Graphs & Inequalities

Topic

AQA

OCR A

OCR MEI

A2 June 2025 Paper 2 Q10

AQACurrent spec13 marksGraphs & Inequalities

10 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{2(x^2 - 4)}{x^2 + 6x + 9}\]

(a) Write down the equations of the asymptotes to the graph of \(y = \mathrm{f}(x)\) [2 marks]

(b) Without using calculus, show that the range of \(\mathrm{f}\) is \(\left\{k : k \geqslant -\dfrac{8}{5}\right\}\) [4 marks]

Without using calculus, find the coordinates of this stationary point. [3 marks]

(d) Sketch the graph of \(y = \mathrm{f}(x)\) on the axes below. [4 marks]

Blank axes: x-axis and y-axis crossing at O

Scheme	Marks	AO
Obtains \(x = -3\) or \(y = 2\)	M1	1.1a
Obtains \(x = -3\) and \(y = 2\) and no other asymptotes.	A1	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
Forms a quadratic equation in \(x\) involving \(k\) or equivalent.	M1	3.1a
Obtains a quadratic equation with \(k\) in a coefficient and sets the discriminant to zero, either in an equation or inequality.	M1	1.1a
Solves their equation or inequality in \(k\)	M1	1.1a
Completes a reasoned argument, without using calculus, to state that the range is \(\left\{k : k \geqslant -\dfrac{8}{5}\right\}\) AG Allow letters other than \(k\) used.	R1	2.1
	(4)

Typical solution

Let \(y = k\)

\[k(x^2 + 6x + 9) = 2x^2 - 8\]\[(k - 2)x^2 + 6kx + (9k + 8) = 0\]\[\Delta \geqslant 0\]\[\therefore (6k)^2 - 4(k - 2)(9k + 8) \geqslant 0\]\[36k^2 - 36k^2 + 40k + 64 \geqslant 0\]\[40k + 64 \geqslant 0\]\[k \geqslant -\frac{8}{5}\]

The range of \(\mathrm{f}\) is \(\left\{k : k \geqslant -\dfrac{8}{5}\right\}\)

Mark scheme (c)

Scheme	Marks	AO
Obtains a quadratic equation by substituting \(k = -\dfrac{8}{5}\)	M1	3.1a
Solves their quadratic equation	M1	1.1a
Deduces \(\left(-\dfrac{4}{3}, -\dfrac{8}{5}\right)\)	A1	2.2a
	(3)

Typical solution

\[k = -\frac{8}{5}\]\[\frac{-18}{5}x^2 + \frac{-48}{5}x + \frac{-32}{5} = 0\]\[9x^2 + 24x + 16 = 0\]\[(3x + 4)^2 = 0\]\[x = -\frac{4}{3}\]\[\text{Stationary point} = \left(-\frac{4}{3}, -\frac{8}{5}\right)\]

Sketch of y = f(x) with asymptotes x = −3 and y = 2; the left branch rises from y = 2 towards +infinity at x = −3; the right branch comes down from +infinity, crosses the x-axis at −2, has a minimum at (−4/3, −8/5), crosses the y-axis at −8/9 and the x-axis at 2, then approaches y = 2 from below — Mark scheme (d)

Scheme	Marks	AO
Draws approximately correct shape of the right-hand branch	B1	1.1b
Draws both branches of the graph approaching the correct asymptotes.	B1	1.1b
Shows at least two correctly labelled axis intercepts.	B1	1.1b
Draws completely correct graph including correctly labelled asymptotes, a stationary point in the third quadrant, and all three correctly labelled axis intercepts. Condone no labelling of stationary point.	B1	1.1b
	(4)
	(13 marks)

AS June 2025 Paper 1 Q9

AQACurrent spec4 marksGraphs & Inequalities

9 Solve the inequality

\[\frac{2x + 1}{3x - 9} \geqslant x - 1\]

[4 marks]

Mark scheme

Scheme	Marks	AO
Multiplies LHS and RHS by \((3x - 9)^2\) or Subtracts one side from the other and correctly combines the fractions. or Cross multiplies and solves for \(x\) PI by obtaining at least one correct region or a correct quadratic factor or both \(\frac{2}{3}\) and 4	M1	3.1a
Obtains a correct quadratic factor eg \(3x^2 - 14x + 8\) PI by obtaining \(\frac{2}{3}\) and 4 or States \(x \neq 3\) PI by \(x \gt 3\)	M1	1.1a
Obtains at least one correct region. Condone \(3 \leqslant x \leqslant 4\)	M1	1.1a
Obtains \(x \leqslant \dfrac{2}{3}\), \(3 \lt x \leqslant 4\)	A1	1.1b
	(4 marks)

Typical solution

\[(2x + 1)(3x - 9) \geqslant (x - 1)(3x - 9)^2\]

but \(x \neq 3\)

\[0 \geqslant (3x - 9)\big((x - 1)(3x - 9) - (2x + 1)\big)\]\[(3x - 9)(3x^2 - 14x + 8) \leqslant 0\]\[(3x - 9)(3x - 2)(x - 4) \leqslant 0\]

Critical values are: \(\frac{2}{3}\), 3, 4

\[x \leqslant \frac{2}{3}, \quad 3 \lt x \leqslant 4\]

A2 June 2025 Paper 1 Q8

AQACurrent spec3 marksGraphs & Inequalities

8 The curve \(C_1\) has equation

\[\frac{x^2}{4} + \frac{y^2}{25} = 1\]

The curve \(C_1\) is translated by the vector \(\begin{bmatrix} 1 \\ 3 \end{bmatrix}\) to give the curve \(C_2\)

Find the coordinates of the points where the curve \(C_2\) intersects the \(x\)-axis. [3 marks]

Mark scheme

Scheme	Marks	AO
Obtains \(\dfrac{(x - 1)^2}{4} + \dfrac{(y - 3)^2}{25} = 1\)	B1	1.2
Substitutes \(y = 0\) into their equation for \(C_2\) and solves to obtain at least one value of \(x\)	M1	1.1a
Obtains \(\left(\dfrac{13}{5}, 0\right)\) and \(\left(-\dfrac{3}{5}, 0\right)\) Must be coordinates.	A1	1.1b
	(3 marks)

Typical solution

\[C_2 : \frac{(x - 1)^2}{4} + \frac{(y - 3)^2}{25} = 1\]\[y = 0 \Rightarrow \frac{(x - 1)^2}{4} + \frac{9}{25} = 1\]\[25(x - 1)^2 + 36 = 100\]\[25(x - 1)^2 = 64\]\[(x - 1)^2 = \frac{64}{25}\]\[x = 1 \pm \frac{8}{5}\]\[\left(\frac{13}{5}, 0\right), \left(-\frac{3}{5}, 0\right)\]

A2 June 2025 Paper 2 Q8

AQACurrent spec5 marksGraphs & Inequalities

8 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = x^3 + x^2 - 12x \qquad (x \in \mathbb{R})\]

Figure 1 shows the graph of \(y = \mathrm{f}(x)\)

Figure 1: graph of the cubic y = f(x), with a local maximum to the left of the y-axis, passing through O, with a local minimum to the right of the y-axis, and crossing the x-axis once on each side of O — Figure 1

(a) The graph of \(y = \mathrm{f}(x)\) is transformed by a stretch, scale factor 2, parallel to the \(x\)-axis with the \(y\)-axis fixed, to give the graph of \(y = \mathrm{g}(x)\)

On Figure 1, sketch the graph of \(y = \mathrm{g}(x)\), showing the values of \(x\) where the graph crosses the \(x\)-axis. [3 marks]

(b) Find the set of values of \(x\) such that the conditions \(\mathrm{f}(x) \gt 0\) and \(\mathrm{g}(x) \lt 0\) are both satisfied. [2 marks]

Sketch showing the original cubic and the stretched cubic y = g(x), which passes through O and crosses the x-axis at −8 and 6, with turning points at about the same heights as the original — Mark scheme (a)

Scheme	Marks	AO
Draws a cubic graph of the correct shape. Condone any horizontal stretch.	B1	1.1b
Draws a cubic graph that crosses \(x\)-axis at \(-8\), 0 and 6 Accept no indication of \(x = 0\) at the origin.	B1	1.1b
Shows local maximum and minimum points which are approximately the same height as in the original graph.	B1	1.1b
	(3)

Mark scheme (b)

Scheme	Marks	AO
Obtains a set of values of the form \(\{x : a \lt x \lt 2a\}\) or \(\{x : 2a \lt x \lt a\}\) Condone non-strict inequality. Condone set notation not used.	M1	2.2a
Obtains \(3 \lt x \lt 6\) Condone set notation not used.	A1	2.2a
	(2)
	(5 marks)

Typical solution

\[\{x : 3 \lt x \lt 6\}\]

AS June 2025 Paper 1 Q3

AQACurrent spec1 markGraphs & Inequalities

3 Find the equations of the asymptotes of the curve with equation

\[y = \frac{(x - 1)(x + 2)}{(x + 1)(x - 2)}\]

Tick (✓) one box. [1 mark]

\(x = 1,\ x = -2,\ y = 1\)
\(x = 1,\ x = -2,\ y = -1\)
\(x = -1,\ x = 2,\ y = 1\)
\(x = -1,\ x = 2,\ y = -1\)

Scheme	Marks	AO
Ticks the 3rd box.	B1	2.2a
	(1 mark)

A2 June 2024 Paper 2 Q16

AQACurrent spec9 marksGraphs & Inequalities Integration

16 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{ax + 5}{x + b}\]

where \(a\) and \(b\) are constants.

The graph of \(y = \mathrm{f}(x)\) has asymptotes \(x = -2\) and \(y = 3\)

(a) Write down the value of \(a\) and the value of \(b\) [2 marks]

(b) The diagram shows the graph of \(y = \mathrm{f}(x)\) and its asymptotes.

The shaded region \(R\) is enclosed by the graph of \(y = \mathrm{f}(x)\), the \(x\)-axis and the \(y\)-axis.

Graph of y = f(x) with dashed asymptotes x = −2 and y = 3; the left branch lies above y = 3, and the right branch rises from below the x-axis near x = −2, crosses the x-axis between −2 and O and approaches y = 3; the region R between this branch, the x-axis and the y-axis is shaded

(i) The shaded region \(R\) is rotated through 360° about the \(x\)-axis to form a solid.

Find the volume of this solid.

Give your answer to three significant figures. [3 marks]

(ii) The shaded region \(R\) is rotated through 360° about the \(y\)-axis to form a solid.

Find the volume of this solid.

