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HL Paper 1

An electric circuit has two power sources. The voltage, $V_{1}$ , provided by the first power source, at time $t$ , is modelled by

$V_{1} = Re (2 e^{3 t i})$ .

The voltage, $V_{2}$ , provided by the second power source is modelled by

$V_{2} = Re (5 e^{(3 t + 4) i})$ .

The total voltage in the circuit, $V_{T}$ , is given by

$V_{T} = V_{1} + V_{2}$ .

Find an expression for $V_{T}$ in the form $A \cos (B t + C)$ , where $A, B$ and $C$ are real constants.

[4]

Hence write down the maximum voltage in the circuit.

[1]

Markscheme

METHOD 1

recognizing that the real part is distributive (M1)

$V_{T} = Re (2 e^{3 t i} + 5 e^{3 t i + 4 i})$

$= Re (e^{3 t i} (2 + 5 e^{4 i}))$ (A1)

(from the GDC) $2 + 5 e^{4 i} = 3.99088 \dots e^{- 1.89418 \dots i}$ (A1)

Note: Accept arguments differing by $2 π$ e.g. $4.38900 \dots)$ .

therefore $V_{T} = 3.99 \cos (3 t - 1.89) (3.99088 \dots \cos (3 t - 1.89418 \dots))$ A1

Note: Award the last A1 for the correct values of $A, B$ and $C$ seen either in the required form or not. If method used is unclear and answer is partially incorrect, assume Method 2 and award appropriate marks eg. (M1)A1A0A0 if only $A$ value is correct.

METHOD 2

converting given expressions to cos form (M1)

$V_{T} = 2 \cos 3 t + 5 \cos (3 t + 4)$

(from graph) $A = 3.99 (3.99088 \dots)$ A1

$V_{T} = 3.99 \cos (B t + C)$

either by considering transformations or inserting points

$B = 3$ A1

$C = - 1.89 (- 1.89418 \dots)$ A1

Note: Accept arguments differing by $2 π$ e.g. $4.38900 \dots$ .

(so, $V_{T} = 3.99 \cos (3 t - 1.89) (3.99088 \dots \cos (3 t - 1.89418 \dots))$ )

Note: It is possible to have $A = 3.99$ , $B = - 3$ with $C = 1.89$ OR $A = - 3.99$ , $B = 3$ with $C = 1.25$ OR $A = - 3.99$ , $B = - 3$ with $C = - 1.25$ due to properties of the cosine curve.

[4 marks]

maximum voltage is $3.99 (3.99088 \dots)$ (units) A1

[1 mark]

Examiners report

The crucial step in this question was to realize that $Re (2 e^{3 t i})+Re (5 e^{(3 t + 4) i})=Re (2 e^{3 t i} + 5 e^{(3 t + 4) i})$ . Candidates who failed to do this step were usually unable to obtain the required result.

[N/A]

The following Argand diagram shows a circle centre $0$ with a radius of $4$ units.

A set of points, $\{z_{θ}\}$ , on the Argand plane are defined by the equation

$z_{θ} = \frac{1}{2} θ e^{θ i}$ , where $θ \geq 0$ .

Plot on the Argand diagram the points corresponding to

Consider the case where $|z_{θ}| = 4$ .

$θ = \frac{π}{2}$ .

[1]

a.i.

$θ = π$ .

[1]

a.ii.

$θ = \frac{3 π}{2}$ .

[1]

a.iii.

Find this value of $θ$ .

[2]

b.i.

For this value of $θ$ , plot the approximate position of $z_{θ}$ on the Argand diagram.

[2]

b.ii.

Markscheme

Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.

[1 mark]

a.i.

Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.

[1 mark]

a.ii.

Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.

[1 mark]

a.iii.

$\frac{1}{2} θ = 4$ (M1)

$\Rightarrow θ = 8$ A1

[2 marks]

b.i.

$z_{8}$ is shown in the diagram above A1A1

Note: Award A1 for a point plotted on the circle and A1 for a point plotted in the second quadrant.

[2 marks]

b.ii.

Examiners report

This question was challenging to many candidates, and some left the answer blank. Those who attempted it often failed to gain any marks. It would have helped examiners credit responses if points that were plotted on the Argand diagram were labelled. Certainly, there was some confusion caused by the appearance of θ both in the modulus and argument of the complex numbers in Euler form. Better use of technology to help visualize the complex numbers by simply getting decimal approximations of values in terms of π or by converting from Euler to Cartesian form would have helped in this question.

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

The sum of the first $n$ terms of a sequence $\{ {u_n}\}$ is given by ${S_n} = 3{n^2} - 2n$ , where $n \in {\mathbb{Z}^ + }$ .

Write down the value of ${u_1}$ .

[1]

Find the value of ${u_6}$ .

[2]

Prove that $\{ {u_n}\}$ is an arithmetic sequence, stating clearly its common difference.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${u_1} = 1$ A1

[1 mark]

${u_6} = {S_6} - {S_5} = 31$ M1A1

[2 marks]

${u_n} = {S_n} - {S_{n - 1}}$ M1

$= (3{n^2} - 2n) - \left( {3{{(n - 1)}^2} - 2(n - 1)} \right)$

$= (3{n^2} - 2n) - (3{n^2} - 6n + 3 - 2n + 2)$

$= 6n - 5$ A1

$d = {u_{n + 1}} - {u_n}$ R1

$= 6n + 6 - 5 - 6n + 5$

$= \left( {6(n + 1) - 5} \right) - (6n - 5)$

$= 6$ (constant) A1

Notes: Award R1 only if candidate provides a clear argument that proves that the difference between ANY two consecutive terms of the sequence is constant. Do not accept examples involving particular terms of the sequence nor circular reasoning arguments (eg use of formulas of APs to prove that it is an AP). Last A1 is independent of R1.

[4 marks]

Examiners report

[N/A]

Let C = $\left( {\begin{array}{*{20}{c}} { - 2}&4 \\ 1&7 \end{array}} \right)$ and D = $\left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$ .

The 2 × 2 matrix Q is such that 3Q = 2C – D

Find Q.

[3]

Find CD.

[4]

Find D^–1.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3Q = $\left( {\begin{array}{*{20}{c}} { - 4}&8 \\ 2&{14} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$ (A1)

3Q = $\left( {\begin{array}{*{20}{c}} { - 9}&6 \\ 3&{14 - a} \end{array}} \right)$ (A1)

Q = $\left( {\begin{array}{*{20}{c}} { - 3}&2 \\ 1&{\frac{{14 - a}}{3}} \end{array}} \right)$ (A1) (N3)

[3 marks]

CD = $\left( {\begin{array}{*{20}{c}} { - 2}&4 \\ 1&7 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5&2 \\ { - 1}&a \end{array}} \right)$

$= \left( {\begin{array}{*{20}{c}} { - 14}&{ - 4 + 4a} \\ { - 2}&{2 + 7a} \end{array}} \right)$ (A1)(A1)(A1)(A1) (N4)

[4 marks]

det D = 5 $a$ + 2 (may be implied) (A1)

D^–1 = $\frac{1}{{5a + 2}}\left( {\begin{array}{*{20}{c}} a&{ - 2} \\ 1&5 \end{array}} \right)$ (A1) (N2)

[2 marks]

Examiners report

[N/A]

Let $w = a{{\text{e}}^{\frac{\pi }{4}{\text{i}}}}$ , where $a \in {\mathbb{R}^ + }$ .

for $a$ = 2,

find the values of ${w^2}$ , ${w^3}$ , and ${w^4}$ .

[2]

a.i.

draw $w$ , ${w^2}$ , ${w^3}$ , and ${w^4}$ on the following Argand diagram.

[3]

a.ii.

Let $z = \frac{w}{{2 - {\text{i}}}}$ .

Find the value of $a$ for which successive powers of $z$ lie on a circle.

[2]

Markscheme

$4{{\text{e}}^{\frac{\pi }{2}{\text{i}}}}$ , $8{{\text{e}}^{\frac{{3\pi }}{4}{\text{i}}}}$ , $16{{\text{e}}^{\pi {\text{i}}}}$ ( $= 4{\text{i}}$ , $- {\text{4}}\sqrt 2 + 4\sqrt 2 {\text{i}}$ , $- 16$ ) (M1)A1

[2 marks]

a.i.

Note: Award A1 for correct arguments, award A1 for $4{\text{i}}$ and −16 clearly indicated, award A1 for | $w$ | < 4 and 4 < | ${w^3}$ | < 16.

[3 marks]

a.ii.

${2^2} + {1^2} = {a^2}$ M1

$a = \sqrt 5 \,\,\left( { = 2.24} \right)$ A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The geometric sequence u₁, u₂, u₃, … has common ratio r.

Consider the sequence $A = \left\{ {{a_n} = {\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right|{\text{:}}\,n \in {\mathbb{Z}^ + }} \right\}$ .

Show that A is an arithmetic sequence, stating its common difference d in terms of r.

[4]

A particular geometric sequence has u₁ = 3 and a sum to infinity of 4.

Find the value of d.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

state that ${u_n} = {u_1}{r^{n - 1}}$ (or equivalent) A1

attempt to consider ${{a_n}}$ and use of at least one log rule M1

${\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right| = {\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right| + \left( {n - 1} \right){\text{lo}}{{\text{g}}_2}\left| r \right|$ A1

(which is an AP) with $d = {\text{lo}}{{\text{g}}_2}\left| r \right|$ (and 1^st term ${\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right|$ ) A1

so A is an arithmetic sequence AG

Note: Condone absence of modulus signs.

Note: The final A mark may be awarded independently.

Note: Consideration of the first two or three terms only will score M0.

[4 marks]

METHOD 2

consideration of $\left( {d = } \right){a_{n + 1}} - {a_n}$ M1

$\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {{u_{n + 1}}} \right| - {\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right|$

$\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right|$ M1

$\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| r \right|$ A1

which is constant R1

Note: Condone absence of modulus signs.

Note: The final A mark may be awarded independently.

Note: Consideration of the first two or three terms only will score M0.

attempting to solve $\frac{3}{{1 - r}} = 4$ M1

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

$d = - \,2$ A1

[3 marks]

Examiners report

[N/A]

Consider $w = i z + 1$ , where $w, z \in ℂ$ .

Find $w$ when

Point $z$ on the Argand diagram can be transformed to point $w$ by two transformations.

$z = 2 i$ .

[2]

a.i.

$z = 1 + i$ .

[1]

a.ii.

Describe these two transformations and give the order in which they are applied.

[3]

Hence, or otherwise, find the value of $z$ when $w = 2 - i$ .

[2]

Markscheme

$i^{2} = - 1$ (M1)

$w = - 2 + 1 = - 1$ A1

[2 marks]

a.i.

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

[1 mark]

a.ii.

