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HL Paper 1

A garden has a triangular sunshade suspended from three points $A (2, 0, 2), B (8, 0, 2)$ and $C (5, 4, 3)$ , relative to an origin in the corner of the garden. All distances are measured in metres.

Find $\vec{CA}$ .

[1]

a.i.

Find $\vec{CB}$ .

[1]

a.ii.

Find $\vec{CA} \times \vec{CB}$ .

[2]

Hence find the area of the triangle $ABC$ .

[2]

Markscheme

$\vec{CA} = (\begin{matrix} - 3 \\ - 4 \\ - 1 \end{matrix})$ A1

[1 mark]

a.i.

$\vec{CB} = (\begin{matrix} 3 \\ - 4 \\ - 1 \end{matrix})$ A1

[1 mark]

a.ii.

$\vec{CA} \times \vec{CB} = (\begin{matrix} 0 \\ - 6 \\ 24 \end{matrix})$ (M1)A1

Note: Do not award (M1) if less than 2 entries are correct.

[2 marks]

area is $\frac{1}{2} \sqrt{6^{2} + 24^{2}} = 12.4 m^{2} (12.3693 \dots, 3 \sqrt{17})$ (M1)A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The position of a helicopter relative to a communications tower at the top of a mountain at time $t$ (hours) can be described by the vector equation below.

$r = (\begin{matrix} 20 \\ - 25 \\ 0 \end{matrix}) + t (\begin{matrix} 4.2 \\ 5.8 \\ - 0.5 \end{matrix})$

The entries in the column vector give the displacements east and north from the communications tower and above/below the top of the mountain respectively, all measured in kilometres.

Find the speed of the helicopter.

[2]

Find the distance of the helicopter from the communications tower at $t = 0$ .

[2]

Find the bearing on which the helicopter is travelling.

[2]

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.

$|v| = \sqrt{4 . 2^{2} + 5 . 8^{2} + 0 . 5^{2}}$ (M1)

$7.18 (7.1784 \dots) ({kmh}^{- 1})$ A1

[2 marks]

$r = (\begin{matrix} 20 \\ - 25 \\ 0 \end{matrix})$

$|r| = \sqrt{20^{2} + 25^{2}}$ (M1)

$= \sqrt{1025} = 32.0 (32.0156 \dots) (km)$ A1

[2 marks]

Bearing is $arctan (\frac{4.2}{5.8})$ or $90 °- arctan (\frac{5.8}{4.2})$ (M1)

$035.9 ° (35.909 \dots)$ A1

[2 marks]

Examiners report

[N/A]

An ant is walking along the edges of a wire frame in the shape of a triangular prism.

The vertices and edges of this frame can be represented by the graph below.

Write down the adjacency matrix, $M$ , for this graph.

[3]

Find the number of ways that the ant can start at the vertex $A$ , and walk along exactly $6$ edges to return to $A$ .

[2]

Markscheme

$M = (\begin{matrix} 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 \end{matrix})$ A1A1A1

Note: Award A1 for each two correct rows.

[3 marks]

calculating $M^{6}$ (M1)

$143$ A1

[2 marks]

Examiners report

[N/A]

The following diagram shows the graph $G$ .

Verify that $G$ satisfies the handshaking lemma.

[3]

Show that $G$ cannot be redrawn as a planar graph.

[3]

State, giving a reason, whether $G$ contains an Eulerian circuit.

[2]

Markscheme

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

METHOD 1

sum of degrees of vertices $= 3 + 5 + 5 + 5 + 4 + 4 = 26$ A1

number of edges $e = 13$ A1

the sum is equal to twice the number of edges which

verifies the handshaking lemma R1

METHOD 2

degrees of vertices $= 3, 5, 5, 5, 4, 4$ A1

there are $4$ vertices of odd order A1

there is an even number of vertices of odd order

which verifies the handshaking lemma R1

[3 marks]

if planar then $e \leq 3 v - 6$ M1

$e = 13, v = 6$ A1

inequality not satisfied R1

therefore $G$ is not planar AG

Note: method explaining that the graph contains $κ_{3, 3}$ is acceptable.

[3 marks]

there are vertices of odd degree R1

hence it does not contain an Eulerian circuit A1

Note: Do not award R0A1.

[2 marks]

Examiners report

[N/A]

An engineer plans to visit six oil rigs $(A - F)$ in the Gulf of Mexico, starting and finishing at $A$ . The travelling time, in minutes, between each of the rigs is shown in the table.

The data above can be represented by a graph $G$ .

Use Prim’s algorithm to find the weight of the minimum spanning tree of the subgraph of $G$ obtained by deleting $A$ and starting at $B$ . List the order in which the edges are selected.

[4]

a.i.

Hence find a lower bound for the travelling time needed to visit all the oil rigs.

[2]

a.ii.

Describe how an improved lower bound might be found.

[1]

Markscheme

use of Prim’s algorithm M1

$BC 46$ A1

$BD 58$ A1

$DE 23$

$EF 47$

Total $174$ A1

Note: Award M0A0A0A1 for $174$ without correct working e.g. use of Kruskal’s, or with no working.
Award M1A0A0A1 for $174$ by using Prim’s from an incorrect starting point.

[4 marks]

a.i.

$AB + AC = 55 + 63 = 118$ (M1)

$174 + 118 = 292$ minutes A1

[2 marks]

a.ii.

delete a different vertex A1

[1 mark]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Points in the plane are subjected to a transformation $T$ in which the point $(x, y)$ is transformed to the point $(x', y')$ where

$[\begin{matrix} x' \\ y' \end{matrix}] = [\begin{matrix} 3 & 0 \\ 0 & 2 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$ .

Describe, in words, the effect of the transformation $T$ .

[1]

Show that the points $A (1, 4), B (4, 8), C (8, 5), D (5, 1)$ form a square.

[3]

b.i.

Determine the area of this square.

[1]

b.ii.

Find the coordinates of $A', B', C', D'$ , the points to which $A, B, C, D$ are transformed under $T$ .

[2]

b.iii.

Show that $A' B' C' D'$ is a parallelogram.

[3]

b.iv.

Determine the area of this parallelogram.

[2]

b.v.

Markscheme

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

a stretch of scale factor $3$ in the $x$ direction

and a stretch of scale factor $2$ in the $y$ direction A1

[1 mark]

the four sides are equal in length $(5)$ A1

$Grad AB = \frac{4}{3}, Grad BC = - \frac{3}{4}$ A1

so product of gradients $= - 1$ , therefore $AB$ is perpendicular to $BC$ A1

therefore $ABCD$ is a square AG

[3 marks]

b.i.

area of square $= 25$ A1

[1 mark]

b.ii.

the transformed points are

$A' = (3, 8)$

$B' = (12, 16)$

$C' = (24, 10)$

$D' = (15, 2)$ A2

Note: Award A1 if one point is incorrect.

[2 marks]

b.iii.

