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More questions on this topic: IGCSE Algebra: Quadratic Inequalities (22) Level 2 FM Algebra: Inequalities (18) AS Pure: Quadratics (179)
21 Solve \((3x - 1)(5x + 2) \lt 0\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-\dfrac{2}{5} \lt x \lt \dfrac{1}{3}\) | B2 | for \(-\dfrac{2}{5} \lt x \lt \dfrac{1}{3}\) oe |
| (B1 | For correct critical values of \(-\dfrac{2}{5}\) and \(\dfrac{1}{3}\) oe) |
May be given as two separate inequalities
Accept \(-0.4\) and 0.33 (or better)
16 Solve \((4x - 3)(x + 5) \lt 0\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-5 \lt x \lt \dfrac{3}{4}\) | M1 | for critical values \(-5\) and \(\dfrac{3}{4}\) oe |
| A1 | oe |
Could be written as two separate expressions,
eg \(x \gt -5\) and \(x \lt \dfrac{3}{4}\) oe
17
| Answer | Mark | Mark scheme |
|---|---|---|
| \((2x + 1)(3x - 4)\) | M1 | for \((2x \pm 1)(3x \pm 4)\) oe or for brackets which when expanded give 2 out of 3 terms correct |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-\dfrac{1}{2} \lt x \lt \dfrac{4}{3}\) | M1 | for two critical values with at least one correct, \(-\dfrac{1}{2}\) or \(\dfrac{4}{3}\) or ft their critical values from (a) |
| A1 | for \(-\dfrac{1}{2} \lt x \lt \dfrac{4}{3}\) oe eg \(x \lt \dfrac{4}{3}, x \gt -\dfrac{1}{2}\) |
24 Find the set of possible values of \(x\) for which
\[4x^2 - 25 \lt 0 \qquad \textbf{and} \qquad 12 - 5x - 3x^2 \gt 0\]You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-2.5 \lt x \lt \dfrac{4}{3}\) | M1 | for method to find the critical values of \(4x^2 - 25 \lt 0\) eg \((2x + 5)(2x - 5)\) or critical values \(-2.5\) and 2.5 oe |
| M1 | (dep on M1) for \(x \gt -2.5\) and \(x \lt 2.5\) or \(x \gt a\) and \(x \lt b\) where \(a\) and \(b\) are their critical values and \(a \lt b\) | |
| M1 | for method to find the critical values of \(12 - 5x - 3x^2 \gt 0\) or \(3x^2 + 5x - 12 \lt 0\) eg \((4 - 3x)(x + 3)\) or \((3x - 4)(x + 3)\) \(\dfrac{--5 \pm \sqrt{(-5)^2 - 4 \times (-3) \times 12}}{2 \times (-3)}\) or \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 3 \times (-12)}}{2 \times 3}\) \(3\left[\left(x + \dfrac{5}{6}\right)^2 - \left(\dfrac{5}{6}\right)^2\right] - 12 = 0\) oe or critical values \(-3\) and \(\dfrac{4}{3}\) oe | |
| M1 | (dep on previous M1) for \(x \gt -3\) and \(x \lt \dfrac{4}{3}\) or \(x \gt c\) and \(x \lt d\) where \(c\) and \(d\) are their critical values and \(c \lt d\) | |
| A1 | for \(-2.5 \lt x \lt \dfrac{4}{3}\) oe eg \(x \lt \dfrac{4}{3}\), \(x \gt -2.5\) |
accept use of = or incorrect inequality symbol for 1st and 3rd M marks
This may be implied by a suitable diagram
A correct answer with no supportive working gets 0 marks
22 Find algebraically the set of values of \(x\) for which
\[x^2 - 49 \gt 0 \qquad \textbf{and} \qquad 5x^2 - 31x - 72 \gt 0\](5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x \lt -7,\ x \gt 8\) | M1 | for method to solve \(x^2 - 49 \gt 0\) eg \((x + 7)(x - 7)\) or 7 and \(-7\) |
| A1 | for \(x \lt -7\) and \(x \gt 7\) | |
| M1 | for method to solve \(5x^2 - 31x - 72 \gt 0\) eg \((5x \pm 9)(x \pm 8)\) or \(\dfrac{--31 \pm \sqrt{(-31)^2 - 4 \times 5 \times (-72)}}{2 \times 5}\) or 8 and \(-1.8\) oe | |