EITHER

rotation of $90 °$ (anticlockwise, centre at the origin) A1A1

Note: Award A1 for “rotation” and A1 for “ $90 °$ ”.

followed by a translation of $(\begin{matrix} 1 \\ 0 \end{matrix})$    A1

OR
translation of $(\begin{matrix} 0 \\ - 1 \end{matrix})$    A1

followed by rotation of $90 °$ (anticlockwise, centre at the origin)    A1A1

Note: Award A1 for “rotation” and A1 for “ $90 °$ ”.

[3 marks]

EITHER

move $1$ to left to $1 - i$ (M1)

then rotate by $- 90 °$ to

$- 1 - i$ A1

$i z + 1 = 2 - i$

$i z = 1 - i$

$z = \frac{1 - i}{i}$ (M1)

$- 1 - i$ A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Let $\left( {\begin{array}{*{20}{c}} b&3 \\ 7&8 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 9&5 \\ { - 2}&7 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4&8 \\ a&{15} \end{array}} \right)$ .

Write down the value of $a$ .

[1]

a.i.

Find the value of $b$ .

[2]

a.ii.

Let $3\left( {\begin{array}{*{20}{c}} { - 4}&8 \\ 2&1 \end{array}} \right) - 5\left( {\begin{array}{*{20}{c}} 2&0 \\ q&{ - 4} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 22}&{24} \\ 9&{23} \end{array}} \right)$ .

Find the value of $q$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$a$ = 5 A1 N1

[1 mark]

a.i.

$b$ + 9 = 4 (M1)

$b$ = −5 A1 N2

[2 marks]

a.ii.

Comparing elements 3(2) − 5( $q$ ) = −9 M1

$q$ = 3 A2 N2

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The diagram shows a sector, $OAB$ , of a circle with centre $O$ and radius $r$ , such that $AÔB = θ$ .

Sam measured the value of $r$ to be $2 cm$ and the value of $θ$ to be $30 °$ .

It is found that Sam’s measurements are accurate to only one significant figure.

Use Sam’s measurements to calculate the area of the sector. Give your answer to four significant figures.

[2]

Find the upper bound and lower bound of the area of the sector.

[3]

Find, with justification, the largest possible percentage error if the answer to part (a) is recorded as the area of the sector.

[3]

Markscheme

$π \times 2^{2} \times \frac{30}{360}$ (M1)

$= 1.047 {cm}^{2}$ A1

Note: Do not award the final mark if the answer is not correct to 4 sf.

[2 marks]

attempt to substitute any two values from $1.5, 2.5, 25$ or $35$ into area of sector formula (M1)

$(upper bound = π \times 2 . 5^{2} \times \frac{35}{360} =) 1.91 {cm}^{2} (1.90895 \dots)$ A1

$(lower bound = π \times 1 . 5^{2} \times \frac{25}{360} =) 0.491 {cm}^{2} (0.490873 \dots)$ A1

Note: Given the nature of the question, accept correctly rounded OR correctly truncated $3$ significant figure answers.

[3 marks]

$(|\frac{1.047 - 1.90895 \dots}{1.90895 \dots}| \times 100 =) 45.2 (%) (45.1532 \dots)$ A1

$(|\frac{1.047 - 0.490873 \dots}{0.490873 \dots}| \times 100 =) 113 (%) (113.293 \dots)$ A1

so the largest percentage error is $113 %$ A1

Note: Accept $45.1 (%)$ ( $45.1428$ ), from use of full accuracy answers. Given the nature of the question, accept correctly rounded OR correctly truncated $3$ significant figure answers. Award A0A1A0 if $113 %$ is the only value found.

[3 marks]

Examiners report

In part (a), the area was almost always found correctly although some candidates gave the answer 1.0472 which is correct to 4 decimal places, not 4 significant figures as required. In part (b), many candidates failed to realize that the upper bounds for r and θ were 2.5 and 35° and lower bounds were 1.5 and 25°. Consequently, the bounds for the area were incorrect. In many cases, the incorrect values in part (b) were followed through into part (c) although in the percentage error calculations, many candidates had 1.047 in the denominator instead of the appropriate bound.

[N/A]

In an arithmetic sequence, the sum of the 3rd and 8th terms is 1.

Given that the sum of the first seven terms is 35, determine the first term and the common difference.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempting to form two equations involving ${u_1}$ and $d$ M1

$\left( {{u_1} + 2d} \right) + \left( {{u_1} + 7d} \right) = 1$ and $\frac{7}{2}\left[ {2{u_1} + 6d} \right] = 35$

$2{u_1} + 9d = 1$

$14{u_1} + 42d = 70\,\,\,\left( {2{u_1} + 6d = 10} \right)$ A1

Note: Award A1 for any two correct equations

attempting to solve their equations: M1

${u_1} = 14$ , $d = - 3$ A1

[4 marks]

Examiners report

[N/A]

Let A, B and C be non-singular 2×2 matrices, I the 2×2 identity matrix and k a scalar. The following statements are incorrect. For each statement, write down the correct version of the right hand side.

(A + B)² = A² + 2AB + B²

[2]

(A – kI)³ = A³ – 3kA² + 3k²A – k³

[2]

CA = B C = $\frac{B}{A}$

[2]

Markscheme

(A + B)² = A² + AB + BA + B² A2

Note: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.

[2 marks]

(A – kI)³ = A³ – 3kA² + 3k²A – k³I A2

Note: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.

[2 marks]

CA = B ⇒ C = BA^–1 A2

Note: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.

[2 marks]

Examiners report

[N/A]

Write down $2 + 5 i$ in exponential form.

[2]

An equilateral triangle is to be drawn on the Argand plane with one of the vertices at the point corresponding to $2 + 5 i$ and all the vertices equidistant from $0$ .

Find the points that correspond to the other two vertices. Give your answers in Cartesian form.

[3]

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.

$5.385 \dots e^{1.1902 \dots i} \approx 5.39 e^{1.19 i}$ A1A1

Note: Accept equivalent answers: $5.39 e^{- 5.09 i}$

[2 marks]

multiply by $e^{\frac{2 π}{3} i}$ (M1)

$- 5.33 - 0.77 i, 3.33 - 4.23 i$ A1A1

[3 marks]

Examiners report

[N/A]

An infinite geometric sequence, with terms $u_{n}$ , is such that $u_{1} = 2$ and $Σ_{k = 1}^{\infty} u_{k} = 10$ .

Find the common ratio, $r$ , for the sequence.

[2]

Find the least value of $n$ such that $u_{n} < \frac{1}{2}$ .

[3]

Markscheme

$10 = \frac{2}{1 - r}$ (M1)

$r = 0.8$ A1

[2 marks]

$2 \times {(0.8)}^{n - 1} < 0.5$ OR $2 \times {(0.8)}^{n - 1} = 0.5$ (M1)

$(n >) 7.212 \dots$ (A1)

$n = 8$ A1

Note: If $n = 7$ is seen, with or without seeing the value $7.212 \dots$ then award M1A1A0.

[3 marks]

Examiners report

A number of candidates did not attempt what should have been a straightforward question. Perhaps because it relied on a part of the syllabus that is restricted to HL and is not in common with SL. Some attempted it but were unaware of the formula for the sum of an infinite geometric sequence, although this is in the formula booklet. By far the biggest error was to fail to recognize that $n$ was the smallest integer value greater than that found from solving the equation. There were disappointingly few candidates who adopted a tabular or graphical approach to this question using technology. Some relied on trial-and-error.

The 1st, 4th and 8th terms of an arithmetic sequence, with common difference $d$ , $d \ne 0$ , are the first three terms of a geometric sequence, with common ratio $r$ . Given that the 1st term of both sequences is 9 find

the value of $d$ ;

[4]

the value of $r$ ;

[1]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

the first three terms of the geometric sequence are $9$ , $9r$ and $9{r^2}$ (M1)

$9 + 3d = 9r( \Rightarrow 3 + d = 3r)$ and $9 + 7d = 9{r^2}$ (A1)

attempt to solve simultaneously (M1)

$9 + 7d = 9{\left( {\frac{{3 + d}}{3}} \right)^2}$

the ${{\text{1}}^{{\text{st}}}}$ , ${{\text{4}}^{{\text{th}}}}$ and ${{\text{8}}^{{\text{th}}}}$ terms of the arithmetic sequence are

$9,{\text{ }}9 + 3d,{\text{ }}9 + 7d$ (M1)

$\frac{{9 + 7d}}{{9 + 3d}} = \frac{{9 + 3d}}{9}$ (A1)

attempt to solve (M1)

THEN

$d = 1$ A1

[4 marks]

$r = \frac{4}{3}$ A1

Note: Accept answers where a candidate obtains $d$ by finding $r$ first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in $r$ .

[1 mark]

Examiners report

[N/A]

Consider $w = 2\left( {{\text{cos}}\frac{\pi }{3} + {\text{i}}\,{\text{sin}}\frac{\pi }{3}} \right)$

These four points form the vertices of a quadrilateral, Q.

Express w² and w³ in modulus-argument form.

[3]

a.i.

Sketch on an Argand diagram the points represented by w⁰ , w¹ , w² and w³.

[2]

a.ii.

Show that the area of the quadrilateral Q is $\frac{{21\sqrt 3 }}{2}$ .

[3]

Let $z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }$ . The points represented on an Argand diagram by ${z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}$ form the vertices of a polygon ${P_n}$ .

Show that the area of the polygon ${P_n}$ can be expressed in the form $a\left( {{b^n} - 1} \right){\text{sin}}\frac{\pi }{n}$ , where $a,\,\,b\, \in \mathbb{R}$ .

[6]

Markscheme

${w^2} = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right){\text{;}}\,\,{w^3} = 8{\text{cis}}\left( \pi \right)$ (M1)A1A1

Note: Accept Euler form.

Note: M1 can be awarded for either both correct moduli or both correct arguments.

Note: Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.

[3 marks]

a.i.

A1A1

[2 marks]

a.ii.

use of area = $\frac{1}{2}ab\,\,{\text{sin}}\,C$ M1

$\frac{1}{2} \times 1 \times 2 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 2 \times 4 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 4 \times 8 \times {\text{sin}}\frac{\pi }{3}$ A1A1

Note: Award A1 for $C = \frac{\pi }{3}$ , A1 for correct moduli.

$= \frac{{21\sqrt 3 }}{2}$ AG

Note: Other methods of splitting the area may receive full marks.

[3 marks]

$\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi }{n} + \, \ldots \, + \frac{1}{2} \times {2^{n - 1}} \times {2^n} \times {\text{sin}}\frac{\pi }{n}$ M1A1

Note: Award M1 for powers of 2, A1 for any correct expression including both the first and last term.

$= {\text{sin}}\frac{\pi }{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n - 2}}} \right)$

identifying a geometric series with common ratio 2²(= 4) (M1)A1

$= \frac{{1 - {2^{2n}}}}{{1 - 4}} \times {\text{sin}}\frac{\pi }{n}$ M1

Note: Award M1 for use of formula for sum of geometric series.

