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Inequalities and inequations questions
Never multiply an inequality by something whose sign you do not know. Move everything to one side, factorise, and let a sign diagram or a sketch read off the answer.
6 original questions · 25 marks · the inequalities and inequations notes · Further Pure 1
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Explain why multiplying both sides of x/(x − 4) < 1 by x − 4 is unsafe, and state the safe first step.
Worked answer
x − 4 is negative when x < 4, and multiplying an inequality by a negative quantity reverses it, so one multiplication silently assumes a sign. The safe move is to subtract everything to one side and combine over a common denominator before factorising. B1 for x − 4 being negative when x < 4, B1 for the safe first step.Solve (x − 1)/(x + 2) > 2.
Worked answer
Subtract: (x − 1 − 2(x + 2))/(x + 2) > 0, that is (−x − 5)/(x + 2) > 0. Critical values −5 and −2. Testing the three intervals gives negative below −5, positive between, negative above −2. So −5 < x < −2, with both endpoints excluded. M1 for collecting on one side over a common denominator, A1 for the critical values, A1 for the solution set.Solve |x − 3| < 2x.
Worked answer
The right side must be positive, so x > 0 throughout. For x ≥ 3 the modulus opens directly and x − 3 < 2x gives x > −3, so all of x ≥ 3 works. For x < 3 it opens with a sign change: 3 − x < 2x gives x > 1. Combining, the solution is x > 1. M1 for opening the modulus on x ≥ 3, A1 for that branch, M1 for opening it with a sign change on x < 3, A1 for the combined solution. Squaring both sides would work here too, since both sides are positive on the region of interest.Solve |x² − 4| > 3x, giving exact values.
Worked answer
Split at x = ±2, where the modulus changes. Where |x| ≥ 2 the modulus opens directly and x² − 3x − 4 > 0, that is (x − 4)(x + 1) > 0, so x < −1 or x > 4; intersecting with |x| ≥ 2 leaves x ≤ −2 or x > 4. Where |x| < 2 the modulus opens with a sign change and 4 − x² > 3x, that is (x + 4)(x − 1) < 0, so −4 < x < 1; intersecting with −2 < x < 2 leaves −2 < x < 1. Taking the union, x < 1 or x > 4. M1 for opening the modulus directly, A1 for (x − 4)(x + 1) > 0, A1 for that branch, M1 for opening it with a sign change, A1 for the second branch, A1 for the union. The two branches are combined with 'or', since a value only has to satisfy the inequality on the region it actually lives in; intersecting the branches instead loses everything below −1.For a weak inequality such as (x + 3)/(x − 2) ≥ 0, explain which critical values may be included in the solution set and which never can.
Worked answer
A zero of the numerator makes the expression exactly 0, which satisfies a weak inequality, so x = −3 is included. A zero of the denominator leaves the expression undefined, so x = 2 is excluded whatever the inequality sign. The solution is x ≤ −3 or x > 2. B1 for including the numerator zero, B1 for excluding the denominator zero, B1 for the solution set.Solve x/(x + 1) ≤ 3/(x − 1), giving exact values.
Worked answer
Subtract and combine: (x(x − 1) − 3(x + 1))/((x + 1)(x − 1)) ≤ 0, that is (x² − 4x − 3)/((x + 1)(x − 1)) ≤ 0. The numerator vanishes at x = 2 ± √7 and the denominator at x = ±1, so there are four critical values. Testing the five intervals gives −1 < x ≤ 2 − √7 or 1 < x ≤ 2 + √7. M1 for collecting over a common denominator, A1 for the single fraction, M1 for solving the numerator quadratic, A1 for 2 ± √7, B1 for the denominator critical values, M1 for testing the intervals, A1 for the solution set. The numerator zeros are included by the weak inequality; both denominator zeros stay out.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise inequalities and inequations one question at a time
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