Mathematics - Microsoft · Mathematics Higher Level – Paper 1 Marking Scheme (300 marks) Section A Concepts and Skills 150 marks Answer all six questions from this section. Question

Page 1 of 68

L.17/20

Pre-Leaving Certificate Examination, 2016

Mathematics Higher Level

Marking Scheme

Paper 1 Pg. 2 Paper 2 Pg. 36

2016.1 L.17/20_MS 2/72 Page 2 of 71 examsDEB


Mathematics

Higher Level – Paper 1 Marking Scheme (300 marks)

Structure of the Marking Scheme

Students’ responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students’ responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table:

Scale label A B C D

No. of categories 2 3 4 5

5 mark scale 0, 2, 5 0, 2, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale 0, 6, 10, 13, 15

A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary.

Marking scales – level descriptors

A-scales (two categories) incorrect response (no credit) correct response (full credit)

B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit)

C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit)

D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions.

Unless otherwise specified, accept correct answer with or without work shown.

Accept students’ work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved.

examsDEB

DEB 2014 LC-H

Scale label A B C DNo of categories 2 3 4 5

5 mark scale 0, 2, 5 0, 2, , 5 0, 2, 3,

10 mark scale 0, 5, 10 0, 3, 7, 10 0, 2, 5,

15 mark scale 0, 7, 15 0, 5, 10,15 0, 4, 7, 1

20 mark scale


Summary of Marks – 2016 LC Maths (Higher Level, Paper 1)

Q.1 (a) (i) 10C* (0, 4, 7, 10) Q.7 (a) (i) 5B* (0, 2, 5) (ii) 5C (0, 2, 4, 5) (ii) 10C (0, 4, 7, 10) (b) 10D (0, 4, 6, 8, 10) (iii) 5B (0, 2, 5) 25 (iv) 10D* (0, 4, 6, 8, 10) (b) (i) 15D (0, 6, 10, 13, 15) Q.2 (a) 5C (0, 2, 4, 5) (ii) 5B (0, 2, 5) (b) (i) 10C (0, 4, 7, 10) 50 (ii) 5B (0, 2, 5) (c) 5B (0, 2, 5) 25 Q.8 (a) 5B (0, 2, 5) Q.3 (a) 10D (0, 4, 6, 8, 10) (b) (i) 5B* (0, 2, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 10C* (0, 4, 7, 10) (ii) 5C (0, 2, 4, 5) (c) (i) 10D* (0, 4, 6, 8, 10) (iii) 5C (0, 2, 4, 5) (ii) 10C* (0, 4, 7, 10) 25 (iii) 5C (0, 2, 4, 5) (iv) 5B (0, 2, 5) 50Q.4 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) 25 Q.9 (a) (i) 5B (0, 2, 5) (ii) 5B* (0, 2, 5) (iii) 5B (0, 2, 5) Q.5 (a) 5B (0, 2, 5) (iv) 10C (0, 4, 7, 10) (b) 10C (0, 4, 7, 10) (b) (i) 10D (0, 4, 6, 8, 10) (c) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (d) 5C (0, 2, 4, 5) (c) 10D* (0, 4, 6, 8, 10) 25 50 Q.6 (a) 5C (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) (c) (i)

10D (0, 4, 6, 8, 10)

(ii) 25

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

General Instructions

There are two sections in this examination paper.

Section A Concepts and Skills 150 marks 6 questions Section B Contexts and Applications 150 marks 3 questions

Answer all questions.

Marks will be lost if all necessary work is not clearly shown.

Answers should include the appropriate units of measurement, where relevant.

Answers should be given in simplest form, where relevant.



Mathematics


Section A Concepts and Skills 150 marks

Answer all six questions from this section.

Question 1 (25 marks)

1(a) Fifty stones are placed in a straight line on level horizontal ground, at equal intervals of 5 m. A bucket is located on the same line 10 m away from the first stone. Tom, who can lift only one stone at a time, wishes to put all of them in the bucket. He starts from the bucket.

10 m 5 m 5 m

Bucket

(i) Calculate the total distance, in km, that Tom travels to put all the stones in the bucket. (10C*)

Stone: 1 2 3 … 50 Distance: 20 30 40 … ... arithmetic series

Sn = 2

n[2a + (n – 1)d ]

a = 20 d = 10

For n = 50

S50 = 2

50[2(20) + 49(10)]

= 25[40 + 490] = 25[530] = 13,250 m = 13·25 km

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down ‘example of arithmetic series’ or relevant correct formula for Tn or Sn of arithmetic series. – Finds a = 20 and d = 10. – Some correct substitution into relevant formula for S50.

High partial credit: (7 marks) – Substitutes correctly into formula for S50, but fails to evaluate or evaluates incorrectly. – Finds S50, but fails to give final answer in the desired form (km).

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘km’) - apply only once in each section (a), (b), (c), etc. of question.

examsDEB


Question 1 (cont’d.)

1(a) (cont’d.)

(ii) More stones are added and placed at the same intervals along the line. Find the number of extra stones added if the total distance that Tom travels to put all of the stones in the bucket is 27 km. (5C)

More stones added and placed at the same interval Sn = 27 km = 27,000 m

Sn = 2

n[2a + (n – 1)d ]

a = 20 d = 10 For n stones in total

Sn = 2

n[2(20) + (n – 1)10]

= 27,000

2

n[40 + 10n – 10] = 27,000

2

n[10n + 30] = 27,000

5n2 + 15n = 27,000 n2 + 3n – 5,400 = 0 (n – 72)(n + 75) = 0 n – 72 = 0 n + 75 = 0 n = 72 n = –75 (not possible)

Number of extra stones added = 72 – 50 = 22

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. Sn formula correctly substituted (in terms of n). – Substitutes correctly into formula for Sn and finds equation in terms in n.

High partial credit: (4 marks) – Solves correctly to find value of n, but fails to finish or finishes incorrectly.



1(b) Tom takes 5 seconds to retrieve the first stone and place it in the bucket. The time it takes him to place each subsequent stone in the bucket increases by 40%. How many stones can Tom retrieve and place in the bucket in 10 minutes? (10D)

Stone: 1 2 3 … 50 Time: 5 5(1·4) 5(1·4)2 … ... geometric series

Total time = 10 × 60 = 600 s

Sn = r

ra n

−−

1

)1(

a = 5 r = 1·4 For n stones in total

Sn = 411

)411(5

⋅−⋅− n

≤ 600

411

)411(5

⋅−⋅− n

≤ 600

40

)411(5

⋅−⋅− n

≤ 600

–12·5(1 – 1·4n) ≤ 600 12·5(1·4n – 1) ≤ 600 1·4n – 1 ≤ 48 1·4n ≤ 48 + 1 ≤ 49 n ≤ log1·4(49) ≤ 11·566542... n = 11 stones

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down ‘example of geometric series’ or relevant correct formula for Tn or Sn of geometric series. – Finds a = 5 and r = 1·4. – Some correct substitution into relevant formula for Sn.