$Grad A'B' = \frac{8}{9}; Grad C'D' = \frac{8}{9}$ A1

therefore $A'B'$ is parallel to $C'D'$ R1

$Grad A'D' = - \frac{6}{12}; Grad B'C' = - \frac{6}{12}$ A1

therefore $A'D'$ is parallel to $B'C'$

therefore $A' B' C' D'$ is a parallelogram AG

[3 marks]

b.iv.

$area of parallelogram = |determinant| \times area of square$

$= 6 \times 25$ (M1)

$= 150$ A1

[2 marks]

b.v.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

b.iv.

[N/A]

b.v.

A geometric transformation $T : (\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x' \\ y' \end{matrix})$ is defined by

$T : (\begin{matrix} x' \\ y' \end{matrix}) = (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ .

Find the coordinates of the image of the point $(6, - 2)$ .

[2]

Given that $T : (\begin{matrix} p \\ q \end{matrix}) \mapsto 2 (\begin{matrix} p \\ q \end{matrix})$ , find the value of $p$ and the value of $q$ .

[3]

A triangle $L$ with vertices lying on the $x y$ plane is transformed by $T$ .

Explain why both $L$ and its image will have exactly the same area.

[2]

Markscheme

$(\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} 6 \\ - 2 \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ (M1)

$= (\begin{matrix} 57 \\ 22 \end{matrix})$ OR $(57, 22)$ A1

[2 marks]

$(\begin{matrix} 2 p \\ 2 q \end{matrix}) = (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} p \\ q \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ (M1)

$7 p - 10 q - 5 = 2 p$

$2 p - 3 q + 4 = 2 q$ (A1)

solve simultaneously:

$p = 13, q = 6$ A1

Note: Award A0 if $13$ and $6$ are not labelled or are labelled the other way around.

[3 marks]

$det (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) = - 1 (OR |det (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix})| = 1)$ A1

scale factor of image area is therefore $(|- 1| =) 1$ (and the translation does not affect the area) A1

[2 marks]

Examiners report

[N/A]

A ship $S$ is travelling with a constant velocity, $v$ , measured in kilometres per hour, where

$v = (\begin{matrix} - 12 \\ 15 \end{matrix})$ .

At time $t = 0$ the ship is at a point $A (300, 100)$ relative to an origin $O$ , where distances are measured in kilometres.

A lighthouse is located at a point $(129, 283)$ .

Find the position vector $\vec{OS}$ of the ship at time $t$ hours.

[1]

Find the value of $t$ when the ship will be closest to the lighthouse.

[6]

An alarm will sound if the ship travels within $20$ kilometres of the lighthouse.

State whether the alarm will sound. Give a reason for your answer.

[2]

Markscheme

$\vec{OS} = (\begin{matrix} 300 \\ 100 \end{matrix}) + t (\begin{matrix} - 12 \\ 15 \end{matrix})$ A1

[1 mark]

attempt to find the vector from $L$ to $S$ (M1)

$\vec{LS} = (\begin{matrix} 171 \\ - 183 \end{matrix}) + t (\begin{matrix} - 12 \\ 15 \end{matrix})$ A1

EITHER

$|\vec{LS}| = \sqrt{{(171 - 12 t)}^{2} + {(15 t - 183)}^{2}}$ (M1)(A1)

minimize to find $t$ on GDC (M1)

$S$ closest when $\vec{LS} \cdot (\begin{matrix} - 12 \\ 15 \end{matrix}) = 0$ (M1)

$((\begin{matrix} 171 \\ - 183 \end{matrix}) + t (\begin{matrix} - 12 \\ 15 \end{matrix})) \cdot (\begin{matrix} - 12 \\ 15 \end{matrix}) = 0$

$- 2052 + 144 t - 2745 + 225 t = 0$ (M1)(A1)

$S$ closest when $\vec{LS} \cdot (\begin{matrix} - 12 \\ 15 \end{matrix}) = 0$ (M1)

$\vec{LS} = (\begin{matrix} 5 k \\ 4 k \end{matrix})$

$\vec{OS} = (\begin{matrix} 129 + 5 k \\ 283 + 4 k \end{matrix})$ (A1)

$(\begin{matrix} 129 + 5 k \\ 283 + 4 k \end{matrix}) = (\begin{matrix} 300 - 12 t \\ 100 + 15 t \end{matrix})$

Solving simultaneously (M1)

THEN

$t = 13$ A1

[6 marks]

the alarm will sound A1

$|\vec{LS}| = 19.2 \dots < 20$ R1

Note: Do not award A1R0.

[2 marks]

Examiners report

Part (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of $t$ by assuming that $\vec{OS}$ and $\vec{OL}$ are perpendicular. Other methods incorrectly applied were equating $\vec{OS}$ and $\vec{OL}$ . Of course, this led to separate values of $t$ for each component. The method of minimizing $|\vec{LS}|$ was not commonly employed.

The diagram below shows a network of roads in a small village with the weights indicating the distance of each road, in metres, and junctions indicated with letters.

Musab is required to deliver leaflets to every house on each road. He wishes to minimize his total distance.

Musab starts and finishes from the village bus-stop at $A$ . Determine the total distance Musab will need to walk.

[5]

Instead of having to catch the bus to the village, Musab’s sister offers to drop him off at any junction and pick him up at any other junction of his choice.

Explain which junctions Musab should choose as his starting and finishing points.

[2]

Markscheme

Odd vertices are $A, B, D, H$ A1

Consider pairings: M1

Note: Award (M1) if there are four vertices not necessarily all correct.

$AB DH$ has shortest route $AD, DE, EB$ and $DE, EH,$
so repeated edges $(19 + 16 + 19) + (16 + 27) = 97$

Note: Condone $AB$ in place of $AD, DE, EB$ giving $56 + (16 + 27) = 99$ .

$AD BH$ has shortest route $AD$ and $BE, EH,$
so repeated edges $19 + (19 + 27) = 65$

$AH BD$ has shortest route $AD, DE, EH$ and $BE, ED$ ,
so repeated edges $(19 + 16 + 27) + (19 + 16) = 97$ A2

Note: Award A1 if only one or two pairings are correctly considered.

so best pairing is $AD, BH$
weight of route is therefore $582 + 65 = 647$ A1

[5 marks]

least value of the pairings is $19$ therefore repeat $AD$ R1

$B$ and $H$ A1

Note: Do not award R0A1.

[2 marks]

Examiners report

[N/A]

Two lines $L_{1}$ and $L_{2}$ are given by the following equations, where $p \in ℝ$ .

$L_{1} : r = (\begin{matrix} 2 \\ p + 9 \\ - 3 \end{matrix}) + λ (\begin{matrix} p \\ 2 p \\ 4 \end{matrix})$

$L_{2} : r = (\begin{matrix} 14 \\ 7 \\ p + 12 \end{matrix}) + μ (\begin{matrix} p + 4 \\ 4 \\ - 7 \end{matrix})$

It is known that $L_{1}$ and $L_{2}$ are perpendicular.

Find the possible value(s) for $p$ .

[3]

In the case that $p < 0$ , determine whether the lines intersect.

