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HL Paper 3

This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.

A power series in $x$ is defined as a function of the form $f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ where the ${a_i} \in \mathbb{R}$ .

It can be considered as an infinite polynomial.

This is an example of a power series, but is only a finite power series, since only a finite number of the ${a_i}$ are non-zero.

We will now attempt to generalise further.

Suppose ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ can be written as the power series ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ .

Expand ${\left( {1 + x} \right)^5}$ using the Binomial Theorem.

[2]

Consider the power series $1 - x + {x^2} - {x^3} + {x^4} - ...$

By considering the ratio of consecutive terms, explain why this series is equal to ${\left( {1 + x} \right)^{ - 1}}$ and state the values of $x$ for which this equality is true.

[4]

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 2}}$ .

[2]

Repeat this process to find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 3}}$ .

[2]

Hence, by recognising the pattern, deduce the first four terms in a power series for ${\left( {1 + x} \right)^{ - n}}$ , $n \in {\mathbb{Z}^ + }$ .

[3]

By substituting $x = 0$ , find the value of ${a_0}$ .

[1]

By differentiating both sides of the expression and then substituting $x = 0$ , find the value of ${a_1}$ .

[2]

Repeat this procedure to find ${a_2}$ and ${a_3}$ .

[4]

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ .

[1]

Write down the power series for $\frac{1}{{1 + {x^2}}}$ .

[2]

Hence, using integration, find the power series for ${\text{arctan}}\,x$ , giving the first four non-zero terms.

[4]

Markscheme

$1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5}$ M1A1

[2 marks]

It is an infinite GP with $a = 1{\text{,}}\,\,r = - x$ R1A1

${S_\infty } = \frac{1}{{1 - \left( { - x} \right)}} = \frac{1}{{1 + x}} = {\left( {1 + x} \right)^{ - 1}}$ M1A1AG

[4 marks]

${\left( {1 + x} \right)^{ - 1}} = 1 - x + {x^2} - {x^3} + {x^4} - ...$

$- 1{\left( {1 + x} \right)^{ - 2}} = - 1 + 2x - 3{x^2} + 4{x^3} - ...$ A1

${\left( {1 + x} \right)^{ - 2}} = 1 - 2x + 3{x^2} - 4{x^3} + ...$ A1

[2 marks]

$- 2{\left( {1 + x} \right)^{ - 3}} = - 2 + 6x - 12{x^2} + 20{x^3}...$ A1

${\left( {1 + x} \right)^{ - 3}} = 1 - 3x + 6{x^2} - 10{x^3}...$ A1

[2 marks]

${\left( {1 + x} \right)^{ - n}} = 1 - nx + \frac{{n\left( {n + 1} \right)}}{{2{\text{!}}}}{x^2} - \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1A1A1

[3 marks]

${1^q} = {a_0} \Rightarrow {a_0} = 1$ A1

[1 mark]

$q{\left( {1 + x} \right)^{q - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ...$ A1

${a_1} = q$ A1

[2 marks]

$q\left( {q - 1} \right){\left( {1 + x} \right)^{q - 2}} = 1 \times 2{a_2} + 2 \times 3{a_3}x + ...$ A1

${a_2} = \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}$ A1

$q\left( {q - 1} \right)\left( {q - 2} \right){\left( {1 + x} \right)^{q - 3}} = 1 \times 2 \times 3{a_3} + ...$ A1

${a_3} = \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}$ A1

[4 marks]

${\left( {1 + x} \right)^q} = 1 + qx + \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}{x^2} + \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1

[1 mark]

$\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} + ...$ M1A1

[2 marks]

${\text{arctan}}\,x + c = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + ...$ M1A1

Putting $x = 0 \Rightarrow c = 0$ R1

So ${\text{arctan}}\,x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + ...$ A1

[4 marks]

Examiners report

[N/A]

This question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree $3$ and $4$ .

The cubic equation $x^{3} + p x^{2} + q x + r = 0$ , where $p, q, r \in ℝ$ , has roots $α, β$ and $γ$ .

Consider the equation $x^{3} - 7 x^{2} + q x + 1 = 0$ , where $q \in ℝ$ .

Noah believes that if $p^{2} \geq 3 q$ then $α, β$ and $γ$ are all real.

Now consider polynomial equations of degree $4$ .

The equation $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ , where $p, q, r, s \in ℝ$ , has roots $α, β, γ$ and $δ$ .

In a similar way to the cubic equation, it can be shown that:

$p = - (α + β + γ + δ)$

$q = α β + α γ + α δ + β γ + β δ + γ δ$

$r = - (α β γ + α β δ + α γ δ + β γ δ)$

$s = α β γ δ$ .

The equation $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = 0$ , has one integer root.

By expanding $(x - α) (x - β) (x - γ)$ show that:

$p = - (α + β + γ)$

$q = α β + β γ + γ α$

$r = - α β γ$ .

[3]

Show that $p^{2} - 2 q = α^{2} + β^{2} + γ^{2}$ .

[3]

b.i.

Hence show that ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} = 2 p^{2} - 6 q$ .

[3]

b.ii.

Given that $p^{2} < 3 q$ , deduce that $α, β$ and $γ$ cannot all be real.

[2]

Using the result from part (c), show that when $q = 17$ , this equation has at least one complex root.

[2]

By varying the value of $q$ in the equation $x^{3} - 7 x^{2} + q x + 1 = 0$ , determine the smallest positive integer value of $q$ required to show that Noah is incorrect.

[2]

e.i.

Explain why the equation will have at least one real root for all values of $q$ .

[1]

e.ii.

Find an expression for $α^{2} + β^{2} + γ^{2} + δ^{2}$ in terms of $p$ and $q$ .

[3]

f.i.

Hence state a condition in terms of $p$ and $q$ that would imply $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ has at least one complex root.

[1]

f.ii.

Use your result from part (f)(ii) to show that the equation $x^{4} - 2 x^{3} + 3 x^{2} - 4 x + 5 = 0$ has at least one complex root.

[1]

State what the result in part (f)(ii) tells us when considering this equation $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = 0$ .

[1]

h.i.

Write down the integer root of this equation.

[1]

h.ii.

By writing $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12$ as a product of one linear and one cubic factor, prove that the equation has at least one complex root.

[4]

h.iii.

Markscheme

attempt to expand $(x - α) (x - β) (x - γ)$ M1

$= (x^{2} - (α + β) x + α β) (x - γ)$ OR $= (x - α) (x^{2} - (β + γ) x + β γ)$ A1

$(x^{3} + p x^{2} + q x + r) = x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ$ A1

comparing coefficients:

$p = - (α + β + γ)$ AG

$q = (α β + β γ + γ α)$ AG

$r = - α β γ$ AG

Note: For candidates who do not include the AG lines award full marks.

[3 marks]

$p^{2} - 2 q = {(α + β + γ)}^{2} - 2 (α β + β γ + γ α)$ (A1)

attempt to expand ${(α + β + γ)}^{2}$ (M1)

$= α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α) - 2 (α β + β γ + γ α)$ or equivalent A1

$= α^{2} + β^{2} + γ^{2}$ AG

Note: Accept equivalent working from RHS to LHS.

[3 marks]

b.i.

EITHER

attempt to expand ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2}$ (M1)

$= (α^{2} + β^{2} - 2 α β) + (β^{2} + γ^{2} - 2 β γ) + (γ^{2} + α^{2} - 2 γ α)$ A1

$= 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α)$

$= 2 (p^{2} - 2 q) - 2 q$ or equivalent A1

$= 2 p^{2} - 6 q$ AG

attempt to write $2 p^{2} - 6 q$ in terms of $α, β, γ$ (M1)

$= 2 (p^{2} - 2 q) - 2 q$

$= 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α)$ A1

$= (α^{2} + β^{2} - 2 α β) + (β^{2} + γ^{2} - 2 β γ) + (γ^{2} + α^{2} - 2 γ α)$ A1

$= {(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2}$ AG

Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.

[3 marks]

b.ii.

$p^{2} < 3 q \Rightarrow 2 p^{2} - 6 q < 0$

$\Rightarrow {(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} < 0$ A1

if all roots were real ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} \geq 0$ R1

Note: Condone strict inequality in the R1 line.
Note: Do not award A0R1.

$\Rightarrow$ roots cannot all be real AG

[2 marks]

$p^{2} = {(- 7)}^{2} = 49$ and $3 q = 51$ A1

so $p^{2} < 3 q \Rightarrow$ the equation has at least one complex root R1

Note: Allow equivalent comparisons; e.g. checking $p^{2} < 6 q$

[2 marks]

use of GDC (eg graphs or tables) (M1)

$q = 12$ A1

[2 marks]

e.i.

complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).

a cubic curve always crosses the $x$ -axis at at least one point. R1

[1 mark]

e.ii.

attempt to expand ${(α + β + γ + δ)}^{2}$ (M1)

${(α + β + γ + δ)}^{2} = α^{2} + β^{2} + γ^{2} + δ^{2} + 2 (α β + α γ + α δ + β γ + β δ + γ δ)$ (A1)

$\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} = {(α + β + γ + δ)}^{2} - 2 (α β + α γ + α δ + β γ + β δ + γ δ)$

$(\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} =) p^{2} - 2 q$ A1

[3 marks]

f.i.

$p^{2} < 2 q$ OR $p^{2} - 2 q < 0$ A1

Note: Allow FT on their result from part (f)(i).

[1 mark]

f.ii.

$4 < 6$ OR $2^{2} - 2 \times 3 < 0$ R1

hence there is at least one complex root. AG

Note: Allow FT from part (f)(ii) for the R mark provided numerical reasoning is seen.

[1 mark]

$(p^{2} > 2 q) (81 > 2 \times 24)$ (so) nothing can be deduced R1

Note: Do not allow FT for the R mark.

[1 mark]

h.i.

$- 1$ A1

[1 mark]

h.ii.

attempt to express as a product of a linear and cubic factor M1

$(x + 1) (x^{3} - 10 x^{2} + 34 x - 12)$ A1A1

Note: Award A1 for each factor. Award at most A1A0 if not written as a product.

since for the cubic, $p^{2} < 3 q (100 < 102)$ R1

there is at least one complex root AG

[4 marks]

h.iii.

Examiners report

The first part of this question proved to be very accessible, with the majority of candidates expanding their brackets as required, to find the coefficients $p, q$ and $r$ .

The first part of this question was usually answered well, though presentation in the second part sometimes left a lot to be desired. The expression $2 (p^{2} - 2 q) - 2 q$ was expected to be seen more often, as a 'pivot' to reaching the required result. Algebra was often lengthy, but untidily so, sometimes leaving examiners to do some mental tidying up on behalf of the candidate.

b.i.