Medium partial credit: (6 marks) – Substitutes correctly into formula for Sn, and stops or continues incorrectly.

High partial credit: (8 marks) – Finds 1·4n ≤ 49 or equivalent, but fails to finish or finishes incorrectly.



The functions f and g are defined for x ∈ ℝ as

f : x 1 →׀ – x and

g : x 2 →׀x2 – 9.

2(a) Given that the function h(x) = g ° f (x), show that h(x) = 2x2 – 4x – 7. (5C)

f (x) = 1 – x g(x) = 2x2 – 9 h(x) = g

° f (x) = g(1 – x) = 2(1 – x)2 – 9 = 2(1 – 2x + x2) – 9 = 2 – 4x + 2x2 – 9 = 2x2 – 4x – 7

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down g(1 – x) and stops.

High partial credit: (4 marks) – Finds g ° f (x) = 2(1 – x)2 – 9, but fails

to finish or finishes incorrectly.

2(b) (i) Express h(x) in the form a(x + b)2 + c, where a, b and c are constants. (10C) h(x) = 2x2 – 4x – 7 = 2[x2 – 2x] – 7 = 2[(x2 – 2x + 1) – 1] – 7 = 2(x2 – 2x + 1) –2 – 7 = 2(x – 1)2 – 9

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. takes 2 as a factor or groups the x terms.

High partial credit: (7 marks) – Completes the square correctly, but fails to finish or finishes incorrectly.



2(b) (cont’d.)

(ii) Hence, or otherwise, find the co-ordinates of the turning point of h(x). (5B)

h(x) = 2(x – 1)2 – 9

the minimum value of h(x) occurs when x = 1 h(x) = 2(1 – 1)2 – 9 = 2(0)2 – 9 = 0 – 9 = –9

Turning point = (1, –9) ... minimum point

or

h(x) = 2(x – 1)2 – 9 h′(x) = 4x – 4 = 0 4x – 4 = 0 4(x – 1) = 0 x – 1 = 0 x = 1

h(x) = 2(x – 1)2 – 9 h(1) = 2(1)2 – 4(1) – 7 = 2 – 4 – 7 = –9

Turning point = (1, –9) ... minimum point

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Finds correct one co-ordinate of the turning point.

2(c) The function k(x) is the image of h(x) under a translation. The co-ordinates of the turning point of k(x) are (–1, –5). Find k(x). (5B)

k (x) is the image of h(x) under a translation Turning point of k (x) is (–1, –5) h(x) = 2(x – 1)2 – 9

k(x) = 2(x – (–1))2 – 5 = 2(x + 1)2 – 5 = 2(x2 + 2x + 1) – 5 = 2x2 + 4x + 2 – 5 = 2x2 + 4x – 3

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Writes down –5 as the constant term or (x + 1)2 or 2 as the co-efficient of x2.



3(a) Solve the inequality:

1

13

+−

x

x ≤

3

5, where x ∈ ℝ and x ≠ –1. (10D)

1

13

+−

x

x(x + 1)2 ≤

3

5(x + 1)2

(3x – 1)(x + 1) ≤ 3

5(x2 + 2x + 1)

3x2 + 2x – 1 ≤ 3

5(x2 + 2x + 1)

3(3x2 + 2x – 1) ≤ 5(x2 + 2x + 1) 9x2 + 6x – 3 ≤ 5x2 + 10x + 5 4x2 – 4x – 8 ≤ 0 x2 – x – 2 ≤ 0 (x – 2)(x + 1) ≤ 0

Consider: (x – 2)(x + 1) = 0 x – 2 = 0 x = 2

x + 1 = 0 x = –1

x2 – x – 2 ≤ 0 x ≤ 2 x > –1 as x ≠ −1

Solution: –1 < x ≤ 2

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant correct step, e.g. multiplies both sides by (x + 1)2. – Finds particular values of x for which the inequality holds. – Some use of quadratic root formula.

Middle partial credit: (6 marks) – Solves the relevant quadratic equation to find the roots, x = 2 and x = –1.

High partial credit: (8 marks) – Wrong shape to graph, but otherwise correct. – Deduces incorrectly from correct values of x. – Deduces correctly for one case only, i.e. –1 < x or x ≤ 2. – Solution set shown on graph only.

21�1

x

y


Question 3 (cont’d.) 3(b) (i) Solve the equation:

x + x

1 =

3

10, where x ∈ ℝ \ {0}. (5C)

x + x

1 =

3

10

3x2 + 3 = 10x 3x2 – 10x + 3 = 0 (3x – 1)(x – 3) = 0 3x – 1 = 0 3x = 1

x = 3

1

x – 3 = 0 x = 3

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. multiples both sides by x.

High partial credit: (4 marks) – Finds correct quadratic equation, but fails to finish or finishes incorrectly.

(ii) Use your answers above to solve the equation:

9x + x9

1 =

3

10, where x ∈ ℝ. (5C)

By substituting 9x for x into x + x

1 =

3

10 from part (b)(i),

the solutions of 9x + x9

1 =

3

10 are:

9x = 3

1

32x = 3–1 2x = –1

x = –2

1

and

9x = 3 32x = 31 2x = 1

x = 2

1

** Accept students’ answers from part (b)(i) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, i.e. makes connection with previous part (replaces 9x for x in equation or equivalent).

High partial credit: (4 marks) – Correct solutions with some manipulation,

i.e. 32x = 3

1 or –3 and 32x = 3 or 31, but

fails to finish or finishes incorrectly. – Finds one correct solution and finishes correctly.



3(b) (cont’d.)