$= \frac{1}{3}\left( {{4^n} - 1} \right){\text{sin}}\frac{\pi }{n}$ A1

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the matrices

A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} 1&3 \\ 2&{ - 2} \end{array}} \right)$ .

Find BA.

[2]

Calculate det (BA).

[2]

Find A(A^–1B + 2A^–1)A.

[3]

Markscheme

BA = $\left( {\left( {\begin{array}{*{20}{c}} 1&3 \\ 2&{ - 2} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} {18}&{ - 14} \\ { - 4}&4 \end{array}} \right)$ A2

Note: Award A1 for one error, A0 for two or more errors.

[2 marks]

det(BA) = (72 – 56) = 16 (M1)A1

[2 marks]

EITHER

A(A^–1B + 2A^–1)A = BA + 2A (M1)A1

$= \left( {\begin{array}{*{20}{c}} {24}&{ - 18} \\ 6&{ - 4} \end{array}} \right)$ A1

A^–1 $= - \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 4}&2 \\ { - 5}&3 \end{array}} \right)$ (A1)

an attempt to evaluate (M1)

A^–1B + 2A^–1 $= - \frac{1}{2}\left( {\begin{array}{*{20}{c}} 0&{ - 16} \\ 1&{ - 21} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 4}&2 \\ { - 5}&3 \end{array}} \right)$

A(A^–1B + 2A^–1)A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 4&6 \\ {4.5}&{7.5} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right)$

$= \left( {\begin{array}{*{20}{c}} 3&3 \\ 2&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ 5&{ - 4} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {24}&{ - 18} \\ 6&{ - 4} \end{array}} \right)$ A1

[3 marks]

Examiners report

[N/A]

The strength of earthquakes is measured on the Richter magnitude scale, with values typically between $0$ and $8$ where $8$ is the most severe.

The Gutenberg–Richter equation gives the average number of earthquakes per year, $N$ , which have a magnitude of at least $M$ . For a particular region the equation is

$\log_{10} N = a - M$ , for some $a \in ℝ$ .

This region has an average of $100$ earthquakes per year with a magnitude of at least $3$ .

The equation for this region can also be written as $N = \frac{b}{10^{M}}$ .

Within this region the most severe earthquake recorded had a magnitude of $7.2$ .

The number of earthquakes in a given year with a magnitude of at least $7.2$ can be modelled by a Poisson distribution, with mean $N$ . The number of earthquakes in one year is independent of the number of earthquakes in any other year.

Let $Y$ be the number of years between the earthquake of magnitude $7.2$ and the next earthquake of at least this magnitude.

Find the value of $a$ .

[2]

Find the value of $b$ .

[2]

Find the average number of earthquakes in a year with a magnitude of at least $7.2$ .

[1]

Find $P (Y > 100)$ .

[3]

Markscheme

$\log_{10} 100 = a - 3$ (M1)

$a = 5$ A1

[2 marks]

EITHER

$N = 10^{5 - M}$ (M1)

$= \frac{10^{5}}{10^{M}} (= \frac{100000}{10^{M}})$

$100 = \frac{b}{10^{3}}$ (M1)

THEN

$b = 100000 (= 10^{5})$ A1

[2 marks]

$N = \frac{10^{5}}{10^{7.2}} = 0.00631 (0.0063095 \dots)$ A1

Note: Do not accept an answer of $10^{- 2.2}$ .

[1 mark]

METHOD 1

$Y > 100 \Rightarrow$ no earthquakes in the first $100$ years (M1)

EITHER

let $X$ be the number of earthquakes of at least magnitude $7.2$ in a year

$X ~ Po (0.0063095 \dots)$

${(P (X = 0))}^{100}$ (M1)

let $X$ be the number of earthquakes in $100$ years

$X ~ Po (0.0063095 \dots \times 100)$ (M1)

$P (X = 0)$

THEN

$0.532 (0.532082 \dots)$ A1

METHOD 2

$Y > 100 \Rightarrow$ no earthquakes in the first $100$ years (M1)

let $X$ be the number of earthquakes in $100$ years

since $n$ is large and $p$ is small

$X ~ B (100, 0.0063095 \dots)$ (M1)

$P (X = 0)$

$0.531 (0.531019 \dots)$ A1

[3 marks]

Examiners report

Parts (a), (b), and (c) were accessible to many candidates who earned full marks with the manipulation of logs and indices presenting no problems. Part (d), however, proved to be too difficult for most and very few correct attempts were seen. As in question 9, most candidates relied on calculator notation when using the Poisson distribution. The discipline of defining a random variable in terms of its distribution and parameters helps to conceptualize the problem in terms that aid a better understanding. Most candidates who attempted this question blindly entered values into the Poisson distribution calculator and were unable to earn any marks. There were a couple of correct solutions using a binomial distribution to approximate the given quantity.

[N/A]

Let A² = 2A + I where A is a 2 × 2 matrix.

Show that A⁴ = 12A + 5I.

[3]

Let B = $\left[ {\begin{array}{*{20}{c}} 4&2 \\ 1&{ - 3} \end{array}} \right]$ .

Given that B² – B – 4I = $\left[ {\begin{array}{*{20}{c}} k&0 \\ 0&k \end{array}} \right]$ , find the value of $k$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1
A⁴ = 4A² + 4AI + I² or equivalent M1A1
= 4(2A + I) + 4A + I A1
= 8A + 4I + 4A + I
= 12A + 5I AG

[3 marks]

METHOD 2
A³ = A(2A + I) = 2A² + AI = 2(2A + I) + A(= 5A + 2I) M1A1
A⁴ = A(5A + 2I) A1
= 5A² + 2A = 5(2A + I) + 2A
= 12A + 5I AG

[3 marks]

B² = $\left[ {\begin{array}{*{20}{c}} {18}&2 \\ 1&{11} \end{array}} \right]$ (A1)

$\left[ {\begin{array}{*{20}{c}} {18}&2 \\ 1&{11} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&2 \\ 1&{ - 3} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&0 \\ 0&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {10}&0 \\ 0&{10} \end{array}} \right]$ (A1)

$\Rightarrow k = 10$ A1

[3 marks]

Examiners report

[N/A]

The matrix A = $\left( {\begin{array}{*{20}{c}} 1&2&0 \\ { - 3}&1&{ - 1} \\ 2&{ - 2}&1 \end{array}} \right)$ has inverse A⁻¹ = $\left( {\begin{array}{*{20}{c}} { - 1}&{ - 2}&{ - 2} \\ 3&1&1 \\ a&6&b \end{array}} \right)$ .

Consider the simultaneous equations

$x + 2y = 7$

$- 3x + y - z = 10$

$2x - 2y + z = - 12$

Write down the value of $a$ .

[1]

a.i.

Write down the value of $b$ .

[1]

a.ii.

Write these equations as a matrix equation.

[1]

Solve the matrix equation.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$a$ = 4 A1 N1

[1 mark]

a.i.

$b$ = 7 A1 N1

[1 mark]

a.ii.

EITHER

A $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7 \\ {10} \\ { - 12} \end{array}} \right)$ A1 N1

$\left( {\begin{array}{*{20}{c}} 1&2&0 \\ { - 3}&1&{ - 1} \\ 2&{ - 2}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7 \\ {10} \\ { - 12} \end{array}} \right)$ A1 N1

[1 mark]

$\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ = A⁻¹ $\left( {\begin{array}{*{20}{c}} 7 \\ {10} \\ { - 12} \end{array}} \right)$ (accept algebraic method) (M1)

$\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 3} \\ 5 \\ 4 \end{array}} \right)$ (accept $x$ = −3, $y$ = 5, $z$ = 4) A2 N3

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Let A = $\left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&{ - 1}&2 \\ 3&{ - 3}&2 \end{array}} \right)$ , D = $\left( {\begin{array}{*{20}{c}} { - 4}&{13}&{ - 7} \\ { - 2}&7&{ - 4} \\ 3&{ - 9}&5 \end{array}} \right)$ , and C = $\left( {\begin{array}{*{20}{c}} 5 \\ 7 \\ {10} \end{array}} \right)$ .

Given matrices A, B, C for which AB = C and det A ≠ 0, express B in terms of A and C.

[2]

Find the matrix DA.

[1]

b.i.

Find B if AB = C.

[2]

b.ii.

Find the coordinates of the point of intersection of the planes $x + 2y + 3z = 5$ , $2x - y + 2z = 7$ , $3x - 3y + 2z = 10$ .

[2]

Markscheme

Since det A ≠ 0, A^–1 exists. (M1)

Hence AB = C ⇒ B = A^–1C (C1)

[2 marks]

DA = $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$ (A1)

[1 mark]

b.i.

B = A^–1C = DC (M1)

$= \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 2 \end{array}} \right)$ (A1)

[2 marks]

b.ii.

The system of equations is $\begin{array}{*{20}{c}} {x + 2y + 3z = 5} \\ {2x - y + 2z = 7} \\ {3x - 3y + 2z = 10} \end{array}$

or A $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) =$ C (M1)

The required point = (1, –1, 2). (A1)

[2 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

The square matrix X is such that X³ = 0. Show that the inverse of the matrix (I – X) is I + X + X².

Markscheme

For multiplying (I – X)(I + X + X²) M1

= I² + IX + IX² – XI – X² – X³ = I + X + X² – X – X² – X³ (A1)(A1)

= I – X³ A1

= I A1

AB = I ⇒ A^–1 = B (R1)

(I – X) (I + X + X²) = I ⇒ (I – X)^–1 = I + X + X² AG N0

[5 marks]

Examiners report

[N/A]

Let A = $\left( {\begin{array}{*{20}{c}} 1&2 \\ 3&{ - 1} \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 3&0 \\ { - 2}&1 \end{array}} \right)$ .

Find A + B.

[2]

Find −3A.

[2]

Find AB.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of addition (M1)

eg at least two correct elements

A + B = $\left( {\begin{array}{*{20}{c}} 4&2 \\ 1&0 \end{array}} \right)$ A1 N2

[2 marks]

evidence of multiplication (M1)

eg at least two correct elements

−3A = $\left( {\begin{array}{*{20}{c}} { - 3}&{ - 6} \\ { - 9}&3 \end{array}} \right)$ A1 N2

[2 marks]

evidence of matrix multiplication (in correct order) (M1)

eg AB = $\left( {\begin{array}{*{20}{c}} {1\left( 3 \right) + 2\left( { - 2} \right)}&{1\left( 0 \right) + 2\left( 1 \right)} \\ {3\left( 3 \right) + \left( { - 1} \right)\left( { - 2} \right)}&{3\left( 0 \right) + \left( { - 1} \right)\left( 1 \right)} \end{array}} \right)$

AB = $\left( {\begin{array}{*{20}{c}} { - 1}&2 \\ {11}&{ - 1} \end{array}} \right)$ A2 N3

[3 marks]

Examiners report

[N/A]

Find a relationship between $a$ and $b$ if the matrices $M = \left( {\begin{array}{*{20}{c}} 1&a \\ 2&3 \end{array}} \right)$ and $N = \left( {\begin{array}{*{20}{c}} 1&b \\ 2&3 \end{array}} \right)$ commute under matrix multiplication.