Give your answer to three significant figures. [4 marks]

Scheme	Marks	AO
Deduces that \(a = 3\)	B1	2.2a
Deduces that \(b = 2\)	B1	2.2a
	(2)

Mark scheme (b)

Scheme	Marks	AO
(i) Deduces that \(x\)-intercept is \(-\dfrac{5}{3}\). PI by correct answer.	B1	2.2a
(i) Uses \(\pi\displaystyle\int y^2\,\mathrm{d}x\) Condone missing \(\mathrm{d}x\) and missing/incorrect limits. PI by correct answer.	M1	1.1a
(i) Obtains AWRT 21.2	A1	1.1b
	(3)
(ii) Deduces that \(y\)-intercept \(= 2.5\) PI correct answer.	B1	2.2a
(ii) Deduces an expression for \(x\) in terms of \(y\) PI correct answer.	M1	2.2a
(ii) Uses \(\pi\displaystyle\int x^2\,\mathrm{d}y\) Condone missing \(\mathrm{d}y\), use of \(\mathrm{d}x\) and missing/incorrect limits. PI by correct answer.	M1	1.1a
(ii) Obtains AWRT 14.1	A1	1.1b
	(4)
	(9 marks)

Typical solution

(i)

\[\mathrm{f}(x) = \frac{3x + 5}{x + 2}\]\[x\text{-intercept} = -\frac{5}{3}\]\[V = \pi\int_{-\frac{5}{3}}^{0} \left(\frac{3x + 5}{x + 2}\right)^2 \mathrm{d}x\]\[= 21.2040\ldots\]\[= 21.2\]

(ii)

\[y\text{-intercept} = 2.5\]\[y = \frac{3x + 5}{x + 2}\]\[yx + 2y = 3x + 5\]\[x(y - 3) = 5 - 2y\]\[x = \frac{5 - 2y}{y - 3}\]\[V = \pi\int_{0}^{2.5} \left(\frac{5 - 2y}{y - 3}\right)^2 \mathrm{d}y\]\[= 14.1360\ldots\]\[= 14.1\]

A2 June 2024 Paper 2 Q15

AQACurrent spec7 marksGraphs & Inequalities

15 The diagram shows the line \(y = 5 - x\)

Axes x and y through O with the straight line y = 5 − x, crossing the positive y-axis and the positive x-axis

(a) On the diagram above, sketch the graph of \(y = \left|x^2 - 4x\right|\), including all parts of the graph where it intersects the line \(y = 5 - x\)

(You do not need to show the coordinates of the points of intersection.) [3 marks]

(b) Find the solution of the inequality\[\left|x^2 - 4x\right| \gt 5 - x\]

Give your answer in an exact form. [4 marks]

Sketch of y = |x² − 4x|: touching the x-axis at O and at 4, with a hump between them, rising steeply outside them, and crossing the line y = 5 − x at four points — Mark scheme (a)

Scheme	Marks	AO
Draws curve with basically correct shape and no negative \(y\)-values.	B1	1.1b
Their graph intersects line at four distinct points.	B1	1.1b
Value of 4 shown at \(x\)-intercept	B1	1.1b
	(3)

Mark scheme (b)

Scheme	Marks	AO
Uses the modulus function to obtain two separate quadratic equations.	M1	3.1a
Obtains four correct \(x\)-values of points of intersection (condone decimal approximations).	A1	1.1a
Uses their graph to obtain at least one subset of the solution set (condone decimal approximations).	M1	2.2a
Deduces a completely correct solution set with exact values.	A1	2.2a
	(4)
	(7 marks)

Typical solution

\[x^2 - 4x = 5 - x\]\[x^2 - 3x - 5 = 0\]\[x = \frac{3 \pm \sqrt{29}}{2}\]\[4x - x^2 = 5 - x\]\[0 = x^2 - 5x + 5\]\[x = \frac{5 \pm \sqrt{5}}{2}\]\[x \lt \frac{3 - \sqrt{29}}{2},\ \frac{5 - \sqrt{5}}{2} \lt x \lt \frac{5 + \sqrt{5}}{2},\ x \gt \frac{3 + \sqrt{29}}{2}\]

AS June 2024 Paper 1 Q10

AQACurrent spec6 marksGraphs & Inequalities

10 The curve \(C\) has equation

\[y = \frac{2x - 10}{3x - 5}\]

Figure 1 shows the curve \(C\) with its asymptotes.

Figure 1: the curve C with a vertical asymptote just right of the y-axis and a horizontal asymptote just above the x-axis; the left branch crosses the y-axis at 2 and rises steeply, the right branch rises from below and crosses the x-axis at 5 — Figure 1

(a) Write down the equations of the asymptotes of \(C\) [2 marks]

(b) The line \(L\) has equation\[y = -\frac{2}{5}x + 2\]

(i) Draw the line \(L\) on Figure 1 [2 marks]

(ii) Hence, or otherwise, solve the inequality\[\frac{2x - 10}{3x - 5} \leqslant -\frac{2}{5}x + 2\] [2 marks]

Mark scheme (a)

Scheme	Marks	AO
States \(x = \dfrac{5}{3}\) or \(y = \dfrac{2}{3}\)	M1	2.2a
States \(x = \dfrac{5}{3}\) and \(y = \dfrac{2}{3}\) and no incorrect equations seen.	A1	2.2a
	(2)

Typical solution

\[x = \frac{5}{3}\]\[y = \frac{2}{3}\]

Figure 1 with the line y = −(2/5)x + 2 drawn through (0, 2) and (5, 0), crossing both branches of the curve — Mark scheme (b)

Scheme	Marks	AO
(i) Draws a straight line with negative gradient passing through \((0, 2)\) Accept freehand if the intention is clear.	M1	1.1a
Draws a straight line passing through \((0, 2)\) and \((5, 0)\)	A1	1.1b
	(2)
(ii) Deduces one of the ranges \(x \leqslant 0\) or \(\dfrac{5}{3} \lt x \leqslant 5\) Condone \(\dfrac{5}{3} \leqslant x \leqslant 5\) or \(\dfrac{5}{3} \lt x \lt 5\) for this mark only. Ignore any incorrect ranges.	M1	2.2a
Deduces the solution \(x \leqslant 0\), \(\dfrac{5}{3} \lt x \leqslant 5\)	A1	2.2a
	(2)
	(6 marks)

A2 June 2024 Paper 1 Q8

AQACurrent spec4 marksGraphs & Inequalities

8 The ellipse \(E\) has equation

\[x^2 + \frac{y^2}{9} = 1\]

The line with equation \(y = mx + 4\) is a tangent to \(E\)

Without using differentiation show that \(m = \pm\sqrt{7}\) [4 marks]

Mark scheme

Scheme	Marks	AO
Forms an equation by eliminating \(y\) or \(x\)	M1	1.1a
Obtains a correct three-term quadratic equation. PI correct discriminant	A1	1.1b
Forms a quadratic equation in \(m\) or \(m^2\) using \(b^2 - 4ac = 0\)	M1	2.2a
Completes fully correct working to obtain \(m = \pm\sqrt{7}\) AG	R1	2.1
	(4 marks)

Typical solution

\[x^2 + \frac{(mx + 4)^2}{9} = 1\]\[9x^2 + m^2x^2 + 8mx + 16 = 9\]\[(9 + m^2)x^2 + 8mx + 7 = 0\]

For a tangent, \(b^2 - 4ac = 0\)

\[64m^2 - 28(9 + m^2) = 0\]\[36m^2 = 252\]\[m^2 = 7\]\[m = \pm\sqrt{7}\]

A2 June 2023 Paper 1 Q14

AQACurrent spec10 marksGraphs & Inequalities Polar Coordinates

14 The curve \(C\) has polar equation

\[r = \frac{4}{5 + 3\cos\theta} \qquad (-\pi \lt \theta \leqslant \pi)\]

(a) Show that \(r\) takes values in the range \(\dfrac{1}{k} \leqslant r \leqslant k\), where \(k\) is an integer. [2 marks]

(b) Find the Cartesian equation of \(C\) in the form \(y^2 = \mathrm{f}(x)\) [4 marks]

Find the transformation that maps the graph of \(E\) onto \(C\) [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Uses the range of \(\cos\theta\) to obtain max and min values of r or obtains \(k = 2\)	M1	2.4
Completes a reasoned argument to obtain the correct inequality	R1	2.1
	(2)

Typical solution

\(r\) is minimum when \(\cos\theta = 1\), and maximum when \(\cos\theta = -1\)

\[\text{So } \frac{1}{2} \leqslant r \leqslant 2\]

Mark scheme (b)

Scheme	Marks	AO
Uses \(x = r\cos\theta\)	M1	3.1a
Uses \(r^2 = x^2 + y^2\) or \(r = \sqrt{x^2 + y^2}\)	M1	3.1a
Rearranges to make \(ky^2\) the subject, having correctly removed any square root in their equation. Equation must be in terms of \(x\) and \(y\) only.	M1	1.1a
Obtains a correct result in required form ISW	A1	1.1b
	(4)

Typical solution

\[\begin{aligned} 5r + 3r\cos\theta &= 4 \\ 5r &= 4 - 3r\cos\theta \\ 5\sqrt{x^2 + y^2} &= 4 - 3x \\ 25(x^2 + y^2) &= (4 - 3x)^2 \\ 25y^2 &= 16 - 24x - 16x^2 \\ y^2 &= \tfrac{1}{25}(16 - 24x - 16x^2) \end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Completes the square to bring together all the \(x\) terms	M1	3.1a
Rearranges equation to obtain a form that enables them to deduce a transformation	M1	1.1a
Deduces that a horizontal translation is required. PI a vector \(\begin{bmatrix} p \\ 0 \end{bmatrix}\)	M1	2.2a
Deduces a translation by \(\begin{bmatrix} -\frac{3}{4} \\ 0 \end{bmatrix}\) as the only transformation	A1	2.2a
	(4)
	(10 marks)

Typical solution

\[\begin{aligned} y^2 &= -\frac{16}{25}\left(x^2 + \frac{3}{2}x - 1\right) \\ &= -\frac{16}{25}\left\{\left(x + \frac{3}{4}\right)^2 - \frac{25}{16}\right\} \\ &= -\frac{16}{25}\left(x + \frac{3}{4}\right)^2 + 1 \end{aligned}\]\[y^2 + \frac{16}{25}\left(x + \frac{3}{4}\right)^2 = 1\]

Starting from \(y^2 + \dfrac{16x^2}{25} = 1\):

The required transformation is a translation by \(\begin{bmatrix} -\frac{3}{4} \\ 0 \end{bmatrix}\)

A2 June 2023 Paper 2 Q14

AQACurrent spec10 marksGraphs & Inequalities Integration

14 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{1}{4x^2 + 16x + 19} \qquad (x \in \mathbb{R})\]

(a) Show, without using calculus, that the graph of \(y = \mathrm{f}(x)\) has a stationary point at \(\left(-2, \dfrac{1}{3}\right)\) [3 marks]

(b) Show that \(\displaystyle\int_{-2}^{-\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x = \frac{\pi\sqrt{3}}{18}\) [5 marks]

Fully justify your answer. [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes the square for denominator. Or Sets \(\mathrm{f}(x) = k\) and forms a quadratic equation in \(x\)	M1	3.1a
Explains that \(\mathrm{f}\) has a stationary point when \(x = -2\) Or Equates the discriminant of the quadratic equation to 0 and solves for \(k\)	E1	2.4
Completes a reasoned argument, without using calculus, to show that the stationary point is at \(\left(-2, \dfrac{1}{3}\right)\) to obtain the required result.	R1	2.1
	(3)

Typical solution

\[\mathrm{f}(x) = \frac{1}{4\left(x^2 + 4x + \frac{19}{4}\right)} = \frac{1}{4\left((x + 2)^2 + \frac{3}{4}\right)}\]

\(\mathrm{f}\) is maximum when the denominator is minimum, that is when \(x = -2\)

\[\text{and } y = \frac{1}{4\left(\frac{3}{4}\right)} = \frac{1}{3}\]

So the graph of \(y = \mathrm{f}(x)\) has a stationary point at \(\left(-2, \dfrac{1}{3}\right)\)

Mark scheme (b)

Scheme	Marks	AO
Expresses the denominator of the integrand in completed square form.	M1	3.1a
Uses inverse tan to integrate their integrand of the form \(\dfrac{1}{(x + k)^2 + a^2}\) Or Makes a correct substitution	M1	3.1a
Integrates to obtain \(A\tan^{-1}\dfrac{2(x + 2)}{\sqrt{3}}\)	A1	1.1b
Substitutes the upper limit correctly into their integrated expression which includes \(\tan^{-1}\)	M1	1.1a
Completes a reasoned argument to obtain the required result. Must see substitution of \(-2\) in the integrated expression.	R1	2.1
	(5)