(iii) Hence, or otherwise, solve the equation:

log3 x + log x

3 = 3

10, where x > 1. (5C)

log x 3 =

x3

3

log

3log

= x3log

1

log3 x +

x3log

1 =

3

10

By substituting log3 x for x into x +

x

1 =

3

10 from part (b)(i),

the solutions of log3 x +

x3log

1 =

3

10 are:

log3 x =

3

1

x = 31

3 or 3 3 or 1·442249...

and

log3 x = 3

x = 33 = 27

or

log3 x +

x3log

1 =

3

10

(log3 x)2 + 1 =

3

10(log3

x)

3(log3 x)2 – 10(log3

x) + 3 = 0 (3(log3

x) – 1)(log3 x – 3) = 0

3log3 x – 1 = 0

log3 x =

3

1

x = 31

3 or 3 3 or 1·442249...

and

log3 x – 3 = 0

log3 x = 3

x = 33 = 27


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. changes

log3 x to

x3log

1 or replaces

x3log

1 for x

in equation or equivalent.

High partial credit: (4 marks) – Finds correct solutions, i.e. log3 x =

3

1

and log3 x = 3, but fails to finish or

finishes incorrectly. – Finds one correct solution and finishes correctly.



4(a) z = – 3 + i is a complex number, where i2 = –1.

(i) Write z in polar form. (5C)

– 3 + i = r(cos + i sin )

r = | z |

= 22 1)3( +−

= 13 +

= 4 = 2

tan = 3

1

= 6

π or 30°

= π – 6

π

= 6

5π

or = 180° – 30° = 150°

– 3 + i = r(cos + i sin )

= 2(cos6

5π + i sin

6

5π) or 2(cos 150° + i sin 150°)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down relevant formula to show complex number in polar form. – Finds correct r or (reference angle). – Plots z on Argand diagram.

High partial credit: (4 marks) – Finds correct both r and , but fails to finish or finishes incorrectly. – Finds r and (reference angle), but complex number in wrong quadrant and finishes correctly.

(ii) Use De Moivre’s theorem to find z4, giving your answer in the form a + bi, where a, b ∈ ℝ. (5C)

z4 = (– 3 + i)4

= 24(cos6

5π + i sin

6

5π)4 or 24(cos 150° + i sin 150°)4

= 16(cos6

20π + i sin

6

20π) or 16(cos 600° + i sin 600°)

= 16(cos3

4π + i sin

3

4π) or 16(cos 240° + i sin 240°)

= 16[–2

1 – i

2

3]

= –8 – i8 3 or –8[1 + i 3]

Im

Re

r1

�

�

3



4(a) (ii) (cont’d.)

** Accept students’ answers from part (a)(i) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. finds r4 = 16 or 4 = 600°. – Correct substitution without manipulation,

i.e. z4 = 24(cos6

5π + i sin

6

5π)4 or

z4 = 24(cos 150° + i sin 150°)4 and stops.

High partial credit: (4 marks) – Correct substitution with manipulation,

i.e. z4 = 16(cos6

20π + i sin

6

20π) or

16(cos 600° + i sin 600°)4, but fails to finish or finishes incorrectly.

(iii) Using the fact that z is one of the roots, or otherwise, find the other roots of z4. (5C)

z4 = –8 – i8 3

z4 = (–8 – i8 41

)3 ... 4 fourth roots

But z = – 3 + i is one of these roots

the 4 roots are on a circle of radius 2 units at 90° intervals

iz, i2z and i3z are the other roots, iz = i(– 3 + i)

= –1 – 3 i

i2z = i2(– 3 + i)

= –1(– 3 + i)

= 3 – i

i3z = i3(– 3 + i)

= –i(– 3 + i)

= 1 + 3 i

** Accept students’ answers from part (a)(i) if not oversimplified. ** Accept students’ answers using other methods to find the roots of z4.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. sketches / plots – 3 + i on Argand diagram with circle of radius 2 units, centre (0,0). – States that other roots are iz, i2z and i3z. – States that –z = 3 – i is also a root. – Finds one other root, i.e. iz, i2z or i3z, correctly.

High partial credit: (4 marks) – Finds two other roots correctly.

Im

Re

i z3

i z2

z

iz



4(b) Verify that 2i is a root of the equation z2 – (1 + 3i)z + (–2 + 2i) = 0 and hence, find the other root of the equation. (10D)

Verify 2i is a root

f (z) = z2 – (1 + 3i)z + (–2 + 2i) = 0

f (2i) = (2i)2 – (1 + 3i)(2i) + (–2 + 2i) = 4i2 – 2i – 6i2 – 2 + 2i = –4 – 2i + 6 – 2 + 2i = 0 z = 2i is a root of the equation

Other root of equation

f (z) = z2 – (1 + 3i)z + (–2 + 2i) f (z) = z2 – (sum of roots)z + (product of roots) = 0

Sum of roots = 1 + 3i 2nd root = a + ib

2i + a + ib = 1 + 3i a + ib = 1 + 3i – 2i = 1 + i

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down ‘2i is a root if f (2i) = 0’ or similar. – Substitutes 2i correctly into f (z), but fails to show that f (2i) = 0.

Middle partial credit: (6 marks) – Substitutes 2i correctly into f (z) and shows correctly that f (2i) = 0 and stops.

High partial credit: (8 marks) – Verifies correctly that 2i is a factor and attempts division without finishing. – Verifies correctly that 2i is a factor and states sum of roots = 1 + 3i or product of roots = –2 + 2i, but fails to finish or finishes incorrectly.



The function f is defined as

f : x ׀→ x2 ln x, for x ∈ ℝ, x > 0.

The diagram shows part of the graph of f.

5(a) State whether or not f is injective. Give a reason for your answer. (5B)

Answer – f is not injective

Reason Any 1: – it fails the horizontal line test // – the graph of the function intersects some horizontal lines

more than once

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct answer, but no reason or incorrect reason given.

5(b) Find f ′(x), the first derivative of f (x). (10C)

y = uv

dx

dy = udx

dv + v

dx

du ... product rule

f (x) = x2 ln x

f ′(x) = x2 (x

1) + ln x(2x)

= x + 2x ln x = x(1 + 2ln x)

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down product rule for differentiation. – Any correct differentiation, e.g.

dx

d (x2) = 2x or dx

d (ln x) = x

1 .