[N/A]

b.ii.

A good number of candidates recognised the reasoning required in this part of the question and were able to score both marks.

Most candidates found applying this specific case to be very straightforward.

Most candidates offered incorrect answers in the first part; despite their working suggested utilisation of the GDC, it was clear that many did not appreciate what the question was asking. The second part was usually answered well, with the idea of complex roots occurring in conjugate pairs being put to good use.

e.i.

[N/A]

e.ii.

Some very dubious algebra was seen here, and often no algebra at all. Despite this, a good number of candidates seemed to make the 'leap' to the correct expression $p^{2} - 2 q$ , perhaps fortuitously so in a number of cases.

f.i.

[N/A]

f.ii.

Of those finding $p^{2} - 2 q$ in part f, a surprising number of answers seen employed the test of checking whether $p^{2} < 3 q$ .

Part i was usually not answered successfully, which may have been due to shortage of time. However, it was pleasing to see a number of candidates reach the end of the paper and successfully factorise the given quartic using a variety of methods. The final part required the $p^{2} < 3 q$ test. Though correct reasoning was sometimes seen, it was rare for this final mark to be gained.

h.i.

[N/A]

h.ii.

[N/A]

h.iii.

This question will explore connections between complex numbers and regular polygons.

The diagram below shows a sector of a circle of radius 1, with the angle subtended at the centre $O$ being $\alpha {\text{,}}\,\,0 < \alpha < \frac{\pi }{2}$ . A perpendicular is drawn from point $P$ to intersect the $x$ -axis at $Q$ . The tangent to the circle at $P$ intersects the $x$ -axis at $R$ .

By considering the area of two triangles and the area of the sector show that ${\text{cos}}\,\alpha \,{\text{sin}}\,\alpha < \alpha < \frac{{{\text{sin}}\,\alpha }}{{{\text{cos}}\,\alpha }}$ .

[5]

Hence show that $\mathop {{\text{lim}}}\limits_{\alpha \to 0} \frac{\alpha }{{{\text{sin}}\,\alpha }} = 1$ .

[2]

Let ${z^n} = 1{\text{,}}\,\,z \in \mathbb{C}{\text{,}}\,\,n \in \mathbb{N}{\text{,}}\,\,n \geqslant 5$ . Working in modulus/argument form find the $n$ solutions to this equation.

[8]

Represent these $n$ solutions on an Argand diagram. Let their positions be denoted by ${P_0}{\text{,}}\,\,{P_1}{\text{,}}\,\,{P_2}{\text{,}}\, \ldots {P_{n - 1}}$ placed in order in an anticlockwise direction round the circle, starting on the positive $x$ -axis. Show the positions of ${P_0}{\text{,}}\,\,{P_1}{\text{,}}\,\,{P_2}$ and ${P_{n - 1}}$ .

[1]

Show that the length of the line segment ${P_0}{P_1}$ is $2\,{\text{sin}}\frac{\pi }{n}$ .

[4]

Hence, write down the total length of the perimeter of the regular $n$ sided polygon ${P_0}{P_1}{P_2} \ldots {P_{n - 1}}{P_0}$ .

[1]

Using part (b) find the limit of this perimeter as $n \to \infty$ .

[2]

Find the total area of this $n$ sided polygon.

[3]

Using part (b) find the limit of this area as $n \to \infty$ .

[2]

Markscheme

Area triangle $OPQ = \frac{1}{2}{\text{cos}}\,\alpha \,{\text{sin}}\,\alpha$ A1

Area sector $= \frac{1}{2}{1^2}\alpha$ A1

Area triangle $OPR = \frac{1}{2}1\,{\text{tan}}\,\alpha$ A1

So looking at the diagram $\frac{1}{2}{\text{cos}}\,\alpha \,{\text{sin}}\,\alpha < \frac{1}{2}\alpha < \frac{1}{2}\frac{{{\text{sin}}\,\alpha }}{{{\text{cos}}\,\alpha }}$ M1

$\Rightarrow {\text{cos}}\,\alpha \,{\text{sin}}\,\alpha < \alpha < \frac{{{\text{sin}}\,\alpha }}{{{\text{cos}}\,\alpha }}$ AG

[5 marks]

Hence ${\text{cos}}\,\alpha < \frac{\alpha }{{{\text{sin}}\,\alpha }} < \frac{1}{{{\text{cos}}\,\alpha }}$ and as $\alpha \to 0{\text{,}}\,\,{\text{cos}}\,\alpha \to 1$ we have M1R1

$\mathop {{\text{lim}}}\limits_{\alpha \to 0} \frac{\alpha }{{{\text{sin}}\,\alpha }} = 1$ AG

[2 marks]

${\left( {r\,cis\,\theta } \right)^n} = 1\,cis\,0 \Rightarrow {r^n}cis\,n\,\theta = 1\,cis\,\theta$ M1A1M1A1

${r^n} = 1 \Rightarrow r = 1$ $n\theta = 0 + 2\pi k{\text{,}}\,\,k \in \mathbb{Z}$ A1A1

$\theta = \frac{{2\pi k}}{n}{\text{,}}\,\,0 \leqslant k \leqslant n - 1$ A1

$z = cis\frac{{2\pi k}}{n}{\text{,}}\,\,0 \leqslant k \leqslant n - 1$ A1

[8 marks]

[1 mark]

Bisecting the triangle $O{P_0}{P_1}$ to form two right angle triangles M1

Length of ${P_0}{P_1} = 2t$ where $t = {\text{sin}}\left( {\frac{{\frac{{2\pi }}{n}}}{2}} \right)$ M1A1A1

So length is $2\,{\text{sin}}\frac{\pi }{n}$ AG

[4 marks]

Length of perimeter is $2n\,{\text{sin}}\frac{\pi }{n}$ A1

[1 mark]

$2n\,{\text{sin}}\frac{\pi }{n} = 2\pi \frac{n}{\pi }{\text{sin}}\frac{\pi }{n} \to 2\pi$ as $n \to \infty$ M1A1

[2 marks]

Area of $O{P_0}{P_1} = \frac{1}{2}1 \times 1\,{\text{sin}}\frac{{2\pi }}{n}$ so total area is $\frac{n}{2}{\text{sin}}\frac{{2\pi }}{n}$ . M1A1A1

[3 marks]

$\frac{n}{2}{\text{sin}}\frac{{2\pi }}{n} = \pi \frac{n}{{2\pi }}{\text{sin}}\frac{{2\pi }}{n} \to \pi$ as $n \to \infty$ M1A1

[2 marks]

Examiners report

[N/A]

In this question you will be exploring the strategies required to solve a system of linear differential equations.

Consider the system of linear differential equations of the form:

$\frac{d x}{d t} = x - y$ and $\frac{d y}{d t} = a x + y$ ,

where $x, y, t \in ℝ^{+}$ and $a$ is a parameter.

First consider the case where $a = 0$ .

Now consider the case where $a = - 1$ .

Now consider the case where $a = - 4$ .

From previous cases, we might conjecture that a solution to this differential equation is $y = F e^{λ t}$ , $λ \in ℝ$ and $F$ is a constant.

By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.

[3]

a.i.

Show that $\frac{d x}{d t} - x = - A e^{t}$ .

[1]

a.ii.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$ .

[4]

a.iii.

By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .

[3]

b.i.

By substituting $Y = \frac{d y}{d t}$ , show that $Y = B e^{2 t}$ where $B$ is a constant.

[3]

b.ii.

Hence find $y$ as a function of $t$ .

[2]

b.iii.

Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.

[3]

b.iv.

Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[3]

c.i.

Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[4]

c.ii.

Let the two values found in part (c)(ii) be $λ_{1}$ and $λ_{2}$ .

Verify that $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

[4]

c.iii.

Markscheme

METHOD 1

$\frac{d y}{d t} = y$

$\int \frac{d y}{y} = \int d t$ (M1)

$\ln y = t + c$ OR $\ln |y| = t + c$ A1A1

Note: Award A1 for $\ln y$ and A1 for $t$ and $c$ .

$y = A e^{t}$ AG

METHOD 2

rearranging to $\frac{d y}{d t} - y = 0$ AND multiplying by integrating factor $e^{- t}$ M1

$y e^{- t} = A$ A1A1

$y = A e^{t}$ AG

[3 marks]

a.i.

substituting $y = A e^{t}$ into differential equation in $x$ M1

$\frac{d x}{d t} = x - A e^{t}$

$\frac{d x}{d t} - x = - A e^{t}$ AG

[1 mark]

a.ii.

integrating factor (IF) is $e^{\int - 1 d t}$ (M1)

$= e^{- t}$ (A1)

$e^{- t} \frac{d x}{d t} - x e^{- t} = - A$

$x e^{- t} = - A t + D$ (A1)

$x = (- A t + D) e^{t}$ A1

Note: The first constant must be $A$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[4 marks]

a.iii.

$\frac{d^{2} y}{d t^{2}} = - \frac{d x}{d t} + \frac{d y}{d t}$ A1

EITHER

$= - x + y + \frac{d y}{d t}$ (M1)

$= \frac{d y}{d t} + \frac{d y}{d t}$ A1

$= - x + y + (- x + y)$ (M1)

$= 2 (- x + y)$ A1

THEN

$= 2 \frac{d y}{d t}$ AG

[3 marks]

b.i.

$\frac{d Y}{d t} = 2 Y$ A1

$\int \frac{d Y}{Y} = \int 2 d t$ M1

$\ln |Y| = 2 t + c$ OR $\ln Y = 2 t + c$ A1

$Y = B e^{2 t}$ AG

[3 marks]

b.ii.

$\frac{d y}{d t} = B e^{2 t}$

$y = \int B e^{2 t} d t$ M1

$y = \frac{B}{2} e^{2 t} + C$ A1

Note: The first constant must be $B$ , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[2 marks]

b.iii.

METHOD 1

substituting $\frac{d y}{d t} = B e^{2 t}$ and their (iii) into $\frac{d y}{d t} = - x + y$ M1(M1)

$B e^{2 t} = - x + \frac{B}{2} e^{2 t} + C$ A1

$x = - \frac{B}{2} e^{2 t} + C$ AG

Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.

