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Inequalities: YIJC 2024 Promo Q3

Attempt the question on paper, then open the worked solution to check your method.

Yishun Innova Junior College2024 PromoPromo Q3●●○ Standard8 marks

What this question tests

Solving inequalities.

(a)By first expressing $x^2-2x+3$ in completed square form, show that $x^2-2x+3$ is positive for all real values of $x$ . [2]

(b)Without the use of a calculator, solve the inequality $\dfrac{(x^2-2x+3)(1-x)}{x^2-x-2}<0$ . [3]

(c)Hence solve the inequality $\dfrac{(x-2\sqrt{x}+3)(1-\sqrt{x})}{x-\sqrt{x}-2}<0$ . [3]

Show full worked solution▾

(a) $x^2-2x+3=(x-1)^2+2$ $(x-1)^2\geq0$ for all $x\in\mathbb R$ , $(x-1)^2+2\geq2>0$ .

$\therefore x^2-2x+3$ is always positive for all values of $x$ .

(b) $\dfrac{(x^2-2x+3)(1-x)}{x^2-x-2}<0$ $x^2-2x+3$ is always positive, $\dfrac{1-x}{x^2-x-2}<0$ .

$\frac{1-x}{x^2-x-2}<0$ $\frac{x-1}{(x+1)(x-2)}>0$

$Yishun Innova Junior College 2024 Promo Q3: solution diagram$

$\therefore -1<x<1$ or $x>2$

(c)Replace $x$ with $\sqrt{x}$ in $\dfrac{(x^2-2x+3)(1-x)}{x^2-x-2}<0$ .

$\therefore -1<\sqrt{x}<1$ or $\sqrt{x}>2$

$0\leq x<1$ or $x>4$