High partial credit: (7 marks) – Applies product rule correctly, i.e.

f ′(x) = x2 (x

1 ) + ln x(2x), but fails to finish

or finishes incorrectly.

y

x

y f x( )�

0



5(c) Find the co-ordinates of the local minimum point of the graph of f , in terms of e. (5C) Turning point: f ′(x) = 0 x + 2x ln x = 0 x(1 + 2ln x) = 0

x = 0 ... not a solution as x > 0

1 + 2ln x = 0

ln x = –2

1

x = 21−

e or e

1

f (x) = x2 ln x

f ( 21−

e ) = ( 21−

e )2ln ( 21−

e )

= e–1 × (–2

1)ln e

= e

1 × (–

2

1)(1)

= –e2

1

∴ Turning point at (e

1, –

e2

1)

f ′(x) = x + 2x ln x

f ″(x) = 1 + 2x(x

1) + (2)ln x

= 1 + 2 + 2ln x = 3 + 2ln x

f ″(e

1) = 3 + 2ln(

e

1)

= 3 + 2ln( 21−

e )

= 3 + 2(–2

1)

= 3 – 1 = 2 > 0

Local min. point @ (e

1, –

e2

1)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down ‘turning point occurs at f ′(x) = 0’ or ‘local minimum point when f ″(x) > 0’ or similar. – Finds x co-ordinate of turning point correctly.

High partial credit: (4 marks) – Finds x and y co-ordinates of turning point correctly, but fails to show or shows incorrectly that turning point is a local minimum point.



5(d) Under what conditions is f bijective? For what domain and codomain does f meet this criteria? (5C) f (x) is bijective if it is both injective and surjective

Domain – for f to be surjective, the domain must be defined

f is defined for x ≥ e

1, x ∈ ℝ or [

e

1, ∞), x ∈ ℝ

Codomain – for f to be injective, the codomain must be defined

f is defined for y ≥ –e2

1, y ∈ ℝ or [–

e2

1, ∞), x ∈ ℝ

** Accept students’ answers from part (c) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. states ‘f is bijective if it is both injective and surjective’ or similar.

High partial credit: (4 marks) – Writes down condition for which f is bijective and finds either correct domain or codomain, i.e. missing one part.



6(a) Given a line segment of length one unit, show clearly how to construct a line segment of length 2 units, using only a compass and a straight edge. Hence, indicate a line segment of length 2 – 1 units. Label each line segment clearly. (5C)

2

12 �1A O BD

C

Construct a line segment of length 2 units

Label end points of given line segment A and O. Place the needle point of the compass at O and swing an arc through A.

Extend the line segment OA until it cuts the arc at B, as shown. Construct the perpendicular bisector of AB as above. Place the needle point of the compass at O and swing an arc through A,

cutting the perpendicular bisector at C. Join A to C: | AC |2 = 12 + 12 = 2 | AC | = 2

Hence, indicate a line segment of length 2 – 1 units

Place the needle point of the compass at A, swing an arc through C and continue the arc until it crosses [ AB ] at D, as shown.

| OD | = | AD | – | OA | = 2 – 1

** Actual size as per examination paper.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. extends line and constructs perpendicular bisector. – Constructs a right-angled triangle with sides of length 1, 1 and 2 indicated. – Using a protractor and ruler, draws a triangle with sides of length 1, 1 and 2 ,

based on cos 45° = 2

1 .

High partial credit: (4 marks) – Constructs correctly line segment of length 2 , but fails to find or finds incorrect line segment of length 2 – 1.



6(b) 2 – 1 is a root of x2 + bx + c = 0, where b, c ∈ ℤ. Find the value of b and the value of c and hence, find the other root. (10D)

2 –1 is a root of x2 + bx + c = 0 ( 2 – 1)2 + b( 2 – 1) + c = 0 2 – 2 2 + 1 + b 2 – b + c = 0 (3 – b + c) + 2 (b – 2) = 0 3 – b + c = 0 and 2 (b – 2) = 0

b – 2 = 0 b = 2

3 – 2 + c = 0 1 + c = 0 c = –1

Equation: x2 + 2x – 1 = 0

x2 + 2x – 1 = 0

x = )1(2

)1)(1(422 2 −−±−

= 2

442 +±−

= 2

82 ±−

= –1 ± 2 1st root = –1 – 2 2nd root = –1 – 2

or

1st root = –1 + 2 2nd root = –1 – 2 ... conjugate root

or

f (x) = x2 – (sum of roots)x + (product of roots) = 0

Sum of roots = –b = –2

2nd root = –2 − 1st root = –2 – (–1 + 2 ) = –2 + 1 – 2 = − 1 − 2

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. substitutes correctly 2 – 1 for x into equation.

Middle partial credit: (6 marks) – Expands ( 2 – 1)2 + b( 2 – 1) + c correctly, but fails to finish or finishes incorrectly.

High partial credit: (8 marks) – Finds correct b and c, but fails to find or finds incorrect other root.



6(c) (i) Explain what it means to say that 2 is not a rational number. (10D)

Answer Any 1:

– it cannot be written in the form q

p, where p, q ∈ ℤ

– it cannot be written / represented as a simple fraction // – it cannot be expressed as a ratio of integers // – it has a decimal expansion that neither terminates nor

becomes periodic // – it cannot be represented as a terminating or a repeating

decimal // etc.

** Accept other appropriate answers.

(ii) Use the method of proof by contradiction to prove that 2 is not a rational number.

Assume that 2 is rational

2 = q

p, where p, q ∈ ℤ and HCF(p, q) = 1

2 = 2

2

q

p

2q2 = p2 Hence, p2 is even and so p is even

Let p = 2k p2 = 4k2 2q2 = 4k2 q2 = 2k2 Hence, q2 is even and so q is even as p and q are both even, it is a false (contradiction) as HCF(p, q) = 1 2 is not a rational number

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. explains why 2 is not a rational number.

Middle partial credit: (6 marks) – Proof substantially correct, but more than one step missing.

High partial credit: (8 marks) – Only one step missing in proof or proof not fully finished.


Section B Contexts and Applications 150 marks

Answer all three questions from this section.


A speed camera is located at P, which is 40 m from the nearest point, A, on a straight level road. The camera monitors a car moving along the road from point B to A, which are 100 m apart. C is a point on the road which is 30 m from A.

P

40 m

CA B

x

s

Let s be the distance from the speed camera to the car after t seconds and let x be the distance the car has travelled from B at that instant. The speed of the car is given by the formula:

dt

dx = 25,

where speed is in metres per second. Both x and s are in metres.

7(a) (i) Find the time it takes the car to reach C. (5B*)

Time = Speed

Distance

= 25

||BC

= 25

30100 −

= 25

70

= 5

14

= 2·8 s

Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down

formula for speed, S = T

D or similar.

– Finds correct value for | BC |.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘s’) - apply only once in each section (a), (b), (c), etc. of question.



7(a) (cont’d.)