Typical solution

\[\int_{-2}^{-\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x = \frac{1}{4}\int_{-2}^{-\frac{1}{2}} \frac{1}{(x + 2)^2 + \frac{3}{4}}\,\mathrm{d}x\]\[\begin{aligned} &= \frac{1}{4} \times \frac{2}{\sqrt{3}}\left[\tan^{-1}\left(\frac{2(x + 2)}{\sqrt{3}}\right)\right]_{-2}^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{3}}\left(\tan^{-1}\sqrt{3} - \tan^{-1}0\right) \\ &= \frac{1}{2\sqrt{3}}\left(\frac{\pi}{3} - 0\right) = \frac{\pi}{6\sqrt{3}} = \frac{\pi\sqrt{3}}{18} \end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Replaces \(\infty\) by a letter (\(N\)) and considers \(\displaystyle\lim_{N \to \infty}\) in the integral or integrated expression.	E1	3.1a
Obtains the correct exact value. ACF	B1	2.2a
	(2)
	(10 marks)

Typical solution

\[\int_{-2}^{\infty} \mathrm{f}(x)\,\mathrm{d}x = \frac{1}{2\sqrt{3}}\lim_{N \to \infty}\left[\tan^{-1}\left(\frac{2(x + 2)}{\sqrt{3}}\right)\right]_{-2}^{N}\]\[= \frac{1}{2\sqrt{3}}\left(\frac{\pi}{2} - 0\right)\]\[= \frac{\pi\sqrt{3}}{12}\]

AS June 2023 Paper 1 Q14

AQACurrent spec4 marksGraphs & Inequalities Using Roots of Polynomials

14 The inequality

\[(x^2 - 5x - 24)(x^2 + 7x + a) \lt 0\]

has the solution set

\[\{x : -9 \lt x \lt -3\} \cup \{x : 2 \lt x \lt b\}\]

Find the values of integers \(a\) and \(b\) [4 marks]

Mark scheme

Scheme	Marks	AO
Factorises \(x^2 - 5x - 24\) to \((x + m)(x + n)\) where \(m + n = -5\) or \(mn = -24\) and identifies their \(m, n \gt 2\) as \(b\) Or uses the coefficients of a quartic to form an equation in \(a\) and/or \(b\) eg \(-(-5 + 7) = -9 + (-3) + 2 + b\) eg \(-24a = -9 \times -3 \times 2 \times b\) Or multiplies two or more of the factors \((x + 9)\), \((x + 3)\), \((x - 2)\) and \((x - b)\)	M1	3.1a
Obtains \(a = -18\) or \(b = 8\)	A1	1.1b
Expands \((x + 9)(x - 2)\) and identifies the constant term as \(a\) Or correctly forms two equations in \(a\) and \(b\) Or divides the expanded quartic by a quadratic or a cubic formed by multiplying two or three of \((x + 9)\), \((x + 3)\), \((x - 2)\) and \((x - b)\) Or compares coefficients in the expansions of \((x^2 - 5x - 24)(x^2 + 7x + a)\) and \((x + 9)(x + 3)(x - 2)(x - b)\)	M1	3.1a
Obtains \(a = -18\) and \(b = 8\)	A1	1.1b
	(4 marks)

Typical solution

\[x^2 - 5x - 24 = (x - 8)(x + 3)\]

\(\therefore\) the critical values include \(-3\) and 8

\[\therefore \ b = 8\]\[(x + 9)(x - 2) = (x^2 + 7x - 18)\]\[\therefore \ a = -18\]

A2 June 2023 Paper 1 Q11

AQACurrent spec7 marksFinding Roots of Polynomials Graphs & Inequalities

11 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = 4x^3 - 8x^2 - 51x - 45 \qquad (x \in \mathbb{R})\]

(a)

(i) Fully factorise \(\mathrm{f}(x)\) [2 marks]

(ii) Hence, solve the inequality \(\mathrm{f}(x) \lt 0\) [2 marks]

(b) The graph of \(y = \mathrm{f}(x)\) is translated by the vector \(\begin{bmatrix} 7 \\ 0 \end{bmatrix}\)

The new graph is then reflected in the \(x\)-axis, to give the graph of \(y = \mathrm{g}(x)\)

Solve the inequality \(\mathrm{g}(x) \leqslant 0\) [3 marks]

Sketch of the cubic y = f(x), touching the x-axis to the left of the y-axis and crossing it to the right — Mark scheme (a)

Scheme	Marks	AO
(i) Obtains one factor of \(\mathrm{f}(x)\)	M1	1.1a
(i) Deduces the correct factorisation	A1	2.2a
	(2)
(ii) Deduces that \(x \lt 5\)	M1	2.2a
(ii) Obtains a completely correct solution to the inequality	A1	1.1b
	(2)

Sketch of y = g(x), coming down to touch the x-axis, rising to a maximum, then falling to cross the x-axis — Mark scheme (b)

Scheme	Marks	AO
Obtains \(\pm 5.5\) or \(\pm 12\) ft 7 + their critical values from (a)(ii)	M1	2.2a
Obtains 5.5 and 12 ft 7 + their critical values from (a)(ii)	A1F	2.2a
Obtains a completely correct solution to the inequality	A1	1.1b
	(3)
	(7 marks)

AS June 2023 Paper 1 Q10

AQACurrent spec9 marksGraphs & Inequalities

10 The curve \(C\) has equation

\[y = \frac{3x^2 + mx + p}{x^2 + px + m}\]

where \(m\) and \(p\) are integers.

The vertical asymptotes of \(C\) are \(x = -4\) and \(x = -1\)

The curve \(C\) is shown in the diagram below.

Graph of C with two vertical asymptotes left of the y-axis and a horizontal asymptote above the x-axis; the left branch rises towards the left asymptote, the middle branch is a downward arch below the x-axis, and the right branch comes down from the right-hand asymptote to a minimum just above the x-axis and then rises slowly

(a) Write down the equation of the horizontal asymptote of \(C\) [1 mark]

(b) Find the value of \(m\) and the value of \(p\) [2 marks]

(d) Without using calculus, show that the line \(y = -1\) does not intersect \(C\) [5 marks]

Scheme	Marks	AO
States \(y = 3\)	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
Identifies the correct factors of the denominator PI	M1	1.1a
Obtains the correct values.	A1	1.1b
	(2)

Typical solution

The denominator is \((x + 4)(x + 1)\)

\[= x^2 + 5x + 4\]\[\therefore \ m = 4 \text{ and } p = 5\]

Mark scheme (c)

Scheme	Marks	AO
Obtains the correct \(y\)-coordinate of the intercept. Follow through their \(\dfrac{p}{m}\)	B1F	1.1b
	(1)

Typical solution

When \(x = 0\), then \(y = \dfrac{p}{m} = \dfrac{5}{4}\)

\(\therefore\) \(y\)-intercept is \(\left(0, \dfrac{5}{4}\right)\)

Mark scheme (d)

Scheme	Marks	AO
Forms an equation to find the intersection point(s) if they exist. Could equate to a letter, eg \(k\) instead of \(-1\)	M1	1.1a
Rearranges into a three-term quadratic equation. Allow one arithmetic error. Could be in terms of \(k\)	M1	1.1a
Obtains a correct quadratic equation. Could be in terms of \(k\) \((k - 3)x^2 + (5k - 4)x + 4k - 5 = 0\)	A1	1.1b
Uses a correct method to deduce that their quadratic equation has no real roots. or Considers the sign of the discriminant in terms of \(k\) \(\Delta = 9k^2 + 28k - 44\)	M1	1.1a
Completes a reasoned argument to conclude that the line \(y = -1\) does not intersect \(C\)	R1	2.1
	(5)
	(9 marks)

Typical solution

\(y = -1\) intersects \(C\) when

\[\frac{3x^2 + 4x + 5}{x^2 + 5x + 4} = -1\]\[\Rightarrow 3x^2 + 4x + 5 = -(x^2 + 5x + 4)\]\[\Rightarrow 4x^2 + 9x + 9 = 0\]\[b^2 - 4ac = 9^2 - 4 \times 4 \times 9 = -63 \lt 0\]

\(\therefore\) there are no real roots

\(\therefore\) \(y = -1\) does not intersect \(C\)

A2 June 2023 Paper 1 Q7

AQACurrent spec5 marksGraphs & Inequalities

7 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \left|\sin x + \frac{1}{2}\right| \qquad (0 \leqslant x \leqslant 2\pi)\]

Find the set of values of \(x\) for which

\[\mathrm{f}(x) \geqslant \frac{1}{2}\]

Give your answer in set notation. [5 marks]

Sketch of y = |sin x + 1/2| for 0 ≤ x ≤ 2π with the line y = 1/2; the curve touches the x-axis twice between π and 2π — Mark scheme

Scheme	Marks	AO
Sketches graph of \(y = \sin x + \tfrac{1}{2}\) PI by graph of \(y = \left\|\sin x + \tfrac{1}{2}\right\|\) or considers one equation or inequality without modulus sign eg \(\sin x + \tfrac{1}{2} = \tfrac{1}{2}\) or \(\left(\sin x + \tfrac{1}{2}\right)^2 = \tfrac{1}{4}\)	M1	3.1a
Obtains the set of values \(0 \leqslant x \leqslant \pi\) Condone \(0 \lt x \lt \pi\)	A1	2.2a
Obtains a graph of \(y = \left\|\sin x + \tfrac{1}{2}\right\|\) with the correct shape or Obtains the other equation or inequality without modulus sign eg \(\sin x + \tfrac{1}{2} = -\tfrac{1}{2}\) or Obtains two critical values from a quadratic in \(\sin x\)	M1	1.1a
Obtains \(3\pi/2\)	A1	1.1b
Obtains a completely correct answer, and expresses it using set notation. eg \([0, \pi] \cup \left\{\dfrac{3\pi}{2}, 2\pi\right\}\) \(\{x : 0 \leqslant x \leqslant \pi\} \cup \left\{x : x = \dfrac{3\pi}{2}\right\} \cup \{x : x = 2\pi\}\) Condone \(\left\{x : 0 \leqslant x \leqslant \pi, \dfrac{3\pi}{2}, 2\pi\right\}\)	A1	2.5
	(5 marks)

A2 June 2023 Paper 2 Q5

AQACurrent spec5 marksGraphs & Inequalities Matrices

5 Josh and Zoe are solving the following mathematics problem:

The curve \(C_1\) has equation

\[\frac{x^2}{16} - \frac{y^2}{9} = 1\]

The matrix \(\mathbf{M} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) maps \(C_1\) onto \(C_2\)

Find the equations of the asymptotes of \(C_2\)

Josh says that to solve this problem you must first carry out the transformation on \(C_1\) to find \(C_2\), and then find the asymptotes of \(C_2\)

Zoe says that you will get the same answer if you first find the asymptotes of \(C_1\), and then carry out the transformation on these asymptotes to obtain the asymptotes of \(C_2\)

Show that Zoe is correct. [5 marks]

Mark scheme

Scheme	Marks	AO
States the correct asymptotes of \(C_1\)	B1	1.1b
States the correct equation of \(C_2\)	B1	3.1a
States the correct asymptotes of \(C_2\)	B1	1.1b
Obtains the asymptotes of \(C_2\) by both methods.	M1	3.1a
Shows that both methods lead to the same answer and concludes that Zoe is correct.	R1	2.3
	(5 marks)

Typical solution

Josh’s method

Reflection in \(y = x\)

\[C_2 \text{ is } \frac{y^2}{16} - \frac{x^2}{9} = 1\]

The asymptotes of \(C_2\) are \(y = \pm\dfrac{4}{3}x\)

Zoe’s method

The asymptotes of \(C_1\) are \(y = \pm\dfrac{3}{4}x\)

The transformation is a reflection in \(y = x\)

The asymptotes of \(C_2\) are \(y = \pm\dfrac{4}{3}x\)

Both answers are the same, so Zoe is correct.

