AQA Further AS Paper 1 2020 June — Question 14

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
TopicInequalities

14
  1. Given $$\frac { x + 7 } { x + 1 } \leq x + 1$$ show that $$\frac { ( x + a ) ( x + b ) } { x + c } \geq 0$$ where \(a , b\), and \(c\) are integers to be found.
    14
  2. Briefly explain why this statement is incorrect. $$\frac { ( x + p ) ( x + q ) } { x + r } \geq 0 \Leftrightarrow ( x + p ) ( x + q ) ( x + r ) \geq 0$$ 14
  3. Solve $$\frac { x + 7 } { x + 1 } \leq x + 1$$