(ii) Express the distance of the car from A in terms of x and hence, show that s, the distance

between the car and the camera at time t, is 600,112002 +− xx . (10C)

the distance of the car from A after t seconds is 100 – x

Using Pythagoras’ theorem s2 = 402 + (100 – x)2 = 1,600 + 10,000 – 200x + x2 = x2 – 200x + 11,600

s = 600,112002 +− xx

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down ‘distance from A is 100 – x’ or similar. – Some correct substitution into Pythagoras’ theorem.

High partial credit: (7 marks) – Substitutes correctly into Pythagoras’ theorem, i.e. s2 = 402 + (100 – x)2, but fails to finish or finishes incorrectly.

(iii) Find dx

ds. (5B)

s = 600,112002 +− xx

s = (x2 – 200x + 11,600 21

)

dx

ds =

2

1(x2 – 200x + 11,600 2

1

)−

× (2x – 200)

= 600,11200

1002 +−

−

xx

x

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down

s in the form (x2 – 200x + 11,600 21

) .

– Some correct differentiation.



7(a) (cont’d.)

(iv) Hence, find the rate of change of s, with respect to t, when the car is at C. (10D*)

dt

ds =

dx

ds ×

dt

dx

= 600,11200

1002 +−

−

xx

x × 25

= 600,11200

)100(252 +−

−

xx

x

when the car is at C, x = 70

dt

ds =

600,11)70(200)70(

)10070(252 +−

−

= 600,11000,14900,4

)30(25

+−−

= 500,2

750−

= 50

750−

= –15 m/s

** Accept students’ answers from part (a)(iii) if not oversimplified.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down

chain rule, dt

ds =

dx

ds ×

dt

dx, and stops.

– Finds 600,11200

)100(252 +−

−

xx

x and stops.

Middle partial credit: (6 marks) – Some correct substitution into dt

ds

and stops or continues.

High partial credit: (8 marks) – Substitutes correctly into

dt

ds, i.e.

dt

ds =

600,11)70(200)70(

)10070(252 +−

−, but

fails to evaluate or evaluates incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘m/s’) - apply only once in each section (a), (b), (c), etc. of question.



7(b) A second car passes B at the same time moving in the same direction towards A. The speed of the car is approximated by the following model:

v = 10e1·05t,

where v is the speed of the car in metres per second, and t is the time in seconds from the instant the car passes B.

(i) Use integration to find the time it takes this car to reach A. Give your answer correct to two decimal places. (15D)

v = dt

ds

= 10e1·05t

dt

ds = 10e1·05t

(dt

ds)dt = 10 e1·05t dt

ds = 10 e1·05t dt

s = 051

10 051

⋅

⋅ te + c

when t = 0, s = 0

0 = 051

10 )0(051

⋅

⋅e + c

c = –051

10 0

⋅e

= –051

10

⋅ or –9·52438...

s = 051

10 051

⋅

⋅ te –

051

10

⋅

∴ s = 051

10

⋅(e1·05t – 1)

When the car is at A, s = 100

051

10

⋅(e1·05t – 1) = 100

10e1·05t – 10 = 100(1·05) = 105 10e1·05t = 105 + 10 = 115 e1·05t = 11·5 ln e1·05t = ln11·5 1·05t = 2·442347... t = 2·326044... ≅ 2·33 seconds

Scale 15D (0, 6, 10, 13, 15) Low partial credit: (6 marks) – Any relevant first step, e.g. replaces

v with dt

ds and stops.

Middle partial credit: (10 marks)– Finds s =

051

10 051

⋅

⋅ te or

051

10 051

⋅

⋅ te + c

and stops or continues.

High partial credit: (13 marks) – Finds correct expression for s, i.e.

s = 051

10 051

⋅

⋅ te –

051

10

⋅ or

051

10

⋅(e1·05t – 1), but

fails to evaluate or incorrectly evaluates t.



7(b) (cont’d.)

(ii) Suggest one limitation of using an exponential function to model the speed of the car and give a possible reason for your answer. (5B)

Limitation – can only apply within limited regions / a limited period of time

Reason Any 1: – e1·05t increases indefinitely for all values of time (t) // – unrealistic for any car to be capable of increasing

its speed indefinitely // etc.


Scale 5B (0, 2, 5) Partial credit: (2 marks) – Limitation correct, but no reason or incorrect reason given.



On July 1st 2014, Meg borrowed €60,000 from her bank. She chose a payment-free option on the loan for 18 months. The bank charges interest that is equivalent to an annual percentage rate (APR) of 4·75%.

8(a) Show that Meg owed the bank €64,325·37 on January 1st 2016. (5B)

Annual rate

F = P(1 + i)t F = 60,000(1 + 0·0475)1⋅5 = 60,000(1·0475)1⋅5 = 60,000(1·072089...) = 64,325·370723... ≅ €64,325·37

or

Monthly rate

r = annual rate (APR) F = P(1 + i)t 1(1+ r) = 1(1 + i)12 1·0475 = (1 + i)12

1 + i = 1·047512

1

i = 12 04751⋅ – 1 = 1·003874... – 1 = 0·003874...

F = P(1 + i)t F = 60,000(1 + 12 04751⋅ – 1)18

= 60,000(12 04751⋅ )18 = 60,000(1·003874...)18 = 60,000(1·072089...) = 64,325·370723... ≅ €64,325·37

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct substitution in compound interest formula [Method ]. – Finds correct monthly interest rate [Method ] with some correct work towards finding amount outstanding.


Question 8 (cont’d.) 8(b) Meg agreed to repay the loan, plus interest, from January 2016 by a series of equal monthly

payments over ten years, payable at the end of each month. By choosing the payment-free option on the initial period of the loan, Meg also agreed to pay a monthly interest rate of 0·425%, fixed for the term of the loan.

(i) Find, correct to five significant figures, the annual percentage rate (APR) that is equivalent to a monthly interest rate of 0·425%. (5B*)

r = annual rate (APR) F = P(1 + i)t 1(1+ r) = 1(1 + i)12 1 + r = (1 + i)12 1 + r = (1 + 0·00425)12 r = 1·0042512 – 1 = 1·052209... – 1 = 0·052209... r (%) = 5·220917... ≅ 5·2209%

Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down correct relevant formula for compound interest. – Some correct substitution into correct formula. – Substitutes correctly into incorrect formula.

* Deduct 1 mark off correct answer only if incorrectly rounded or is not rounded - apply only once in each section (a), (b), (c), etc. of question.

* No deduction applied for the omission of or incorrect use of units (‘%’).

