Inequalities: Sketch & solve \(\frac{1}{(x-a)^2}\) vs \(|x-a|\) — JPJC 2025 H2 Math Prelim Paper 1
Jurong Pioneer Junior College2025 PrelimPaper 1●●● Challenging6 marks
What this question tests
Sketch & solve \(\frac{1}{(x-a)^2}\) vs \(|x-a|\).
Question
- Find, in terms of \(a\), the roots of the equation \[ \frac{1}{(x-a)^{2}} = |x-a|. \]
- On the same axes, sketch the curves with equations \[ y=\frac{1}{(x-a)^{2}} \qquad \text{and} \qquad y=|x-a|, \] where \(a>1\). Hence solve the inequality \[ \frac{1}{(x-a)^{2}} > |x-a|. \]
Show full worked solution▾
(i) [2 marks]
\[ \frac{1}{(x-a)^{2}} = |x-a| \]| \(\dfrac{1}{(x-a)^{2}} = x-a\) | or \(\dfrac{1}{(x-a)^{2}} = -(x-a)\) |
|---|---|
| \((x-a)^{3} = 1\) | \((x-a)^{3} = -1\) |
| \(x-a = 1\) | \(x-a = -1\) |
| \(x = a+1\) | \(x = a-1\) |
(ii) [4 marks]
For \(\dfrac{1}{(x-a)^{2}} > |x-a|\): \[ a-1 < x < a+1, \quad x \neq a. \]
Or equivalently: \[ a-1 < x < a \quad\text{or}\quad a < x < a+1. \]
Answer: (i) \(x=a+1\) or \(x=a-1\). (ii) \(a-1<x<a+1,\ x\neq a\).