[4]

Find the value of $a$ if the determinant of matrix $M$ is −1.

[2]

b.i.

Write down ${M^{ - 1}}$ for this value of $a$ .

[1]

b.ii.

Markscheme

$\left( {\begin{array}{*{20}{c}} 1&a \\ 2&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&b \\ 2&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {1 + 2a}&{b + 3a} \\ 8&{2b + 9} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} 1&b \\ 2&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&a \\ 2&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {1 + 2b}&{a + 3b} \\ 8&{2a + 9} \end{array}} \right)$ M1A1

So require $a = b$ M1A1

[4 marks]

$\left| {\begin{array}{*{20}{c}} 1&a \\ 2&3 \end{array}} \right| = 3 - 2a = - 1 \Rightarrow a = 2$ M1A1

[2 marks]

b.i.

${\left( {\begin{array}{*{20}{c}} 1&2 \\ 2&3 \end{array}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}} { - 3}&2 \\ 2&{ - 1} \end{array}} \right)$ A1

[1 mark]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

If A = $\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right)$ , find 2 values of $x$ and $y$ , given that AB = BA.

Markscheme

AB = $\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2x + 32}&{xy + 16} \\ {24}&{4y + 8} \end{array}} \right)$ (A1)

BA = $\left( {\begin{array}{*{20}{c}} 2&y \\ 8&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x&4 \\ 4&2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2x + 4y}&{2y + 8} \\ {8x + 16}&{40} \end{array}} \right)$ (A1)

AB = BA ⇒ 8 $x$ + 16 = 24 and 4 $y$ + 8 = 40

This gives $x = 1$ and $y = 8$ . (A1) (C3)

[3 marks]

Examiners report

[N/A]

The equation of the line $y = m x + c$ can be expressed in vector form $r = a + λ b$ .

The matrix $M$ is defined by $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix})$ .

The line $y = m x + c$ (where $m \neq - 2$ ) is transformed into a new line using the transformation described by matrix $M$ .

Find the vectors $a$ and $b$ in terms of $m$ and/or $c$ .

[2]

Find the value of $det M$ .

[1]

Show that the equation of the resulting line does not depend on $m$ or $c$ .

[4]

Markscheme

(one vector to the line is $(\begin{matrix} 0 \\ c \end{matrix})$ therefore) $a = (\begin{matrix} 0 \\ c \end{matrix})$ A1

the line goes $m$ up for every $1$ across

(so the direction vector is) $b = (\begin{matrix} 1 \\ m \end{matrix})$ A1

Note: Although these are the most likely answers, many others are possible.

[2 marks]

(from GDC OR $6 \times 2 - 4 \times 3$ ) $|M| = 0$ A1

[1 mark]

METHOD 1

$(\begin{matrix} X \\ Y \end{matrix}) = (\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} x \\ m x + c \end{matrix}) = (\begin{matrix} 6 x + 3 m x + 3 c \\ 4 x + 2 m x + 2 c \end{matrix})$ M1A1

$= (\begin{matrix} 3 (2 x + m x + c) \\ 2 (2 x + m x + c) \end{matrix})$ A1

therefore the new line has equation $3 Y = 2 X$ A1

which is independent of $m$ or $c$ AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

METHOD 2

take two points on the line, e.g $(0, c)$ and $(1, m + c)$ M1

these map to $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} 0 \\ c \end{matrix}) = (\begin{matrix} 3 c \\ 2 c \end{matrix})$

and $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} 1 \\ m + c \end{matrix}) = (\begin{matrix} 6 + 3 m + 3 c \\ 4 + 2 m + 2 c \end{matrix})$ A1

therefore a direction vector is $(\begin{matrix} 6 + 3 m \\ 4 + 2 m \end{matrix}) = (2 + m) (\begin{matrix} 3 \\ 2 \end{matrix})$

(since $m \neq - 2$ ) a direction vector is $(\begin{matrix} 3 \\ 2 \end{matrix})$

the line passes through $(\begin{matrix} 3 c \\ 2 c \end{matrix}) - c (\begin{matrix} 3 \\ 2 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ therefore it always has the origin as a jump-on vector A1

the vector equation is therefore $r = μ (\begin{matrix} 3 \\ 2 \end{matrix})$ A1

which is independent of $m$ or $c$ AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

METHOD 3

$r = (\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) ((\begin{matrix} 0 \\ c \end{matrix}) + λ (\begin{matrix} 1 \\ m \end{matrix})) = (\begin{matrix} 3 c \\ 2 c \end{matrix}) + λ (\begin{matrix} 6 + 3 m \\ 4 + 2 m \end{matrix})$ M1A1

$= c (\begin{matrix} 3 \\ 2 \end{matrix}) + (2 + m) λ (\begin{matrix} 3 \\ 2 \end{matrix})$ A1

$= μ (\begin{matrix} 3 \\ 2 \end{matrix})$

where $μ = c + (2 + m) λ$ is an arbitrary parameter. A1

which is independent of $m$ or $c$ (as $μ$ can take any value) AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

[4 marks]

Examiners report

In part (a), most candidates were unable to convert the Cartesian equation of a line into its vector form. In part (b), almost every candidate showed that the value of the determinant was zero. In part (c), the great majority of candidates failed to come up with any sort of strategy to solve the problem.

[N/A]

$A$ and $B$ are 2 × 2 matrices, where $A = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 2&0 \end{array}} \right]$ and $BA = \left[ {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right]$ . Find $B$

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$B = \left( {BA} \right){A^{ - 1}}$ (M1)

$= - \frac{1}{4}\left( {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&{ - 2} \\ { - 2}&5 \end{array}} \right)$ (M1)

$= - \frac{1}{4}\left( {\begin{array}{*{20}{c}} { - 4}&{ - 12} \\ { - 16}&{ - 48} \end{array}} \right)$ (A1)

$= \left( {\begin{array}{*{20}{c}} 1&3 \\ 4&{12} \end{array}} \right)$ (A1)

$\left( {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5&2 \\ 2&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {11}&2 \\ {44}&8 \end{array}} \right)$ (M1)

$\left. { \Rightarrow \begin{array}{*{20}{c}} {5a + 2b = 11} \\ {2a\,\,\,\,\,\,\,\,\,\,\,\,\, = 2} \end{array}} \right\}$

$\Rightarrow a = 1$ , $b = 3$ (A1)

$\left. {\begin{array}{*{20}{c}} {5c + 2d = 44} \\ {2c{\text{ }} = 8} \end{array}} \right\}$

$\Rightarrow c = 4$ , $d = 12$ (A1)

$B = \left( {\begin{array}{*{20}{c}} 1&3 \\ 4&{12} \end{array}} \right)$ (A1) (C4)

Note: Correct solution with inversion (ie AB instead of BA) earns FT marks, (maximum [3 marks]).

[4 marks]

Examiners report

[N/A]

Let A = $\left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&1&2 \\ 2&0&1 \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} {18} \\ {23} \\ {13} \end{array}} \right)$ , and X = $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ .

Consider the equation AX = B.

Write down the inverse matrix A⁻¹.

[2]

Express X in terms of A⁻¹ and B.

[1]

b.i.

Hence, solve for X.

[3]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

A⁻¹ = $\left( {\begin{array}{*{20}{c}} { - \frac{1}{3}}&{\frac{2}{3}}&{ - \frac{1}{3}} \\ { - \frac{1}{3}}&{\frac{5}{3}}&{ - \frac{7}{3}} \\ {\frac{2}{3}}&{ - \frac{4}{3}}&{\frac{5}{3}} \end{array}} \right)$ or $\left( {\begin{array}{*{20}{c}} { - 0.333}&{0.667}&{ - 0.333} \\ { - 0.333}&{1.67}&{ - 2.33} \\ {0.667}&{ - 1.33}&{1.67} \end{array}} \right)$ A2 N2

[2 marks]

X = A⁻¹B A1 N1

[1 mark]

b.i.

X = $\left( {\begin{array}{*{20}{c}} 5 \\ 2 \\ 3 \end{array}} \right)$ A3 N3

[3 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

The sum of an infinite geometric sequence is $9$ .

The first term is $4$ more than the second term.

Find the third term. Justify your answer.

Markscheme

METHOD 1

$\frac{u_{1}}{1 - r} = 9$ A1

therefore $u_{1} = 9 - 9 r$

$u_{1} = 4 + u_{1} r$ A1

substitute or solve graphically: M1

$9 - 9 r = 4 + (9 - 9 r) r$ OR $\frac{4}{{(1 - r)}^{2}} = 9$

$9 r^{2} - 18 r + 5 = 0$

$r = \frac{1}{3}$ or $r = \frac{5}{3}$

only $r = \frac{1}{3}$ is possible as the sum to infinity exists R1

then $u_{1} = 9 - (9 \times \frac{1}{3}) = 6$

$u_{3} = 6 \times {\frac{1}{3}}^{2} = \frac{2}{3}$ A1

METHOD 2

$\frac{u_{1}}{1 - r} = 9$ A1

$r = \frac{u_{1} - 4}{u_{1}}$ A1

attempt to solve M1

$\frac{u_{1}}{1 - (\frac{u_{1} - 4}{u_{1}})} = 9$

$\frac{u_{1}}{(\frac{4}{u_{1}})} = 9$

${(u_{1})}^{2} = 36$

$u_{1} = \pm 6$

attempting to solve both possible sequences

$6, 2, \dots$ or $- 6, - 10 \dots$

$r = \frac{1}{3}$ or $r = \frac{5}{3}$

only $r = \frac{1}{3}$ is possible as the sum to infinity exists R1

$u_{3} = 6 \times {(\frac{1}{3})}^{2} = \frac{2}{3}$ A1

[5 marks]

Examiners report

Many candidates submitted quite poor attempts at this question. Many managed to state the equation $u_{1} = 9 (1 - r)$ obtained by considering the sum to infinity but few managed to find the second equation $u_{1} (1 - r) = 4$ . Common errors in failing to obtain this equation were that “four more” meant multiplied by four or thinking that the second term was four more than the first term. Even those candidates who obtained both equations were often unable to solve them. Attempted solutions often filled the page with algebra going nowhere. Most of those candidates who actually found the third term correctly then failed to realize that there were two solutions to the equations, one of which had to be rejected. Consequently, the final “reasoning” mark was seldom awarded.

Consider the following system of equations where $a \in \mathbb{R}$ .

$2x + 4y - z = 10$

$x + 2y + az = 5$

$5x + 12y = 2a$ .