(ii) Calculate, correct to the nearest cent, the amount of each of Meg’s monthly repayments. (10C*)

A = 1)1(

)1(

−++

t

t

i

iiP

t = 10 × 12 = 120 i = 0·00425 P = 64,325·37

A = 1)0042501(

)0042501)(004250(37325,64120

120

−⋅+⋅+⋅⋅

= 1)004251(

)004251)(004250(37325,64120

120

−⋅⋅⋅⋅

= 685·418859... ≅ €685·42

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down correct relevant formula for amortisation.– Some correct substitution into formula.

High partial credit: (7 marks) – Correct substitution into amortisation formula or equivalent geometric series, but fails to evaluate or evaluates incorrectly.


* No deduction applied for the omission of or incorrect use of units involving currency.



8(c) Meg’s bank has agreed that, from January 1st 2022, to reduce the APR to 4·4% for the remainder of the loan if she makes all repayments up to that date on time and in full. The outstanding amount would be repaid in equal monthly repayments over the remaining term of the loan.

(i) By finding the present values of all her remaining repayments on January 1st 2022 and using the sum of a geometric series, calculate how much Meg still owes the bank on that date, correct to the nearest cent. (10D*)

01/2016 to 01/2022 = 6 years = 6 × 12 = 72 payments

Outstanding 01/2022 = 120 – 72 = 48 payments

F = P(1 + i)t

P = ti

F

)1( +

A = Month 1 + Month 2 + ... + Month 48

= 1)004251(

42685

⋅⋅

+ 2)004251(

42685

⋅⋅

+ ... + 48)004251(

42685

⋅⋅

Geometric series with n = 48, a = 004251

42685

⋅⋅

and r = 004251

1

⋅

Sn = r

ra n

−−

1

)1(

A = S48

=

004251

11

004251

11

004251

4268548

⋅−

⋅−

⋅⋅

= 1004251

...)81581401(42685

−⋅⋅−⋅

= €29,704·497801... ≅ €29,704·50

** Accept students’ answers from part (b)(ii) if not oversimplified.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. finds 48 outstanding payments. – Finds present value of first payment

correctly, i.e. 1)004251(

42685

⋅⋅

.

Middle partial credit: (6 marks) – Correct expression for geometric series,

i.e. A = 1)004251(

42685

⋅⋅

+ 2)004251(

42685

⋅⋅

+ ...

– States ‘Sum of a geometric series with

n = 48, a = 004251

42685

⋅⋅

and r = 004251

1

⋅’

or similar. – Some substitution into formula for S48 , but with errors.

High partial credit: (8 marks) – Correct substitution into formula for S48 , but fails to finish or finishes incorrectly.





8(c) (cont’d.)

(ii) Calculate, correct to the nearest cent, the new amount that Meg will have to repay monthly for the remaining term of the loan. (10C*)

r = annual percentage rate (APR) F = P(1 + i)t 1(1+ r) = 1(1 + i)12 1·044 = (1 + i)12

1 + i = 1·04412

1

i = 12 0441⋅ – 1 = 1·003594... – 1 = 0·003594...

A = 1)1(

)1(

−++

t

t

i

iiP

t = 48 i = 12 0441⋅ – 1 or 0·003594... P = 29,704·50

A = 1)104411(

)104411)(10441(50704,294812

481212

−−⋅+−⋅+−⋅⋅

= 1)0441(

)0441)(10441(50704,294812

481212

−⋅⋅−⋅⋅

= 1...)0035941(

...)0035941)(1...0035941(50704,2948

48

−⋅⋅−⋅⋅

= 1...)1879601(

...)1879601...)(0035940(50704,29

−⋅⋅⋅⋅

= 674·877170... ≅ €674·88

** Accept students’ answers from part (c)(i) if not oversimplified.

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. calculates new monthly interest rate correctly. – Some correct substitution into amortisation formula, but with errors.

High partial credit: (7 marks) – Correct substitution into amortisation formula or equivalent geometric series, but fails to evaluate or evaluates incorrectly.





8(c) (cont’d.)

(iii) Calculate the total amount of interest that Meg will pay over the entire term of the loan. (5C)

Total interest = Total amount repayable – amount borrowed = [(72 × 685·42) + (48 × 674·88)] – 60,000 = [49,350·24 + 32,394·24] – 60,000 = 81,744·48 – 60,000 = €21,744·48

** Accept students’ answers from parts (b)(ii) and (c)(ii) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. calculates (based on student’s answers) ‘Total amount repayable’ for first six years or for the remaining four years.

High partial credit: (4 marks) – Correct substitution into ‘Total interest, i.e. [(72 × 685·42) + (48 × 674·88)] – 60,000, but fails to evaluate or evaluates incorrectly.


(iv) What advice would you offer Meg? Justify your answer by calculation. (5B)

Advice – to make all payments on time and in full so that she can avail of the lower rate from 2022 // etc.


Justification – savings over the term of the loan = (48 × 685·42) – (48 × 674·88) = 32,900·16 – 32,394·24 = €505·92

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct advice, but no justification or incorrect justification given.



When a heavy chain hangs freely under the effect of gravity, and is supported only at each end, the shape that the chain forms is called a catenary, as shown.

Using a suitable choice of co-ordinates, the equation of a catenary can be written as:

y =

+ − a

xax

eea

2, where a is a constant.

To demonstrate a catenary, a heavy chain is let hang freely between two points, P and Q, on opposite walls on level ground. The co-ordinates of P and Q are (–2, k) and (4, l) respectively, where k and l are constants. Both x and y are in metres.

9(a) (i) Given that the point R(0, 2) lies on the path of the chain, show that the value of a is 2. (5B)

R(0, 2) ∈ curve

+ − a

xax

eea

2 = y

+ − aa ee

a 00

2 = 2

2

a(e0 + e0) = 2

2

a(1 + 1) = 2

2

a(2) = 2

a = 2

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct substitution into equation, but fails to finish or finishes incorrectly.

(ii) Find the height of the chain above ground level when the value of the x co-ordinate is 3, correct to one decimal place. (5B*)

y =

+ − a

xax

eea

2

=

+

−22

2

2 xx

ee

= 2x

e + 2x

e−

@ x = 3

y = 23

e + 23−

e = 4·481689... + 0·223130... = 4·704819.... ≅ 4·7 m


Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Correct substitution into equation, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only if incorrectly rounded or is not rounded or for the omission of or incorrect use of units (metres) - apply only once in each section (a), (b), (c), etc. of question.

P k( 2, )�

Q l(4, )

R(0, 2)



9(a) (cont’d.)

