2017 H2 Mathematics Paper 1 Question 2

Equations and Inequalities

Answers

x<a{x<a } or x>a+1b{x > a + \sqrt{\frac{1}{b}}}

Full solutions

(i)

(ii)

At intersection between the two graphs,
1xa=bxa\frac{1}{x-a} = b\left|x-a\right|
Since x>a{x>a} (from the graph),
1xa=b(xa)b(xa)2=1(xa)2=1b\begin{gather*} \frac{1}{x-a} = b(x-a) \\ b (x-a)^2 = 1 \\ (x-a)^2 = \frac{1}{b} \\ \end{gather*}
Since x>a,{x>a,}
x=a+1bx = a + \sqrt{\frac{1}{b}}
From the graphs, solution to 1xa<bxa:{\displaystyle \frac{1}{x-a} < b|x-a|: }
x<a   or   x>a+1b  x<a \; \textrm{ or } \; x > a + \sqrt{\frac{1}{b}} \; \blacksquare