[4]

Markscheme

setting a dot product of the direction vectors equal to zero (M1)

$(\begin{matrix} p \\ 2 p \\ 4 \end{matrix}) \cdot (\begin{matrix} p + 4 \\ 4 \\ - 7 \end{matrix}) = 0$

$p (p + 4) + 8 p - 28 = 0$ (A1)

$p^{2} + 12 p - 28 = 0$

$(p + 14) (p - 2) = 0$

$p = - 14, p = 2$ A1

[3 marks]

$p = - 14 \Rightarrow$

$L_{1} : r = (\begin{matrix} 2 \\ - 5 \\ - 3 \end{matrix}) + λ (\begin{matrix} - 14 \\ - 28 \\ 4 \end{matrix})$

$L_{2} : r = (\begin{matrix} 14 \\ 7 \\ - 2 \end{matrix}) + μ (\begin{matrix} - 10 \\ 4 \\ - 7 \end{matrix})$

a common point would satisfy the equations

$2 - 14 λ = 14 - 10 μ$

$- 5 - 28 λ = 7 + 4 μ$ (M1)

$- 3 + 4 λ = - 2 - 7 μ$

METHOD 1

solving the first two equations simultaneously

$λ = - \frac{1}{2}, μ = \frac{1}{2}$ A1

substitute into the third equation: M1

$- 3 + 4 (- \frac{1}{2}) \neq - 2 + \frac{1}{2} (- 7)$

so lines do not intersect. R1

Note: Accept equivalent methods based on the order in which the equations are considered.

METHOD 2

attempting to solve the equations using a GDC M1

GDC indicates no solution A1

so lines do not intersect R1

[4 marks]

Examiners report

[N/A]

The following diagram shows a corner of a field bounded by two walls defined by lines $L_{1}$ and $L_{2}$ . The walls meet at a point $A$ , making an angle of $40 °$ .

Farmer Nate has $7 m$ of fencing to make a triangular enclosure for his sheep. One end of the fence is positioned at a point $B$ on $L_{2}$ , $10 m$ from $A$ . The other end of the fence will be positioned at some point $C$ on $L_{1}$ , as shown on the diagram.

He wants the enclosure to take up as little of the current field as possible.

Find the minimum possible area of the triangular enclosure $ABC$ .

Markscheme

METHOD 1

attempt to find $AC$ using cosine rule M1

$7^{2} = 10^{2} + {AC}^{2} - 2 \times 10 \times AC \times \cos 40 °$ (A1)

attempt to solve a quadratic equation (M1)

$AC = 4.888 \dots$ AND $10.432 \dots$ (A1)

Note: At least $AC = 4.888 \dots$ must be seen, or implied by subsequent working.

minimum area $= \frac{1}{2} \times 10 \times 4.888 \dots \times \sin (40 °)$ M1

Note: Do not award M1 if incorrect value for minimizing the area has been chosen.

$= 15.7 m^{2}$ A1

METHOD 2

attempt to find $A \hat{C} B$ using the sine Rule M1

$\frac{\sin C}{10} = \frac{\sin 40}{7}$ (A1)

$C = 66.674 \dots °$ OR $113.325 \dots °$ (A1)

EITHER

$B = 180 - 40 - 113.325 \dots$

$B = 26.675 \dots °$ (A1)

area $= \frac{1}{2} \times 10 \times 7 \times \sin (26.675 \dots °)$ M1

sine rule or cosine rule to find $AC = 4.888 \dots$ (A1)

minimum area $= \frac{1}{2} \times 10 \times 4.888 \dots \times \sin (40 °)$ M1

THEN

$= 15.7 m^{2}$ A1

Note: Award A0M1A0 if the wrong length $AC$ or the wrong angle $B$ selected but used correctly finding a value of $33.5 m^{2}$ for the area.

[6 marks]

Examiners report

As has often been the case in the past, trigonometry is a topic that is poorly understood and candidates are poorly prepared for. Approaches to this question required the use of the cosine or sine rules. Some candidates tried to use right-angled trigonometry instead. A minority of candidates used the cosine rule approach and were more likely to be successful, navigating the roots of the quadratic equation formed. When using the sine rule the method involved the ambiguous case as the required angle was obtuse. Few candidates realized this and this was the most common mistake. In a few instances, the word “minimum” led candidates to attempt an approach using calculus.

The following diagram shows a frame that is made from wire. The total length of wire is equal to $15 cm$ . The frame is made up of two identical sectors of a circle that are parallel to each other. The sectors have angle $θ$ radians and radius $r cm$ . They are connected by $1 cm$ lengths of wire perpendicular to the sectors. This is shown in the diagram below.

The faces of the frame are covered by paper to enclose a volume, $V$ .

Show that $r = \frac{6}{2 + θ}$ .

[2]

Find an expression for $V$ in terms of $θ$ .

[2]

b.i.

Find the expression $\frac{d V}{d θ}$ .

[3]

b.ii.

Solve algebraically $\frac{d V}{d θ} = 0$ to find the value of $θ$ that will maximize the volume, $V$ .

[2]

b.iii.

Markscheme

$15 = 3 + 4 r + 2 r θ$ M1

$12 = 2 r (2 + θ)$ A1

Note: Award A1 for any reasonable working leading to expected result e,g, factorizing $r$ .

$r = \frac{6}{2 + θ}$ AG

[2 marks]

attempt to use sector area to find volume (M1)

volume $= \frac{1}{2} r^{2} θ \times 1$

$= \frac{1}{2} \times \frac{36}{{(2 + θ)}^{2}} \times θ (= \frac{18 θ}{{(2 + θ)}^{2}})$ A1

[2 marks]

b.i.

$\frac{d V}{d θ} = \frac{{(2 + θ)}^{2} \times 18 - 36 θ (2 + θ)}{{(2 + θ)}^{4}}$ M1A1A1

$\frac{d V}{d θ} = \frac{36 - 18 θ}{{(2 + θ)}^{3}}$

[3 marks]

b.ii.

$\frac{d V}{d θ} = \frac{36 - 18 θ}{{(2 + θ)}^{3}} = 0$ M1

Note: Award this M1 for simplified version equated to zero. The simplified version may have been seen in part (b)(ii).

$θ = 2$ A1

[2 marks]

b.iii.

Examiners report

Several candidates missed that the angle $θ$ was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate $\frac{18 θ}{{(2 + θ)}^{2}}$ as $18 θ {(2 + θ)}^{- 2}$ rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.

b.i.

b.ii.

b.iii.

At $1 : 00 pm$ a ship is $1 km$ east and $4 km$ north of a harbour. A coordinate system is defined with the harbour at the origin. The position vector of the ship at $1 : 00 pm$ is given by $(\begin{matrix} 1 \\ 4 \end{matrix})$ .

The ship has a constant velocity of $(\begin{matrix} 1.2 \\ - 0.6 \end{matrix})$ kilometres per hour ( ${km h}^{- 1}$ ).

Write down an expression for the position vector $r$ of the ship, $t$ hours after $1 : 00 pm$ .