METHOD 2

$\frac{d x}{d t} = x - \frac{B}{2} e^{2 t} - C$

$\frac{d x}{d t} - x = - \frac{B}{2} e^{2 t} - C$

$\frac{d (x e^{- t})}{d t} = - \frac{B}{2} e^{t} - C e^{- t}$ M1

$x e^{- t} = \int - \frac{B}{2} e^{t} - C e^{- t} d t$

$x e^{- t} = - \frac{B}{2} e^{t} - C e^{- t} + D$ A1

$x = - \frac{B}{2} e^{2 t} + C + D e^{t}$

$\frac{d y}{d t} = - x + y \Rightarrow B e^{2 t} = \frac{B}{2} e^{2 t} - C - D e^{t} + \frac{B}{2} e^{2 t} + C \Rightarrow D = 0$ M1

$x = - \frac{B}{2} e^{2 t} + C$ AG

[3 marks]

b.iv.

$\frac{d y}{d t} = - 4 x + y$

$\frac{d^{2} y}{d t^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t}$ seen anywhere M1

METHOD 1

$\frac{d^{2} y}{d t^{2}} = - 4 (x - y) + \frac{d y}{d t}$

attempt to eliminate $x$ M1

$= - 4 (\frac{1}{4} (y - \frac{d y}{d t}) - y) + \frac{d y}{d t}$

$= 2 \frac{d y}{d t} + 3 y$ A1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ AG

METHOD 2

rewriting LHS in terms of $x$ and $y$ M1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = (- 8 x + 5 y) - 2 (- 4 x + y) - 3 y$ A1

$= 0$ AG

[3 marks]

c.i.

$\frac{d y}{d t} = F λ e^{λ t}, \frac{d^{2} y}{d t^{2}} = F λ^{2} e^{λ t}$ (A1)

$F λ^{2} e^{λ t} - 2 F λ e^{λ t} - 3 F e^{λ t} = 0$ (M1)

$λ^{2} - 2 λ - 3 = 0$ (since $e^{λ t} \neq 0$ ) A1

$λ_{1}$ and $λ_{2}$ are $3$ and $- 1$ (either order) A1

[4 marks]

c.ii.

METHOD 1

$y = F e^{3 t} + G e^{- t}$

$\frac{d y}{d t} = 3 F e^{3 t} - G e^{- t}, \frac{d^{2} y}{d t^{2}} = 9 F e^{3 t} - G e^{- t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 9 F e^{3 t} + G e^{- t} - 2 (3 F e^{3 t} - G e^{- t}) - 3 (F e^{3 t} - G e^{- t})$ M1

$= 9 F e^{3 t} + G e^{- t} - 6 F e^{3 t} + 2 G e^{- t} - 3 F e^{3 t} - 3 G e^{- t}$ A1

$= 0$ AG

METHOD 2

$y = F e^{λ_{1} t} + G e^{λ_{2} t}$

$\frac{d y}{d t} = F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}, \frac{d^{2} y}{d t^{2}} = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t} - 2 (F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}) - 3 (F e^{λ_{1} t} + G e^{λ_{2} t})$ M1

$= F e^{λ_{1} t} (λ^{2} - 2 λ - 3) + G e^{λ_{2} t} (λ^{2} - 2 λ - 3)$ A1

$= 0$ AG

[4 marks]

c.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

b.iv.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

A Gaussian integer is a complex number, $z$ , such that $z = a + b i$ where $a, b \in ℤ$ . In this question, you are asked to investigate certain divisibility properties of Gaussian integers.

Consider two Gaussian integers, $α = 3 + 4 i$ and $β = 1 - 2 i$ , such that $γ = α β$ for some Gaussian integer $γ$ .

Now consider two Gaussian integers, $α = 3 + 4 i$ and $γ = 11 + 2 i$ .

The norm of a complex number $z$ , denoted by $N (z)$ , is defined by $N (z) = {|z|}^{2}$ . For example, if $z = 2 + 3 i$ then $N (2 + 3 i) = 2^{2} + 3^{2} = 13$ .

A Gaussian prime is a Gaussian integer, $z$ , that cannot be expressed in the form $z = α β$ where $α, β$ are Gaussian integers with $N (α), N (β) > 1$ .

The positive integer $2$ is a prime number, however it is not a Gaussian prime.

Let $α, β$ be Gaussian integers.

The result from part (h) provides a way of determining whether a Gaussian integer is a Gaussian prime.

Find $γ$ .

[2]

Determine whether $\frac{γ}{α}$ is a Gaussian integer.

[3]

On an Argand diagram, plot and label all Gaussian integers that have a norm less than $3$ .

[2]

Given that $α = a + b i$ where $a, b \in ℤ$ , show that $N (α) = a^{2} + b^{2}$ .

[1]

By expressing the positive integer $n = c^{2} + d^{2}$ as a product of two Gaussian integers each of norm $c^{2} + d^{2}$ , show that $n$ is not a Gaussian prime.

[3]

Verify that $2$ is not a Gaussian prime.

[2]

Write down another prime number of the form $c^{2} + d^{2}$ that is not a Gaussian prime and express it as a product of two Gaussian integers.

[2]

Show that $N (α β) = N (α) N (β)$ .

[6]

Hence show that $1 + 4 i$ is a Gaussian prime.

[3]

Use proof by contradiction to prove that a prime number, $p$ , that is not of the form $a^{2} + b^{2}$ is a Gaussian prime.

[6]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$(3 + 4 i) (1 - 2 i) = 11 - 2 i$ (M1)A1

[2 marks]

$\frac{γ}{α} = \frac{41}{25} - \frac{38}{25} i$ (M1)A1

(Since $Re \frac{γ}{α} (= \frac{41}{25})$ and/or $Im \frac{γ}{α} (= - \frac{38}{25})$ are not integers)

$\frac{γ}{α}$ is not a Gaussian integer R1

Note: Award R1 for correct conclusion from their answer.

[3 marks]

$\pm 1, \pm i, 0$ plotted and labelled A1

$1 \pm i, - 1 \pm i$ plotted and labelled A1

Note: Award A1A0 if extra points to the above are plotted and labelled.

[2 marks]

$|z| = \sqrt{a^{2} + b^{2}}$ (and as $N (z) = {|z|}^{2}$ ) A1

then $N (α) = a^{2} + b^{2}$ AG

[1 mark]

$c^{2} + d^{2} = (c + d i) (c - d i)$ A1

and $N (c + d i) = N (c - d i) = c^{2} + d^{2}$ R1

$N (c + d i), N (c - d i) > 1$ (since $c, d$ are positive) R1

so $c^{2} + d^{2}$ is not a Gaussian prime, by definition AG

[3 marks]

$2 (= 1^{2} + 1^{2}) = (1 + i) (1 - i)$ (A1)

$N (1 + i) = N (1 - i) = 2$ A1

so $2$ is not a Gaussian prime AG

[2 marks]

For example, $5 (= 1^{2} + 2^{2}) = (1 + 2 i) (1 - 2 i)$ (M1)A1

[2 marks]

METHOD 1

Let $α = m + n i$ and $β = p + q i$

LHS:

$α β = (m p - n q) + (m q + n p) i$ M1

$N (α β) = {(m p - n q)}^{2} + {(m q + n p)}^{2}$ A1

${(m p)}^{2} - 2 m n p q + {(n q)}^{2} + {(m q)}^{2} + 2 m n p q + {(n p)}^{2}$ A1

${(m p)}^{2} + {(n q)}^{2} + {(m q)}^{2} + {(n p)}^{2}$ A1

RHS:

$N (α) N (β) = (m^{2} + n^{2}) (p^{2} + q^{2})$ M1

${(m p)}^{2} + {(m q)}^{2} + {(n p)}^{2} + {(n q)}^{2}$ A1

LHS = RHS and so $N (α β) = N (α) N (β)$ AG

METHOD 2

Let $α = m + n i$ and $β = p + q i$

LHS

$N (α β) = (m^{2} + n^{2}) (p^{2} + q^{2})$ M1

$= (m + n i) (m - n i) (p + q i) (p - q i)$ A1

$= (m + n i) (p + q i) (m - n i) (p - q i)$

$= ((m p - n q) + (m q + n p) i) ((m p - n q) - (m q + n p) i)$ M1A1

$= {(m p - n q)}^{2} + {(m q + n p)}^{2}$ A1

$N = ((m p - n q) + (m q + n p) i)$ A1

$= N (α) N (β)$ (= RHS) AG

[6 marks]

$N (1 + 4 i) = 17$ which is a prime (in $ℤ$ ) R1

if $1 + 4 i = α β$ then $17 = N (α β) = N (α) N (β)$ R1

we cannot have $N (α), N (β) > 1$ R1

Note: Award R1 for stating that $1 + 4 i$ is not the product of Gaussian integers of smaller norm because no such norms divide $17$

so $1 + 4 i$ is a Gaussian prime AG

[3 marks]

Assume $p$ is not a Gaussian prime

$\Rightarrow p = α β$ where $α, β$ are Gaussian integers and $N (α), N (β) > 1$ M1

$\Rightarrow N (p) = N (α) N (β)$ M1

$p^{2} = N (α) N (β)$ A1

It cannot be $N (α) = 1, N (β) = p^{2}$ from definition of Gaussian prime R1

hence $N (α) = p, N (β) = p$ R1

If $α = a + b i$ then $N (α) = a^{2} + b^{2} = p$ which is a contradiction R1

hence a prime number, $p$ , that is not of the form $a^{2} + b^{2}$ is a Gaussian prime AG

[6 marks]

Examiners report

[N/A]

In this question you will explore some of the properties of special functions $f$ and $g$ and their relationship with the trigonometric functions, sine and cosine.

Functions $f$ and $g$ are defined as $f (z) = \frac{e^{z} + e^{- z}}{2}$ and $g (z) = \frac{e^{z} - e^{- z}}{2}$ , where $z \in ℂ$ .

Consider $t$ and $u$ , such that $t, u \in ℝ$ .

Using $e^{i u} = \cos u + i \sin u$ , find expressions, in terms of $\sin u$ and $\cos u$ , for

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point ( $\cos θ, \sin θ$ ) defines points on the unit circle with equation $x^{2} + y^{2} = 1$ .

The functions $f (x)$ and $g (x)$ are known as hyperbolic functions, as the general point ( $f (θ), g (θ)$ ) defines points on a curve known as a hyperbola with equation $x^{2} - y^{2} = 1$ . This hyperbola has two asymptotes.

Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .

[2]

Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .

[3]

$f (i u)$ .

[3]

c.i.

$g (i u)$ .

[2]

c.ii.

Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

[2]

Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .

[4]

Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.

[4]

The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .

