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Inequalities: CJC 2024 Promo Q2

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Catholic Junior College2024 PromoPromo Q2●●○ Standard5 marks

What this question tests

Inequalities.

Question

By expressing $\dfrac{3x - 7}{(x+3)(x-2)} - 1$ as a single simplified fraction, solve exactly the inequality $\frac{3x - 7}{(x+3)(x-2)} < 1. \quad\text{([3])}$ Hence, solve exactly the inequality $\dfrac{3\mathrm{e}^x - 7}{(\mathrm{e}^x + 3)(\mathrm{e}^x - 2)} < 1$ . [2]

Show full worked solution▾

$\begin{aligned} \frac{3x - 7}{(x+3)(x-2)} - 1 &= \frac{3x - 7 - (x+3)(x-2)}{(x+3)(x-2)}\\ &= \frac{-x^2-x+6+3x-7}{(x+3)(x-2)}\\ &= \frac{-x^2 + 2x - 1}{(x+3)(x-2)}. \end{aligned}$ $\begin{aligned} \frac{3x - 7}{(x+3)(x-2)} &< 1\\ \frac{3x - 7}{(x+3)(x-2)} - 1 &< 0\\ \frac{-x^2 + 2x - 1}{(x+3)(x-2)} &< 0\\ \frac{x^2 - 2x + 1}{(x+3)(x-2)} &> 0\\ \frac{(x - 1)^2}{(x+3)(x-2)} &> 0. \end{aligned}$

$Catholic Junior College 2024 Promo Q2: solution diagram$

$\begin{aligned} \therefore\ x&<-3\text{ or }x>2. \end{aligned}$ Replace $x$ by $\mathrm{e}^x$ , $\begin{aligned} \mathrm{e}^x&<-3\ \text{(reject $\because\mathrm{e}^x>0$) or }\mathrm{e}^x>2\\ x&>\ln2 \end{aligned}$

Answers

$x < -3$ or $x > 2$ ; $x > \ln 2$

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