AS June 2022 Paper 1 Q14

AQACurrent spec15 marksGraphs & Inequalities

14 The function f is defined by

\[\mathrm{f}(x) = \frac{x^2 - 3}{x^2 + px + 7} \qquad x \in \mathbb{R}\]

where \(p\) is a constant.

The graph of \(y = \mathrm{f}(x)\) has only one asymptote.

(a) Write down the equation of the asymptote. [1 mark]

(b) Find the set of possible values of \(p\) [4 marks]

(d) A curve \(C\) has equation\[y = \frac{x^2 - 3}{x^2 - 3x + 7}\]

The curve \(C\) has a local minimum at the point \(M\) as shown in the diagram.

Curve C: coming down from the left, crossing the x-axis left of O, dipping to a local minimum M just below the x-axis to the right of O, then crossing the x-axis and rising, levelling off to the right

The line \(y = k\) intersects curve \(C\)

(i) Show that\[19k^2 - 16k - 12 \leqslant 0\] [5 marks]

(ii) Hence, find the \(y\)-coordinate of point \(M\) [2 marks]

Scheme	Marks	AO
Writes the correct equation.	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
Indicates that the denominator cannot be equal to zero. PI by use of the discriminant of the denominator.	M1	3.1a
Obtains a relevant inequality for \(p\) eg use of the discriminant of the denominator.	M1	1.1a
Obtains a correct inequality for \(p\)	A1	1.1b
Obtains the correct set of values for \(p\) Accept \(-\sqrt{28} \lt p \lt \sqrt{28}\) Condone \(-5.29 \lt p \lt 5.29\) or better.	A1	3.2a
	(4)

Typical solution

\[x^2 + px + 7 \neq 0\]\[\therefore \ p^2 - 4 \times 1 \times 7 \lt 0\]\[p^2 \lt 28\]\[-2\sqrt{7} \lt p \lt 2\sqrt{7}\]

Mark scheme (c)

Scheme	Marks	AO
Obtains the correct \(y\)-intercept.	B1	1.1b
Solves \(x^2 - 3 = 0\)	M1	1.1a
Obtains the correct coordinates for all three intercepts. Must be written as coordinates.	A1	1.1b
	(3)

Typical solution

\[x = 0 \ \Rightarrow \ y = \frac{0 - 3}{0 + 0 + 7} = -\frac{3}{7}\]\[y = 0 \ \Rightarrow \ x^2 - 3 = 0\]\[\Rightarrow x = \pm\sqrt{3}\]\[\left(0, -\frac{3}{7}\right), \left(\sqrt{3}, 0\right), \left(-\sqrt{3}, 0\right)\]

Mark scheme (d)

Scheme	Marks	AO
(i) Multiplies by the denominator and forms a quadratic equation in \(x\)	M1	1.1a
Obtains a correct quadratic equation in \(x\) in the form \(ax^2 + bx + c = 0\) PI by a correct discriminant.	A1	1.1b
Selects a method to demonstrate the required inequality. Substitutes \(k\) for \(y\) and uses the discriminant to form an inequality in \(k\)	M1	3.1a
Obtains a correct quadratic inequality in \(k\)	A1	1.1b
Completes a rigorous proof to show that \(19k^2 - 16k - 12 \leqslant 0\)	R1	2.1
	(5)
(ii) Selects a method to find the \(y\)-coordinate of the minimum point. Obtains at least one correct root of the given quadratic. PI by \(-0.48\) or \(1.32\) or better	M1	3.1a
Obtains the correct \(y\)-coordinate.	A1	1.1b
	(2)
	(15 marks)

Typical solution

(i)

\[k = \frac{x^2 - 3}{x^2 - 3x + 7}\]\[k(x^2 - 3x + 7) = x^2 - 3\]\[(k - 1)x^2 - 3kx + 7k + 3 = 0\]

At least one solution, so \(b^2 - 4ac \geqslant 0\)

\[(-3k)^2 - 4(k - 1)(7k + 3) \geqslant 0\]\[9k^2 - 4(7k^2 - 4k - 3) \geqslant 0\]\[-19k^2 + 16k + 12 \geqslant 0\]\[19k^2 - 16k - 12 \leqslant 0\]

(ii)

\[k = \frac{8 \pm 2\sqrt{73}}{19}\]\[y = \frac{8 - 2\sqrt{73}}{19}\]

AS June 2022 Paper 1 Q13

AQACurrent spec10 marksGraphs & Inequalities

13 A curve \(C_1\) has equation

\[y = \frac{2x + 7}{3x + 5}\]

(a) Write down the equations of the asymptotes of curve \(C_1\) [2 marks]

(b) On the axes below, sketch the graph of curve \(C_1\)

Indicate the values of the intercepts of the curve with the axes. [3 marks]

(d) Curve \(C_2\) is a reflection of curve \(C_1\) in the line \(y = -x\)

Find an equation for curve \(C_2\) in the form \(y = \mathrm{f}(x)\) [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains at least one correct asymptote.	M1	1.1a
Obtains two correct asymptotes and no incorrect asymptotes.	A1	1.1b
	(2)

Typical solution

\[y = \frac{2}{3} \quad \text{and} \quad x = -\frac{5}{3}\]

Sketch of y = (2x + 7)/(3x + 5) with dashed asymptotes x = −5/3 and y = 2/3; the right branch crosses the y-axis at 7/5 and the left branch crosses the x-axis at −7/2 — Mark scheme (b)

Scheme	Marks	AO
Sketches a curve asymptotic to either a horizontal asymptote or a vertical asymptote. Only one branch required for this mark.	B1	1.1b
Sketches a curve with two branches asymptotic to their horizontal asymptote and vertical asymptote.	M1	1.1a
Deduces the shape of the curve and sketches it correctly with the correct axis intercepts. Must draw the asymptotes – condone solid lines.	A1	2.2a
	(3)

Mark scheme (c)

Scheme	Marks	AO
Identifies both critical values and no others. Or identifies one correct set of \(x\)-values. FT their asymptote or \(x\)-intercept.	M1	3.1a
Obtains the correct set of \(x\)-values.	A1	3.2a
	(2)

Typical solution

\[x \leqslant -\frac{7}{2}, \quad x \gt -\frac{5}{3}\]

Mark scheme (d)

Scheme	Marks	AO
Selects a method to find the equation of the reflected curve. Swaps \(-x\) for \(y\) and \(-y\) for \(x\) or any two of: \(x = -\dfrac{2}{3}\) \(\quad y = \dfrac{5}{3}\) \(x\)-intercept \(= -1.4\) \(\quad y\)-intercept \(= 3.5\) Follow through their asymptotes and/or intercepts from parts (a) and (b)	M1	3.1a
Multiplies an equation of the form \(\pm x = \dfrac{\pm 2y + 7}{\pm 3y + 5}\) throughout by the denominator and attempts to isolate \(y\) or any three of: \(x = -\dfrac{2}{3}\) \(\quad y = \dfrac{5}{3}\) \(x\)-intercept \(= -1.4\) \(\quad y\)-intercept \(= 3.5\) Follow through their asymptotes and/or intercepts from parts (a) and (b)	M1	1.1a
Obtains a correct equation in the correct format.	A1	3.2a
	(3)
	(10 marks)

Typical solution

\[-x = \frac{2(-y) + 7}{3(-y) + 5}\]\[-x(-3y + 5) = -2y + 7\]\[3xy - 5x = -2y + 7\]\[3xy + 2y = 5x + 7\]\[y(3x + 2) = 5x + 7\]\[y = \frac{5x + 7}{3x + 2}\]

A2 June 2022 Paper 2 Q10

AQACurrent spec7 marksGraphs & Inequalities

10 The curve \(C_1\) has equation

\[\frac{x^2}{25} - \frac{y^2}{4} = 1\]

The curve \(C_2\) has equation

\[x^2 - 25y^2 - 6x - 200y - 416 = 0\]

(a) Find a sequence of transformations that maps the graph of \(C_1\) onto the graph of \(C_2\) [4 marks]

(b) Find the equations of the asymptotes to \(C_2\)

Give your answers in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers. [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Deduces that a stretch, only along the \(y\)-axis, is required	M1	2.2a
Deduces that a translation with two non-zero components is needed (or a translation parallel to the \(x\)-axis and a translation parallel to the \(y\)-axis)	M1	2.2a
Obtains correct stretch and/or correct translation (accept \(\begin{bmatrix}3 \\ -8\end{bmatrix}\))	A1	1.1b
States correct order of completely correct transformations (accept any correct sequence)	E1	2.4
	(4)

Typical solution

\[\frac{x^2}{25} - \frac{y^2}{4} = 1\]

Stretch s.f. ½ along \(y\)-axis gives

\[\frac{x^2}{25} - \frac{(2y)^2}{4} = 1\]\[\frac{x^2}{25} - y^2 = 1\]

Translation by vector \(\begin{bmatrix}3 \\ -4\end{bmatrix}\) gives

\[\frac{(x - 3)^2}{25} - (y + 4)^2 = 1\]

which is equivalent to the equation for \(C_2\)

Stretch, scale factor ½, along \(y\)-axis followed by translation by vector \(\begin{bmatrix}3 \\ -4\end{bmatrix}\)

Mark scheme (b)

Scheme	Marks	AO
Obtains the correct asymptotes of \(C_1\) (PI) or uses a correct method to obtain both asymptotes of \(C_2\) directly PI by an equation of the form \(5y \pm x + k = 0\)	M1	1.1a
Correctly applies their sequence of at least two different types of transformations to one asymptote or obtains an equation of the form \(5y \pm x + k = 0\) where \(k\) is a non-zero integer constant	M1	3.1a
Obtains both correct results	A1	1.1b
	(3)
	(7 marks)

Typical solution

Asymptotes of \(C_1\): \(y = \pm\dfrac{2x}{5}\)

\[y = \frac{2x}{5}\]

Stretch \(\Rightarrow 2y = \dfrac{2x}{5} \Rightarrow 5y = x\)

Translation \(\Rightarrow 5(y + 4) = x - 3\)

\[5y - x + 23 = 0\]\[y = -\frac{2x}{5}\]

Stretch \(\Rightarrow 2y = -\dfrac{2x}{5} \Rightarrow 5y = -x\)

Translation \(\Rightarrow 5(y + 4) = -x - 3\)

\[5y + x + 17 = 0\]

A2 June 2022 Paper 2 Q7

AQACurrent spec8 marksGraphs & Inequalities

7 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{ax - 5}{2x + b} \qquad x \in \mathbb{R},\ x \neq \frac{9}{2}\]

where \(a\) and \(b\) are integers.

