Questions on inequalities and inequations

Question 1

Explain why multiplying both sides of x/(x − 4) < 1 by x − 4 is unsafe, and state the safe first step.

Accepted 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.

Question 2

Solve (x − 1)/(x + 2) > 2.

Accepted 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.

Question 3

Solve |x − 2| ≤ 3, and describe the solution set as a set of distances on the number line.

Accepted Answer

|x − 2| ≤ 3 means −3 ≤ x − 2 ≤ 3, so −1 ≤ x ≤ 5. B1 for the double inequality, B1 for the closed interval. On the number line |x − 2| is the distance from x to 2, so the solution is every point within 3 of 2, endpoints included by the weak inequality.

Question 4

Solve x² > 9, and explain with a sketch of y = x² − 9 why the answer is not x > ±3.

Accepted Answer

x² − 9 = (x − 3)(x + 3) is a parabola opening upwards with roots at ±3, so it is above the axis outside the roots: x < −3 or x > 3. B1 for the sketch with the roots, B1 for the two outer intervals. Writing x > ±3 keeps −3 < x < 3 on the wrong side, and x = 0, where x² = 0, is the one-number check that shows it.

Question 5

Solve |x − 3| < 2x.

Accepted 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.

Question 6

Solve |x² − 4| > 3x, giving exact values.

Accepted 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 leaves no solutions at all, since the two regions do not overlap.

Question 7

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.

Accepted 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.

Question 8

On the same axes, sketch y = |2x − 1| and y = x + 1, and hence solve |2x − 1| ≤ x + 1.

Accepted Answer

The modulus graph is a V with vertex at (1/2, 0), and the line crosses it twice. On the right branch 2x − 1 = x + 1 gives x = 2, and on the reflected branch 1 − 2x = x + 1 gives x = 0. The V sits on or below the line exactly between the crossings, so 0 ≤ x ≤ 2. B1 for the two sketches with the vertex at x = 1/2, M1 for equating on each branch, A1 for x = 0 and x = 2, A1 for the closed interval.

Question 9

Solve |x + 1| ≥ 2|x − 2| by squaring both sides, explaining why squaring is legitimate here.

Accepted Answer

Both sides are moduli, so both are non-negative and squaring preserves the inequality's direction: (x + 1)² ≥ 4(x − 2)². Expanding, x² + 2x + 1 ≥ 4x² − 16x + 16, so 3x² − 18x + 15 ≤ 0, that is 3(x − 1)(x − 5) ≤ 0. A quadratic with positive leading coefficient is at most zero between its roots, giving 1 ≤ x ≤ 5. B1 for the justification that both sides are non-negative, M1 for squaring and collecting on one side, A1 for 3(x − 1)(x − 5) ≤ 0, A1 for the closed interval. Case-splitting would need the sign of both moduli tracked at once; squaring turns the whole question into one quadratic.

Question 10

A student solves (x + 3)/(x − 1) > 2 by multiplying both sides by x − 1, obtaining x + 3 > 2x − 2 and so x < 5. Sketch y = (x + 3)/(x − 1) and y = 2 on the same axes, use the sketch to find the correct solution, and explain what the student's method assumed.

Accepted Answer

y = (x + 3)/(x − 1) = 1 + 4/(x − 1) is a hyperbola with vertical asymptote x = 1 and horizontal asymptote y = 1, positive and falling to the right of x = 1, and below y = 1 and also falling on the left. It meets y = 2 where 4/(x − 1) = 1, at x = 5. B1 for the sketch with both asymptotes and the crossing at x = 5. The curve is above the line only between the asymptote and the crossing, so 1 < x < 5. M1 for reading the region from the sketch, A1. Algebraically, (x + 3 − 2(x − 1))/(x − 1) > 0 gives (5 − x)/(x − 1) > 0 and the same interval.
Multiplying by x − 1 assumed it was positive; for x < 1 it is negative and the inequality should have reversed, so the student's answer wrongly includes every x below 1. At x = 0 the left side is −3, not greater than 2. B1 for the explanation.