[5]

Markscheme

$f' (t) = \frac{e^{t} - e^{- t}}{2}$ A1

$f'' (t) = \frac{e^{t} + e^{- t}}{2}$ A1

$= f (t)$ AG

[2 marks]

METHOD 1

${(f (t))}^{2} + {(g (t))}^{2}$

substituting $f$ and $g$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$

$= \frac{{(e^{t})}^{2} + 2 + {(e^{- t})}^{2} + {(e^{t})}^{2} - 2 + {(e^{- t})}^{2}}{4}$ (M1)

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2} (= \frac{e^{2 t} + e^{- 2 t}}{2})$ A1

$= f (2 t)$ AG

METHOD 2

$f (2 t) = \frac{e^{2 t} + e^{- 2 t}}{2}$

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2}$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$ M1A1

$= {(f (t))}^{2} + {(g (t))}^{2}$ AG

Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.

[3 marks]

substituting $e^{i u} = \cos u + i \sin u$ into the expression for $f$ (M1)

obtaining $e^{-i u} = \cos u - i \sin u$ (A1)

$f (i u) = \frac{\cos u + i \sin u + \cos u - i \sin u}{2}$

Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.

$= \frac{2 \cos u}{2}$

$= \cos u$ A1

[3 marks]

c.i.

$g (i u) = \frac{\cos u + i \sin u - \cos u + i \sin u}{2}$

substituting and attempt to simplify (M1)

$= \frac{2 i \sin u}{2}$

$= i \sin u$ A1

[2 marks]

c.ii.

METHOD 1

${(f (i u))}^{2} + {(g (i u))}^{2}$

substituting expressions found in part (c) (M1)

$= \cos^{2} u - \sin^{2} u (= \cos 2 u)$ A1

METHOD 2

$f (2 i u) = \frac{e^{2 i u} + e^{- 2 i u}}{2}$

$= \frac{\cos 2 u + i \sin 2 u + \cos 2 u - i \sin 2 u}{2}$ M1

$= \cos 2 u$ A1

Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg $2 \cos^{2} u - 1$ etc

[2 marks]

${(f (t))}^{2} - {(g (t))}^{2} = \frac{{(e^{t} + e^{- t})}^{2} - {(e^{t} - e^{- t})}^{2}}{4}$ M1

$= \frac{(e^{2 t} + e^{- 2 t} + 2) - (e^{2 t} + e^{- 2 t} - 2)}{4}$ A1

$= \frac{4}{4} = 1$ A1

Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.

${(f (i u))}^{2} - {(g (i u))}^{2} = \cos^{2} u + \sin^{2} u$ M1

$= 1$ (hence ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ ) AG

Note: Award full marks for showing that ${(f (z))}^{2} - {(g (z))}^{2} = 1, \forall z \in ℂ$ .

[4 marks]

A1A1A1A1

Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct $x$ -intercepts of $(- 1, 0)$ and $(1, 0)$ (condone $x = - 1$ and $1$ ), A1 for $y = x$ and $y = - x$ .

[4 marks]

attempt to rotate by $45 °$ in either direction (M1)

Note: Evidence of an attempt to relate to a sketch of $x y = k$ would be sufficient for this (M1).

attempting to rotate a particular point, eg $(1, 0)$ (M1)

$(1, 0)$ rotates to $(\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}})$ (or similar) (A1)

hence $k = \pm \frac{1}{2}$ A1A1

[5 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

This question asks you to investigate and prove a geometric property involving the roots of the equation $z^{n} = 1$ where $z \in ℂ$ for integers $n$ , where $n \geq 2$ .

The roots of the equation $z^{n} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}, \dots, ω^{n - 1}$ , where $ω = e^{\frac{2 πi}{n}}$ . Each root can be represented by a point $P_{0}, P_{1}, P_{2}, \dots, P_{n - 1}$ , respectively, on an Argand diagram.

For example, the roots of the equation $z^{2} = 1$ where $z \in ℂ$ are $1$ and $ω$ . On an Argand diagram, the root $1$ can be represented by a point $P_{0}$ and the root $ω$ can be represented by a point $P_{1}$ .

Consider the case where $n = 3$ .

The roots of the equation $z^{3} = 1$ where $z \in ℂ$ are $1$ , $ω$ and $ω^{2}$ . On the following Argand diagram, the points $P_{0}, P_{1}$ and $P_{2}$ lie on a circle of radius $1$ unit with centre $O (0, 0)$ .

Line segments $[P_{0} P_{1}]$ and $[P_{0} P_{2}]$ are added to the Argand diagram in part (a) and are shown on the following Argand diagram.

$P_{0} P_{1}$ is the length of $[P_{0} P_{1}]$ and $P_{0} P_{2}$ is the length of $[P_{0} P_{2}]$ .

Consider the case where $n = 4$ .

The roots of the equation $z^{4} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}$ and $ω^{3}$ .

On the following Argand diagram, the points $P_{0}, P_{1}, P_{2}$ and $P_{3}$ lie on a circle of radius $1$ unit with centre $O (0, 0)$ . $[P_{0} P_{1}]$ , $[P_{0} P_{2}]$ and $[P_{0} P_{3}]$ are line segments.

For the case where $n = 5$ , the equation $z^{5} = 1$ where $z \in ℂ$ has roots $1, ω, ω^{2}, ω^{3}$ and $ω^{4}$ .

It can be shown that $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} \times P_{0} P_{4} = 5$ .

Now consider the general case for integer values of $n$ , where $n \geq 2$ .

The roots of the equation $z^{n} = 1$ where $z \in ℂ$ are $1, ω, ω^{2}, \dots, ω^{n - 1}$ . On an Argand diagram, these roots can be represented by the points $P_{0}, P_{1}, P_{2}, \dots, P_{n - 1}$ respectively where $[P_{0} P_{1}], [P_{0} P_{2}], \dots, [P_{0} P_{n - 1}]$ are line segments. The roots lie on a circle of radius $1$ unit with centre $O (0, 0)$ .

$P_{0} P_{1}$ can be expressed as $| 1 - ω |$ .

Consider $z^{n} - 1 = (z - 1) (z^{n - 1} + z^{n - 2} + \dots + z + 1)$ where $z \in ℂ$ .

Show that $(ω - 1) (ω^{2} + ω + 1) = ω^{3} - 1$ .

[2]

a.i.

Hence, deduce that $ω^{2} + ω + 1 = 0$ .

[2]

a.ii.

Show that $P_{0} P_{1} \times P_{0} P_{2} = 3$ .

[3]

By factorizing $z^{4} - 1$ , or otherwise, deduce that $ω^{3} + ω^{2} + ω + 1 = 0$ .

[2]

Show that $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ .

[4]

Suggest a value for $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}$ .

[1]

Write down expressions for $P_{0} P_{2}$ and $P_{0} P_{3}$ in terms of $ω$ .

[2]

f.i.

Hence, write down an expression for $P_{0} P_{n - 1}$ in terms of $n$ and $ω$ .

[1]

f.ii.

Express $z^{n - 1} + z^{n - 2} + \dots + z + 1$ as a product of linear factors over the set $ℂ$ .

[3]

g.i.

Hence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).

[4]

g.ii.

Markscheme

METHOD 1

attempts to expand $(ω - 1) (ω^{2} + ω + 1)$ (M1)

$= ω^{3} + ω^{2} + ω - ω^{2} - ω - 1$ A1

$= ω^{3} - 1$ AG

METHOD 2

attempts polynomial division on $\frac{ω^{3} - 1}{ω - 1}$ M1

$= ω^{2} + ω + 1$ A1

so $(ω - 1) (ω^{2} + ω + 1) = ω^{3} - 1$ AG

Note: In part (a), award marks as appropriate where $ω$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

a.i.

(since $ω$ is a root of $z^{3} = 1$ ) $\Rightarrow ω^{3} - 1 = 0$ R1

and $ω \neq 1$ R1

$\Rightarrow ω^{2} + ω + 1 = 0$ AG

Note: In part (a), award marks as appropriate where $ω$ has been converted into Cartesian, modulus-argument (polar) or Euler form.

[2 marks]

a.ii.

METHOD 1

attempts to find either $P_{0} P_{1}$ or $P_{0} P_{2}$ (M1)

accept any valid method

e.g. $2 \sin \frac{π}{3}, 1^{2} + 1^{2} - 2 \cos \frac{2 π}{3}, \frac{1}{\sin \frac{π}{6}} = \frac{P_{0} P_{1}}{\sin \frac{2 π}{3}}$ from either ${ΔOP}_{0} P_{1}$ or ${ΔOP}_{0} P_{2}$

e.g. use of Pythagoras’ theorem

e.g. $|1 - e^{i \frac{2 π}{3}}|, |1 - (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)|$ by calculating the distance between $2$ points

$P_{0} P_{1} = \sqrt{3}$ A1

$P_{0} P_{2} = \sqrt{3}$ A1

Note: Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either $P_{0} P_{1}$ or $P_{0} P_{2}$ or both.

so $P_{0} P_{1} \times P_{0} P_{2} = 3$ AG

METHOD 2

attempts to find $P_{0} P_{1} \times P_{0} P_{2} = |1 - ω| |1 - ω^{2}|$ (M1)

$P_{0} P_{1} \times P_{0} P_{2} = |ω^{3} - ω^{2} - ω + 1|$ A1

$= |1 - (ω^{2} + ω + 1) + 2|$ and since $ω^{2} + ω + 1 = 0$ R1

so $P_{0} P_{1} \times P_{0} P_{2} = 3$ AG

[3 marks]

METHOD 1

$z^{4} - 1 = (z - 1) (z^{3} + z^{2} + z + 1)$ A1

( $ω$ is a root hence) $ω^{4} - 1 = 0$ and $ω \neq 1$ R1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

Note: Condone the use of $ω$ throughout.

METHOD 2

considers the sum of roots of $z^{4} - 1 = 0$ (M1)

the sum of roots is zero (there is no $z^{3}$ term) A1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

METHOD 3

substitutes for $ω$ (M1)

e.g. $LHS = e^{i \frac{3 π}{2}} + e^{πi} + e^{i \frac{π}{2}} + 1$

$= - i - 1 + i + 1$ A1

Note: This can be demonstrated geometrically or by using vectors. Accept Cartesian or modulus-argument (polar) form.

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

METHOD 4

$ω^{3} + ω^{2} + ω + 1 = \frac{ω^{4} - 1}{ω - 1}$ A1

$= \frac{0}{ω - 1} = 0$ as $ω \neq 1$ R1

$\Rightarrow ω^{3} + ω^{2} + ω + 1 = 0$ AG

[2 marks]

METHOD 1

$P_{0} P_{2} = 2$ A1

attempts to find either $P_{0} P_{1}$ or $P_{0} P_{3}$ (M1)

Note: For example $P_{0} P_{1} = |1 - i|$ and $P_{0} P_{3} = |1 + i|$ .
Various geometric and trigonometric approaches can be used by candidates.

