Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Inequalities. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Find the set of values of \(x\) for which \(4(x + 5) \lt 7x + 20\)[2]
(b) Find the set of values of \(x\) for which \(x^{2} - 25 \lt 0\)[3]
(c) Hence find the set of values of \(x\) for which both inequalities are satisfied.[1]
Show the answer and mark scheme
(a)Answer: \(x \gt 0\)
M1 for expanding the bracket and collecting the \(x\) terms: \(-3x \lt 0\)
A1 for \(x \gt 0\)
Worked solution: \(4x + 20 \lt 7x + 20\), so \(-3x \lt 0\); dividing by \(-3\) reverses the inequality: \(x \gt 0\).
(b)Answer: \(-5 \lt x \lt 5\)
M1 for finding the critical values, e.g. \((x + 5)(x - 5) = 0\)
A1 for \(x = -5\) and \(x = 5\)
A1ft for choosing the inside region: \(-5 \lt x \lt 5\)
Worked solution: \(x^{2} - 25 = (x + 5)(x - 5)\), so the critical values are \(x = -5\) and \(x = 5\). The graph of \(y = x^{2} - 25\) is \(\cup\)-shaped, so it is below the \(x\)-axis between the roots: \(-5 \lt x \lt 5\).
(c)Answer: \(0 \lt x \lt 5\)
B1ft for \(0 \lt x \lt 5\)
Worked solution: Combining \(x \gt 0\) with \(-5 \lt x \lt 5\) (a number line helps) gives \(0 \lt x \lt 5\). In set notation: \(\{x : 0 \lt x \lt 5\}\).
Question 2Medium6 marks
A student is asked to find the set of values of \(x\) for which \[\frac{x - 1}{x + 2} \le 2\]The student writes: \(x - 1 \le 2(x + 2)\) \(x - 1 \le 2x + 4\) \(-5 \le x\), so the answer is \(x \ge -5\).
(a) Show that \(x = -3\) satisfies the student's answer but does not satisfy the original inequality.[1]
(b) Explain the error in the student's method.[1]
(c) Find the correct set of values of \(x\) for which \(\dfrac{x - 1}{x + 2} \le 2\).[4]
Show the answer and mark scheme
(a)Answer: \(-3 \ge -5\), but \(\frac{-3 - 1}{-3 + 2} = 4\), which is not \(\le 2\).
B1 for evaluating \(\frac{-4}{-1} = 4 \gt 2\) and noting \(-3 \ge -5\)
Worked solution: \(-3 \ge -5\), so \(x = -3\) is in the student's answer. But \(\dfrac{-3 - 1}{-3 + 2} = \dfrac{-4}{-1} = 4\), and \(4 \gt 2\).
(b)Answer: The student multiplied by \(x + 2\), which is negative when \(x \lt -2\); multiplying by a negative number reverses the inequality.
B1 for the student multiplied both sides by \(x + 2\), which can be negative (reversing the inequality), without considering its sign
Worked solution: Multiplying an inequality by a negative number reverses it. The student multiplied by \(x + 2\), which is negative for \(x \lt -2\), so the method fails for those values.
(c)Answer: \(x \le -5\) or \(x \gt -2\)
M1 for multiplying both sides by \((x + 2)^2\) (or forming a single fraction \(\frac{-x - 5}{x + 2} \le 0\))
A1 for critical values \(-5\) and \(-2\), e.g. from \((x + 2)(x + 5) \ge 0\)
M1 for choosing the outside region
A1 for \(x \le -5\) or \(x \gt -2\) (strict at \(-2\))
Worked solution: Multiply both sides by \((x + 2)^2 \gt 0\) (for \(x \ne -2\)): \((x - 1)(x + 2) \le 2(x + 2)^2\). So \(0 \le 2(x + 2)^2 - (x - 1)(x + 2) = (x + 2)(2x + 4 - x + 1) = (x + 2)(x + 5)\). \((x + 2)(x + 5) \ge 0\) for \(x \le -5\) or \(x \ge -2\), and \(x = -2\) is excluded, so \(x \le -5\) or \(x \gt -2\).
Question 3Hard5 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The diagram shows a sketch of the curve \(C\) with equation \(y = x(x - 1)(x + 1)\) and the line \(l\) with equation \(y = 8x\).
[object Object]
Using algebra and the diagram, find the set of values of \(x\) for which \(x(x - 1)(x + 1) \lt 8x\)[5]
Show the answer and mark scheme
Answer: \(x \lt -3\) or \(0 \lt x \lt 3\)
M1 for equating, collecting terms and factorising out \(x\): \(x(x^{2} - 9) = 0\)
A1 for the critical values \(x = -3\), \(x = 0\) and \(x = 3\)
M1 for using the diagram to identify where \(C\) is below \(l\)
A1 for one correct interval
A1 for \(x \lt -3\) or \(0 \lt x \lt 3\) and no other values
Worked solution: \(x(x - 1)(x + 1) = 8x\) gives \(x^{3} - 9x = 0\), i.e. \(x(x^{2} - 9) = 0\), so \(x = 0\) or \((x + 3)(x - 3) = 0\). From the diagram, \(C\) is below \(l\) when \(x \lt -3\) or \(0 \lt x \lt 3\).