| A1 | for \(x \lt -1.8\) and \(x \gt 8\) | |
| A1 | cao |
accept use of = or incorrect inequality symbol for both the M marks
This may be implied by a suitable diagram
23 The diagram shows a parallelogram.
The area of the parallelogram is greater than 15 cm\(^2\)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | C1 | for a method to find the area of half of the parallelogram or of the whole parallelogram, eg \(\frac{1}{2}(2x - 1)(10 - x)\sin 150\) or \(\frac{1}{2}(2x - 1)(10 - x) \times \frac{1}{2}\) oe or \((2x - 1)(10 - x)\sin 150\) or \((2x - 1)(10 - x) \times \frac{1}{2}\) oe |
| C1 | for a correct expansion of the whole area eg \(\frac{1}{2}(20x - 10 - 2x^2 + x)\) or \(\frac{1}{2}(-2x^2 + 21x - 10)\) or \(-x^2 + 10.5x - 5\) | |
| C1 | complete chain of reasoning with fully correct algebra dealing with the inequality eg \(x^2 - 10.5x + 5 \lt -15\) or \(x^2 - 10.5x + 20 \lt 0\) or \(2x^2 - 21x + 10 \lt -30\) which lead to \(2x^2 - 21x + 40 \lt 0\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2.5 \lt x \lt 8\) | M1 | for factorising, \((2x - 5)(x - 8)\) |
| A1 | for critical values, 2.5, 8 | |
| A1 | for any statement that \(x\) is greater than 2.5 and \(x\) is less than 8 |
Could use the formula
Need not be given as an inequality statement
18
| Answer | Mark | Mark scheme |
|---|---|---|
| \(6x^3 + 35x^2 + 58x + 21\) | M1 | for a method to find the product of two linear expressions, 3 correct terms out of 4 terms e.g. \(2x^2 + x + 6x + 3\) or \(3x^2 + 7x + 9x + 21\) or \(6x^2 + 14x + 3x + 7\) |
| M1 | for a complete method to obtain all terms, at least half of which are correct (ft their first product) e.g. \(6x^3 + 32x^2 + 42x + 3x^2 + 16x + 21\) | |
| A1 | cao |
Note that, for example, \(7x + 3\) is regarded as three terms in the expansion of \((2x + 1)(x + 3)\)
First product must be a 3 or 4 term quadratic but need not be simplified or may be simplified incorrectly
Accept \(a = 6\), \(b = 35\), \(c = 58\), \(d = 21\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{2}{5} \lt x \lt 1\dfrac{3}{5}\) | M1 | for first step of finding the square root of both sides eg \(1 - x \lt \pm\dfrac{3}{5}\) OR for writing in the form \(ax^2 + bx + c\ (\lt 0)\) eg \(x^2 - 2x + \dfrac{16}{25}\ (\lt 0)\) or \(25x^2 - 50x + 16\ (\lt 0)\) |
| M1 | for showing critical values \(\dfrac{2}{5}\ (= 0.4)\) and \(1\dfrac{3}{5}\ (= 1.6)\) oe | |
| A1 | for \(\dfrac{2}{5} \lt x \lt 1\dfrac{3}{5}\) oe |
Condone use of an “=” sign; accept one square root (eg \(\tfrac{3}{5}\)) only shown.
Critical values can be stated, or shown in an expression (which may have incorrect inequality symbols)
Could be written as two separate expressions eg \(x \gt \dfrac{2}{5}\) and \(x \lt 1\dfrac{3}{5}\) oe
19 Solve \(22 \lt \dfrac{m^2 + 7}{4} \lt 32\)
Show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(9 \lt m \lt 11\) \(-11 \lt m \lt -9\) | M1 | for a correct method to begin rearranging to solve for \(m^2\) eg \(88 \lt m^2 + 7\) or \(m^2 + 7 \lt 128\) or \(81 \lt m^2 \lt 121\) |
| M1 | for a complete method to \(m^2 = 81\) or \(m^2 = 121\) or better | |
| M1 | for a set of critical values: at least two out of 9, 11, \(-9\), \(-11\) | |
| M1 | for selecting a correct inequality for one set of critical values eg \(9 \lt m\) and \(m \lt -9\) or \(m \lt 11\) and \(-11 \lt m\) or \(9 \lt m\) and \(m \lt 11\) or a set of inequalities with some error eg \(9\ ?\ m\ ?\ 11\) and \(-11\ ?\ m\ ?\ -9\) where ? is an incorrect inequality symbol like \(9 \lt m \leqslant 11\) or \(9 \geqslant m \geqslant 11\) or answer given as \(\pm 9 \lt m \lt \pm 11\) | |
| A1 | \(9 \lt m \lt 11\) and \(-11 \lt m \lt -9\) given as boundaries of \(m\) |
It is insufficient to just multiply all three elements by 4; some rearrangement must occur such as showing as two separate inequalities or isolating \(m^2\)
Accept an inequality used in place of “=”.