Find the value of $a$ for which the system of equations does not have a unique solution.

[2]

Find the solution of the system of equations when $a = 2$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

an attempt at a valid method eg by inspection or row reduction (M1)

$2 \times {R_2} = {R_1} \Rightarrow 2a = - 1$

$\Rightarrow a = - \frac{1}{2}$ A1

[2 marks]

using elimination or row reduction to eliminate one variable (M1)

correct pair of equations in 2 variables, such as

$\left. {\begin{array}{*{20}{c}} {5x + 10y = 25} \\ {5x + 12y = 4} \end{array}} \right\}$ A1

Note: Award A1 for $z$ = 0 and one other equation in two variables.

attempting to solve for these two variables (M1)

$x = 26$ , $y = - 10.5$ , $z = 0$ A1A1

Note: Award A1A0 for only two correct values, and A0A0 for only one.

Note: Award marks in part (b) for equivalent steps seen in part (a).

[5 marks]

Examiners report

[N/A]

An arithmetic sequence ${u_1}{\text{, }}{u_2}{\text{, }}{u_3} \ldots$ has ${u_1} = 1$ and common difference $d \ne 0$ . Given that ${u_2}{\text{, }}{u_3}$ and ${u_6}$ are the first three terms of a geometric sequence

Given that ${u_N} = - 15$

find the value of $d$ .

[4]

determine the value of $\sum\limits_{r = 1}^N {{u_r}}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

use of ${u_n} = {u_1} + (n - 1)d$ M1

${(1 + 2d)^2} = (1 + d)(1 + 5d)$ (or equivalent) M1A1

$d = - 2$ A1

[4 marks]

$1 + (N - 1) \times - 2 = - 15$

$N = 9$ (A1)

$\sum\limits_{r = 1}^9 {{u_r}} = \frac{9}{2}(2 + 8 \times - 2)$ (M1)

$= - 63$ A1

[3 marks]

Examiners report

[N/A]

Find the determinant of A, where A = $\left( {\begin{array}{*{20}{c}} 3&1&2 \\ 9&5&8 \\ 7&4&6 \end{array}} \right)$ .

Markscheme

det A = −2 A2

[2 marks]

Examiners report

[N/A]

It is given that ${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$ .

Show that ${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$ where $r,\,x \in {\mathbb{R}^ + }$ .

[2]

Express $y$ in terms of $x$ . Give your answer in the form $y = p{x^q}$ , where p , q are constants.

[5]

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines $x = 1$ and $x = \alpha$ where $\alpha > 1$ . The area of R is $\sqrt 2$ .

Find the value of $\alpha$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)$ M1A1

$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 2

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}$ M1

$= \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}$ A1

$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y = - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}$

${\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)$ M1A1

$y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$ .

[5 marks]

METHOD 2

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0$ M1

$\sqrt 2 xy = 1$ A1

$y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$ .

[5 marks]

the area of R is $\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x$ M1

$= \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha$ A1

$= \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha$ A1

$\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2$ M1

$\alpha = {{\text{e}}^2}$ A1

Note: Only follow through from part (b) if $y$ is in the form $y = p{x^q}$

[5 marks]

Examiners report

[N/A]

Consider the system of equations A $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ { - 3} \end{array}} \right)$ where A $= \left( {\begin{array}{*{20}{c}} {k + 1}&{ - k} \\ 2&{k - 1} \end{array}} \right)$ and $k \in \mathbb{R}$ .

Find det A.

[2]

Find the set of values of $k$ for which the system has a unique solution.

[4]

Markscheme

Attempting to find det A (M1)

det A $= {k^2} + 2k - 1$ A1 N2

[2 marks]

System has a unique solution provided det A ≠ 0 (R1)

${k^2} + 2k - 1 \ne 0$ (A1)

Solving ${k^2} + 2k - 1 \ne 0$ or equivalent for $k$ M1

$k \in \mathbb{R}\,{\text{\ }}\left\{ { - 1 \pm \,\sqrt 2 } \right\}\,\,\left( {{\text{accept}}\,\,k \ne - 1 \pm \,\sqrt 2 {\text{,}}\,\,\,k \ne - 2.41{\text{,}}\,\,0.414} \right)$ A1 N3

[4 marks]

Examiners report

[N/A]

The matrix A is given by A $= \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]$ .

The matrix B is given by B $= \left[ {\begin{array}{*{20}{c}} 3&2 \\ 2&3 \end{array}} \right]$ .

Show that the eigenvalues of A are real if ${\left( {a - d} \right)^2} + 4bc \geqslant 0$ .

[4]

a.i.

Deduce that the eigenvalues are real if A is symmetric.

[2]

a.ii.

Determine the eigenvalues of B.

[2]

b.i.

Determine the corresponding eigenvectors.

[4]

b.ii.

Markscheme

the eigenvalues satisfy

$\left| {\begin{array}{*{20}{c}} {a - \lambda }&b \\ c&{d - \lambda } \end{array}} \right| = 0$ M1

$\left( {a - \lambda } \right)\left( {d - \lambda } \right) - bc = 0$ A1

${\lambda ^2} - \left( {a + d} \right)\lambda + ad - bc = 0$ A1

the condition for real roots is

${\left( {a + d} \right)^2} - 4\left( {ad - bc} \right) \geqslant 0$ M1

${\left( {a - d} \right)^2} + 4bc \geqslant 0$ AG

[4 marks]

a.i.

if the matrix is symmetric, b = c. In this case, M1

${\left( {a - d} \right)^2} + 4bc = {\left( {a - d} \right)^2} + 4{b^2} \geqslant 0$

because each square term is non-negative R1AG

[2 marks]

a.ii.

the characteristic equation is

${\lambda ^2} - 6\lambda + 5 = 0$ M1

$\lambda = 1,5$ A1

[2 marks]

b.i.

taking $\lambda = 1$

$\left[ {\begin{array}{*{20}{c}} 2&2 \\ 2&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right]$ M1

giving eigenvector $= \left[ {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right]$ A1

taking $\lambda = 5$

$\left[ {\begin{array}{*{20}{c}} { - 2}&2 \\ 2&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right]$ M1

giving eigenvector $= \left[ {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right]$ A1

[4 marks]

b.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

Solve ${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)$

EITHER

${\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}$ M1

$= \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}$ A1

$\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)$ M1A1

THEN

${\text{ln}}\,x = 2\,{\text{ln}}\,2$ or $- {\text{ln}}\,2$ A1

$\Rightarrow x = 4$ or $x = \frac{1}{2}$ (M1)A1

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is $\frac{1}{2} < x < 4$ A1

[6 marks]

Examiners report

[N/A]

Matrices A, B and C are defined as

A = $\left( {\begin{array}{*{20}{c}} 1&5&1 \\ 3&{ - 1}&3 \\ { - 9}&3&7 \end{array}} \right)$ , B = $\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)$ , C = $\left( {\begin{array}{*{20}{c}} 8 \\ 0 \\ { - 4} \end{array}} \right)$ .

Given that AB = $\left( {\begin{array}{*{20}{c}} a&0&0 \\ 0&a&0 \\ 0&0&a \end{array}} \right)$ , find $a$ .

[1]

Hence, or otherwise, find A^–1.

[2]

Find the matrix X, such that AX = C.

[2]

Markscheme

$a = 16$ A1

[1 mark]

A^–1 = $\frac{1}{{16}}\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)$ (M1)A1

[2 marks]

AX = C ⇒ X = A^–1C (M1)

$= \frac{1}{{16}}\left( {\begin{array}{*{20}{c}} 1&2&{ - 1} \\ 3&{ - 1}&0 \\ 0&3&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 8 \\ 0 \\ { - 4} \end{array}} \right)$

$= \frac{1}{{16}}\left( {\begin{array}{*{20}{c}} {12} \\ {24} \\ { - 4} \end{array}} \right)\,\,\,\left( { = \left( {\begin{array}{*{20}{c}} {0.75} \\ {1.5} \\ { - 0.25} \end{array}} \right)} \right)$ A1

[2 marks]

Examiners report

[N/A]

Let M = $\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right)$ where $a$ and $b$ are non-zero real numbers.

Show that M is non-singular.

[2]

Calculate M².

[2]

Show that det(M²) is positive.

[2]

Markscheme

finding det M $= {a^2} + {b^2}$ A1

${a^2} + {b^2} > 0$ , therefore M is non-singular or equivalent statement R1

[2 marks]

M² = $\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right)\left( {\begin{array}{*{20}{c}} a&b \\ { - b}&a \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a^2} - {b^2}}&{2ab} \\ { - 2ab}&{{a^2} - {b^2}} \end{array}} \right)$ M1A1

[2 marks]

EITHER

det(M²) $= \left( {{a^2} - {b^2}} \right)\left( {{a^2} - {b^2}} \right) + \left( {2ab} \right)\left( {2ab} \right)$ A1

det(M²) $= {\left( {{a^2} - {b^2}} \right)^2} + {\left( {2ab} \right)^2}$ $\left( { = {{\left( {{a^2} + {b^2}} \right)}^2}} \right)$

since the first term is non-negative and the second is positive R1

therefore det(M2) > 0

Note: Do not penalise first term stated as positive.

det(M²) = (det M)² A1

since det M is positive so too is det (M²) R1

[2 marks]

Examiners report

[N/A]

Given the matrix A = $\left( {\begin{array}{*{20}{c}} 3&2 \\ { - 1}&0 \end{array}} \right)$ find the values of the real number $k$ for which ${\text{det}}\left( {A - kI} \right) = 0$ where $I = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)$

Markscheme

${\text{det}}\left( {A - kI} \right) = 0$

$\Rightarrow \left| {\begin{array}{*{20}{c}} {3 - k}&2 \\ { - 1}&{ - k} \end{array}} \right| = 0$ (M1)

$\Rightarrow {k^2} - 3k + 2 = 0$ (M1)

$\Rightarrow \left( {k - 2} \right)\left( {k - 1} \right) = 0$

$\Rightarrow k = 1{\text{,}}\,\,2$ (A2) (C4)

[4 marks]

Examiners report

[N/A]

The matrix $M = (\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix})$ has eigenvalues $- 5$ and $1$ .

A switch has two states, $A$ and $B$ . Each second it either remains in the same state or moves according to the following rule: If it is in state $A$ it will move to state $B$ with a probability of $0.8$ and if it is in state $B$ it will move to state $A$ with a probability of $0.7$ .

Find an eigenvector corresponding to the eigenvalue of $1$ . Give your answer in the form $(\begin{matrix} a \\ b \end{matrix})$ , where $a, b \in ℤ$ .

[3]

Using your answer to (a), or otherwise, find the long-term probability of the switch being in state $A$ . Give your answer in the form $\frac{c}{d}$ , where $c, d \in ℤ^{+}$ .