The graph of \(y = \mathrm{f}(x)\) has asymptotes \(x = \dfrac{9}{2}\) and \(y = 3\)

(a) Find the value of \(a\) and the value of \(b\) [2 marks]

(b) Solve the inequality\[\mathrm{f}(x) \leqslant x + 2\]

Fully justify your answer. [6 marks]

Scheme	Marks	AO
Deduces correct value of \(b\)	B1	2.2a
Deduces correct value of \(a\)	B1	2.2a
	(2)

Mark scheme (b)

Scheme	Marks	AO
Selects a suitable method to solve the inequality, for example Multiplies by square of denominator or sketches graphs of their \(y = \mathrm{f}(x)\) and \(y = x + 2\)	M1	3.1a
Simplifies their inequality or equation or Indicates points of intersection of the two graphs	M1	1.1a
Obtains at least two critical values of their inequality or equation	A1F	1.1b
Excludes \(x = \dfrac{9}{2}\) (PI by final answer)	A1	2.2a
Deduces correct solution set for their inequality or graph. Condone inclusion of \(x = \dfrac{9}{2}\) Follow through their answers to part a)	A1F	2.2a
Obtains completely correct solution OE with each step clearly shown	R1	2.1
	(6)
	(8 marks)

Typical solution

\[x + 2 \geqslant \frac{6x - 5}{2x - 9}\]\[(x + 2)(2x - 9)^2 \geqslant (6x - 5)(2x - 9)\]\[(2x - 9)\{(x + 2)(2x - 9) - (6x - 5)\} \geqslant 0\]\[(2x - 9)\{2x^2 - 5x - 18 - 6x + 5\} \geqslant 0\]\[(2x - 9)\{2x^2 - 11x - 13\} \geqslant 0\]\[(2x - 9)(x + 1)(2x - 13) \geqslant 0\]

Change of sign occurs at

\[x = -1,\ x = \frac{9}{2},\ x = \frac{13}{2}\]

Exclude \(x = \dfrac{9}{2}\) (f not defined)

\[-1 \leqslant x \lt \frac{9}{2} \text{ or } x \geqslant \frac{13}{2}\]

A2 June 2022 Paper 1 Q5

AQACurrent spec6 marksFinding Roots of Polynomials Graphs & Inequalities

5 It is given that \(z = -\dfrac{3}{2} + \mathrm{i}\dfrac{\sqrt{11}}{2}\) is a root of the equation

\[z^4 - 3z^3 - 5z^2 + kz + 40 = 0\]

where \(k\) is a real number.

(a) Find the other three roots. [5 marks]

(b) Given that \(x \in \mathbb{R}\), solve\[x^4 - 3x^3 - 5x^2 + kx + 40 \lt 0\] [1 mark]

Mark scheme (a)

Scheme	Marks	AO
States complex conjugate	B1	1.1b
Obtains a quadratic factor or Obtains sum and product of their two complex roots or Substitutes given root or their conjugate into the quartic	M1	1.1a
Obtains \(z^2 + 3z + 5\) or Obtains \(k = -6\)	A1	1.1b
Forms a second quadratic or Solves the quartic equation with their value of \(k\)	M1	1.1a
Obtains 2 and 4	A1	1.1b
	(5)

Typical solution

\(z = -\dfrac{3}{2} - \mathrm{i}\dfrac{\sqrt{11}}{2}\) is another root

Sum of complex roots = −3

Product of complex roots = 5

So \(z^2 + 3z + 5\) is a factor of \(z^4 - 3z^3 - 5z^2 + kz + 40\)

The other quadratic factor is

\[z^2 - 6z + 8 = (z - 2)(z - 4)\]

The other roots are

\[2,\ 4,\ -\frac{3}{2} - \mathrm{i}\frac{\sqrt{11}}{2}\]

Scheme	Marks	AO
Deduces correct solution (ft from their exactly two distinct real roots)	B1F	2.2a
	(1)
	(6 marks)

AS June 2021 Paper 1 Q16

AQACurrent spec8 marksGraphs & Inequalities

16 Curve \(C\) has equation \(y = \dfrac{ax}{x + b}\) where \(a\) and \(b\) are constants.
The equations of the asymptotes to \(C\) are \(x = -2\) and \(y = 3\)

Graph of y = ax/(x + b) with a dashed vertical asymptote left of the y-axis and a dashed horizontal asymptote above the x-axis; the left branch lies above the horizontal asymptote and rises steeply towards the vertical asymptote; the right branch rises from below through O towards the horizontal asymptote

(a) Write down the value of \(a\) and the value of \(b\) [2 marks]

(b) The gradient of \(C\) at the origin is \(\dfrac{3}{2}\)

With reference to the graph, explain why there is exactly one root of the equation

\[\frac{ax}{x + b} = \frac{3x}{2}\]

[2 marks]

[4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(a = 3\) Accept \(y = \dfrac{3x}{x + b}\) with any non-zero value of \(b\)	B1	2.2a
Obtains \(b = 2\) Accept \(y = \dfrac{ax}{x + 2}\) with any non-zero value of \(a\)	B1	2.2a
	(2)

Typical solution

\[y = \frac{3x}{x + 2}\]\[a = 3 \quad \text{and} \quad b = 2\]

Mark scheme (b)

Scheme	Marks	AO
Explains that \(y = \frac{3x}{2}\) is a tangent to the curve. Condone ‘tangent’ only.	B1	2.4
Explains that a tangent to a hyperbola meets the curve exactly once. Accept ‘conic’ instead of ‘hyperbola’. Condone no conclusion given.	B1	2.4
	(2)

Typical solution

\(y = \frac{3x}{2}\) is a tangent to the hyperbola.

Any tangent to a hyperbola only intersects once.

So there is exactly one root of the equation.

Mark scheme (c)

Scheme	Marks	AO
Forms a simplified quadratic equation from \(ax - (1 - x)(x + b) = 0\) Accept any inequality sign instead of = PI by \(x^2 + 4x - 2\) or \(-x^2 - 4x + 2\) PI by \(-2 + \sqrt{6}\) or \(-2 - \sqrt{6}\) Or forms a simplified cubic equation from \(ax(x + 2) - (1 - x)(x + b)(x + 2) = 0\)	M1	1.1a
Identifies \(-2 + \sqrt{6}\) or \(-2 - \sqrt{6}\) as a critical value. Follow through their \(a\) and \(b\)	A1F	1.1b
Identifies \(-2\) as a critical value.	B1	1.1b
Obtains the correct set of values of \(x\)	A1	3.2a
	(4)
	(8 marks)

Typical solution

\[\frac{3x}{x + 2} = 1 - x\]\[3x = (1 - x)(x + 2)\]\[3x = -x^2 - x + 2\]\[x^2 + 4x - 2 = 0\]\[x = \frac{-4 \pm \sqrt{16 + 4 \times 2}}{2}\]\[x = -2 \pm \sqrt{6}\]\[x \leqslant -2 - \sqrt{6},\]\[-2 \lt x \leqslant -2 + \sqrt{6}\]

AS June 2021 Paper 1 Q14

AQACurrent spec9 marksGraphs & Inequalities

14 Curve \(C_1\) has equation

\[\frac{x^2}{16} + \frac{y^2}{4} = 1\]

(a) Curve \(C_2\) is a reflection of \(C_1\) in the line \(y = x\)

Write down an equation of \(C_2\) [1 mark]

(b) Curve \(C_3\) is a circle of radius 4, centred at the origin.

Describe a single transformation which maps \(C_1\) onto \(C_3\) [2 marks]

(c) Curve \(C_4\) is a translation of \(C_1\)
The positive \(x\)-axis and the positive \(y\)-axis are tangents to \(C_4\)

(i) Sketch the graphs of \(C_1\) and \(C_4\) on the axes below. Indicate the coordinates of the \(x\) and \(y\) intercepts on your graphs. [2 marks]

(ii) Determine the translation vector. [2 marks]

(iii) The line \(y = mx + c\) is a tangent to both \(C_1\) and \(C_4\)
Find the value of \(m\) [2 marks]

Scheme	Marks	AO
Writes down a correct equation.	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
Indicates a stretch. Condone a stretch in any direction.	M1	3.1a
Identifies the correct transformation.	A1	1.1b
	(2)

Typical solution

\[y \longrightarrow \frac{y}{2}\]

stretch, parallel to the \(y\)-axis, scale factor 2

Sketch: C1, an ellipse centred on O crossing the x-axis at −4 and 4 and the y-axis at 2 and −2; C4, the same ellipse moved into the first quadrant so that it touches the y-axis at 2 and the x-axis at 4 — Mark scheme (c)

The same sketch with an arrow drawn from (0, −2) on C1 to (4, 0) on C4 — Mark scheme (c)

A2 June 2021 Paper 1 Q7

AQACurrent spec7 marksGraphs & Inequalities

7 The diagram below shows the graph of \(y = \mathrm{f}(x)\) \((-4 \leqslant x \leqslant 4)\)

The graph meets the \(x\)-axis at \(x = 1\) and \(x = 3\)

The graph meets the \(y\)-axis at \(y = 2\)

Graph of y = f(x) for x from −4 to 4: it decreases from the left, crosses the y-axis at 2 and the x-axis at 1, has a minimum just to the right of 1, then increases through the x-axis at 3 up to x = 4

(a) Sketch the graph of \(y = |\mathrm{f}(x)|\) on the axes below.

Show any axis intercepts. [2 marks]

Blank axes with x- and y-axes crossing at O

(b) Sketch the graph of \(y = \dfrac{1}{\mathrm{f}(x)}\) on the axes below.

Show any axis intercepts and asymptotes. [3 marks]

Show any axis intercepts. [2 marks]

Graph of y = |f(x)|: as the original but with the part between x = 1 and x = 3 reflected above the x-axis; it touches the x-axis at 1 and 3 and meets the y-axis at 2 — Mark scheme (a)

Graph of y = 1/f(x): vertical asymptotes x = 1 and x = 3 (dashed); to the left of x = 1 the curve is above the x-axis, increasing through (0, 1/2) towards the asymptote; between 1 and 3 it is below the x-axis with a local maximum; to the right of 3 it is above the x-axis and decreasing — Mark scheme (b)

Graph of y = f(|x|): symmetric about the y-axis, with maximum 2 on the y-axis, crossing the x-axis at −3, −1, 1 and 3 with minima between −3 and −1 and between 1 and 3 — Mark scheme (c)

A2 June 2021 Paper 2 Q6

AQACurrent spec8 marksGraphs & Inequalities

6 The ellipse \(E_1\) has equation

\[x^2 + \frac{y^2}{4} = 1\]

\(E_1\) is translated by the vector \(\begin{bmatrix} 3 \\ 0 \end{bmatrix}\) to give the ellipse \(E_2\)

(a) Write down the equation of \(E_2\) [1 mark]

(b) The ellipse \(E_3\) has equation\[\frac{x^2}{4} + (y - 3)^2 = 1\]

Describe the transformation that maps \(E_2\) to \(E_3\) [1 mark]

\(L_A\) is closer to the origin than \(L_B\)

\(E_2\) and \(E_3\) both lie between \(L_A\) and \(L_B\)

Sketch and label \(E_2\), \(E_3\), \(L_A\) and \(L_B\) on the axes below.