\(m^2\) must be isolated at this stage.
Do not award if other values are also given eg 10
Could be shown as \(9 \lt m \lt 11\) or \(-11 \lt m \lt -9\) or \(-11 \lt m \lt 11\)
Accept with an “and” or an “or” or neither
20 \(n\) is an integer such that \(3n + 2 \leqslant 14\) and \(\dfrac{6n}{n^2 + 5} \gt 1\)
Find all the possible values of \(n\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2, 3, 4 | M1 | for method to solve \(3n + 2 \leqslant 14\) eg \(n \leqslant (14 - 2) \div 3\) oe |
| M1 | for complete method to rearrange \(\dfrac{6n}{n^2 + 5} \gt 1\) to the form \(an^2 + bn + c\ (\lt 0)\) | |
| M1 | for method to begin to solve \(n^2 - 6n + 5\ (\lt 0)\) eg \((n \pm 5)(n \pm 1)\ (\lt 0)\) | |
| M1 | (dep on previous M2) for \(n \gt 1\) and \(n \leqslant 4\) or \(1 \lt n \lt 5\) | |
| A1 | (dep M4) cao |
This could be shown within an equation rather than an inequality at this stage
For the 2nd and 3rd M marks condone no ‘\(\lt 0\)’ and condone use of incorrect inequality signs or ‘=’
Accept \(\dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 5}}{2 \times 1}\) (condone one sign error)
Must come from correct working
Could be shown on a number line
| Answer | Mark | Mark scheme |
|---|---|---|
| 2, 3, 4 | M1 | for method to solve \(3n + 2 \leqslant 14\) eg \(n \leqslant (14 - 2) \div 3\) oe OR for \(3 \times 4 + 2 = 14\) |
| M3 | for trials with 1, 2, 3 and 4 in the quadratic inequality, correctly evaluated | |
| (M2 | for trials with three of 1, 2, 3 and 4, correctly evaluated) | |
| (M1 | for trials with two of 1, 2, 3 and 4, correctly evaluated) | |
| A1 | (dep M4) cao |
This could be shown within an equation rather than an inequality at this stage
The values from the trials may be given as improper fractions eg \(\dfrac{24}{21}\), \(\dfrac{18}{14}\), \(\dfrac{12}{9}\), \(\dfrac{6}{6}\)
23 Here is a rectangle and a right-angled triangle.
All measurements are in centimetres.
The area of the rectangle is greater than the area of the triangle.
Find the set of possible values of \(x\). (5)
| Answer | Mark | Notes |
|---|---|---|
| \(x \gt 2\) | P1 | for process to derive algebraic expressions for area of both rectangle and triangle eg \((x - 1)(3x - 2)\) and \((2x \times x) \div 2\) (condone missing brackets) |
| M1 | for method to rearrange inequality to \(2x^2-5x+2 \gt 0\) oe providing in the form \(ax^2 + bx + c \gt 0\) | |
| M1 | for a correct method to solve \(2x^2-5x+2\gt 0\) | |
| M1 | for establishing critical values 2 and \(\dfrac{1}{2}\) | |
| A1 | \(x \gt 2\) |
19 Solve \(2x^2 + 3x - 2 \gt 0\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(x \lt -2\), \(x \gt \dfrac{1}{2}\) | M1 | for a first step to solve the quadratic e.g. factorisation: \((2x + 4)\left(x - \tfrac{1}{2}\right)\) or \((2x - 1)(x + 2)\) or using the formula \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2)}}{2 \times 2}\) |
| A1 | for \(-2\) and \(\tfrac{1}{2}\) | |
| A1 | for \(x \lt -2\), \(x \gt \tfrac{1}{2}\) |
22 Solve \(\quad 2x^2 \gt 12 - 5x\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2x^2 + 5x - 12\ (\gt 0)\) or \(12 - 5x - 2x^2\ (\lt 0)\) | B1 | oe may be implied by 1.5 and \(-4\) |
| Correct method to solve their three-term quadratic | M1 | eg \((2x - 3)(x + 4)\ (\gt 0)\) or \(2\left(x + \dfrac{5}{4}\right)^2 - \dfrac{121}{8}\ (\gt 0)\) or \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 2 \times -12}}{2 \times 2}\) any of these imply B1M1 |
| 1.5 and \(-4\) | A1ft | oe ft their three-term quadratic |