[2]

Markscheme

$λ = 1$

$(\begin{matrix} - 0.8 & 0.7 \\ 0.8 & - 0.7 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ OR $(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ y \end{matrix})$ (M1)

$0.8 x = 0.7 y$ (A1)

an eigenvector is $(\begin{matrix} 7 \\ 8 \end{matrix})$ (or equivalent with integer values) A1

[3 marks]

EITHER

(the long-term probability matrix is given by the eigenvector corresponding to the eigenvalue equal to $1$ , scaled so that the sum of the entries is $1$ )

$8 + 7 = 15$ (M1)

$(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix}) (\begin{matrix} p \\ 1 - p \end{matrix}) = (\begin{matrix} p \\ 1 - p \end{matrix})$ (M1)

considering high powers of the matrix e.g. ${(\begin{matrix} 0.2 & 0.7 \\ 0.8 & 0.3 \end{matrix})}^{50}$ (M1)

$(\begin{matrix} \frac{7}{15} & \frac{7}{15} \\ \frac{8}{15} & \frac{8}{15} \end{matrix})$

THEN

probability of being in state $A$ is $\frac{7}{15}$ A1

[2 marks]

Examiners report

In part (a), some candidates could correctly use either $(A - λ I) x = 0$ or $A x = λ x$ to find an eigenvector but many did not pay attention to the fact that integer values of the eigenvector were required. Some candidates used the method of finding the steady state by finding $A^{n}$ for some high value of n in part (b) but ignored the fact that they needed to express their answer in rational form. Some did try to convert their calculated answer of 0.467 to $\frac{467}{1000}$ but this could only receive partial credit as an exact answer was required.

[N/A]

Consider the complex numbers ${z_1} = 1 + \sqrt 3 {\text{i, }}{z_2} = 1 + {\text{i}}$ and $w = \frac{{{z_1}}}{{{z_2}}}$ .

By expressing ${z_1}$ and ${z_2}$ in modulus-argument form write down the modulus of $w$ ;

[3]

a.i.

By expressing ${z_1}$ and ${z_2}$ in modulus-argument form write down the argument of $w$ .

[1]

a.ii.

Find the smallest positive integer value of $n$ , such that ${w^n}$ is a real number.

[2]

Markscheme

${z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)$ and ${z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)$ A1A1

Note: Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.

$\left| w \right| = \sqrt 2$ A1

[3 marks]

a.i.

${z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)$ and ${z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)$ A1A1

Note: Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.

$\arg w = \frac{\pi }{{12}}$ A1

Notes: Allow FT from incorrect answers for ${z_1}$ and ${z_2}$ in modulus-argument form.

[1 mark]

a.ii.

EITHER

$\sin \left( {\frac{{\pi n}}{{12}}} \right) = 0$ (M1)

$\arg ({w^n}) = \pi$ (M1)

$\frac{{n\pi }}{{12}} = \pi$

THEN

$\therefore n = 12$ A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the matrix A $= \left( {\begin{array}{*{20}{c}} 5&{ - 2} \\ 7&1 \end{array}} \right)$ .

B, C and X are also 2 × 2 matrices.

Write down the inverse, A^–1.

[2]

Given that XA + B = C, express X in terms of A^–1, B and C.

[2]

b.i.

Given that B $= \left( {\begin{array}{*{20}{c}} 6&7 \\ 5&{ - 2} \end{array}} \right)$ , and C $= \left( {\begin{array}{*{20}{c}} { - 5}&0 \\ { - 8}&7 \end{array}} \right)$ , find X.

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

det A = 5(1) − 7(−2) = 19

A^–1 $= \frac{1}{{19}}\left( {\begin{array}{*{20}{c}} 1&2 \\ { - 7}&5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{{19}}}&{\frac{2}{{19}}} \\ {\frac{{ - 7}}{{19}}}&{\frac{5}{{19}}} \end{array}} \right)$ (A2)

Note: Award (A1) for $\left( {\begin{array}{*{20}{c}} 1&2 \\ { - 7}&5 \end{array}} \right)$ , (A1) for dividing by 19.

A^–1 $= \left( {\begin{array}{*{20}{c}} {0.0526}&{0.105} \\ { - 0.368}&{0.263} \end{array}} \right)$ (G2)

[2 marks]

XA + B = C ⇒ XA = C – Β (M1)

X = (C – Β)Α^–1 (A1)

X = (C – B)A^–1 (A2)

[2 marks]

b.i.

(C – Β)Α^–1 = $\left( {\begin{array}{*{20}{c}} { - 11}&{ - 7} \\ { - 13}&9 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\frac{1}{{19}}}&{\frac{2}{{19}}} \\ {\frac{{ - 7}}{{19}}}&{\frac{5}{{19}}} \end{array}} \right)$ (A1)

⇒ X = $\left( {\begin{array}{*{20}{c}} {\frac{{38}}{{19}}}&{\frac{{ - 57}}{{19}}} \\ {\frac{{ - 76}}{{19}}}&{\frac{{19}}{{19}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 4}&1 \end{array}} \right)$ (A1)

X = $\left( {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 4}&1 \end{array}} \right)$ (G2)

Note: If premultiplication by A^–1 is used, award (M1)(M0) in part (i) but award (A2) for $\left( {\begin{array}{*{20}{c}} {\frac{{ - 37}}{{19}}}&{\frac{{11}}{{19}}} \\ {\frac{{12}}{{19}}}&{\frac{{94}}{{19}}} \end{array}} \right)$ in part (ii).

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Find the values of $a$ and $b$ given that the matrix $A = \left( {\begin{array}{*{20}{c}} a&{ - 4}&{ - 6} \\ { - 8}&5&7 \\ { - 5}&3&4 \end{array}} \right)$ is the inverse of the matrix $B = \left( {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ 3&b&1 \\ { - 1}&1&{ - 3} \end{array}} \right)$ .

[2]

For the values of $a$ and $b$ found in part (a), solve the system of linear equations

$\begin{array}{*{20}{c}} {x + 2y - 2z = 5} \\ {3x + by + z = 0} \\ { - x + y - 3z = a - 1.} \end{array}$

[2]

Markscheme

AB = I

(AB)₁₁ = 1 ⇒ $a$ – 12 + 6 = 1, giving $a$ = 7 (A1) (C1)

(AB)₂₂ = 1 ⇒ –16 + 5 $b$ + 7 = 1, giving $b$ = 2 (A1) (C1)

[2 marks]

the system is $BX = \left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right)$ where $X = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ .

Then, $X = A\left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7&{ - 4}&{ - 6} \\ { - 8}&5&7 \\ { - 5}&3&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 6 \end{array}} \right).$ (M1)

Thus $x = - 1$ , $y = 2$ , $z = - 1$ (A1) (C2)

[2 marks]

Examiners report

[N/A]

The function f : M → M where M is the set of 2 × 2 matrices, is given by f(X) = AX where A is a 2 × 2 matrix.

Given that A is non-singular, prove that f is a bijection.

[7]

It is now given that A is singular.

By considering appropriate determinants, prove that f is not a bijection.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

suppose f(X) = f(Y) , ie AX = AY (M1)

then A⁻¹AX = A⁻¹AY A1

X = Y A1

since f(X) = f(Y) ⇒ X = Y, f is an injection R1

now suppose C ∈ M and consider f(D) = C , ie AD = C M1

then D = A⁻¹ C (A⁻¹ exists since A is non- singular) A1

since given C ∈ M, there exists D ∈ M such that f(D) = C , f is a surjection R1

therefore f is a bijection AG

[7 marks]

suppose f(X) = Y, ie AX = Y (M1)

then det(A)det(X) = det(Y) A1

since det(A) = 0, it follows that det(Y) = 0 A1

it follows that f is not surjective since the function cannot reach non-singular matrices R1

therefore f is not a bijection AG

[4 marks]

Examiners report

[N/A]

Let $z = 1 - \cos 2\theta - {\text{i}}\sin 2\theta ,{\text{ }}z \in \mathbb{C},{\text{ }}0 \leqslant \theta \leqslant \pi$ .

Solve $2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ$ .

[5]

Show that $\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$ .

[3]

Find the modulus and argument of $z$ in terms of $\theta$ . Express each answer in its simplest form.

[9]

c.i.

Hence find the cube roots of $z$ in modulus-argument form.

[5]

c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2\sin (x + 60^\circ ) = \cos (x + 30^\circ )$

$2(\sin x\cos 60^\circ + \cos x\sin 60^\circ ) = \cos x\cos 30^\circ - \sin x\sin 30^\circ$ (M1)(A1)

$2\sin x \times \frac{1}{2} + 2\cos x \times \frac{{\sqrt 3 }}{2} = \cos x \times \frac{{\sqrt 3 }}{2} - \sin x \times \frac{1}{2}$ A1

$\Rightarrow \frac{3}{2}\sin x = - \frac{{\sqrt 3 }}{2}\cos x$

$\Rightarrow \tan x = - \frac{1}{{\sqrt 3 }}$ M1

$\Rightarrow x = 150^\circ$ A1

[5 marks]

EITHER

choosing two appropriate angles, for example 60° and 45° M1

$\sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$ and

$\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$ (A1)

$\sin 105^\circ + \cos 105^\circ = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}$ A1

$= \frac{1}{{\sqrt 2 }}$ AG

attempt to square the expression M1

${(\sin 105^\circ + \cos 105^\circ )^2} = {\sin ^2}105^\circ + 2\sin 105^\circ \cos 105^\circ + {\cos ^2}105^\circ$

${(\sin 105^\circ + \cos 105^\circ )^2} = 1 + \sin 210^\circ$ A1

$= \frac{1}{2}$ A1

$\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$ AG

[3 marks]

EITHER

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$\left| z \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{(\sin 2\theta )}^2}}$ M1

$\left| z \right| = \sqrt {1 - 2\cos 2\theta + {{\cos }^2}2\theta + {{\sin }^2}2\theta }$ A1

$= \sqrt 2 \sqrt {(1 - \cos 2\theta )}$ A1

$= \sqrt {2(2{{\sin }^2}\theta )}$

$= 2\sin \theta$ A1

let $\arg (z) = \alpha$

$\tan \alpha = - \frac{{\sin 2\theta }}{{1 - \cos 2\theta }}$ M1

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

$= - \cot \theta$ A1

$\arg (z) = \alpha = - \arctan \left( {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right)$ A1

$= \theta - \frac{\pi }{2}$ A1

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$= 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta$ M1A1

$= 2\sin \theta (\sin \theta - {\text{i}}\cos \theta )$ (A1)

$= - 2{\text{i}}\sin \theta (\cos \theta + {\text{i}}\sin \theta )$ M1A1

$= 2\sin \theta \left( {\cos \left( {\theta - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\theta - \frac{\pi }{2}} \right)} \right)$ M1A1

$\left| z \right| = 2\sin \theta$ A1

$\arg (z) = \theta - \frac{\pi }{2}$ A1

[9 marks]

c.i.

attempt to apply De Moivre’s theorem M1

${(1 - \cos 2\theta - {\text{i}}\sin 2\theta )^{\frac{1}{3}}} = {2^{\frac{1}{3}}}{(\sin \theta )^{\frac{1}{3}}}\left[ {\cos \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right)} \right]$ A1A1A1

Note: A1 for modulus, A1 for dividing argument of $z$ by 3 and A1 for $2n\pi$ .