[1]

Find the time at which the bearing of the ship from the harbour is $045 ˚$ .

[4]

Markscheme

$(r =) (\begin{matrix} 1 \\ 4 \end{matrix}) + t (\begin{matrix} 1.2 \\ - 0.6 \end{matrix})$ A1

Note: Do not condone the use of $λ$ or any other variable apart from $t$ .

[1 mark]

when the bearing from the port is $045 ˚$ , the distance east from the port is equal to the distance north from the port (M1)

$1 + 1.2 t = 4 - 0.6 t$ (A1)

$1.8 t = 3$

$t = \frac{5}{3}$ ( $1.6666 \dots,$ $1$ hour $40$ minutes) (A1)

time is $2 : 40 pm (14 : 40)$ A1

[4 marks]

Examiners report

Most candidates were able to answer part (a) correctly but there were some very poor examples of vector notation. The question asked for an expression of $r$ in terms of $t$ and although a failure to write $r$ was condoned the use of $λ$ or some other variable was penalized. In part (b) few candidates recognized that the eastern and northern distances would be equal with a bearing of 045°. Those who correctly obtained a value of $t = \frac{5}{3}$ often did not use this to find the time as required.

[N/A]

Consider the following directed network.

Write down the adjacency matrix for this network.

[2]

Determine the number of different walks of length $5$ that start and end at the same vertex.

[3]

Markscheme

$(\begin{matrix} 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 \end{matrix})$ A2

Note: Award A2 for the transposed matrix. Presentation in markscheme assumes columns/rows ordered A-E; accept a matrix with rows and/or columns in a different order only if appropriately communicated. Do not FT from part (a) into part (b).

[2 marks]

raising their matrix to a power of $5$ (M1)

$M^{5} = (\begin{matrix} 17 & 9 & 2 & 3 & 5 \\ 17 & 10 & 3 & 4 & 4 \\ 13 & 6 & 2 & 2 & 4 \\ 8 & 5 & 1 & 2 & 2 \\ 18 & 11 & 2 & 4 & 5 \end{matrix})$ (A1)

Note: The numbers along the diagonal are sufficient to award M1A1.

(the required number is $17 + 10 + 2 + 2 + 5 =$ ) $36$ A1

[3 marks]

Examiners report

This was well answered by the majority of candidates with most writing down the correct adjacency matrix and then raising it to the power 5.

[N/A]

A vertical pole stands on a sloped platform. The bottom of the pole is used as the origin, $O$ , of a coordinate system in which the top, $F$ , of the pole has coordinates $(0, 0, 5.8)$ . All units are in metres.

The pole is held in place by ropes attached at $F$ .

One of these ropes is attached to the platform at point $A (3.2, 4.5, - 0.3)$ . The rope forms a straight line from $A$ to $F$ .

Find $\vec{AF}$ .

[1]

Find the length of the rope.

[2]

Find $FÂO$ , the angle the rope makes with the platform.

[5]

Markscheme

$(\begin{matrix} - 3.2 \\ - 4.5 \\ 6.1 \end{matrix})$ A1

[1 mark]

$\sqrt{{(- 3.2)}^{2} + {(- 4.5)}^{2} + 6 . 1^{2}}$ (M1)

$8.22800 \dots \approx 8.23 m$ A1

[2 marks]

EITHER

$\vec{AO} = (\begin{matrix} - 3.2 \\ - 4.5 \\ 0.3 \end{matrix})$ A1

$\cos θ = \frac{\vec{AO} \cdot \vec{AF}}{|\vec{AO}| |\vec{AF}|}$

$\vec{AO} \cdot \vec{AF} = {(- 3.2)}^{2} + {(- 4.5)}^{2} + (0.3 \times 6.1) (= 32.32)$ (A1)

$\cos θ = \frac{32.32}{\sqrt{3 . 2^{2} + 4 . 5^{2} + 0 . 3^{2}} \times 8.22800 \dots}$ (M1)

$= 0.710326 \dots$ (A1)

Note: If $\vec{OA}$ is used in place of $\vec{AO}$ then $\cos θ$ will be negative.
Award A1(A1)(M1)(A1) as above. In order to award the final A1, some justification for changing the resulting obtuse angle to its supplementary angle must be seen.

$AO = \sqrt{3 . 2^{2} + 4 . 5^{2} + 0 . 3^{2}} (= 5.52991 \dots)$ (A1)

$\cos θ = \frac{8.22800 \dots^{2} + 5.52991 \dots^{2} - 5 . 8^{2}}{2 \times 8.22800 \dots \times 5.52991 \dots}$ (M1)(A1)

$= 0.710326 \dots$ (A1)

THEN

$θ = 0.780833 \dots \approx 0.781$ OR $44.7384 \dots ° \approx 44.7 °$ A1

[5 marks]

Examiners report

Many candidates appeared unfamiliar with the notation $\vec{AF}$ for a vector and interpreted it to mean the magnitude of the vector. This notation is defined in the notation list in the subject guide. Candidates are recommended to be aware of this list and to use it throughout the course; this list indicates notation that will be used in the examination questions without introduction and thus it is important that candidates are familiar. It was also common to see $\vec{FA}$ in place of $\vec{AF}$ . Often, candidates were able to find the magnitude correctly even though they failed to write the vector. Correct answers to part (c) were more elusive. The use of scalar product or cosine rule was common although some incorrectly assumed that $F \hat{A} O$ was a right angle. Other errors were to find one of the other angles of the triangle. Some candidates using the scalar product found the obtuse angle between $\vec{OA}$ and $\vec{AF}$ , but failed to find its supplementary angle. In a question like this, it is important that a suitable level of accuracy is maintained throughout so that early rounding will not affect subsequent parts of the question. Candidates must also learn to final round answers and not to truncate them.

[N/A]

The position vector of a particle, $P$ , relative to a fixed origin $O$ at time $t$ is given by

$\vec{OP} = (\begin{matrix} \sin (t^{2}) \\ \cos (t^{2}) \end{matrix})$ .

Find the velocity vector of $P$ .

[2]

Show that the acceleration vector of $P$ is never parallel to the position vector of $P$ .

[5]

Markscheme

attempt at chain rule (M1)

$(v = \frac{d O P}{d t} =) (\begin{matrix} 2 t \cos t^{2} \\ - 2 t \sin t^{2} \end{matrix})$ A1

[2 marks]

attempt at product rule (M1)

$a = (\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \end{matrix})$ A1

METHOD 1

let $S = \sin t^{2}$ and $C = \cos t^{2}$

finding $\cos θ$ using

$a \cdot \vec{OP} = 2 S C - 4 t^{2} S^{2} - 2 S C - 4 t^{2} C^{2} = - 4 t^{2}$ M1

$|\vec{OP}| = 1$

$|a| = \sqrt{{(2 C - 4 t^{2} S)}^{2} + {(- 2 S - 4 t^{2} C)}^{2}}$

$= \sqrt{4 + 16 t^{4}} > 4 t^{2}$

if $θ$ is the angle between them, then

$\cos θ = - \frac{4 t^{2}}{\sqrt{4 + 16 t^{4}}}$ A1

so $- 1 < \cos θ < 0$ therefore the vectors are never parallel R1

METHOD 2

solve

$(\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \end{matrix}) = k (\begin{matrix} \sin t^{2} \\ \cos t^{2} \end{matrix})$ M1

then

$(k =) \frac{2 \cos t^{2} - 4 t^{2} \sin t^{2}}{\sin t^{2}} = \frac{- 2 \sin t^{2} - 4 t^{2} \cos t^{2}}{\cos t^{2}}$

Note: Condone candidates not excluding the division by zero case here. Some might go straight to the next line.