(iii) Find k, the value of the y co-ordinate of P, in terms of e. (5B)

P(–2, k) ∈ curve

y = 2x

e + 2x

e−

Substituting (–2, k) into equation

k = 22−

e + 22

e = e–1 + e1

= e

1 + e

** Accept students’ answers from part (a)(ii) if not oversimplified.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct substitution into equation, but fails to finish or finishes incorrectly.

(iv) Use your answer to part (a)(iii) above to show that l, the value of the y co-ordinate of Q, is k2 – 2. (10C)

Q(4, l) ∈ curve

y = 2x

e + 2x

e−

Substituting (4, l) into equation

l = 24

e + 24−

e = e2 + e–2

= e2 + 2

1

e

k = e + e

1

k2 = e2 + 2(e)(e

1) +

2

1

e

= e2 + 2

1

e + 2

k2 – 2 = e2 + 2

1

e

= l

** Accept students’ answers from part (a)(ii) and (a)(iii) if not oversimplified.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. replaces x with or finds k2. – Correct substitution of Q into equation, but fails to finish or finishes incorrectly.

– Finds k2 = e2 + 2

1

e + 2 or k2 – 2 = e2 +

2

1

e

without attempting to find l.

High partial credit: (7 marks) – Finds l = e2 + 2

1

e and attempts to find

k2 – 2.



9(b) (i) Show that R is the minimum point on the path of the chain. (10D)

y = 2x

e + 2x

e−

dx

dy =

2

1 2x

e – 2

1 2x

e−

Turning point occurs at dx

dy = 0

2

1 2x

e – 2

1 2x

e−

= 0

2

1 2x

e = 2

1 2x

e−

2x

e = 2x

e−

2

x = –

2

x

x = –x x = 0

@ x = 0

y = 20

e + 20−

e = e0 + e0 = 1 + 1 = 2

R(0, 2) is a turning point

2

2

dx

yd =

4

1 2x

e + 4

1 2x

e−

@ x = 0

2

2

dx

yd =

4

1 20

e + 4

1 20−

e

= 4

1 +

4

1

= 2

1 > 0

2

2

dx

yd > 0

minimum point at R(0, 2)

** Accept students’ answers from part(a)(ii) if not oversimplified.

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. differentiates

dx

dy correctly.

Middle partial credit: (6 marks) – Equates correctly dx

dy to 0, but fails

to find or find incorrect value of x.

High partial credit: (8 marks) – Shows that turning point is R(0, 2), but fails to verify or verifies incorrectly that it is a minimum point.



9(b) (cont’d.)

(ii) Show, by calculation, that the slope of the tangent to the curve is always increasing from R to Q. (5C)

dx

dy = slope of a tangent to the curve

2

2

dx

yd = rate of change of slope of tangent

2

2

dx

yd =

4

1 2x

e + 4

1 2x

e−

= 4

1

+

−22xx

ee

> 0

as 2

2

dx

yd > 0 for 0 < x < 4, the slope of tangent is always increasing from R to Q

** Accept students’ answers from previous parts if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down ‘slope of a tangent (to the curve) is equal

to dx

dy ’ or similar.

High partial credit: (4 marks) – Finds

2

2

dx

yd > 0, but fails to conclude

or concludes incorrectly.



9(c) Use integration to find the average height of the chain above ground level. Give your answer in metres, correct to one decimal place. (10D*)

Average value of f (x) in the interval [a, b]

= ab −

1

b

a

f (x) dx

Average height = )2(4

1

−− −

4

2

( 2x

e + 2x

e−

) dx

= 6

1(2 2

x

e – 2 2x

e−

)4

2−

= 6

1(2 2

4

e – 2 24−

e ) – 6

1(2 2

2−e – 2 2

2

e )

= 3

1[(e2 – e–2) – (e–1 – e1)]

= 3

1[e2 – e–2 – e–1 + e1]

= 3

1[9·604123...]

= 3·2013744... ≅ 3·2 m


Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down relevant formula for the average value of a function. – Integrates one term correctly.

Middle partial credit: (6 marks) – Integrates both terms correctly, but

excludes ab −

1 from calculation.

High partial credit: (8 marks) – Integrates correctly, i.e. average height

= 6

1(2 2

x

e – 2 2x

e−

) or 6

1(2 2

x

e – 2 2x

e−

)4

2−,

but fails to evaluate or evaluates incorrectly or evaluates using incorrect limits. – Finds area under the curve and evaluates correctly.



.


Mathematics


Structure of the Marking Scheme

Students’ responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students’ responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table:

Scale label A B C D

No. of categories 2 3 4 5

5 mark scale 0, 2, 5 0, 2, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale

A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary.

Marking scales – level descriptors

A-scales (two categories) incorrect response (no credit) correct response (full credit)

B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit)

C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit)

D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions.

Unless otherwise specified, accept correct answer with or without work shown.

Accept students’ work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved.

examsDEB

DEB 2014 LC-H

Scale label A B C D No of categories 2 3 4 5

5 mark scale 0, 2, 5 0, 2, , 5 0, 2, 3, 4

10 mark scale 0, 5, 10 0, 3, 7, 10 0, 2, 5, 8,

15 mark scale 0, 7, 15 0, 5, 10,15 0, 4, 7, 11

20 mark scale


Summary of Marks – 2016 LC Maths (Higher Level, Paper 2)

Q.1 (a) (i) 5B (0, 2, 5) Q.7 (a) (i) 5B (0, 2, 5) × 2 (ii) 5C (0, 2, 4, 5) (ii) 5B (0, 2, 5) (b) 10D (0, 4, 6, 8, 10) (iii) 5B (0, 2, 5) (c) 5C* (0, 2, 4, 5) (b) (i) 5C (0, 2, 4, 5) 25 (ii) 5C (0, 2, 4, 5) (iii) 10D (0, 4, 6, 8, 10) (iv) 5C (0, 2, 4, 5) Q.2 (a) (i) 10D (0, 4, 6, 8, 10) (c) (i) 10D (0, 4, 6, 8, 10) (b) (i) 10D (0, 4, 6, 8, 10) (ii) 5C (0, 2, 4, 5) (ii) 5B (0, 2, 5) 60 25 Q.3 (a) 5B (0, 2, 5) Q.8 (a) (i) 5B (0, 2, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 10D (0, 4, 6, 8, 10) (ii) 5B (0, 2, 5) (b) 10C* (0, 4, 7, 10) (c) 10D (0, 4, 6, 8, 10) (c) (i) 5C (0, 2, 4, 5) 25 (ii) 10D* (0, 4, 6, 8, 10) 40 Q.4 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (b) (i) 5C (0, 2, 4, 5) Q.9 (a) (i) 10C* (0, 4, 7, 10) (ii) 10C (0, 4, 7, 10) (ii) 5C* (0, 2, 4, 5) 25 (b) (i) 5C* (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) (iii) 5C* (0, 2, 4, 5) Q.5 (a) (i) 5B (0, 2, 5) (c) (i) 5B* (0, 2, 5) (ii) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) (iii) 10D* (0, 4, 6, 8, 10) (b) (i)

10D (0, 4, 6, 8, 10) 50

(ii) 25 Q.6 (a) (i)

10D (0, 4, 6, 8, 10)

(ii) (iii) 5B (0, 2, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) 25

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

General Instructions

There are two sections in this examination paper.