Hence cube roots are the above expression when $n = - 1,{\text{ }}0,{\text{ }}1$ . Equivalent forms are acceptable. A1

[5 marks]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Find the solution of ${\log _2}x - {\log _2}5 = 2 + {\log _2}3$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\log _2}x - {\log _2}5 = 2 + {\log _2}3$

collecting at least two log terms (M1)

eg $\,\,\,\,\,$ ${\log _2}\frac{x}{5} = 2 + {\log _2}3{\text{ or }}{\log _2}\frac{x}{{15}} = 2$

obtaining a correct equation without logs (M1)

eg $\,\,\,\,\,$ $\frac{x}{5} = 12$ $\,\,\,$ OR $\,\,\,$ $\frac{x}{{15}} = {2^2}$ (A1)

$x = 60$ A1

[4 marks]

Examiners report

[N/A]

Write down the inverse of the matrix

A = $\left( {\begin{array}{*{20}{c}} 1&{ - 3}&1 \\ 2&2&{ - 1} \\ 1&{ - 5}&3 \end{array}} \right)$

[2]

Hence, find the point of intersection of the three planes.

$\begin{array}{*{20}{c}} {x - 3y + z = 1} \\ {2x + 2y - z = 2} \\ {x - 5y + 3z = 3} \end{array}$

[3]

A fourth plane with equation $x + y + z = d$ passes through the point of intersection. Find the value of $d$ .

[1]

Markscheme

A^–1 = $\left( {\begin{array}{*{20}{c}} {0.1}&{0.4}&{0.1} \\ { - 0.7}&{0.2}&{0.3} \\ { - 1.2}&{0.2}&{0.8} \end{array}} \right)$ A2 N2

[2 marks]

For attempting to calculate $\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)$ = A⁻¹ $\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right)$ M1

$x = 1.2{\text{,}}\,\,y = 0.6{\text{,}}\,\,z = 1.6$ (so the point is (1.2, 0.6, 1.6)) A2 N2

[3 marks]

(1.2, 0.6, 1.6) lies on $x + y + z = d$

$\therefore d = 3.4$ A1 N1

[1 mark]

Examiners report

[N/A]

Let A = $\left( {\begin{array}{*{20}{c}} 3&2 \\ k&4 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&2 \\ 1&3 \end{array}} \right)$ . Find, in terms of $k$ ,

2A − B.

[3]

det (2A − B).

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

2A = $\left( {\begin{array}{*{20}{c}} 6&4 \\ {2k}&8 \end{array}} \right)$ (A1)

2A − B = $\left( {\begin{array}{*{20}{c}} 4&2 \\ {2k - 1}&5 \end{array}} \right)$ A2 N3

[3 marks]

Evidence of using the definition of determinant (M1)

Correct substitution (A1)

eg 4(5) − 2(2 $k$ − 1), 20 − 2(2 $k$ − 1), 20 − 4 $k$ + 2

det (2A − B) = 22 − 4 $k$ A1 N3

[3 marks]

Examiners report

[N/A]

The square matrix X is such that X³ = 0. Show that the inverse of the matrix (I – X) is I + X + X².

Markscheme

For multiplying (I – X)(I + X + X²) M1

= I² + IX + IX² – XI – X² – X³ = I + X + X² – X – X² – X³ (A1)(A1)

= I – X³ A1

= I A1

AB = I ⇒ A^–1 = B (R1)

(I – X)(I + X + X²) = I ⇒ (I – X)^–1 = I + X + X² AG N0

[6 marks]

Examiners report

[N/A]

It is given that $z_{1} = 3 cis (\frac{3 π}{4})$ and $z_{2} = 2 cis (\frac{n π}{16}), n \in ℤ^{+}$ .

In parts (a)(i) and (a)(ii), give your answers in the form $r e^{i θ}, r \geq 0, - π < θ \leq π$ .

Find the value of ${z_{1}}^{3}$ .

[2]

a.i.

Find the value of ${(\frac{z_{1}}{z_{2}})}^{4}$ for $n = 2$ .

[3]

a.ii.

Find the least value of $n$ such that $z_{1} z_{2} \in ℝ^{+}$ .

[3]

Markscheme

${z_{1}}^{3} = 27 e^{\frac{iπ}{4}} (= 27 e^{0.785398 \dots i})$ A1A1

Note: Award A1 for $27$ and A1 for the angle in the correct form.

[2 marks]

a.i.

${(\frac{z_{1}}{z_{2}})}^{4} = (\frac{81}{16}) e^{\frac{iπ}{2}} (= 5.0625 e^{1.57079 \dots i})$ A1A2

Note: Award A1 for $\frac{81}{16}$ , A2 for the angle in the correct form and A1 for the angle in incorrect form e.g. $cis \frac{π}{2}$ and/or $\frac{5 π}{2}$ . Award A1 if $i$ is given in place of $cis \frac{π}{2}$ .

[3 marks]

a.ii.

$z_{1} z_{2} = 6 cis (\frac{3 π}{4} + \frac{n π}{16})$ (M1)

$= 6 cis (\frac{12 π + n π}{16})$

$12 π + n π = 32 π$ (M1)

$n = 20$ A1

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Given that A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ { - 3}&4 \end{array}} \right)$ and I = $\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)$ , find the values of $\lambda$ for which (A – $\lambda$ I) is a singular matrix.

Markscheme

singular matrix ⇒ det = 0 (R1)

$\left| {\begin{array}{*{20}{c}} {3 - \lambda }&{ - 2} \\ { - 3}&{4 - \lambda } \end{array}} \right|$ (A1)

$\left( {3 - \lambda } \right)\left( {4 - \lambda } \right) - 6 = 0$ (M1)

$\Rightarrow {\lambda ^2} - 7\lambda + 6 = 0$ (A1)

$\lambda = 1$ or 6 (A1)(A1) (C6)

Note: Award (C2) for one correct answer with no working.

[6 marks]

Examiners report

[N/A]

Find the inverse of the matrix $\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 1&1&2 \\ 2&1&4 \end{array}} \right)$ .

[2]

Hence solve the system of equations

$x + 2y + z = 0$

$x + y + 2z = 7$

$2x + y + z = 17$

[4]

Markscheme

${\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 1&1&2 \\ 2&1&4 \end{array}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}} 2&{ - 7}&3 \\ 0&2&{ - 1} \\ { - 1}&3&{ - 1} \end{array}} \right)$ A2 N2

[2 marks]

In matrix form A $x$ = B or $x$ = A⁻¹ B M1

$x = 2$ , $y = -3$ , $z = 4$ A1A1A1 N0

[4 marks]

Examiners report

[N/A]

Given that A = $\left( {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 2} \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&{ - 3} \end{array}} \right)$ , find X if BX = A – AB.

Markscheme

METHOD 1

A – AB = $\left( {\begin{array}{*{20}{c}} 2&3 \\ 1&{ - 2} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&{ - 9} \\ 2&6 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (M1)(A1)

X = B^–1(A – AB) = B^–1 $\left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (M1)

$= - \frac{1}{6}\left( {\begin{array}{*{20}{c}} { - 3}&0 \\ 0&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} { - 2}&{12} \\ { - 1}&{ - 8} \end{array}} \right)$ (A1)

$= \left( {\begin{array}{*{20}{c}} { - 1}&6 \\ {\frac{1}{3}}&{\frac{8}{3}} \end{array}} \right)$ (A2) (C6)

METHOD 2

Attempting to set up a matrix equation (M2)

X = B^–1(A – AB) (A2)

$= \left( {\begin{array}{*{20}{c}} { - 1}&6 \\ {\frac{1}{3}}&{\frac{8}{3}} \end{array}} \right)$ (from GDC) (A2) (C6)

[6 marks]

Examiners report

[N/A]

The matrix $A$ is given by $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ .

By considering the determinant of a relevant matrix, show that the eigenvalues, $λ$ , of $A$ satisfy the equation

$λ^{2} - α λ + β = 0$ ,

where $α$ and $β$ are functions of $a, b, c, d$ to be determined.

[4]

Verify that

$A^{2} - α A + β I =$ 0.

[5]

b.i.

Assuming that $A$ is non-singular, use the result in part (b)(i) to show that

$A^{- 1} = \frac{1}{β} (α I - A)$ .

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$|\begin{matrix} a - λ & b \\ c & d - λ \end{matrix}| = 0$ M1

$(a - λ) (d - λ) - b c = 0$ M1A1

$λ^{2} - (a + d) λ + a d - b c = 0$ A1

$α = (a + d); β = a d - b c$

[4 marks]

$A^{2} = [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} a^{2} + b c & a b + b d \\ a c + c d & b c + d^{2} \end{matrix}]$ (M1)A1

$A^{2} - (a + d) A + (a d - b c) I =$

$[\begin{matrix} a^{2} + b c & a b + b d \\ a c + c d & b c + d^{2} \end{matrix}] - (a + d) [\begin{matrix} a & b \\ c & d \end{matrix}] + (a d - b c) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ M1

$= [\begin{matrix} a^{2} + b c - a (a + d) + a d - b c & a b + b d - b (a + d) \\ a c + c d - c (a + d) & b c + d^{2} - d (a + d) + a d - b c \end{matrix}]$ A2

$=$ 0 AG

Note: Award A1A0 for a single error.

[5 marks]

b.i.

multiply throughout by $A^{- 1}$ giving M1

$A - α I + β A^{- 1} =$ 0 A1

$A^{- 1} = \frac{1}{β} (α I - A)$ AG

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Roger buys a new laptop for himself at a cost of $£ 495$ . At the same time, he buys his daughter Chloe a higher specification laptop at a cost of $£ 2200$ .

It is anticipated that Roger’s laptop will depreciate at a rate of $10 %$ per year, whereas Chloe’s laptop will depreciate at a rate of $15 %$ per year.

Roger and Chloe’s laptops will have the same value $k$ years after they were purchased.

Estimate the value of Roger’s laptop after $5$ years.

[2]

Find the value of $k$ .

[2]

Comment on the validity of your answer to part (b).

[1]

Markscheme

$£ 495 \times 0 . 9^{5} = £ 292 (£ 292.292 \dots)$ (M1)A1

[2 marks]

$£ 495 \times 0 . 9^{k} = 2200 \times 0 . 85^{k}$ (M1)

$k = 26.1 (26.0968 \dots)$ A1

Note: Award M1A0 for $k - 1$ in place of $k$ .