$P_{0} P_{1} = \sqrt{2}, P_{0} P_{3} = \sqrt{2}$ A1A1

Note: Award a maximum of A1M1A1A0 if labels such as $P_{0} P_{1}$ are not clearly shown.
Award full marks if the lengths are shown on a clearly labelled diagram.
Award a maximum of A1M1A1A0 for any decimal approximation seen in the calculation of either $P_{0} P_{1}$ or $P_{0} P_{3}$ or both.

$P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

METHOD 2

attempts to find $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = |1 - ω| |1 - ω^{2}| |1 - ω^{3}|$ M1

$P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = |- ω^{6} + ω^{5} + ω^{4} - ω^{2} - ω + 1|$ A1

$= |- (- 1) + ω^{5} + 1 - (- 1) - ω + 1|$ since $ω^{6} = ω^{2} = - 1$ and $ω^{4} = 1$ A1

$= |ω^{5} - ω + 4|$ and since $ω^{5} = ω$ R1

so $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

METHOD 3

$P_{0} P_{2} = 2$ A1

attempts to find $P_{0} P_{1} \times P_{0} P_{3} = |1 - ω| |1 - ω^{3}|$ M1

$P_{0} P_{1} \times P_{0} P_{3} = |ω^{4} - ω^{3} - ω + 1|$ A1

$= |2 - (- ω) - ω|$ since $ω^{4} = 1$ and $ω^{3} = - ω$ R1

so $P_{0} P_{1} \times P_{0} P_{2} \times P_{0} P_{3} = 4$ AG

[4 marks]

$(P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1}) = n$ A1

[1 mark]

$P_{0} P_{2} = |1 - ω^{2}|, P_{0} P_{3} = |1 - ω^{3}|$ A1A1

[2 marks]

f.i.

$P_{0} P_{n - 1} = |1 - ω^{n - 1}|$ A1A1

Note: Accept $|1 - ω|$ from symmetry.

[1 mark]

f.ii.

$z^{n} - 1 = (z - 1) (z^{n - 1} + z^{n - 2} + \dots + z + 1)$

considers the equation $z^{n - 1} + z^{n - 2} + \dots + z + 1 = 0$ (M1)

the roots are $ω, ω^{2}, \dots, ω^{n - 1}$ (A1)

so $(z - ω) (z - ω^{2}) \dots (z - ω^{n - 1})$ A1

[3 marks]

g.i.

METHOD 1

substitutes $z = 1$ into $(z - ω) (z - ω^{2}) \dots (z - ω^{n - 1}) \equiv z^{n - 1} + z^{n - 2} + \dots + z + 1$ M1

$(1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1}) = n$ (A1)

takes modulus of both sides M1

$|(1 - ω) (1 - ω^{2}) \dots (1 - ω^{n - 1})| = |n|$

$|1 - ω| |1 - ω^{2}| \dots |1 - ω^{n - 1}| = n$ A1

so $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1} = n$ AG

Note: Award a maximum of M1A1FTM1A0 from part (e).

METHOD 2

$(1 - ω), (1 - ω^{2}), \dots, (1 - ω^{n - 1})$ are the roots of ${(1 - v)}^{n - 1} + {(1 - v)}^{n - 2} + \dots + (1 - v) + 1 = 0$ M1

coefficient of $v^{n - 1}$ is ${(- 1)}^{n - 1}$ and the coefficient of $1$ is $n$ A1

product of the roots is $\frac{{(- 1)}^{n - 1} n}{{(- 1)}^{n - 1}} = n$ A1

$|1 - ω| |1 - ω^{2}| \dots |1 - ω^{n - 1}| = n$ A1

so $P_{0} P_{1} \times P_{0} P_{2} \times \dots \times P_{0} P_{n - 1} = n$ AG

[4 marks]

g.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

g.i.

[N/A]

g.ii.

This question asks you to explore some properties of polygonal numbers and to determine and prove interesting results involving these numbers.

A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers.

For example, a triangular number is a number that can be arranged in the shape of an equilateral triangle. The first five triangular numbers are $1, 3, 6, 10$ and $15$ .

The following table illustrates the first five triangular, square and pentagonal numbers respectively. In each case the first polygonal number is one represented by a single dot.

For an $r$ -sided regular polygon, where $r \in ℤ^{+}, r \geq 3$ , the $n$ th polygonal number $P_{r} (n)$ is given by

$P_{r} (n) = \frac{(r - 2) n^{2} - (r - 4) n}{2}$ , where $n \in ℤ^{+}$ .

Hence, for square numbers, $P_{4} (n) = \frac{(4 - 2) n^{2} - (4 - 4) n}{2} = n^{2}$ .

The $n$ th pentagonal number can be represented by the arithmetic series

$P_{5} (n) = 1 + 4 + 7 + \dots + (3 n - 2)$ .

For triangular numbers, verify that $P_{3} (n) = \frac{n (n + 1)}{2}$ .

[2]

a.i.

The number $351$ is a triangular number. Determine which one it is.

[2]

a.ii.

Show that $P_{3} (n) + P_{3} (n + 1) \equiv {(n + 1)}^{2}$ .

[2]

b.i.

State, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.

[1]

b.ii.

For $n = 4$ , sketch a diagram clearly showing your answer to part (b)(ii).

[1]

b.iii.

Show that $8 P_{3} (n) + 1$ is the square of an odd number for all $n \in ℤ^{+}$ .

[3]

Hence show that $P_{5} (n) = \frac{n (3 n - 1)}{2}$ for $n \in ℤ^{+}$ .

[3]

By using a suitable table of values or otherwise, determine the smallest positive integer, greater than $1$ , that is both a triangular number and a pentagonal number.

[5]

A polygonal number, $P_{r} (n)$ , can be represented by the series

$Σ_{m = 1}^{n} (1 + (m - 1) (r - 2))$ where $r \in ℤ^{+}, r \geq 3$ .

Use mathematical induction to prove that $P_{r} (n) = \frac{(r - 2) n^{2} - (r - 4) n}{2}$ where $n \in ℤ^{+}$ .

[8]

Markscheme

$P_{3} (n) = \frac{(3 - 2) n^{2} - (3 - 4) n}{2}$ OR $P_{3} (n) = \frac{n^{2} - (- n)}{2}$ A1

$P_{3} (n) = \frac{n^{2} + n}{2}$ A1

Note: Award A0A1 if $P_{3} (n) = \frac{n^{2} + n}{2}$ only is seen.

Do not award any marks for numerical verification.

so for triangular numbers, $P_{3} (n) = \frac{n (n + 1)}{2}$ AG

[2 marks]

a.i.

METHOD 1

uses a table of values to find a positive integer that satisfies $P_{3} (n) = 351$ (M1)

for example, a list showing at least $3$ consecutive terms $(\dots 325, 351, 378 \dots)$

Note: Award (M1) for use of a GDC’s numerical solve or graph feature.

$n = 26$ ( $26$ th triangular number) A1

Note: Award A0 for $n = - 27, 26$ . Award A0 if additional solutions besides $n = 26$ are given.

METHOD 2

attempts to solve $\frac{n (n + 1)}{2} = 351 (n^{2} + n - 702 = 0)$ for $n$ (M1)

$n = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 702)}}{2}$ OR $(n - 26) (n + 27) = 0$

$n = 26$ ( $26$ th triangular number) A1

Note: Award A0 for $n = - 27, 26$ . Award A0 if additional solutions besides $n = 26$ are given.

[2 marks]

a.ii.

attempts to form an expression for $P_{3} (n) + P_{3} (n + 1)$ in terms of $n$ M1

EITHER

$P_{3} (n) + P_{3} (n + 1) \equiv \frac{n (n + 1)}{2} + \frac{(n + 1) (n + 2)}{2}$

$\equiv \frac{(n + 1) (2 n + 2)}{2} (\equiv \frac{2 (n + 1) (n + 1)}{2})$ A1

$P_{3} (n) + P_{3} (n + 1) \equiv (\frac{n^{2}}{2} + \frac{n}{2}) + (\frac{{(n + 1)}^{2}}{2} + \frac{n + 1}{2})$

$\equiv (\frac{n^{2} + n}{2}) + (\frac{n^{2} + 2 n + 1 + n + 1}{2}) (\equiv n^{2} + 2 n + 1)$ A1

THEN

$\equiv {(n + 1)}^{2}$ AG

[2 marks]

b.i.

the sum of the $n$ th and $(n + 1)$ th triangular numbers

is the $(n + 1)$ th square number A1

[1 mark]

b.ii.

Note: Accept equivalent single diagrams, such as the one above, where the $4$ th and $5$ th triangular numbers and the $5$ th square number are clearly shown.
Award A1 for a diagram that show $P_{3} (4)$ (a triangle with $10$ dots) and $P_{3} (5)$ (a triangle with $15$ dots) and $P_{4} (5)$ (a square with $25$ dots).

[1 mark]

b.iii.

METHOD 1

$8 P_{3} (n) + 1 = 8 (\frac{n (n + 1)}{2}) + 1 (= 4 n (n + 1) + 1)$ A1

attempts to expand their expression for $8 P_{3} (n) + 1$ (M1)

$= 4 n^{2} + 4 n + 1$

$= {(2 n + 1)}^{2}$ A1

and $2 n + 1$ is odd AG

METHOD 2

$8 P_{3} (n) + 1 = 8 ({(n + 1)}^{2} - P_{3} (n + 1)) + 1 (= 8 ({(n + 1)}^{2} - \frac{(n + 1) (n + 2)}{2}) + 1)$ A1

attempts to expand their expression for $8 P_{3} (n) + 1$ (M1)

$8 (n^{2} + 2 n + 1) - 4 (n^{2} + 3 n + 2) + 1 (= 4 n^{2} + 4 n + 1)$

$= {(2 n + 1)}^{2}$ A1

and $2 n + 1$ is odd AG

Method 3

$8 P_{3} (n) + 1 = 8 (\frac{n (n + 1)}{2}) + 1 (= {(A n + B)}^{2})$ (where $A, B \in ℤ^{+}$ ) A1

attempts to expand their expression for $8 P_{3} (n) + 1$ (M1)

$4 n^{2} + 4 n + 1 (= A^{2} n^{2} + 2 A B n + B^{2})$

now equates coefficients and obtains $B = 1$ and $A = 2$

$= {(2 n + 1)}^{2}$ A1

and $2 n + 1$ is odd AG

[3 marks]

EITHER

$u_{1} = 1$ and $d = 3$ (A1)

substitutes their $u_{1}$ and their $d$ into $P_{5} (n) = \frac{n}{2} (2 u_{1} + (n - 1) d)$ M1