You do not need to show the values of the axis intercepts for \(L_A\) and \(L_B\) [4 marks]

(d) Explain, without doing any calculations, why \(L_A\) has an equation of the form\[x + y = c\]

where \(c\) is a constant. [2 marks]

Sketch: ellipse E3 centred on the y-axis between y = 2 and y = 4, ellipse E2 centred on the x-axis between x = 2 and x = 4, and two parallel lines of gradient −1, LA nearer the origin and LB further out, each touching both ellipses — Mark scheme (c)

Mark scheme (d)

Scheme	Marks	AO
Uses the fact that \(E_3\) is a reflection of \(E_2\) in the line \(y = x\)	E1	2.4
Explains that the tangent is perpendicular to the line \(y = x\) and concludes that its equation is \(x + y = c\)	E1	2.4
	(2)
	(8 marks)

Typical solution

Points on \(E_2\) and \(E_3\) joined by \(L_A\) are symmetrical about \(y = x\), therefore the line is perpendicular to \(y = x\) and has a gradient of \(-1\) and is of the form \(y = -x + c\) or \(x + y = c\)

AS June 2020 Paper 1 Q14

AQACurrent spec7 marksGraphs & Inequalities

(a) Given\[\frac{x + 7}{x + 1} \leqslant x + 1\]

show that

\[\frac{(x + a)(x + b)}{x + c} \geqslant 0\]

where \(a\), \(b\), and \(c\) are integers to be found. [4 marks]

(b) Briefly explain why this statement is incorrect.\[\frac{(x + p)(x + q)}{x + r} \geqslant 0 \Leftrightarrow (x + p)(x + q)(x + r) \geqslant 0\]

[1 mark]

[2 marks]

Mark scheme (a)

Scheme	Marks	AO
Writes both terms on one side of the inequality, with a common denominator.	M1	1.1a
Correctly combines their two fractions into one fraction.	A1F	1.1b
Obtains a single fraction in which the numerator is a three-term quadratic.	M1	1.1a
Completes a rigorous argument to show the correct inequality in the required form. Accept \(0 \leqslant \frac{(x + -2)(x + 3)}{x + 1}\).	R1	2.1

Typical solution

\[0 \leqslant \frac{(x + 1)^2}{x + 1} - \frac{x + 7}{x + 1}\]\[0 \leqslant \frac{x^2 + 2x + 1 - x - 7}{x + 1}\]\[0 \leqslant \frac{x^2 + x - 6}{x + 1}\]\[\frac{(x + 3)(x - 2)}{x + 1} \geqslant 0\]

Mark scheme (b)

Scheme	Marks	AO
Explains that \(x = -r\) is a solution of the inequality on the RHS, but not the one of the LHS.	B1	2.4

Typical solution

\(x = -r\) is a solution of the inequality on the right, but not the one on the left.

Number line with critical values −3, −1 and 2; the sign of the expression is − for x < −3, + between −3 and −1, − between −1 and 2, and + for x > 2 — Mark scheme (c)

AS June 2020 Paper 1 Q10

AQACurrent spec8 marksGraphs & Inequalities

(a) Show that the equation\[y = \frac{3x - 5}{2x + 4}\]

can be written in the form

\[(x + a)(y + b) = c\]

where \(a\), \(b\) and \(c\) are integers to be found. [3 marks]

(b) Write down the equations of the asymptotes of the graph of\[y = \frac{3x - 5}{2x + 4}\]

[2 marks]

[3 marks]

Mark scheme (a)

Scheme	Marks	AO
Selects a method to find the values of \(a, b, c\). e.g. by expanding \((x + a)(y + b) = c\) or by multiplying the equation by \((2x + 4)\) and expanding or by dividing the numerator of the equation by its denominator.	M1	3.1a
Expresses the original equation in a form that allows comparison with \((x + a)(y + b) = c\).	M1	1.1a
Completes a rigorous argument to show that \(y = \frac{3x - 5}{2x + 4}\) can be written as \((x + 2)\left(y - \frac{3}{2}\right) = -\frac{11}{2}\).	R1	2.1

Typical solution

\[y(2x + 4) = 3x - 5\]\[2xy + 4y = 3x - 5\]\[xy + 2y - \tfrac{3}{2}x + \tfrac{5}{2} = 0\]\[xy + 2y - \tfrac{3}{2}x - 3 = -\tfrac{11}{2}\]\[(x + 2)\left(y - \tfrac{3}{2}\right) = -\tfrac{11}{2}\]

Sketch of y = (3x − 5)/(2x + 4) with dashed asymptotes x = −2 and y = 3/2; the left branch lies above y = 3/2 and rises towards x = −2; the right branch rises from below, crossing the y-axis at −5/4 and the x-axis at 5/3, towards y = 3/2 — Mark scheme (c)

A2 June 2020 Paper 1 Q9

AQACurrent spec13 marksGraphs & Inequalities

9 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{x(x + 3)}{x + 4} \qquad (x \in \mathbb{R},\ x \neq -4)\]

(a) Find the interval \((a, b)\) in which \(\mathrm{f}(x)\) does not take any values.

Fully justify your answer. [5 marks]

(b) Find the coordinates of the two stationary points of the graph of \(y = \mathrm{f}(x)\) [2 marks]

(d) Sketch the graph of \(y = \mathrm{f}(x)\) on the axes below. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms an equation \(\dfrac{x(x + 3)}{x + 4} = {}\)‘\(k\)’ or Differentiates using quotient rule.	M1	3.1a
Rearranges their equation into a quadratic in \(x\). or Obtains the correct \(\mathrm{f}^{\prime}(x) = \dfrac{(x + 4) \times (2x + 3) - (x^2 + 3x) \times 1}{(x + 4)^2}\).	M1	1.1a
Explains that the discriminant of this quadratic \(\lt 0\) or Explains that because there is a vertical asymptote the minimum is higher up the graph than the maximum and the two turning points lie on different branches of the graph.	E1	2.4
Forms a quadratic equation or inequality in ‘\(k\)’ from their discriminant. or Equates their \(\mathrm{f}^{\prime}(x)\) or its numerator to 0 and solves	M1	1.1a
Completes a rigorous argument to show that \(\mathrm{f}(x)\) does not take any values in the interval \((-9, -1)\). Condone \(-9 \lt k \lt -1\)	R1	2.1

Typical solution

Assume \(y = k\) meets \(y = \dfrac{x(x + 3)}{x + 4}\)

Then

\[\begin{aligned} k(x + 4) &= x(x + 3) \\ kx + 4k &= x^2 + 3x \\ x^2 + (3 - k)x - 4k &= 0 \end{aligned}\]\[\Delta \lt 0 \therefore (3 - k)^2 - 4(1)(-4k) \lt 0\]\[\begin{aligned} k^2 + 10k + 9 &\lt 0 \\ (k + 9)(k + 1) &\lt 0 \\ -9 \lt k &\lt -1 \end{aligned}\]

So \(\mathrm{f}(x)\) does not take any values in the interval \((-9, -1)\)

Mark scheme (b)

Scheme	Marks	AO
Substitutes their \(-9\) or \(-1\) into \(y = f(x)\) and forms a quadratic in \(x\) or Differentiates using the quotient rule and equates their \(\mathrm{f}^{\prime}(x)\) or its numerator to 0	M1	1.1a
Finds the coordinates of both stationary points	A1F	1.1b

Typical solution

Stationary points \((-6, -9)\) and \((-2, -1)\)

Mark scheme (c)

Scheme	Marks	AO
Divides the numerator by \(x + 4\) and obtains \(\mathrm{f}(x) = x + \cdots\)	M1	3.1a
Obtains the correct equation of the asymptote \(y = x - 1\)	A1	1.1b

Typical solution

\[\begin{aligned} \mathrm{f}(x) = \frac{x^2 + 3x}{x + 4} &= \frac{(x + 4)(x - 1) + 4}{x + 4} \\ &= x - 1 + \frac{4}{x + 4} \end{aligned}\]

Asymptote is \(y = x - 1\)

Sketch of y = f(x) with asymptotes x = −4 and y = x − 1: the right branch comes down beside x = −4, crosses the x-axis at −3, has a minimum at (−2, −1), passes through O and approaches y = x − 1; the left branch rises to a maximum at (−6, −9) and falls beside x = −4 — Mark scheme (d)

A2 June 2020 Paper 1 Q5

AQACurrent spec9 marksGraphs & Inequalities

5 \(H_1\) is the locus of points such that the distance from the point \((5, 0)\) is twice the distance from the line \(x = 2\)

(a) Show that the equation of \(H_1\) can be written in the form\[(x - 1)^2 - \frac{y^2}{q} = r\]

where \(q\) and \(r\) are integers. [5 marks]

(b) \(H_2\) is the hyperbola\[x^2 - y^2 = 4\]

Describe fully a sequence of two transformations which maps the graph of \(H_2\) onto the graph of \(H_1\) [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms an expression for the distance of a general point from the line \(x = 2\). Condone \((x - 2)\) or \((2 - x)\)	B1	1.1b
Forms an expression for the (squared) distance of a general point from \((5, 0)\)	B1	1.1b
Forms an equation in \(x\) and \(y\) for the locus, of the form: distance from \((5, 0)\) = 2 x distance from \(x = 2\)	M1	3.1a
Simplifies their equation to a quadratic in \(x\) and \(y\) with coefficient of \(x\) equal to 1	M1	1.1a
Completes a rigorous argument to show \((x - 1)^2 - \frac{y^2}{3} = 4\) AG	R1	2.1

Typical solution

\[\begin{aligned} \sqrt{(x - 5)^2 + y^2} &= 2|x - 2| \\ (x - 5)^2 + y^2 &= 4(x - 2)^2 \\ x^2 - 10x + 25 + y^2 &= 4x^2 - 16x + 16 \\ 3x^2 - 6x - y^2 &= 9 \\ x^2 - 2x - \frac{y^2}{3} &= 3 \\ (x - 1)^2 - 1 - \frac{y^2}{3} &= 3 \\ (x - 1)^2 - \frac{y^2}{3} &= 4 \end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
States that there is a horizontal (PI by vector with no \(y\)-component) translation.	B1	2.2a
Describes translation as \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) OE	B1	2.5
States that there is a stretch parallel to the \(y\)-axis.	B1	2.2a
States that the scale factor is \(\sqrt{3}\) FT their \(q\) (or \(\sqrt{q}\) if no value of \(q\) given.)	B1F	2.5
	(9 marks)

Typical solution

Translation by \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\)

Stretch parallel to \(y\)-axis.
Scale factor \(\sqrt{3}\)

A2 June 2020 Paper 2 Q5

AQACurrent spec5 marksGraphs & Inequalities

5 Solve the inequality

\[\frac{2x + 3}{x - 1} \leqslant x + 5\]

[5 marks]

Mark scheme

Scheme	Marks	AO
Selects a correct approach which would lead to solving the inequality eg Multiplies the inequality by \((x - 1)^2\) or Rearranges to an inequality with 0 on LHS or RHS or Replaces “\(\leqslant\)” with “\(=\)” and multiplies by \((x - 1)\)	M1	1.1a
Manipulates their equation/inequality to allow the critical values to be found	M1	1.1a
Obtains critical values of \(-4\), \(1\) and \(2\)	M1	1.1a
Gives one correct region from \(x \geqslant 2\), \(-4 \leqslant x \lt 1\) Condone \(-4 \leqslant x \leqslant 1\) Must have three critical values.	A1	1.1b
Obtains correct solution \(x \geqslant 2\), \(-4 \leqslant x \lt 1\)	A1	1.1b
	(5 marks)

Typical solution

\[\frac{(x - 1)^2(2x + 3)}{x - 1} \leqslant (x - 1)^2(x + 5)\]\[(x - 1)(2x + 3) \leqslant (x - 1)^2(x + 5)\]\[(x - 1)\{(x - 1)(x + 5) - (2x + 3)\} \geqslant 0\]\[(x - 1)\{x^2 + 2x - 8\} \geqslant 0\]\[(x - 1)(x + 4)(x - 2) \geqslant 0\]

Considering cubic curve:
\(x \geqslant 2\) or between \(-4\) and \(1\)

But \(x \neq 1\)

So \(x \geqslant 2\) or \(-4 \leqslant x \lt 1\)

AS June 2020 Paper 1 Q3

AQACurrent spec1 markGraphs & Inequalities

3 Given \((x - 1)(x - 2)(x - a) \lt 0\) and \(a \gt 2\)

Find the set of possible values of \(x\).