[2 marks]

depreciation rates unlikely to be constant (especially over a long time period) R1

Note: Accept reasonable answers based on the magnitude of $k$ or the fact that “value” depends on factors other than time.

[1 mark]

Examiners report

[N/A]

Let A = $\left( {\begin{array}{*{20}{c}} 1&x&{ - 1} \\ 3&1&4 \end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}} 3 \\ x \\ 2 \end{array}} \right)$ .

Find AB.

[3]

The matrix C = $\left( {\begin{array}{*{20}{c}} {20} \\ {28} \end{array}} \right)$ and 2AB = C. Find the value of $x$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

Attempting to multiply matrices (M1)

$\left( {\begin{array}{*{20}{c}} 1&x&{ - 1} \\ 3&1&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3 \\ x \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {3 + {x^2} - 2} \\ {9 + x + 8} \end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}} {1 + {x^2}} \\ {17 + x} \end{array}} \right)} \right)$ A1A1 N3

[3 marks]

Setting up equation M1

eg $2\left( {\begin{array}{*{20}{c}} {1 + {x^2}} \\ {17 + x} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {20} \\ {28} \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} {2 + 2{x^2}} \\ {34 + 2x} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {20} \\ {28} \end{array}} \right)$ , $\left( {\begin{array}{*{20}{c}} {1 + {x^2}} \\ {17 + x} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {10} \\ {14} \end{array}} \right)$

$\begin{array}{*{20}{c}} {2 + 2{x^2} = 20} \\ {34 + 2x = 28} \end{array}\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} {1 + {x^2} = 10} \\ {17 + x = 14} \end{array}} \right)$ (A1)

$x = - 3$ A1 N2

[3 marks]

Examiners report

[N/A]

A meteorologist models the height of a hot air balloon launched from the ground. The model assumes the balloon travels vertically upwards and travels $450$ metres in the first minute.

Due to the decrease in temperature as the balloon rises, the balloon will continually slow down. The model suggests that each minute the balloon will travel only $82 %$ of the distance travelled in the previous minute.

Find how high the balloon will travel in the first $10$ minutes after it is launched.

[3]

The balloon is required to reach a height of at least $2520$ metres.

Determine whether it will reach this height.

[2]

Suggest a limitation of the given model.

[1]

Markscheme

recognition of geometric sequence eg $r = 0.82$ (M1)

$S_{10} = \frac{450 (1 - 0 . 82^{10})}{1 - 0.82}$ (A1)

$= 2160 m (2156.37 \dots)$ A1

[3 marks]

$S_{\infty} = \frac{450}{1 - 0.82}$ (M1)

$= 2500 < 2520$ so the balloon will not reach the required height. A1

[2 marks]

horizontal motion not taken into account,

rate of cooling will not likely be linear,

balloon is considered a point mass / size of balloon not considered,

effects of wind/weather unlikely to be consistent,

a discrete model has been used, whereas a continuous one may offer greater accuracy R1

Note: Accept any other sensible answer.

[1 mark]

Examiners report

[N/A]

Consider the matrix A = $\left( {\begin{array}{*{20}{c}} 0&2 \\ a&{ - 1} \end{array}} \right)$ .

Find the matrix A².

[2]

If det A² = 16, determine the possible values of $a$ .

[3]

Markscheme

A² = $\left( {\begin{array}{*{20}{c}} {2a}&{ - 2} \\ { - a}&{2a + 1} \end{array}} \right)$ (M1)A1

[2 marks]

METHOD 1

det A² = $4{a^2} + 2a - 2a = 4{a^2}$ M1

$a$ = ±2 A1A1 N2

METHOD 2

det A = $- 2a$ M1

det A = ±4

$a$ = ±2 A1A1 N2

[3 marks]

Examiners report

[N/A]

Consider the matrix A $= \left( {\begin{array}{*{20}{c}} {{{\text{e}}^x}}&{{{\text{e}}^{ - x}}} \\ {2 + {{\text{e}}^x}}&1 \end{array}} \right)$ , where $x \in \mathbb{R}$ .

Find the value of $x$ for which A is singular.

Markscheme

finding det A = ${{\text{e}}^x} - {{\text{e}}^{ - x}}\left( {2 + {{\text{e}}^x}} \right)$ or equivalent A1

A is singular ⇒ det A = 0 (R1)

${{\text{e}}^x} - {{\text{e}}^{ - x}}\left( {2 + {{\text{e}}^x}} \right) = 0$

${{\text{e}}^{2x}} - {{\text{e}}^x} - 2 = 0$ A1

solving for ${{\text{e}}^x}$ (M1)

${{\text{e}}^x}$ > 0 (or equivalent explanation) (R1)

${{\text{e}}^x} = 2$

$x =$ ln 2 (only) A1 N0

[6 marks]

Examiners report

[N/A]

The matrices A, B, C and X are all non-singular 3 × 3 matrices.

Given that A^–1XB = C, express X in terms of the other matrices.

Markscheme

AA^–1XB = ΑC (M1)(A1)

IXBB^–1 = ACB^–1 (M1)(A1)

X = ACB^–1 (M1)(A1) (C6)

[6 marks]

Examiners report

[N/A]

If A = $\left( {\begin{array}{*{20}{c}} 1&2 \\ k&{ - 1} \end{array}} \right)$ and A² is a matrix whose entries are all 0, find $k$ .

Markscheme

A² = $\left( {\begin{array}{*{20}{c}} 1&2 \\ k&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&2 \\ k&{ - 1} \end{array}} \right)$ M1

$= \left( {\begin{array}{*{20}{c}} {1 + 2k}&0 \\ 0&{2k + 1} \end{array}} \right)$ A2

Note: Award A2 for 4 correct, A1 for 2 or 3 correct.

$1 + 2k = 0$ M1

$k = - \frac{1}{2}$ A1

[5 marks]

Examiners report

[N/A]

Given that M = $\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right)$ and that M² – 6M + kI = 0 find k.

Markscheme

M² = $\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&{ - 1} \\ { - 3}&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 7&{ - 6} \\ { - 18}&{19} \end{array}} \right)$ M1A1

$\Rightarrow \left( {\begin{array}{*{20}{c}} 7&{ - 6} \\ { - 18}&{19} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {12}&{ - 6} \\ { - 18}&{24} \end{array}} \right)$ + kI = 0 (M1)

$\Rightarrow \left( {\begin{array}{*{20}{c}} { - 5}&0 \\ 0&{ - 5} \end{array}} \right)$ + kI = 0 (A1)

⇒ k = 5 A1

[5 marks]

Examiners report

[N/A]

In the following Argand diagram the point A represents the complex number $- 1 + 4{\text{i}}$ and the point B represents the complex number $- 3 + 0{\text{i}}$ . The shape of ABCD is a square. Determine the complex numbers represented by the points C and D.

M17/5/MATHL/HP1/ENG/TZ2/05

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

C represents the complex number $1 - 2{\text{i}}$ A2

D represents the complex number $3 + 2{\text{i}}$ A2

[4 marks]

Examiners report

[N/A]

Find the values of the real number $k$ for which the determinant of the matrix $\left( {\begin{array}{*{20}{c}} {k - 4}&3 \\ { - 2}&{k + 1} \end{array}} \right)$ is equal to zero.

Markscheme

$\left| {\begin{array}{*{20}{c}} {k - 4}&3 \\ { - 2}&{k + 1} \end{array}} \right| = 0$

$\Rightarrow \left( {k - 4} \right)\left( {k + 1} \right) + 6 = 0$ (M1)

$\Rightarrow {k^2} - 3k + 2 = 0$ (M1)

$\Rightarrow \left( {k - 2} \right)\left( {k - 1} \right) = 0$

$\Rightarrow k = 2$ or $k = 1$ (A1) (C3)

[3 marks]

Examiners report

[N/A]

If $A = \left( {\begin{array}{*{20}{c}} {2p}&3 \\ { - 4p}&p \end{array}} \right)$ and det $A = 14$ , find the possible values of $p$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2{p^2} + 12p = 14$ (M1)(A1)

${p^2} + 6p - 7 = 0$

$\left( {p + 7} \right)\left( {p - 1} \right) = 0$ (A1)

$p = - 7$ or $p = 1$ (A1) (C4)

Note: Both answers are required for the final (A1).

[4 marks]

Examiners report

[N/A]

Solve the equation ${4^x} + {2^{x + 2}} = 3$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to form a quadratic in ${2^x}$ M1

${({2^x})^2} + 4 \bullet {2^x} - 3 = 0$ A1

${2^x} = \frac{{ - 4 \pm \sqrt {16 + 12} }}{2}{\text{ }}\left( { = - 2 \pm \sqrt 7 } \right)$ M1

${2^x} = - 2 + \sqrt 7 {\text{ }}\left( {{\text{as }} - 2 - \sqrt 7 < 0} \right)$ R1

$x = {\log _2}\left( { - 2 + \sqrt 7 } \right){\text{ }}\left( {x = \frac{{\ln \left( { - 2 + \sqrt 7 } \right)}}{{\ln 2}}} \right)$ A1

Note: Award R0 A1 if final answer is $x = {\log _2}\left( { - 2 + \sqrt 7 } \right)$ .

[5 marks]

Examiners report

[N/A]

Solve the equation ${\log _2}(x + 3) + {\log _2}(x - 3) = 4$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\log _2}(x + 3) + {\log _2}(x - 3) = 4$

${\log _2}({x^2} - 9) = 4$ (M1)

${x^2} - 9 = {2^4}{\text{ }}( = 16)$ M1A1

${x^2} = 25$

$x = \pm 5$ (A1)

$x = 5$ A1

[5 marks]

Examiners report

[N/A]

The rate, $A$ , of a chemical reaction at a fixed temperature is related to the concentration of two compounds, $B$ and $C$ , by the equation

$A = k{B^x}{C^y}$ , where $x$ , $y$ , $k \in \mathbb{R}$ .

A scientist measures the three variables three times during the reaction and obtains the following values.

Find $x$ , $y$ and $k$ .

Markscheme

${\text{log}}\,A = x\,{\text{log}}\,B + y\,{\text{log}}\,C + {\text{log}}\,k$ (M1)

${\text{log}}\,5.74 = x\,{\text{log}}\,2.1 + y\,{\text{log}}\,3.4 + {\text{log}}\,k$

${\text{log}}\,2.88 = x\,{\text{log}}\,1.5 + y\,{\text{log}}\,2.4 + {\text{log}}\,k$

${\text{log}}\,0.980 = x\,{\text{log}}\,0.8 + y\,{\text{log}}\,1.9 + {\text{log}}\,k$ M1A1

Note: Allow any consistent base, allow numerical equivalents.

attempting to solve their system of equations (M1)

$x$ = 1.53, $y$ = 0.505 A1

$k$ = 0.997 A1

[6 marks]

Examiners report

[N/A]