$2 \cos^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2} = - 2 \sin^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2}$

$2 \cos^{2} t^{2} + 2 \sin^{2} t^{2} = 0$

$2 = 0$ A1

this is never true so the two vectors are never parallel R1

METHOD 3

embedding vectors in a 3d space and taking the cross product: M1

$(\begin{matrix} \sin t^{2} \\ \cos t^{2} \\ 0 \end{matrix}) \times (\begin{matrix} 2 \cos t^{2} - 4 t^{2} \sin t^{2} \\ - 2 \sin t^{2} - 4 t^{2} \cos t^{2} \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ - 2 \sin^{2} t^{2} - 4 t^{2} \cos t^{2} \sin t^{2} - 2 \cos^{2} t^{2} + 4 t^{2} \cos t^{2} \sin t^{2} \end{matrix})$

$= (\begin{matrix} 0 \\ 0 \\ - 2 \end{matrix})$ A1

since the cross product is never zero, the two vectors are never parallel R1

[5 marks]

Examiners report

In part (a), many candidates found the velocity vector correctly. In part (b), however, many candidates failed to use the product rule correctly to find the acceleration vector. To show that the acceleration vector is never parallel to the position vector, a few candidates put $\overset{. .}{r} = k r$ presumably hoping to show that no value of the constant k existed for any t but this usually went nowhere.

[N/A]

A submarine is located in a sea at coordinates $(0.8, 1.3, - 0.3)$ relative to a ship positioned at the origin $O$ . The $x$ direction is due east, the $y$ direction is due north and the $z$ direction is vertically upwards.

All distances are measured in kilometres.

The submarine travels with direction vector $(\begin{matrix} - 2 \\ - 3 \\ 1 \end{matrix})$ .

The submarine reaches the surface of the sea at the point $P$ .

Assuming the submarine travels in a straight line, write down an equation for the line along which it travels.

[2]

Find the coordinates of $P$ .

[3]

b.i.

Find $OP$ .

[2]

b.ii.

Markscheme

$r = (\begin{matrix} 0.8 \\ 1.3 \\ - 0.3 \end{matrix}) + λ (\begin{matrix} - 2 \\ - 3 \\ 1 \end{matrix})$ A1A1

Note: Award A1 for each correct vector. Award A0A1 if their “ $r =$ ” is omitted.

[2 marks]

$- 0.3 + λ = 0$ (M1)

$\Rightarrow λ = 0.3$

$r = (\begin{matrix} 0.8 \\ 1.3 \\ - 0.3 \end{matrix}) + 0.3 (\begin{matrix} - 2 \\ - 3 \\ 1 \end{matrix}) = (\begin{matrix} 0.2 \\ 0.4 \\ 0 \end{matrix})$ (M1)

$P$ has coordinates $(0.2, 0.4, 0)$ A1

Note: Accept the coordinates of $P$ in vector form.

[3 marks]

b.i.

$\sqrt{0 . 2^{2} + 0 . 4^{2}}$ (M1)

$= 0.447 km (= 447 m)$ A1

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

The cost adjacency matrix below represents the distance in kilometres, along routes between bus stations.

All the values in the matrix are positive, distinct integers.

It is decided to electrify some of the routes, so that it will be possible to travel from any station to any other station solely on electrified routes. In order to achieve this with a minimal total length of electrified routes, Prim’s algorithm for a minimal spanning tree is used, starting at vertex A.

The algorithm adds the edges in the following order:

AB AC CD DE.

There is only one minimal spanning tree.

Find with a reason, the value of $x$ .

[2]

If the total length of the minimal spanning tree is 14, find the value of $s$ .

[2]

Hence, state, with a reason, what can be deduced about the values of $p$ , $q$ , $r$ .

[2]

Markscheme

AB must be the length of the smallest edge from A so $x = 1$ . R1A1

[2 marks]

$1 + 2 + 3 + s = 14 \Rightarrow s = 8$ M1A1

[2 marks]

The last minimal edge chosen must connect to E , so since $s = 8$ each of $p$ , $q$ , $r$ must be ≥ 9. R1A1

[2 marks]

Examiners report

[N/A]

A particle P moves with velocity v = $\left( {\begin{array}{*{20}{c}} { - 15} \\ 2 \\ 4 \end{array}} \right)$ in a magnetic field, B = $\left( {\begin{array}{*{20}{c}} 0 \\ d \\ 1 \end{array}} \right)$ , $d \in \mathbb{R}$ .

Given that v is perpendicular to B, find the value of $d$ .

[2]

The force, F, produced by P moving in the magnetic field is given by the vector equation F = $a$ v × B, $a \in {\mathbb{R}^ + }$ .

Given that | F | = 14, find the value of $a$ .

[4]

Markscheme

$15 \times 0 + 2d + 4 = 0$ (M1)

$d = - 2$ A1

[2 marks]

$a\left( {\begin{array}{*{20}{c}} { - 15} \\ 2 \\ 4 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 0 \\ { - 2} \\ 1 \end{array}} \right)$ (M1)

$= a\left( {\begin{array}{*{20}{c}} {10} \\ {15} \\ {30} \end{array}} \right)\left( { = 5a\left( {\begin{array}{*{20}{c}} 2 \\ 3 \\ 6 \end{array}} \right)} \right)$ A1

magnitude is $5a\sqrt {{2^2} + {3^2} + {6^2}} = 14$ M1

$a = \frac{{14}}{{35}}\,\left( { = 0.4} \right)$ A1

[4 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]

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]

The above diagram shows the weighted graph G.

Write down the adjacency matrix for G.

[1]

a.i.

Find the number of distinct walks of length 4 beginning and ending at A.

[3]

a.ii.

Starting at A, use Prim’s algorithm to find and draw the minimum spanning tree for G.

Your solution should indicate clearly the way in which the tree is constructed.

[5]

Markscheme

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

[1 mark]

a.i.

We require the (A, A) element of M⁴ which is 13. M1A2

[3 marks]

a.ii.

A1A1A1A1A1

[5 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Sameer is trying to design a road system to connect six towns, A, B, C, D, E and F.

The possible roads and the costs of building them are shown in the graph below. Each vertex represents a town, each edge represents a road and the weight of each edge is the cost of building that road. He needs to design the lowest cost road system that will connect the six towns.

Name an algorithm that will allow Sameer to find the lowest cost road system.

