence, without the use of a calculator, find the set of values of $x$ for which $\frac{3\left| x \right|-1}{2{{x}^{2}}-\left| x \right|+2}\ge \frac{1}{\left| x \right|}$. [2]

Accepted Answer

$-\sqrt{2} \leq x < 0 \text{ or } x \geq \sqrt{2}; \text{ hence } x \geq \sqrt{2} \text{ or } x \leq -\sqrt{2}$

2013 NJC P1 Q2

2013 NJC P1 Q2

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