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N2023 P2 Q1

N2023 P2 Q1

Junior College 2
6 marks
Free
  1. Find the set of values of \(x\) for which \(3|x-2|<|5-2x|\).[2]
  2. Express \(\frac{x+25}{x^2-4x-5}+3\) as a single simplified fraction. Hence, without using a calculator, solve exactly the inequality \(\frac{x+25}{x^2-4x-5}>-3\).[4]

Default solution

  1. On the same axes, compare \(y=3|x-2|\) with \(y=|5-2x|\). Their intersections are \((1,3)\) and \(\left(\frac{11}{5},\frac{3}{5}\right)\).
    The graph of \(y=3|x-2|\) lies strictly below the other graph between these intersections.
    \(\boxed{1<x<\frac{11}{5}}\)
  2. The original denominator requires \(x\ne-1,5\). Combining and factorising gives
    \(\frac{x+25}{x^2-4x-5}+3\)\({}=\frac{x+25+3(x^2-4x-5)}{x^2-4x-5}\)\({}=\frac{3x^2-11x+10}{x^2-4x-5}\)\({}=\boxed{\frac{(3x-5)(x-2)}{(x-5)(x+1)}}\)
    The required inequality is equivalent to this fraction being positive. Its zeros are \(\frac{5}{3},2\), and its excluded values are \(-1,5\). The signs of the four linear factors give the following sign diagram without a calculator.
    \(\boxed{x<-1\quad\text{or}\quad\frac{5}{3}<x<2\quad\text{or}\quad x>5}\)
    The zeros are excluded because the inequality is strict; the poles are outside the original domain.
Answer:(a) \(1<x<\frac{11}{5}\). (b) \(\frac{(3x-5)(x-2)}{(x-5)(x+1)}\), \(x\ne-1,5\); solution \(x<-1\) or \(\frac{5}{3}<x<2\) or \(x>5\).

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