[1]

Find the lowest cost road system and state the cost of building it. Show clearly the steps of the algorithm.

[7]

Markscheme

EITHER

Prim’s algorithm A1

Kruskal’s algorithm A1

[1 mark]

EITHER

using Prim’s algorithm, starting at A

A1A1A1A1A1

lowest cost road system contains roads AC, CD, CF, FE and AB A1

cost is 20 A1

using Kruskal’s algorithm

A1A1A1A1A1

lowest cost road system contains roads CD, CF, FE, AC and AB A1

cost is 20 A1

Note: Accept alternative correct solutions.

[7 marks]

Examiners report

[N/A]

The diagram below shows a weighted graph.

Use Prim’s algorithms to find a minimal spanning tree, starting at J. Draw the tree, and find its total weight.

Markscheme

(C4)

Total weight = 17 (A2)

Note: There are other possible spanning trees.

[6 marks]

Examiners report

[N/A]

Let G be a weighted graph with 6 vertices L, M, N, P, Q, and R. The weight of the edges joining the vertices is given in the table below:

For example the weight of the edge joining the vertices L and N is 3.

Use Prim’s algorithm to draw a minimum spanning tree starting at M.

[5]

What is the total weight of the tree?

[1]

Markscheme

M → Q (M1)

Q → L (A1)

M → P (A1)

P → N → R (A1)

(A1)

Note: There are other correct answers.

[5 marks]

The total weight is 2 + 1 + 3 + 2 + 3 = 11. (A1)

[1 mark]

Examiners report

[N/A]

In the following diagram, $\overrightarrow {{\text{OA}}}$ = a, $\overrightarrow {{\text{OB}}}$ = b. C is the midpoint of [OA] and $\overrightarrow {{\text{OF}}} = \frac{1}{6}\overrightarrow {{\text{FB}}}$ .

N17/5/MATHL/HP1/ENG/TZ0/09

It is given also that $\overrightarrow {{\text{AD}}} = \lambda \overrightarrow {{\text{AF}}}$ and $\overrightarrow {{\text{CD}}} = \mu \overrightarrow {{\text{CB}}}$ , where $\lambda ,{\text{ }}\mu \in \mathbb{R}$ .

Find, in terms of a and b $\overrightarrow {{\text{OF}}}$ .

[1]

a.i.

Find, in terms of a and b $\overrightarrow {{\text{AF}}}$ .

[2]

a.ii.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\lambda$ ;

[2]

b.i.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\mu$ .

[2]

b.ii.

Show that $\mu = \frac{1}{{13}}$ , and find the value of $\lambda$ .

[4]

Deduce an expression for $\overrightarrow {{\text{CD}}}$ in terms of a and b only.

[2]

Given that area $\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$ , find the value of $k$ .

[5]

Markscheme

$\overrightarrow {{\text{OF}}} = \frac{1}{7}$ b A1

[1 mark]

a.i.

$\overrightarrow {{\text{AF}}} = \overrightarrow {{\text{OF}}} - \overrightarrow {{\text{OA}}}$ (M1)

$= \frac{1}{7}$ b – a A1

[2 marks]

a.ii.

$\overrightarrow {{\text{OD}}} =$ a $+ \lambda \left( {\frac{1}{7}b -a} \right){\text{ }}\left( { = (1 - \lambda )a + \frac{\lambda }{7}b} \right)$ M1A1

[2 marks]

b.i.

$\overrightarrow {{\text{OD}}} = \frac{1}{2}$ a $+ \mu \left( { - \frac{1}{2}a + b} \right){\text{ }}\left( { = \left( {\frac{1}{2} - \frac{\mu }{2}} \right)a + \mu b} \right)$ M1A1

[2 marks]

b.ii.

equating coefficients: M1

$\frac{\lambda }{7} = \mu ,{\text{ }}1 - \lambda = \frac{{1 - \mu }}{2}$ A1

solving simultaneously: M1

$\lambda = \frac{7}{{13}},{\text{ }}\mu = \frac{1}{{13}}$ A1AG

[4 marks]

$\overrightarrow {{\text{CD}}} = \frac{1}{{13}}\overrightarrow {{\text{CB}}}$

$= \frac{1}{{13}}\left( {b - \frac{1}{2}a} \right){\text{ }}\left( { = - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)$ M1A1

[2 marks]

METHOD 1

${\text{area }}\Delta {\text{ACD}} = \frac{1}{2}{\text{CD}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{area }}\Delta {\text{ACB}} = \frac{1}{2}{\text{CB}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{ratio }}\frac{{{\text{area }}\Delta {\text{ACD}}}}{{{\text{area }}\Delta {\text{ACB}}}} = \frac{{{\text{CD}}}}{{{\text{CB}}}} = \frac{1}{{13}}$ A1

$k = \frac{{{\text{area }}\Delta {\text{OAB}}}}{{{\text{area }}\Delta {\text{CAD}}}} = \frac{{13}}{{{\text{area }}\Delta {\text{CAB}}}} \times {\text{area }}\Delta {\text{OAB}}$ (M1)

$= 13 \times 2 = 26$ A1

METHOD 2

${\text{area }}\Delta {\text{OAB}} = \frac{1}{2}\left| {a \times b} \right|$ A1

${\text{area }}\Delta {\text{CAD}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CD}}} } \right|$ or $\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{AD}}} } \right|$ M1

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)} \right|$

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a} \right) + \frac{1}{2}a \times \frac{1}{{13}}b} \right|$ (M1)

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{{13}}\left| {a \times b} \right|{\text{ }}\left( { = \frac{1}{{52}}\left| {a \times b} \right|} \right)$ A1

${\text{area }}\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$

$\frac{1}{2}\left| {a \times b} \right| = k\frac{1}{{52}}\left| {a \times b} \right|$

$k = 26$ A1

[5 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Consider the lines ${l_1}$ and ${l_2}$ defined by

${l_1}:$ r $= \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$ and ${l_2}:\frac{{6 - x}}{3} = \frac{{y - 2}}{4} = 1 - z$ where $a$ is a constant.

Given that the lines ${l_1}$ and ${l_2}$ intersect at a point P,

find the value of $a$ ;

[4]

determine the coordinates of the point of intersection P.

[2]

Markscheme

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

METHOD 1

${l_1}:$ r $= \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = - 3 + \beta } \\ {y = - 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.$ M1

$\frac{{6 - ( - 3 + \beta )}}{3} = \frac{{( - 2 + 4\beta ) - 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3$ M1A1

$\frac{{6 - ( - 3 + \beta )}}{3} = 1 - (a + 2\beta ) \Rightarrow 2 = - 5 - a \Rightarrow a = - 7$ A1

METHOD 2

$\left\{ {\begin{array}{*{20}{l}} { - 3 + \beta = 6 - 3\lambda } \\ { - 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 - \lambda } \end{array}} \right.$ M1

attempt to solve M1

$\lambda = 2,{\text{ }}\beta = 3$ A1

$a = 1 - \lambda - 2\beta = - 7$ A1

[4 marks]

$\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ { - 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$ (M1)

$= \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 1} \end{array}} \right)$ A1

$\therefore {\text{P}}(0,{\text{ 10, }} - 1)$

[2 marks]

Examiners report

[N/A]

Nymphenburg Palace in Munich has extensive grounds with $9$ points of interest (stations) within them.