Question 11

Solve x³ − 4x > 0, using a sketch to confirm the sign of the expression on each interval.

Accepted Answer

x³ − 4x = x(x − 2)(x + 2), with roots at −2, 0 and 2. M1 for the factorisation. The cubic has positive leading coefficient, so it comes from below on the left, crosses at each simple root, and finishes above on the right: negative below −2, positive on (−2, 0), negative on (0, 2), positive beyond 2. B1 for the sign pattern from the sketch. So −2 < x < 0 or x > 2. A1. A test value in each interval does the same job: x = −1 gives 3, x = 1 gives −3.

Question 12

Solve x/(x + 1) ≤ 3/(x − 1), giving exact values.

Accepted 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.

Question 13

Solve |x/(x − 2)| < 3.

Accepted Answer

The modulus inequality says −3 < x/(x − 2) < 3, and each side is handled separately without multiplying by x − 2. For x/(x − 2) < 3, subtracting gives (6 − 2x)/(x − 2) < 0, with critical values 2 and 3, so x < 2 or x > 3. For x/(x − 2) > −3, adding gives (4x − 6)/(x − 2) > 0, with critical values 3/2 and 2, so x < 3/2 or x > 2. A value must satisfy both conditions, so the two sets are intersected, not united: x < 3/2 or x > 3, with x = 2 excluded throughout since the fraction is undefined there. M1 for rewriting the modulus as a double inequality, M1 for collecting each side over a common denominator, A1 for the first branch, A1 for the second, M1 for intersecting the two sets, A1 for the solution.

Question 14

Solve |x² − 1| > 2(x + 1) by writing the left side as |x − 1||x + 1| and considering the sign of x + 1.

Accepted Answer

|x² − 1| = |x − 1||x + 1|. M1 for factorising inside the modulus.
If x + 1 < 0, that is x < −1, the right side 2(x + 1) is negative while the left side is a modulus, never negative, so the inequality holds for every such x. B1 for the whole of x < −1.
If x = −1 both sides are 0 and the strict inequality fails.
If x + 1 > 0, that is x > −1, divide both sides by the positive quantity |x + 1| = x + 1: |x − 1| > 2. M1 for dividing by x + 1 on the branch where it is positive, A1 for |x − 1| > 2. That gives x − 1 > 2 or x − 1 < −2, so x > 3 or x < −1; on this branch only x > 3 survives. A1.
Combining the branches, x < −1 or x > 3. A1. Checks: x = −2 gives 3 > −2, x = 0 gives 1 > 2 which fails, x = 4 gives 15 > 10. Opening the modulus by cases on x² − 1 works too but needs the three regions x ≤ −1, −1 < x < 1 and x ≥ 1 handled separately; factorising lets the sign of one factor do the sorting.

Question 15

Solve x + 1/x ≥ 2 for x ≠ 0 algebraically, and sketch y = x + 1/x to check the answer. Explain what the sketch shows about where the curve can lie.

Accepted Answer

Put everything over x without multiplying through: x + 1/x − 2 = (x² − 2x + 1)/x = (x − 1)²/x ≥ 0. M1 for the single fraction, A1 for (x − 1)²/x. The numerator is a square, never negative and zero only at x = 1, so the sign of the fraction is the sign of x, and the inequality holds exactly when x > 0, with x = 1 included as the equality case. A1.
Sketch: the curve has two branches, one in each of the first and third quadrants, with asymptotes x = 0 and y = x. Differentiating, 1 − 1/x² = 0 at x = ±1, giving a minimum of 2 at (1, 2) on the right branch and a maximum of −2 at (−1, −2) on the left. B1 for the sketch with both turning points. So the right branch never drops below y = 2 and the left branch never rises above y = −2: the curve never enters the strip −2 < y < 2 at all, which confirms that y ≥ 2 holds for every positive x and no negative one. B1 for the explanation. Multiplying by x to get x² − 2x + 1 ≥ 0, true for all x, would have returned every x including the negative ones, and the sketch is what exposes that answer as wrong.