Inequalities: ASRJC 2024 Promo Q4

Anderson Serangoon Junior College2024 PromoPromo Q4●●● Challenging7 marks

What this question tests

Inequalities & Modulus.

Question

Sketch the graph of $y = \dfrac{4 - x}{2 + 3x - 2x^2}$ . Give the equations of the asymptotes and the coordinates of the point where the curve intersects the $x$ -axis. [4]
Solve the inequality $\dfrac{4 - x}{2 + 3x - 2x^2} > 0$ . [1]
Hence solve the inequality $\dfrac{4 - |x|}{2 + 3|x| - 2x^2} > 0$ . [2]

Show full worked solution▾

(a)

$y=\frac{4-x}{2+3x-2x^2}=\frac{4-x}{-(2-x)(1+2x)} \Rightarrow\text{ Asymptotes are }y=0,\ x=2,\ x=-\frac12$

$Anderson Serangoon Junior College 2024 Promo Q4: solution diagram$

(b)

From the graph, $-0.5<x<2$ or $x>4$

OR: $\frac{4-x}{2+3x-2x^2}>0\Rightarrow-(4-x)(x-2)(2x+1)>0$

$Anderson Serangoon Junior College 2024 Promo Q4: solution diagram$

$\Rightarrow-0.5<x<2\text{ or }x>4$

(c)

Replace $x$ with $|x|$ , $-0.5<|x|<2\qquad\text{or}\qquad |x|>4$ Since $|x|\geq0>-0.5$ for all real values of $x$ , $|x|<2\qquad\text{or}\qquad |x|>4$ $-2<x<2\qquad\text{or}\qquad x<-4\text{ or }x>4$

Answers

(a)asymptotes $y = 0$ , $x = 2$ , $x = -\tfrac{1}{2}$ ; intersects $x$ -axis at $(4, 0)$
(b) $-\dfrac{1}{2} < x < 2$ or $x > 4$ (c) $-2 < x < 2$ or $x < -4$ or $x > 4$

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