| Both \(x \gt 1.5\) and \(x \lt -4\) stated | B1ft | oe ft B0M1 or B1M1A0 ft their two solutions for their three-term quadratic correct for their inequality in the form \(ax^2 + bx + c \gt 0\) or \(ax^2 + bx + c \lt 0\), which must be seen if it is not the correct inequality |
| For the first B1 and the M1 condone = 0 in place of \(\gt 0\) or \(\lt 0\) | |
| Trial and Improvement is 0, 3 (for 1.5 and \(-4\)) or 4 marks | |
| Incorrectly joined inequalities cannot score the final B1 eg \(1.5 \lt x \lt -4\) | B1M1A1B0 |
| Example where 1.5 and \(-4\) do not imply first B1: \(2x^2 + 5x - 12 \lt 0\), \((2x - 3)(x + 4) \lt 0\), \(x = 1.5\) and \(x = -4\), \(-4 \lt x \lt 1.5\) final B1ft is only awarded if B0M1 or B1M1A0 is scored | B0M1A1B1ft |
| \(2x^2 - 5x - 12 \lt 0\), \((2x + 3)(x - 4) \lt 0\), \(x = -1.5\) and \(x = 4\), \(-1.5 \lt x \lt 4\) joined inequality is correct for their three-term quadratic inequality | B0M1A1ftB1ft |
20 David has solved the inequality \(\quad x^2 \lt 16 \quad\) and represented his solution on the number line below.
Give one reason why David’s solution is wrong. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Correct reason indicating that the circles should be unshaded or \((x =)\ 4\) is not a solution or \((x =)\ -4\) is not a solution | B1 | eg the circles should be white or \((x =)\ 4\) doesn’t satisfy \(x^2 \lt 16\) |
| Ignore non-contradictory statements alongside a correct reason eg 4 is not a solution. Squaring a number is always \(\geqslant 0\) | B1 |
| Incorrect statement with a correct reason is choice eg Circles should be white and should be at \(-3\) and 3 | B0 |
| Condone \(-4^2\) if recovered to 16 | |
| Circles should be blank | B1 |
| Circles crossed out and replaced with open circles above \(-4\) and 4 | B1 |
| \(-4^2\) is 16 so \(-4\) shouldn’t be a solution | B1 |
| One circle should be white | B0 |
| Coloured dot means equal to | B0 |
| The circles are the wrong colour | B0 |
| Both circles are black so he is doing \(x\) squared is 16 | B0 |
| 4 squared is 16 | B0 |
| His solution is \(-4 \leqslant x \leqslant 4\) | B0 |
27 Write down the solution of \(\quad x^2 \lt 16\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(-4 \lt x \lt 4\) or \(4 \gt x \gt -4\) | B1 | condone the inequality in two parts eg \(x \gt -4\) and \(x \lt 4\) |
25
| Answer | Mark | Comments |
|---|---|---|
| \((x + 1)(x - 6)\) or \(\dfrac{5 \pm \sqrt{(-5)^2 - 4\ (\times 1) \times (-6)}}{2\ (\times 1)}\) or \(2.5 \pm \sqrt{12.25}\) or \(-1\) and 6 identified | M1 | oe do not accept missing bracket on \((-5)^2\) unless recovered |
| \(-1 \lt x \lt 6\) | A1 | condone \(-1 \lt x\) and \(x \lt 6\) |
| Answer | Mark | Comments |
|---|---|---|
| Open circles at \(-1\) and 6 joined by line | B1ft | ft their double-sided inequality in (a) if the bounds are within the number line condone ft an inequality given in two parts if the bounds are within the number line condone ft a single-sided inequality if the bound is within the number line |
17 \(m^2 \gt 9\)
Circle the possible value of \(m\). [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(-\dfrac{7}{2}\) | B1 |
24 Here is a sketch of the curve \(\quad y = x^2 + 4x - 12\)
Work out the values of \(x\) for which \(\quad x^2 + 4x - 12 < 0\)
Give your answer as an inequality. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x + 6)(x - 2)\) or \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times -12}}{2 \times 1}\) or \(-2 \pm \sqrt{16}\) | M1 | oe |