These nine points, along with the palace, are shown as the vertices in the graph below. The weights on the edges are the walking times in minutes between each of the stations and the total of all the weights is $105$ minutes.

Anders decides he would like to walk along all the paths shown beginning and ending at the Palace (vertex A).

Use the Chinese Postman algorithm, clearly showing all the stages, to find the shortest time to walk along all the paths.

Markscheme

Odd vertices are B, F, H and I (M1)A1

Pairing the vertices M1

BF and HI    $9 + 3 = 12$
BH and FI    $4 + 11 = 15$
BI and FH    $3 + 8 = 11$ A2

Note: award A1 for two correct totals.

Shortest time is $105 + 11 = 116$ (minutes) M1A1

[7 marks]

Examiners report

[N/A]

Find the value of $\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}$ .

[2]

Show that $\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[2]

Use the principle of mathematical induction to prove that

$\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[9]

Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .

[6]

Markscheme

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

$\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}$ (M1)A1

Note: Award M1 for 5 equal terms with \) + \) or $-$ signs.

[2 marks]

$\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \frac{{1 - (1 - 2{{\sin }^2}x)}}{{2\sin x}}$ M1

$\equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}$ A1

$\equiv \sin x$ AG

[2 marks]

let ${\text{P}}(n):\sin x + \sin 3x + \ldots + \sin (2n - 1)x \equiv \frac{{1 - \cos 2nx}}{{2\sin x}}$

if $n = 1$

${\text{P}}(1):\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x$ which is true (as proved in part (b)) R1

assume ${\text{P}}(k)$ true, $\sin x + \sin 3x + \ldots + \sin (2k - 1)x \equiv \frac{{1 - \cos 2kx}}{{2\sin x}}$ M1

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let $n = k$ ” only. Subsequent marks are independent of this M1.

consider ${\text{P}}(k + 1)$ :

${\text{P}}(k + 1):\sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

$LHS = \sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x$ M1

$\equiv \frac{{1 - \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x$ A1

$\equiv \frac{{1 - \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}$

$\equiv \frac{{1 - \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}$ M1

$\equiv \frac{{1 - \left( {(1 - 2{{\sin }^2}x)\cos 2kx - \sin 2x\sin 2kx} \right)}}{{2\sin x}}$ M1

$\equiv \frac{{1 - (\cos 2x\cos 2kx - \sin 2x\sin 2kx)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos (2kx + 2x)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

so if true for $n = k$ , then also true for $n = k + 1$

as true for $n = 1$ then true for all $n \in {\mathbb{Z}^ + }$ R1

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

[9 marks]

EITHER

$\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 - \cos 4x}}{{2\sin x}} = \cos x$ M1

$\Rightarrow 1 - \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)$ A1

$\Rightarrow 1 - (1 - 2{\sin ^2}2x) = \sin 2x$ M1

$\Rightarrow \sin 2x(2\sin 2x - 1) = 0$ M1

$\Rightarrow \sin 2x = 0$ or $\sin 2x = \frac{1}{2}$ A1

$2x = \pi ,{\text{ }}2x = \frac{\pi }{6}$ and $2x = \frac{{5\pi }}{6}$

$\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x$ M1A1

$\Rightarrow (2\sin 2x - 1)\cos x = 0,{\text{ }}(\sin x \ne 0)$ M1A1

$\Rightarrow \sin 2x = \frac{1}{2}$ of $\cos x = 0$ A1

$2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}$ and $x = \frac{\pi }{2}$

THEN

$\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}$ and $x = \frac{{5\pi }}{{12}}$ A1

Note: Do not award the final A1 if extra solutions are seen.

[6 marks]

Examiners report

[N/A]

The weights of the edges in a simple graph G are given in the following table.

Use Prim’s Algorithm, starting with vertex F, to find and draw the minimum spanning tree for G. Your solution should indicate the order in which the edges are introduced.

Markscheme

The edges are introduced in the following order:

FD, FC, CB, BA, CE A2A2A2A2A2

[12 marks]

Examiners report

[N/A]

Apply Prim’s algorithm to the weighted graph given below to obtain the minimal spanning tree starting with the vertex A.

Find the weight of the minimal spanning tree.

Markscheme

We start with point A and write S as the set of vertices and T as the set of edges.
The weights on each edge will be used in applying Prim’s algorithm.
Initially, S = {A}, T = Φ. In each subsequent stage, we shall update S and T.

Step 1: Add edge h:   So S = {A, D}, T = {h}
Step 2: Add edge e:   So S = {A, D, E}   T = {h, e}
Step 3: Add edge d:   Then S = {A, D, E, F}   T = {h, e, d}
Step 4: Add edge a:   Then S = {A, D, E, F, B}   T = {h, e, d, a}
Step 5: Add edge i:    Then S = {A, D, E, F, B, G} T = {h, e, d, a, i}
Step 6: Add edge g:   Then S = (A, D, E, F, B, G, C}   T = {h, e, d, a, I, g} (M4)(A3)

Notes: Award (M4)(A3) for all 6 correct,
(M4)(A2) for 5 correct;
(M3)(A2) for 4 correct,
(M3)(A1) for 3 correct;
(M1)(A1) for 2 correct,
(M1)(AO) for 1 correct

OR
(M2) for the correct definition of Prim’s algorithm,
(M2) for the correct application of Prim’s algorithm,
(A3) for the correct answers at the last three stages.

Now S has all the vertices and the minimal spanning tree is obtained.

The weight of the edges in T is 5 + 3 + 5 + 7 + 5 + 6

= 31 (A1)

[8 marks]

Examiners report

[N/A]

In this part, marks will only be awarded if you show the correct application of the required algorithms, and show all your working.

In an offshore drilling site for a large oil company, the distances between the planned wells are given below in metres.

It is intended to construct a network of paths to connect the different wells in a way that minimises the sum of the distances between them.

Use Prim’s algorithm, starting at vertex 3, to find a network of paths of minimum total length that can span the whole site.

Markscheme

(R2)(A4)(M1)

(A1)

Note: Award (R2) for correct algorithms, (R1) for 1 error, (R0) for 2 or more errors.
Award (A4) for correct calculations, (A3) for 1 error, (A2) for 2 errors, (A1) for 3 errors, (A0) for 4 or more errors.
Award (M1) for tree/table/method.
Award (A1) for minimum weight.

[8 marks]

Examiners report

[N/A]

Let a = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$ , $k \in \mathbb{R}$ .

Given that a and b are perpendicular, find the possible values of $k$ .