$P_{5} (n) = \frac{n}{2} (2 + 3 (n - 1)) (= \frac{n}{2} (2 + 3 n - 3))$ A1

$u_{1} = 1$ and $u_{n} = 3 n - 2$ (A1)

substitutes their $u_{1}$ and their $u_{n}$ into $P_{5} (n) = \frac{n}{2} (u_{1} + u_{n})$ M1

$P_{5} (n) = \frac{n}{2} (1 + 3 n - 2)$ A1

$P_{5} (n) = (3 (1) - 2) + (3 (2) - 2) + (3 (3) - 2) + \dots 3 n - 2$

$P_{5} (n) = (3 (1) + 3 (2) + 3 (3) + \dots + 3 n) - 2 n (= 3 (1 + 2 + 3 + \dots + n) - 2 n)$ (A1)

substitutes $\frac{n (n + 1)}{2}$ into their expression for $P_{5} (n)$ M1

$P_{5} (n) = 3 (\frac{n (n + 1)}{2}) - 2 n$

$P_{5} (n) = \frac{n}{2} (3 (n + 1) - 4)$ A1

attempts to find the arithmetic mean of $n$ terms (M1)

$= \frac{1 + (3 n - 2)}{2}$ A1

multiplies the above expression by the number of terms $n$

$P_{5} (n) = \frac{n}{2} (1 + 3 n - 2)$ A1

THEN

so $P_{5} (n) = \frac{n (3 n - 1)}{2}$ AG

[3 marks]

METHOD 1

forms a table of $P_{3} (n)$ values that includes some values for $n > 5$ (M1)

forms a table of $P_{5} (m)$ values that includes some values for $m > 5$ (M1)

Note: Award (M1) if at least one $P_{3} (n)$ value is correct. Award (M1) if at least one $P_{5} (m)$ value is correct. Accept as above for $(n^{2} + n)$ values and $(3 m^{2} - m)$ values.

$n = 20$ for triangular numbers (A1)

$m = 12$ for pentagonal numbers (A1)

Note: Award (A1) if $n = 20$ is seen in or out of a table. Award (A1) if $m = 12$ is seen in or out of a table. Condone the use of the same parameter for triangular numbers and pentagonal numbers, for example, $n = 20$ for triangular numbers and $n = 12$ for pentagonal numbers.

$210$ (is a triangular number and a pentagonal number) A1

Note: Award all five marks for $210$ seen anywhere with or without working shown.

METHOD 2

EITHER

attempts to express $P_{3} (n) = P_{5} (m)$ as a quadratic in $n$ (M1)

$n^{2} + n + (m - 3 m^{2}) (= 0)$ (or equivalent)

attempts to solve their quadratic in $n$ (M1)

$n = \frac{- 1 \pm \sqrt{12 m^{2} - 4 m + 1}}{2} (= \frac{- 1 \pm \sqrt{1^{2} - 4 (m - 3 m^{2})}}{2})$

attempts to express $P_{3} (n) = P_{5} (m)$ as a quadratic in $m$ (M1)

$3 m^{2} - m - (n^{2} + n) (= 0)$ (or equivalent)

attempts to solve their quadratic in $m$ (M1)

$m = \frac{1 \pm \sqrt{12 n^{2} - 12 n + 1}}{6} (= \frac{1 \pm \sqrt{{(- 1)}^{2} + 12 (n^{2} + n)}}{6})$

THEN

$n = 20$ for triangular numbers (A1)

$m = 12$ for pentagonal numbers (A1)

$210$ (is a triangular number and a pentagonal number) A1

METHOD 3

$\frac{n (n + 1)}{2} = \frac{m (3 m - 1)}{2}$

let $n = m + k (n > m)$ and so $3 m^{2} - m = (m + k) (m + k + 1)$ M1

$2 m^{2} - 2 (k + 1) m - (k^{2} + k) = 0$ A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square M1

$Δ = 4 {(k + 1)}^{2} + 8 (k^{2} + k)$

$N^{2} = 4 {(k + 1)}^{2} + 8 (k^{2} + k) (= 4 (k + 1) (3 k + 1))$

determines that $k = 8$ leading to $2 m^{2} - 18 m - 72 = 0 \Rightarrow m = - 3, 12$ and so $m = 12$ A1

$210$ (is a triangular number and a pentagonal number) A1

METHOD 4

$\frac{n (n + 1)}{2} = \frac{m (3 m - 1)}{2}$

let $m = n - k (m < n)$ and so $n^{2} + n = (n - k) (3 (n - k) - 1)$ M1

$2 n^{2} - 2 (3 k + 1) n + (3 k^{2} + k) = 0$ A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square M1

$Δ = 4 {(3 k + 1)}^{2} - 8 (3 k^{2} + k)$

$N^{2} = 4 {(3 k + 1)}^{2} - 8 (3 k^{2} + k) (= 4 (k + 1) (3 k + 1))$

determines that $k = 8$ leading to $2 n^{2} - 50 n + 200 = 0 \Rightarrow n = 5, 20$ and so $n = 20$ A1

$210$ (is a triangular number and a pentagonal number) A1

[5 marks]

Note: Award a maximum of R1M0M0A1M1A1A1R0 for a ‘correct’ proof using $n$ and $n + 1$ .

consider $n = 1 : P_{r} (1) = 1 + (1 - 1) (r - 2) = 1$ and $P_{r} (1) = \frac{(r - 2) (1^{2}) - (r - 4) (1)}{2} = 1$

so true for $n = 1$ R1

Note: Accept $P_{r} (1) = 1$ and $P_{r} (1) = \frac{(r - 2) (1^{2}) - (r - 4) (1)}{2} = 1$ .
Do not accept one-sided considerations such as ' $P_{r} (1) = 1$ and so true for $n = 1$ '.
Subsequent marks after this R1 are independent of this mark can be awarded.

Assume true for $n = k$ , ie. $P_{r} (k) = \frac{(r - 2) k^{2} - (r - 4) k}{2}$ M1

Note: Award M0 for statements such as “let $n = k$ ”. The assumption of truth must be clear.
Subsequent marks after this M1 are independent of this mark and can be awarded.

Consider $n = k + 1 :$

( $P_{r} (k + 1)$ can be represented by the sum

$Σ_{m = 1}^{k + 1} (1 + (m - 1) (r - 2)) = Σ_{m = 1}^{k} (1 + (m - 1) (r - 2)) + (1 + k (r - 2))$ and so

$P_{r} (k + 1) = \frac{(r - 2) k^{2} - (r - 4) k}{2} + (1 + k (r - 2)) (P_{r} (k + 1) = P_{r} (k) + (1 + k (r - 2)))$ M1

$= \frac{(r - 2) k^{2} - (r - 4) k + 2 + 2 k (r - 2)}{2}$ A1

$= \frac{(r - 2) (k^{2} + 2 k) - (r - 4) k + 2}{2}$

$= \frac{(r - 2) (k^{2} + 2 k + 1) - (r - 2) - (r - 4) k + 2}{2}$ M1

$= \frac{(r - 2) {(k + 1)}^{2} - (r - 4) k - (r - 4)}{2}$ (A1)

$= \frac{(r - 2) {(k + 1)}^{2} - (r - 4) (k + 1)}{2}$ A1

hence true for $n = 1$ and $n = k$ true $\Rightarrow n = k + 1$ true R1

therefore true for all $n \in ℤ^{+}$

Note: Only award the final R1 if the first five marks have been awarded. Award marks as appropriate for solutions that expand both the LHS and (given) RHS of the equation.

[8 marks]

Examiners report

Part (a) (i) was generally well done. Unfortunately, some candidates adopted numerical verification. Part (a) (ii) was generally well done with the majority of successful candidates using their GDC judiciously and disregarding n = −27 as a possible solution. A few candidates interpreted the question as needing to deal with P₃(351).

Although part (b) (i) was generally well done, a significant number of candidates laboured unnecessarily to show the required result. Many candidates set their LHS to equal the RHS throughout the solution. Part (b) (ii) was generally not well done with many candidates unable to articulate clearly in words and symbols what the given identity shows for the sum of two consecutive triangular numbers. In part (b) (iii), most candidates were unable to produce a clear diagram illustrating the identity stated in part (b) (i).

Part (c) was reasonably well done. Most candidates were able to show algebraically that $8 P_{3} (n) + 1 = 4 n^{2} + 4 n + 1$ . A good number of candidates were then able to express $4 n^{2} + 4 n + 1$ as ${(2 n + 1)}^{2}$ and conclude that $(2 n + 1)$ is odd. Rather than making the connection that $4 n^{2} + 4 n + 1$ is a perfect square, many candidates attempted instead to analyse the parity of either $4 n (n + 1) + 1$ or $4 n^{2} + 4 n + 1$ . As with part (b) (i), many candidates set their LHS to equal the RHS throughout the solution. A number of candidates unfortunately adopted numerical verification.

Part (d) was not answered as well as anticipated with many candidates not understanding what was
required. Instead of using the given arithmetic series to show that $P_{5} (n) = \frac{n (3 n - 1)}{2}$ , a large number of
candidates used $P_{5} (n) = \frac{(5 - 2) n^{2} - (5 - 4) n}{2}$ . Unfortunately, a number of candidates adopted numerical verification.

In part (e), the overwhelming majority of candidates who successfully determined that 210 is the smallest positive integer greater than 1 that is both triangular and pentagonal used a table of values. Unfortunately, a large proportion of these candidates seemingly spent quite a few minutes listing the first 20 triangular numbers and the first 12 pentagonal numbers. And it can be surmised that a number of these candidates constructed their table of values either without the use of a GDC or with the arithmetic functionality of a GDC rather than with a GDC's table of values facility. Candidates should be aware that a relevant excerpt from a table of values is sufficient evidence of correct working. A number of candidates started constructing a table of values but stopped before identifying 210. Disappointingly, a significant number of candidates attempted to solve $P_{3} (n) = P_{5} (n)$ for $n$ .

Part (f) proved beyond the reach of most with only a small number of candidates successfully proving the given result. A significant number of candidates were unable to show that the result is true for $n = 1$ . A number of candidates established the validity of the base case for the RHS only while a number of other candidates attempted to prove the base case for r = 3. A large number of candidates did not state the inductive step correctly with the assumption of truth not clear. A number of candidates then either attempted to work backwards from the given result or misinterpreted the question and attempted to prove the result stated in the question stem rather than the result stated in the question. Some candidates who were awarded the first answer mark when considering the $n = k + 1$ case were unable to complete the square or equivalent simplification correctly. Disappointingly, a significant number listed the steps involved in an induction proof without engaging in the actual proof.