| \(-6\) and 2 | A1 | may be seen in inequalities or as intersections with \(x\)-axis on the graph must be selected if appearing in a list of values or a table |
| \(-6 < x < 2\) or \(2 > x > -6\) | A1ft | ft M1A0 and two values must be a single inequality |
| To award A1ft the values must be used to give a continuous interval eg1 \((x + 6)(x - 2)\) followed by \((x =)\ 6\) and \((x =)\ {-2}\) Answer \(-2 < x < 6\) eg2 \((x + 6)(x - 2)\) followed by \((x =)\ 6\) and \((x =)\ {-2}\) Answer \(6 < x < -2\) | M1A0A1ft M1A0A0ft |
| \(x < 2\) and \(x > -6\) | M1A1A0 |
| \(-6 < x > 2\) | M1A1A0 |
| \(-6 \leqslant x < 2\) | M1A1A0 |
| \(-6 < x < 2\) in working with different answer on answer line | M1A1A0 |
| \(-6 < x < 2\) in working with integers on answer line | M1A1A0 |
29 Here is the graph of \(\;y = \mathrm{f}(x)\;\) where \(\mathrm{f}(x)\) is a quadratic function.
Write down all the integer solutions of \(\;\mathrm{f}(x) \geqslant 0\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| −3 −2 −1 0 1 2 | B2 | B1 for 5 correct and 0 incorrect or 6 correct and 1 incorrect |
Do not accept coordinates
21 Here is a sketch of \(\quad y = \mathrm{f}(x) \quad\) where \(\mathrm{f}(x)\) is a quadratic function.
The graph intersects the \(x\)-axis where \(\quad x = -2.5 \quad\) and \(\quad x = 1\)
Not drawn accurately
Circle the solution of \(\;\mathrm{f}(x) \gt 0\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(-2.5 \lt x \lt 1\) | B1 |
13 Solve the inequality.
\(x^2 - 100 \geqslant 0\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Both inequalities \(x \geqslant 10\) , \(x \leqslant -10\) as final answer | 3 | Allow 3 marks for \(-10 \geqslant x \geqslant 10\) final answer | |
| B2 for answer one of \(x \geqslant 10\) or \(x \leqslant -10\) | B2 implied by answer e.g. \(-10 \leqslant x \geqslant 10\) or \(x \geqslant \pm 10\) as one element of inequality correct | ||
| or M1 for \(x^2 \geqslant 100\) or for \((x + 10)(x - 10)\) | M1 implied by 10 or –10 seen in answer | ||
20 Solve this inequality.
\[x^2 + 4x - 12 \leqslant 0\]Give your answer using set notation.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\{x : -6 \leqslant x \leqslant 2\}\) final answer and with correct working | 5 | “correct working” requires at least M2 e.g. \((x + 6)(x - 2)\) | |
| B4 for \(-6 \leqslant x \leqslant 2\) with correct working and not written separately | |||
| OR | |||
| M2 for \((x + 6)(x - 2)\) or \(\frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times -12}}{2}\) or M1 for brackets which give two correct terms or the formula with at most two errors B1 −6 and 2 | condone \(x(x - 2) + 6(x - 2)\) for M2 could be seen as roots on a sketch of graph or with incorrect inequality symbols completing the square : allow \((x + 2)^2 - 16\) for M2 or \((x + 2)^2 + k\) for M1 | ||
| If 0 or 1 scored award instead SC2 for \(\{x : -6 \leqslant x \leqslant 2\}\) If 0 scored SC1 for \(-6 \leqslant x \leqslant 2\) | |||
17 Solve the inequality.
\(x^2 - 5x - 6 \leqslant 0\) [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(-1 \leqslant x \leqslant 6\) oe | 4 | M2 for \((x - 6)(x + 1)\) [\(\leqslant\) 0 ] oe | For M2 or M1, condone [= 0] M2 for correct formula or complete square condone 1 error M1 for \((x - 2.5)^2\) oe seen or for correct formula with 2 errors |
| Or M1 for \((x + a)(x + b)\) [\(\leqslant\) 0 ] where \(ab = -6\) or \(a + b = -5\) | |||
| B1 for –1 and 6 soi | Could be seen as roots on sketch of graph or in wrong inequality | ||
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