Markscheme

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

a • b = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$

$= - 6 + k\left( {k + 2} \right) - k$ A1

a • b = 0 (M1)

${k^2} + k - 6 = 0$

attempt at solving their quadratic equation (M1)

$\left( {k + 3} \right)\left( {k - 2} \right) = 0$

$k = - 3{\text{,}}\,\,2$ A1

Note: Attempt at solving using |a||b| = |a × b| will be M1A0A0A0 if neither answer found M1(A1)A1A0
for one correct answer and M1(A1)A1A1 for two correct answers.

[4 marks]

Examiners report

[N/A]

The lengths of two of the sides in a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The triangle has an area of $\frac{{5\sqrt {15} }}{2}$ cm².

Show that ${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ .

[1]

Find the two possible values for the length of the third side.

[6]

Markscheme

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

EITHER

$\frac{{5\sqrt {15} }}{2} = \frac{1}{2} \times 4 \times 5\,{\text{sin}}\,\theta$ A1

height of triangle is $\frac{{5\sqrt {15} }}{4}$ if using 4 as the base or ${\sqrt {15} }$ if using 5 as the base A1

THEN

${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ AG

[1 mark]

let the third side be $x$

${x^2} = {4^2} + {5^2} - 2 \times 4 \times 5 \times {\text{cos}}\,\theta$ M1

valid attempt to find ${\text{cos}}\,\theta$ (M1)

Note: Do not accept writing ${\text{cos}}\left( {{\text{arcsin}}\left( {\frac{{\sqrt {15} }}{4}} \right)} \right)$ as a valid method.

${\text{cos}}\,\theta = \pm \sqrt {1 - \frac{{15}}{{16}}}$

$= \frac{1}{4}{\text{,}}\,\, - \frac{1}{4}$ A1A1

${x^2} = 16 + 25 - 2 \times 4 \times 5 \times \pm \frac{1}{4}$

$x = \sqrt {31}$ or $\sqrt {51}$ A1A1

[6 marks]

Examiners report

[N/A]

Solve the equation ${\sec ^2}x + 2\tan x = 0,{\text{ }}0 \leqslant x \leqslant 2\pi$ .

Markscheme

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

METHOD 1

use of ${\sec ^2}x = {\tan ^2}x + 1$ M1

${\tan ^2}x + 2\tan x + 1 = 0$

${(\tan x + 1)^2} = 0$ (M1)

$\tan x = - 1$ A1

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$ A1A1

METHOD 2

$\frac{1}{{{{\cos }^2}x}} + \frac{{2\sin x}}{{\cos x}} = 0$ M1

$1 + 2\sin x\cos x = 0$

$\sin 2x = - 1$ M1A1

$2x = \frac{{3\pi }}{2},{\text{ }}\frac{{7\pi }}{2}$

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$ A1A1

Note: Award A1A0 if extra solutions given or if solutions given in degrees (or both).

[5 marks]

Examiners report

[N/A]

Three points in three-dimensional space have coordinates A(0, 0, 2), B(0, 2, 0) and C(3, 1, 0).

Find the vector $\overrightarrow {{\text{AB}}}$ .

[1]

a.i.

Find the vector $\overrightarrow {{\text{AC}}}$ .

[1]

a.ii.

Hence or otherwise, find the area of the triangle ABC.

[4]

Markscheme

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.i.

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

Note: Accept row vectors or equivalent.

[1 mark]

a.ii.

METHOD 1

attempt at vector product using $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{AC}}}$ . (M1)

±(2i + 6j +6k) A1

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|$ M1

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

METHOD 2

attempt to use $\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = \left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|{\text{cos}}\,\theta$ M1

$\left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right) = \sqrt {{0^2} + {2^2} + {{\left( { - 2} \right)}^2}} \sqrt {{3^2} + {1^2} + {{\left( { - 2} \right)}^2}} {\text{cos}}\,\theta$

$6 = \sqrt 8 \sqrt {14} \,{\text{cos}}\,\theta$ A1

${\text{cos}}\,\theta = \frac{6}{{\sqrt 8 \sqrt {14} }} = \frac{6}{{\sqrt {112} }}$

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|{\text{sin}}\,\theta$ M1

$= \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {1 - \frac{{36}}{{112}}} \,\left( { = \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {\frac{{76}}{{112}}} } \right)$

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

The weights of the edges of a complete graph G are shown in the following table.

Starting at B, use Prim’s algorithm to find and draw a minimum spanning tree for G. Your solution should indicate the order in which the vertices are added. State the total weight of your tree.

Markscheme

Different notations may be used but the edges should be added in the following order.

Using Prim’s Algorithm, (M1)

BD A1

DF A1

FA A1

FE A1

EC A1

Total weight = 12 A2

[10 marks]

Examiners report

[N/A]

A sector of a circle with radius $r$ cm , where $r$ > 0, is shown on the following diagram.
The sector has an angle of 1 radian at the centre.

Let the area of the sector be $A$ cm² and the perimeter be $P$ cm. Given that $A = P$ , find the value of $r$ .

Markscheme

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

$A = P$

use of the correct formula for area and arc length (M1)

perimeter is $r\theta + 2r$ (A1)

Note: A1 independent of previous M1.

$\frac{1}{2}{r^2}\left( 1 \right) = r\left( 1 \right) + 2r$ A1

${r^2} - 6r = 0$

$r = 6$ (as $r$ > 0) A1

Note: Do not award final A1 if $r = 0$ is included.

[4 marks]

Examiners report

[N/A]

The vectors a and b are defined by a = $\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ t \end{array}} \right)$ , b = $\left( {\begin{array}{*{20}{c}} 0 \\ { - t} \\ {4t} \end{array}} \right)$ , where $t \in \mathbb{R}$ .

Find and simplify an expression for a • b in terms of $t$ .

[2]

Hence or otherwise, find the values of $t$ for which the angle between a and b is obtuse .

[4]

Markscheme

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

a • b = $\left( {1 \times 0} \right) + \left( {1 \times - t} \right) + \left( {t \times 4t} \right)$ (M1)

= $- t + 4{t^2}$ A1

[2 marks]

recognition that a • b = |a||b|cos θ (M1)

a • b < 0 or $- t + 4{t^2}$ < 0 or cos θ < 0 R1

Note: Allow ≤ for R1.

attempt to solve using sketch or sign diagram (M1)

$0 < t < \frac{1}{4}$ A1

[4 marks]

Examiners report

[N/A]

Helen is building a cabin using cylindrical logs of length 2.4 m and radius 8.4 cm. A wedge is cut from one log and the cross-section of this log is illustrated in the following diagram.

Find 50° in radians.

[1]

Find the volume of this log.

[4]

Markscheme

$\frac{{50 \times \pi }}{{180}} = 0.873\,\,\left( {0.872664 \ldots } \right)$ A1

[1 mark]

volume $= 240\left( {\pi \times {{8.4}^2} - \frac{1}{2} \times {{8.4}^2} \times 0.872664 \ldots } \right)$ M1M1M1

Note: Award M1 240 × area, award M1 for correctly substituting area sector formula, award M1 for subtraction of the angles or their areas.

= 45800 (= 45811.96071) A1

[4 marks]

Examiners report

[N/A]