Tick (✓) one box. [1 mark]

\(\{x : x \lt 1\} \cup \{x : 2 \lt x \lt a\}\)
\(\{x : 1 \lt x \lt 2\} \cup \{x : x \gt a\}\)
\(\{x : x \lt -a\} \cup \{x : -2 \lt x \lt -1\}\)
\(\{x : -a \lt x \lt -2\} \cup \{x : x \gt -1\}\)

AS June 2019 Paper 1 Q11

AQACurrent spec8 marksGraphs & Inequalities

(a) Curve \(C\) has equation\[y = \frac{x^2 + px - q}{x^2 - r}\]

where \(p\), \(q\) and \(r\) are positive constants.

Write down the equations of its asymptotes. [2 marks]

(b) Find the set of possible \(y\)-coordinates for the graph of\[y = \frac{x^2 + x - 6}{x^2 - 1}, \quad x \neq \pm 1\]

giving your answer in exact form.

No credit will be given for solutions based on differentiation. [6 marks]

Mark scheme (a)

Scheme	Marks	AO
Gives \(x = \sqrt{r}\) or \(x = -\sqrt{r}\) or \(y = 1\) as an asymptote. Condone other incorrect asymptotes.	B1	1.1b
Gives \(x = \pm\sqrt{r}\) and \(y = 1\) as asymptotes, with no incorrect asymptotes given.	B1	1.1b

Typical solution

\[x^2 - r = 0\]\[x^2 = r\]\[x = \pm\sqrt{r}\]\[y = 1\]

Mark scheme (b)

Scheme	Marks	AO
Rearranges \(k = \dfrac{x^2 + x - 6}{x^2 - 1}\) into a non-fractional form. Accept any sensible alternative for \(k\), e.g. \(y\) or f	M1	3.1a
Rearranges their equation into a correct three-term quadratic equation in \(x\) Condone missing \(= 0\) Possibly implied by a correct discriminant.	A1	1.1b
Correctly substitutes their coefficients into \(b^2 - 4ac\) to obtain an expression in \(k\) only. Accept any sensible alternative for \(k\), e.g. \(y\) or f	M1	3.1a
Obtains a correct quadratic equation/inequality in \(k\) – may be unsimplified. Or obtains the correct critical values of \(k\). Accept any sensible alternative for \(k\), e.g. \(y\) or f	A1	1.1b
Obtains the correct critical values. Accept non-exact values to at least 3 sig figs, e.g. 1.05 and 5.95	A1	1.1b
Gives a correct range in terms of \(y\) using exact values. Condone ‘and’. Follow through their critical values if M2 scored and quadratic inequality seen. Do not accept an alternative for \(y\) Accept any equivalent expressions for \(\frac{7 - 2\sqrt{6}}{2}\) and \(\frac{7 + 2\sqrt{6}}{2}\) NMS scores 0/6	A1F	2.2a
	(8 marks)

Typical solution

Let \(k = \dfrac{x^2 + x - 6}{x^2 - 1}\)

\[k(x^2 - 1) = x^2 + x - 6\]\[(k - 1)x^2 - x + 6 - k = 0\]\[1 - 4(k - 1)(6 - k) \geqslant 0\]\[1 - 4(6k - k^2 - 6 + k) \geqslant 0\]\[4k^2 - 28k + 25 \geqslant 0\]\[y \leqslant \frac{7 - 2\sqrt{6}}{2}, \quad y \geqslant \frac{7 + 2\sqrt{6}}{2}\]

A2 June 2019 Paper 2 Q8

AQACurrent spec9 marksGraphs & Inequalities Integration

8 A parabola \(P_1\) has equation \(y^2 = 4ax\) where \(a \gt 0\)

\(P_1\) is translated by the vector \(\begin{bmatrix} b \\ 0 \end{bmatrix}\), where \(b \gt 0\), to give the parabola \(P_2\)

(a) The line \(y = mx\) is a tangent to \(P_2\)

Prove that \(m = \pm\sqrt{\dfrac{a}{b}}\)

Solutions using differentiation will be given no marks. [4 marks]

(b) The line \(y = \sqrt{\dfrac{a}{b}}\,x\) meets \(P_2\) at the point \(D\).

The finite region \(R\) is bounded by the \(x\)-axis, \(P_2\) and a line through \(D\) perpendicular to the \(x\)-axis.

The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid.

Find, in terms of \(a\) and \(b\), the volume of this solid.

Fully justify your answer. [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains the equation of \(P_2\)	B1	1.2
Combines equations to give a quadratic equation in \(x\)	M1	1.1a
Sets the discriminant to zero, with working	M1	1.1a
Completes a rigorous argument to show the required result.	R1	2.1

Typical solution

\[y^2 = 4a(x - b)\]\[m^2x^2 = 4a(x - b)\]\[m^2x^2 - 4ax + 4ab = 0\]

For equal roots \(\Delta = 0\)

\[16a^2 - 4m^2(4ab) = 0\]\[16a^2 = 16m^2ab\]\[m = \pm\sqrt{\frac{a}{b}}\]

Sketch: the parabola P2 with vertex at (b, 0), the line through the origin touching it, and the shaded region R between P2, the x-axis and the vertical line x = 2b — Mark scheme (b)

AS June 2019 Paper 1 Q5

AQACurrent spec8 marksGraphs & Inequalities Integration

5 A hyperbola \(H\) has the equation

\[\frac{x^2}{a^2} - \frac{y^2}{4a^2} = 1\]

where \(a\) is a positive constant.

(a) Write down the equations of the asymptotes of \(H\). [1 mark]

(b) Sketch the hyperbola \(H\) on the axes below, indicating the coordinates of any points of intersection with the coordinate axes.
The asymptotes have already been drawn. [2 marks]

Axes crossing at O with two dashed asymptotes drawn through O

(c) The finite region bounded by \(H\), the positive \(x\)-axis, the positive \(y\)-axis and the line \(y = a\) is rotated through \(360^\circ\) about the \(y\)-axis.
Show that the volume of the solid generated is \(ma^3\), where \(m = 3.40\) correct to three significant figures. [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Writes the correct equations with \(a\) removed. Accept any equivalent equations, e.g. \(x = \pm\frac{1}{2}y\)	B1	1.1b

Typical solution

\[\frac{x}{a} = \pm\frac{y}{2a}\]\[y = \pm 2x\]

Sketch of the hyperbola H: two branches approaching the dashed asymptotes, crossing the x-axis at −a and a — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
Formulates an expression for a volume generated by rotating the hyperbola about an axis – must be a clear intent to integrate. A volume expression must be of the form \(\int \mathrm{f}(x)\) or \(\int \mathrm{f}(y)\) where f is a polynomial function of degree 2. Condone missing limits and/or \(\pi\) and/or \(\mathrm{d}y\) (or \(\mathrm{d}x\)).	M1	3.1a
Expresses the volume as \(\pi\displaystyle\int\left(\frac{y^2}{4} + a^2\right)\mathrm{d}y\) or equivalent. Must include \(\pi\) – may be seen later. Condone missing limits and/or \(\mathrm{d}y\).	A1	1.1b
Correctly integrates their expression. Their expression must be of the form \(cy^2 + d\) or \(cx^2 + d\) where \(c\) and \(d\) are constants.	M1	1.1a
Substitutes correct limits into \(py^3 + qy\), where \(p\) and \(q\) are positive constants. May be unsimplified. Accept substitution of 0 not seen.	A1	1.1b
Completes a rigorous mathematical argument, including either \(ka^3\) where \(k \in [3.4005, 3.4045]\) or \(\frac{13}{12}\pi a^3\) (or equivalent), that the volume can be expressed as \(3.40a^3\) to 3 significant figures. Must use \(\mathrm{d}y\) correctly throughout. Must include an appropriate reference to 3 significant figures, e.g. \(\frac{13\pi}{12} = 3.40\) (3sf) Accept substitution of 0 not seen. This mark can only be awarded if M2A2 scored. NMS scores 0/5	R1	2.1
	(8 marks)

Typical solution

\[\begin{aligned}\text{Volume} &= \pi\int x^2\,\mathrm{d}y \\ &= \pi\int_0^a \left(\frac{y^2}{4} + a^2\right)\mathrm{d}y \\ &= \pi\left[\frac{y^3}{12} + a^2y\right]_0^a \\ &= \pi\left(\frac{a^3}{12} + a^3 - 0\right) \\ &= \frac{13\pi}{12}a^3 \\ &= 3.403a^3 \\ &= 3.40a^3 \text{ (3 sig figs)}\end{aligned}\]

A2 June 2019 Paper 2 Q3

AQACurrent spec1 markGraphs & Inequalities

3 The set \(\mathcal{A}\) is defined by \(\mathcal{A} = \{x : -\sqrt{2} \lt x \lt 0\} \cup \{x : 0 \lt x \lt \sqrt{2}\}\)

Which of the inequalities given below has \(\mathcal{A}\) as its solution?

Circle your answer. [1 mark]

\(|x^2 - 1| \gt 1\)
\(|x^2 - 1| \geqslant 1\)
\(|x^2 - 1| \lt 1\)
\(|x^2 - 1| \leqslant 1\)

A2 June 2019 Paper 2 Q2

AQACurrent spec1 markGraphs & Inequalities

2 Which of the straight lines given below is an asymptote to the curve

\[y = \frac{ax^2}{x - 1}\]

where \(a\) is a non-zero constant?

Circle your answer. [1 mark]

\(y = ax + a\)
\(y = ax\)
\(y = ax - a\)
\(y = a\)

AS June 2018 Paper 1 Q13

AQACurrent spec9 marksGraphs & Inequalities

13 The graph of the rational function \(y = \mathrm{f}(x)\) intersects the \(x\)-axis exactly once at \((-3, 0)\)

The graph has exactly two asymptotes, \(y = 2\) and \(x = -1\)

(a) Find \(\mathrm{f}(x)\) [2 marks]

(b) Sketch the graph of the function. [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Writes any rational function with a horizontal asymptote of \(y = 2\) or one vertical asymptote of \(x = -1\), e.g. \(y = \frac{ax + b}{x + 1}\) or \(y = \frac{2x + b}{x + c}\) or \(y = \frac{ax^n + bx^{n-1} + cx^{n-2} + \ldots\ldots}{dx^n + ex^{n-1} + fx^{n-2} + \ldots\ldots}\) where \(\frac{a}{d} = 2\) Accept any correct rearrangement of \(y = \mathrm{f}(x)\), where \(\mathrm{f}(x)\) is a function as described above.	M1	3.1a
Obtains a fully correct answer.	A1	1.1b

Typical solution

\[y = \frac{2x + c}{x + 1}\]

but \(x = -3\) when \(y = 0\)

\[\therefore 0 = \frac{2 \times -3 + c}{-3 + 1}\]\[-6 + c = 0\]\[c = 6\]\[\therefore y = \frac{2x + 6}{x + 1}\]

(Corrected from the printed mark scheme: its first line is printed as \(y = \frac{2x + m}{x + 1}\), but the working that follows uses \(c\).)