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

This question asks you to explore cubic polynomials of the form $(x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ and corresponding cubic equations with one real root and two complex roots of the form $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ .

In parts (a), (b) and (c), let $r = 1, a = 4$ and $b = 1$ .

Consider the equation $(z - 1) (z^{2} - 8 z + 17) = 0$ for $z \in ℂ$ .

Consider the function $f (x) = (x - 1) (x^{2} - 8 x + 17)$ for $x \in ℝ$ .

Consider the function $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

The equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ has roots $r$ and $a \pm b i$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

On the Cartesian plane, the points $C_{1} (a, \sqrt{g' (a)})$ and $C_{2} (a, - \sqrt{g' (a)})$ represent the real and imaginary parts of the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ .

The following diagram shows a particular curve of the form $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ and the tangent to the curve at the point $A (a, 80)$ . The curve and the tangent both intersect the $x$ -axis at the point $R (- 2, 0)$ . The points $C_{1}$ and $C_{2}$ are also shown.

Consider the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a \neq r, b > 0$ . The points $A (a, g (a))$ and $R (r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$ .

Consider the special case where $a = r$ and $b > 0$ .

Given that $1$ and $4 + i$ are roots of the equation, write down the third root.

[1]

a.i.

Verify that the mean of the two complex roots is $4$ .

[1]

a.ii.

Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .

[4]

Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.

[2]

Show that $g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ .

[2]

d.i.

Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .

[6]

d.ii.

Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .

[1]

Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .

[4]

f.i.

State the coordinates of $C_{2}$ .

[1]

f.ii.

Show that the $x$ -coordinate of $P$ is $\frac{1}{3} (2 a + r)$ .

You are not required to demonstrate a change in concavity.

[2]

g.i.

Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$ .

[1]

g.ii.

Sketch the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a = r = 1$ and $b = 2$ .

[2]

h.i.

For $a = r$ and $b > 0$ , state in terms of $r$ , the coordinates of points $P$ and $A$ .

[1]

h.ii.

Markscheme

$4 - i$ A1

[1 mark]

a.i.

mean $= \frac{1}{2} (4 + i + 4 - i)$ A1

$= 4$ AG

[1 mark]

a.ii.

METHOD 1

attempts product rule differentiation (M1)

Note: Award (M1) for attempting to express $f (x)$ as $f (x) = x^{3} - 9 x^{2} + 25 x - 17$

$f' (x) = (x - 1) (2 x - 8) + x^{2} - 8 x + 17 (f' (x) = 3 x^{2} - 18 x + 25)$ A1

$f' (4) = 1$ A1

Note: Where $f' (x)$ is correct, award A1 for solving $f' (x) = 1$ and obtaining $x = 4$ .

EITHER

$y - 3 = 1 (x - 4)$ A1

$y = x + c$

$3 = 4 + c \Rightarrow c = - 1$ A1

states the gradient of $y = x - 1$ is also $1$ and verifies that $(4, 3)$ lies on the line $y = x - 1$ A1

THEN

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f' (x)$ .

METHOD 2

sets $f (x) = x - 1$ to form $x - 1 = (x - 1) (x^{2} - 8 x + 17)$ (M1)

EITHER

$(x - 1) (x^{2} - 8 x + 16) = 0 (x^{3} - 9 x^{2} + 24 x - 16 = 0)$ A1

attempts to solve a correct cubic equation (M1)

$(x - 1) {(x - 4)}^{2} = 0 \Rightarrow x = 1, 4$

recognises that $x \neq 1$ and forms $x^{2} - 8 x + 17 = 1 (x^{2} - 8 x + 16 = 0)$ A1

attempts to solve a correct quadratic equation (M1)

${(x - 4)}^{2} = 0 \Rightarrow x = 4$

THEN

$x = 4$ is a double root R1

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Candidates using this method are not required to verify that $y = 3$ .

[4 marks]

a positive cubic with an $x$ -intercept $(x = 1)$ , and a local maximum and local minimum in the first quadrant both positioned to the left of $A$ A1

Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as $A, (4, 3)$ or the point labelled from both axes. Coordinates are not required.

a correct sketch of the tangent passing through $A$ and crossing the $x$ -axis at the same point $(x = 1)$ as the curve A1

Note: Award A1A0 if both graphs cross the $x$ -axis at distinctly different points.

[2 marks]

EITHER

$g' (x) = (x - r) (2 x - 2 a) + x^{2} - 2 a x + a^{2} + b^{2}$ (M1)A1

$g (x) = x^{3} - (2 a + r) x^{2} + (a^{2} + b^{2} + 2 a r) x - (a^{2} + b^{2}) r$

attempts to find $g' (x)$ M1

$g' (x) = 3 x^{2} - 2 (2 a + r) x + a^{2} + b^{2} + 2 a r$

$= 2 x^{2} - 2 (a + r) x + 2 a r + x^{2} - 2 a x + a^{2} + b^{2}$ A1

$(= 2 (x^{2} - a x - r x + a r) + x^{2} - 2 a x + a^{2} + b^{2})$

THEN

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ AG

[2 marks]

d.i.

METHOD 1

$g (a) = b^{2} (a - r)$ (A1)

$g' (a) = b^{2}$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y - g (a) = g' (a) (x - a)$ M1

$y - b^{2} (a - r) = b^{2} (x - a)$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $- b^{2} (a - r) = b^{2} (x - a)$ M1

$b > 0$ OR $b \neq 0 \Rightarrow - (a - r) = x - a$ R1

$x = r$ A1

THEN

so the tangent intersects the $x$ -axis at the point $R (r, 0)$ AG

METHOD 2

$g' (a) = b^{2}$ (A1)

$g (a) = b^{2} (a - r)$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y = g' (a) x + c$ and attempts to find $c$ M1

$c = - b^{2} r$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0$ OR $b \neq 0 \Rightarrow x - r = 0$ R1

$x = r$ A1

METHOD 3

$g' (a) = b^{2}$ (A1)

the line through $R (r, 0)$ parallel to the tangent at $A$ has equation
$y = b^{2} (x - r)$ A1

sets $g (x) = b^{2} (x - r)$ to form $b^{2} (x - r) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ M1

$b^{2} = x^{2} - 2 a x + a^{2} + b^{2}, (x \neq r)$ A1

${(x - a)}^{2} = 0$ A1

since there is a double root $(x = a)$ , this parallel line through $R (r, 0)$ is the required tangent at $A$ R1

[6 marks]

d.ii.

EITHER

$g' (a) = b^{2} \Rightarrow b = \sqrt{g' (a)}$ (since $b > 0$ ) R1

Note: Accept $b = \pm \sqrt{g' (a)}$ .

$(a \pm b i=) a \pm i \sqrt{b^{2}}$ and $g' (a) = b^{2}$ R1

THEN

hence the complex roots can be expressed as $a \pm i \sqrt{g' (a)}$ AG

[1 mark]

$b = 4$ (seen anywhere) A1

EITHER

attempts to find the gradient of the tangent in terms of $a$ and equates to $16$ (M1)

OR

substitutes $r = - 2, x = a$ and $y = 80$ to form $80 = (a - (- 2)) (a^{2} - 2 a^{2} + a^{2} + 16)$ (M1)

substitutes $r = - 2, x = a$ and $y = 80$ into $y = 16 (x - r)$ (M1)

THEN

$\frac{80}{a + 2} = 16 \Rightarrow a = 3$

roots are $- 2$ (seen anywhere) and $3 \pm 4 i$ A1A1

Note: Award A1 for $- 2$ and A1 for $3 \pm 4 i$ . Do not accept coordinates.

[4 marks]

f.i.

$(3, - 4)$ A1

Note: Accept “ $x = 3$ and $y = - 4$ ”.
Do not award A1FT for $(a, - 4)$ .

[1 mark]

f.ii.

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$

attempts to find $g'' (x)$ M1

$g'' (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a (= 6 x - 2 r - 4 a)$

sets $g'' (x) = 0$ and correctly solves for $x$ A1

for example, obtaining $x - r + 2 (x - a) = 0$ leading to $3 x = 2 a + r$

so $x = \frac{1}{3} (2 a + r)$ AG

Note: Do not award A1 if the answer does not lead to the AG.

[2 marks]

g.i.

point $P$ is $\frac{2}{3}$ of the horizontal distance (way) from point $R$ to point $A$ A1

Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ $P$ is between $R$ and $A$ , closer to $A$ ”.

[1 mark]

g.ii.

$y = (x - 1) (x^{2} - 2 x + 5)$ (A1)

a positive cubic with no stationary points and a non-stationary point of inflexion at $x = 1$ A1

Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of $x = 1$ is apparent.
Coordinates are not required and the $y$ -intercept need not be indicated.

[2 marks]

h.i.

$(r, 0)$ A1

[1 mark]

h.ii.

Examiners report

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4 - i$ as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of $4 + i$ and $4 - i$ .

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f' (x)$ and a good number of those candidates were able to determine that $f' (4) = 1$ . Candidates that did not determine the equation of the tangent had to state that the gradient of $y = x - 1$ is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find $f' (x)$ .

Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.

Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding $g (x)$ , then differentiating to find $g' (x)$ and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that $g' (a) = b^{2}$ and/or $g (a) = b^{2} (a - r)$ . Only the very best candidates obtained full marks by concluding that as $b > 0$ or $b \neq 0$ , then $x = r$ when $y = 0$ .

In general, only the best candidates were able to use the result $g' (a) = b^{2}$ to deduce that the complex roots of the equation can be expressed as $a \pm i \sqrt{g' (a)}$ . Although given the complex roots $a \pm b i$ , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation $z^{2} - 2 a z + a^{2} + b^{2} = 0$ .

In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state $- 2$ as a root. Some candidates determined that $b = 4$ but were unable to use the diagram to determine that $a = 3$ . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C₂ . The most frequent error was to give the y-coordinate as $3 - 4 i$ .

Of the candidates who attempted part (g) (i), most were able to find an expression for $g'' (x)$ and a reasonable number of these were then able to convincingly show that $x = \frac{1}{3} (2 a + r)$ . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated $x = \frac{1}{3} (2 a + r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y = (x - 1) (x^{2} - 2 x + 5)$ . However, most graphs were poorly drawn with many showing a change in concavity at $x = 0$ rather than at $x = 1$ . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).

a.i.

[N/A]

a.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

g.i.

[N/A]

g.ii.

[N/A]

h.i.

[N/A]

h.ii.

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{{{\text{d}}P}}{{{\text{d}}t}} = kP$ , where $k \in \mathbb{R}$ , $t$ is the time (in years) and $P$ is the population

The initial population is 1000.

Given that $k = 0.003$ , use your answer from part (a) to find

Consider now the situation when $k$ is not a constant, but a function of time.

Given that $k = 0.003 + 0.002t$ , find

Another model for population growth assumes

there is a maximum value for the population, $L$ .
that $k$ is not a constant, but is proportional to $\left( {1 - \frac{P}{L}} \right)$ .

Show that the general solution of this differential equation is $P = A{{\text{e}}^{kt}}$ , where $A \in \mathbb{R}$ .

[5]

the population after 10 years

[2]

b.i.

the number of years it will take for the population to triple.

[2]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P$

[1]

b.iii.

the solution of the differential equation, giving your answer in the form $P = f\left( t \right)$ .

[5]

c.i.

the number of years it will take for the population to triple.

[4]

c.ii.

Show that $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , where $m \in \mathbb{R}$ .

[2]

Solve the differential equation $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , giving your answer in the form $P = g\left( t \right)$ .

[10]

Given that the initial population is 1000, $L = 10000$ and $m = 0.003$ , find the number of years it will take for the population to triple.

[4]

Markscheme

$\int {\frac{1}{P}} {\text{d}}P = \int {k{\text{d}}t}$ M1A1

${\text{ln}}\,P = kt + c$ A1A1

$P = {e^{kt + c}}$ A1

$P = A{e^{kt}}$ , where $A = {e^c}$ AG

[5 marks]

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow A = 1000$ A1

$P\left( {10} \right) = 1000{e^{0.003\left( {10} \right)}} = 1030$ A1

[2 marks]

b.i.

$3000 = 1000{e^{0.003t}}$ M1

$t = \frac{{{\text{ln}}\,3}}{{0.003}} = 366$ years A1

[2 marks]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P = \infty$ A1

[1 mark]

b.iii.

$\int {\frac{1}{P}} {\text{d}}P = \int {\left( {0.003 + 0.002t} \right){\text{d}}t}$ M1

${\text{ln}}\,P = 0.003t + 0.001{t^2} + c$ A1A1

$P = {e^{0.003t + 0.001{t^2} + c}}$ A1

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow {e^c} = 1000$ M1

$P = 1000{e^{0.003t + 0.001{t^2}}}$

[5 marks]

c.i.

$3000 = 1000{e^{0.003t + 0.001{t^2}}}$ M1

${\text{ln}}\,3 = 0.003t + 0.001{t^2}$ A1

Use of quadratic formula or GDC graph or GDC polysmlt M1

$t = 31.7$ years A1

[4 marks]

c.ii.

$k = m\left( {1 - \frac{P}{L}} \right)$ , where $m$ is the constant of proportionality A1

So $\frac{{{\text{d}}P}}{{{\text{d}}t}} = m\left( {1 - \frac{P}{L}} \right)P$ A1

$\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ AG

[2 marks]

$\int {\frac{1}{{P\left( {L - P} \right)}}} {\text{d}}P = \int {\frac{m}{L}{\text{d}}t}$ M1

$\frac{1}{{P\left( {L - P} \right)}} = \frac{A}{P} + \frac{B}{{L - P}}$ M1

$1 \equiv A\left( {L - P} \right) + BP$ A1

$A = \frac{1}{L}{\text{,}}\,\,B = \frac{1}{L}$ A1

$\frac{1}{L}\int {\left( {\frac{1}{P} + \frac{1}{{L - P}}} \right){\text{d}}P} = \int {\frac{m}{L}{\text{d}}t}$

$\frac{1}{L}\left( {{\text{ln}}\,P - {\text{ln}}\left( {L - P} \right)} \right) = \frac{m}{L}t + c$ A1A1

${\text{ln}}\left( {\frac{P}{{L - P}}} \right) = mt + d$ , where $d = cL$ M1

$\frac{P}{{L - P}} = C{e^{mt}}$ , where $C = {e^d}$ A1

$P\left( {1 + C{e^{mt}}} \right) = CL{e^{mt}}$ M1

$P = \frac{{CL{e^{mt}}}}{{\left( {1 + C{e^{mt}}} \right)}}{\text{ }}\left( { = \frac{L}{{\left( {D{e^{ - mt}} + 1} \right)}}{\text{,}}\,\,{\text{where}}\;D = \frac{1}{C}} \right)$ A1

[10 marks]

$1000 = \frac{{10000}}{{D + 1}}$ M1

$D = 9$ A1

$3000 = \frac{{10000}}{{9{e^{ - 0.003t}} + 1}}$ M1

$t = 450$ years A1

[4 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

In parts (b) and (c), ${\left( {abc \ldots } \right)_n}$ denotes the number ${abc \ldots }$ written in base $n$ , where $n \in {\mathbb{Z}^ + }$ . For example, ${\left( {359} \right)_n} = 3{n^2} + 5n + 9$ .

State Fermat’s little theorem.

[2]

a.i.

Find the remainder when ${15^{1207}}$ is divided by $13$ .

[5]

a.ii.

Convert ${\left( {7A2} \right)_{16}}$ to base $5$ , where ${\left( A \right)_{16}} = {\left( {10} \right)_{10}}$ .

[4]

Consider the equation ${\left( {1251} \right)_n} + {\left( {30} \right)_n} = {\left( {504} \right)_n} + {\left( {504} \right)_n}$ .

Find the value of $n$ .

[4]

Markscheme

EITHER

${a^p} \equiv a\left( {{\text{mod}}\,p} \right)$ A1

where $p$ is prime A1

${a^{p - 1}} \equiv 1\left( {{\text{mod}}\,p} \right)$ A1

where $p$ is prime and $p$ does not divide $a$ (or equivalent statement) A1

[2 marks]

a.i.

${15^{1207}} \equiv {2^{1207}}\left( {{\text{mod}}\,13} \right)$

${2^{12}} \equiv 1\left( {{\text{mod}}\,13} \right)$ (M1)(A1)

${2^{1207}} = {\left( {{2^{12}}} \right)^{100}}{2^7}$ (M1)

${2^{1207}}\left( { \equiv {2^7}} \right) \equiv 11\left( {{\text{mod}}\,13} \right)$ (M1)A1

the remainder is $11$

Note: Award as above for using $15$ instead of $2$ .

[5 marks]

a.ii.

${\left( {7A2} \right)_{16}} = 7 \times {16^2} + 10 \times 16 + 2$ M1

$= 1954$ A1

EITHER

$5\left| \!{\underline {\, {1954} \,}} \right.$

$390\,r\,4$

$78\,r\,0$

$15\,r\,3$ M1

$3\,r\,0$

$0\,r\,3$

$1954 = 3 \times {5^4} + 0 \times {5^3} + 3 \times {5^2} + 0 \times {5^1} + 4$ M1

THEN

${\left( {7A2} \right)_{16}} = {\left( {30304} \right)_5}$ A1

[4 marks]

the equation can be written as

${n^3} + 2{n^2} + 5n + 1 + 3n = 2\left( {5{n^2} + 4} \right)$ M1A1

$\Rightarrow {n^3} - 8{n^2} + 8n - 7 = 0$ (M1)

Note: The (M1) is for an attempt to solve the original equation.

$n = 7$ A1

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Write down the remainder when $14^{2022}$ is divided by $7$ .

[1]

a.i.

Use Fermat’s little theorem to find the remainder when $14^{2022}$ is divided by $17$ .

[4]

a.ii.

Prove that a number in base $13$ is divisible by $6$ if, and only if, the sum of its digits is divisible by $6$ .

[4]

b.i.

The base $13$ number $1 y 93 y 25$ is divisible by $6$ . Find the possible values of the digit $y$ .

[4]

b.ii.

Markscheme

the remainder is $0$ A1

[1 mark]

a.i.

$14^{16} \equiv 1 (mod 17)$ (from Fermat’s little theorem) (A1)

$14^{2022} = 14^{16 \times 126 + 6}$ (M1)

Note: Award M1 for a $RHS$ exponent consistent with the correct use of Fermat’s little theorem.

$14^{2022} \equiv 14^{6} (mod 17) (\equiv 15 (mod 17))$ A1

the remainder is $15$ A1

[4 marks]

a.ii.

METHOD 1

let $N = a_{n} 13^{n} + a_{n - 1} 13^{n - 1} + \dots + a_{1} 13 + a_{0}$ M1

Note: The above M1 is independent of the A marks below.

$13 \equiv 1 (mod 6)$ A1

EITHER

$13^{x} \equiv 1 (mod 6)$ (for all $x \in ℕ$ ) A1

$N \equiv a_{n} 1^{n} + a_{n - 1} 1^{n - 1} + \dots + a_{1} 1 + a_{0} (mod 6) (N \equiv a_{n} + a_{n - 1} + \dots + a_{1} + a_{0} (mod 6))$ A1

THEN

so $N \equiv 0 (mod 6)$ if and only if $a_{n} + a_{n - 1} + \dots + a_{1} + a_{0} \equiv 0 (mod 6)$ R1

so $6 N$ if and only if $6 (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0})$ AG

METHOD 2

let $N = a_{n} 13^{n} + a_{n - 1} 13^{n - 1} + \dots + a_{1} 13 + a_{0}$ (M1)

$N = (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0}) + (13 - 1) (a_{n} (13^{n - 1} + \dots + 13^{0}) + a_{n - 1} (13^{n - 2} + \dots + 13^{0}) + \dots + a_{1} 13^{0})$ M1A1

Note: Award M1 for attempting to express $N$ in the form $N = (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0}) + (13 - 1) M$ .

as $6 (13 - 1) M$ R1

so $6 N$ if and only if $6 (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0})$ AG

[4 marks]

b.i.

METHOD 1

the sum of the digits is $2 y + 20$ (A1)

uses $2 y + 20 = 6 k$ (or equivalent) to attempt to find a valid value of $y$ (M1)

$y = 2, 5, 8, 11 (B)$ A1A1

Note: Award A1 for $y = 2, 5, 8$ and A1 for $y = 11 (B)$ .

METHOD 2

${(1 y 93 y 25)}_{13} = 1 \times 13^{6} + y \times 13^{5} + 9 \times 13^{4} + 3 \times 13^{3} + y \times 13^{2} + 2 \times 13^{1} + 5 \times 13^{0}$ (A1)

$= 371462 y + 5090480$

attempts to find a valid value of $y$ such that

$371462 y + 5090480 \equiv 0 (mod 6)$ (M1)

$y = 2, 5, 8, 11 (B)$ A1A1

Note: Award A1 for $y = 2, 5, 8$ and A1 for $y = 11 (B)$ .

[4 marks]

b.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.