Alternative

\[(x + 1)(y - 2) = n\]

but \(x = -3\) when \(y = 0\)

\[\therefore (-3 + 1)(0 - 2) = n\]\[4 = n\]\[\therefore (x + 1)(y - 2) = 4\]\[y - 2 = \frac{4}{x + 1}\]\[y = 2 + \frac{4}{x + 1}\]

Sketch of y = (2x + 6)/(x + 1) with dashed asymptotes x = −1 and y = 2; the right branch comes down from the top near x = −1, crosses the y-axis at 6 and levels off above y = 2; the left branch lies below y = 2, crosses the x-axis at −3 and falls steeply towards x = −1 — Mark scheme (b)

Number line for x with critical values −1 and 1/3; the sign is + for x < −1, − between −1 and 1/3, and + for x > 1/3 — Mark scheme (c)

AS June 2018 Paper 1 Q12

AQACurrent spec6 marksGraphs & Inequalities Matrices

(a) Show that the matrix \(\begin{bmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{bmatrix}\) is singular when \(k = 1\). [1 mark]

(b) Find the values of \(k\) for which the matrix \(\begin{bmatrix} 5 - k & 2 \\ k^3 + 1 & k \end{bmatrix}\) has a negative determinant.
Fully justify your answer. [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Substitutes \(k = 1\) and correctly calculates the determinant, and concludes that the matrix is singular.	B1	2.2a

Typical solution

\[\text{determinant} = 4 \times 1 - 2 \times 2 = 0\]

\(\therefore\) the matrix is singular

Number line for k with critical values −2, 1/2 and 1; the sign is − for k < −2, + between −2 and 1/2, − between 1/2 and 1, and + for k > 1 — Mark scheme (b)

AS June 2018 Paper 1 Q9

AQACurrent spec6 marksGraphs & Inequalities Integration

(a) Sketch the graph of \(y^2 = 4x\) [1 mark]

(b) Ben is using a 3D printer to make a plastic bowl which holds exactly \(1000\,\text{cm}^3\) of water.
Ben models the bowl as a region which is rotated through \(2\pi\) radians about the \(x\)-axis.
He uses the finite region enclosed by the lines \(x = d\) and \(y = 0\) and the curve with equation \(y^2 = 4x\) for \(y \geqslant 0\)

(i) Find the depth of the bowl to the nearest millimetre. [4 marks]

(ii) What assumption has Ben made about the bowl? [1 mark]

Mark scheme (b)

Scheme	Marks	AO
(i) Obtains \(\pi\displaystyle\int 4x\,\mathrm{d}x\) Limits not required for this mark. Condone missing \(\mathrm{d}x\)	M1	3.3
Obtains \(2x^2\) and uses limits of \(d\) and \(0\) (oe).	B1	1.1b
Forms an equation of the form \(kd^2 = \text{volume}\) (oe)	M1	3.4
Correct depth to nearest millimetre. Condone 126 or 12.6 without units. NMS: 126 or 12.6 scores 4/4. Using \(1000000\,\text{mm}^3\) leads to a correct answer of 399 mm for 4/4.	A1	3.2a
(ii) States appropriate assumption	B1	3.5b
	(6 marks)

Typical solution

(i)

\[\text{volume} = \pi\int_0^d y^2\,\mathrm{d}x\]\[\therefore 1000 = \pi\int_0^d 4x\,\mathrm{d}x\]\[\frac{1000}{\pi} = \left[\frac{4x^2}{2}\right]_0^d\]\[\frac{1000}{\pi} = 2d^2 - 0\]\[2\pi d^2 = 1000\]\[d = \sqrt{\frac{500}{\pi}}\]

depth = 12.6 cm

(ii)

The thickness of the plastic is negligible

A2 June 2023 Paper 1 Q4

OCR ACurrent spec11 marksGraphs & Inequalities Matrices

4 The transformations \(\mathrm{T_A}\) and \(\mathrm{T_B}\) are represented by the matrices \(\mathbf{A}\) and \(\mathbf{B}\) respectively, where

\[\mathbf{A} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \text{ and } \mathbf{B} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\]

(a) Describe geometrically the single transformation consisting of \(\mathrm{T_A}\) followed by \(\mathrm{T_B}\). [2]

(b) By considering the transformation \(\mathrm{T_A}\), determine the matrix \(\mathbf{A}^{423}\). [3]

The transformation \(\mathrm{T_C}\) is represented by the matrix \(\mathbf{C}\), where

\[\mathbf{C} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{pmatrix}.\]

The region \(R\) is defined by the set of points \((x, y)\) satisfying the inequality \(x^2 + y^2 \leqslant 36\).

The region \(R^{\prime}\) is defined as the image of \(R\) under \(\mathrm{T_C}\).

(c)

(i) Find the exact area of the region \(R^{\prime}\). [2]

(ii) Sketch the region \(R^{\prime}\), specifying all the points where the boundary of \(R^{\prime}\) intersects the coordinate axes. [4]

Mark scheme (a)

Scheme	Marks
\(\mathbf{BA} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)	M1
i.e. a reflection in the \(x\)-axis	A1
	[2]

Notes

M1: Multiplication in correct order

A1: Or reflection in the line \(y = 0\)
Answer with no working B2

Alternative method

Scheme	Marks
A represents a rotation anti-clockwise of \(90^\circ\) B represents a reflection in the line \(y = x\)	M1
Taken one after the other gives a reflection in the \(x\)- axis	A1

Mark scheme (b)

Scheme	Marks
\(\mathrm{T_A}\) is a rotation 90 degrees (anti-clockwise about \(O\))	B1
(423 has remainder 3 when divided by 4) so \(\mathbf{A}^{423} = \mathbf{A}^3\)	M1
So \(\mathbf{A}^{423} = \mathbf{A}^3 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\)	A1
	[3]

Notes

M1: For \(\mathbf{A}^4 = \mathbf{I}\) or \(\mathrm{T_A}\) repeated four times is a 360 degree rotation. Condone clockwise instead of anticlockwise for \(\mathrm{T_A}\) so notes that 423 is divisible by 4 with remainder 3 so \(\mathbf{A}^{423} = \mathbf{A}^3\)

A1: As \(\mathbf{A}^3\) represents a 270 degrees rotation anti-clockwise (or 90 degrees clockwise) or by direct calculation of \(\mathbf{A}^3\).

Sketch: ellipse centred at the origin, shaded and labelled R prime, crossing the axes at (3,0), (-3,0), (0,2) and (0,-2) — Mark scheme (c)

A2 October 2021 Paper 2 Q10

OCR ACurrent spec6 marksDifferentiation & Maclaurin Graphs & Inequalities

10 In this question you must show detailed reasoning.

(a) By using an appropriate Maclaurin series prove that if \(x \gt 0\) then \(\mathrm{e}^x \gt 1 + x\). [2]

(b) Hence, by using a suitable substitution, deduce that \(\mathrm{e}^t \gt \mathrm{e}t\) for \(t \gt 1\). [1]

(c) Using the inequality in part (b), and by making a suitable choice for \(t\), determine which is greater, \(\mathrm{e}^\pi\) or \(\pi^\mathrm{e}\). [3]

Mark scheme (a)

Scheme	Marks	AO
DR \(\mathrm{e}^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ldots = (1 + x) + \left(\dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ldots\right)\)	M1	1.1
\(x \gt 0 \Rightarrow \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ldots \gt 0\) \(\Rightarrow \mathrm{e}^x \gt 1 + x\)	A1	2.2a
	[2]

Notes

M1: Quoting and using the Maclaurin series

A1: AG. Result with sufficient justification

Mark scheme (c)

Scheme	Marks	AO
DR \(t = \dfrac{\pi}{\mathrm{e}} \gt 1\) since \(2 \lt \mathrm{e} \lt 3\) and \(\pi \gt 3\)	B1	3.1a
\(\mathrm{e}^{\frac{\pi}{\mathrm{e}}} \gt \mathrm{e} \times \dfrac{\pi}{\mathrm{e}}\ (= \pi)\)	M1	3.1a
\(\Rightarrow \mathrm{e}^\pi \gt \pi^\mathrm{e}\) (ie \(\mathrm{e}^\pi\) is greater)	A1	1.1
	[3]

Notes

B1: Some justification that \(t \gt 1\) is required

M1: Substituting their choice into the inequality

A1: Answer without use of inequality in part (b) scores M0A0

Alternative method

Scheme	Marks
\(t = \ln\pi\)	B1
\(\mathrm{e}^{\ln\pi} \gt \mathrm{e}\ln\pi\)	M1
\(\pi \gt \ln\left(\pi^\mathrm{e}\right)\) \(\mathrm{e}^\pi \gt \pi^\mathrm{e}\)	A1
	[3]

B1: Some justification that \(t \gt 1\) is required

A2 June 2023 Paper 1 Q13

OCR MEICurrent spec14 marksComplex Numbers Graphs & Inequalities

(a) On the separate Argand diagrams below, show the set of points representing each of the following inequalities.

(i) \(|z| \leqslant \sqrt{5}\)

Argand diagram grid from the Printed Answer Booklet: Re axis from -8 to 8 and Im axis from -8i to 8i, with unit grid squares

[3]

(ii) \(|z + 2 - 4\mathrm{i}| \geqslant |z - 2 - 6\mathrm{i}|\)

[3]

(b) Show that there is a unique value of \(z\), which should be determined, for which both \(|z| \leqslant \sqrt{5}\) and \(|z + 2 - 4\mathrm{i}| \geqslant |z - 2 - 6\mathrm{i}|\). [8]

Argand diagram: shaded disc centred at O with radius root 5, crossing the axes just beyond plus and minus 2 and plus and minus 2i — Mark scheme (a)

Argand diagram: points -2 + 4i and 2 + 6i joined by a segment; their perpendicular bisector, a line through 5i with gradient -2 crossing the real axis at 2.5; the region to the right of the line is shaded — Mark scheme (a)

Mark scheme (b)

Scheme	Marks	AO
gradient \(= -2\)	M1	1.1
passing through \((0, 5)\)	B1	3.1a
equation \(y = -2x + 5\)	A1	1.1
circle is \(x^2 + y^2 = 5\)	B1	3.1a
\(x^2 + (5 - 2x)^2 = 5\)	M1	2.1
\(\Rightarrow 5x^2 - 20x + 20 = 0\)	M1	1.1
\(\Rightarrow x = 2\) [only]	A1*	2.2a
[unique solution is] \(z = 2 + \mathrm{i}\)	A1dep	3.2a
	[8]

Notes

A1: (1st) oe. Allow inequality. Could be obtained from diagram in (a).

B1: (2nd) allow inequality

M1: (2nd) or \(\left(\frac{5}{2} - \frac{y}{2}\right)^2 + y^2 = 5\). Must be an equation.

M1: (3rd) simplifying to a three-term quadratic equation

A1*: or \(y = 1\) [only]

Alternative method

Scheme	Marks
\((x + 2)^2 + (y - 4)^2 = (x - 2)^2 + (y - 6)^2\)	M1 M1
\(y = -2x + 5\)	A1
circle is \(x^2 + y^2 = 5\)	B1
\(x^2 + (5 - 2x)^2 = 5\)	M1
\(\Rightarrow 5x^2 - 20x + 20 = 0\)	M1
\(\Rightarrow x = 2\) [only]	A1*
[unique solution is] \(z = 2 + \mathrm{i}\)	A1dep

M1: (1st) squaring both sides of equations or inequality

M1: (2nd) expanding all four sets of brackets

A1: oe. Allow inequality.

B1: allow inequality

M1: (3rd) or \(\left(\frac{5}{2} - \frac{y}{2}\right)^2 + y^2 = 5\). Must be an equation.

M1: (4th) simplifying to a three term quadratic equation

A1*: or \(y = 1\) [only]