Section A Concepts and Skills 150 marks 6 questions Section B Contexts and Applications 150 marks 3 questions

Answer all questions.

Marks will be lost if all necessary work is not clearly shown.

Answers should include the appropriate units of measurement, where relevant.

Answers should be given in simplest form, where relevant.



Mathematics


Section A Concepts and Skills 150 marks

Answer all six questions from this section.


The points A, B, C, and D are the vertices of a square, as shown. The equation of AC is 3x – 7y + 2 = 0 and the co-ordinates of B are (2, –3).

1(a) (i) Find the slope of AC. (5B) Line AC:

3x – 7y + 2 = 0 7y = 3x + 2

y = 7

3x +

7

2

y = mx + c

mAC = 7

3

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down relevant line formula with some manipulation of the equation of AC. – Effort at finding slope of AB, e.g. finds

y = 7

3x +

7

2, but fails to identify mAC.

examsDEB



1(a) (cont’d.)

(ii) Hence, find the equation of BD. (5C)

BD ⊥ AC ... diagionals of a square intersect at 90°

slope of BD, mBD = –3

7

Equation of BD:

Point (2, –3), mBD = –3

7

y – y1 = m(x – x1)

y – (–3) = –3

7(x – 2)

3(y + 3) = –7(x – 2) 3y + 9 = –7x + 14 7x + 3y + 9 – 14 = 0 7x + 3y – 5 = 0


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down the relevant formula for the equation of a line and stops. – Finds correct slope of BD. – Some correct substitution into the formula for the equation of the line BD.

High partial credit: (4 marks) – Correct substitution into the formula for the equation of the line BD (including correct slope, mBD), but fails to finish or finishes incorrectly.



1(b) Find the co-ordinates of the point of intersection of the two diagonals, [ AC ] and [ BD ], and hence, find the co-ordinates of D. (10D)

AC ∩ BD

3x – 7y = –2 (×3) 7x + 3y = 5 (×7)

9x – 21y = –6 49x + 21y = 35 58x = 29

x = 58

29

x = 2

1 or 0·5

3x – 7y = –2 –7y = –2 – 3x

= –2 – 3(2

1)

= –2

7

y = 2

1 or 0·5

or 7x + 3y = 5 3y = 5 – 7x

= 5 – 7(2

1)

= 2

3

y = 2

1 or 0·5

AC ∩ BD = (2

1,

2

1) or (0·5, 0·5)

Co-ordinates of D

let T = AC ∩ BD

B → T = (2, −3) → (2

1,

2

1)

2 → 2

1 x: ↓1

2

1

–3 → 2

1 y: ↑3

2

1

T → D = (2

1,

2

1) → (

2

1 – 1

2

1,

2

1 + 3

2

1)

Co-ordinates of D = (–1, 4)


Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. multiplying equation by appropriate constant to facilitate cancellation of x or y term. – Finds 58x = 29 or –58y = –29, but fails to finish or finishes incorrectly. – Finds either variable (x or y) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

Middle partial credit: (6 marks) – Finds first variable (x or y) correctly, but fails to find second variable or finds it incorrectly. – Finds both variables (x and y) correctly with no work shown. – Finds both variables (x and y) by graphical means. – Finds both variables (x and y) by trial and error, but does not verify in both equations.

High partial credit: (8 marks) – Finds both variables (x and y) correctly, but fails to find or finds incorrect co-ordinates of D.



1(c) Hence, or otherwise, find the area of the square ABCD. (5C*)

Area of ABCD = | BC | × | CD | = | BC | × | BC | | BC |2

| BC |2 + | CD |2 = | BD |2 | BC |2 + | BC |2 = | BD |2 2| BC |2 = | BD |2

| BC |2 = 2

1| BD |2

| BC |2 = 2

1| BD |2

B(2, –3), D(–1, 4)

| BD | = 212

212 )()( yyxx −+−

= 22 ))3(4()21( −−+−−

= 22 )7()3( +−

= 499 +

= 58

Area of ABCD = 2

1| BD |2

= 2

1( 58)2

= 2

1(58)

= 29 units2

** Accept students’ answers from part (b) if not oversimplified.

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down

area of ABCD equal to | BC |2 or 2

1| BD |2

or similar (without any calculations). – Correct use of distance formula.

– Finds correct value for | BD | or 2

1| BD |

and stops.

High partial credit: (4 marks) – Finds correct value for | BD | or 2

1| BD |

and attempts to find an area (triangle or square), but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘units2’) - apply only once in each section (a), (b), (c), etc. of question.



2(a) The function f : x 3 – 1– →׀cos 6x is defined for x ∈ ℝ. Write down the period and range of f and hence, sketch the graph of f in the domain 0 ≤ x ≤ π. (10D)

Period = 6

2π

= 3

π or 60°

Range = [–1 – 3, –1 + 3] = [–4, 2]

Graph

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Correct period or range and stops. – Plots at least three points on graph correctly.

Middle partial credit: (6 marks) – Correct period and range and stops. – Correct graph of f , but period and/or range incorrect.

High partial credit: (8 marks) – Correct period and range, but graph of f incomplete.

�

2

�

f x( )

x

y

4

2

�2

�4

4

�

4

3�



2(b) A triangle has sides of length a, b and c. The angle opposite the side of length a is A.

(i) Prove that a2 = b2 + c2 – 2bc cos A. (10D)

s is a circle with centre A(0, 0) and radius b

Let B be any point on the x-axis with co-ordinates (c, 0)

| AB | = c

Let C be any point on the circle such that [ AC ] makes an angle A with the positive sense of the x-axis

| AC | = b C = (b cos A, b sin A) | BC | = a

= 22 )sin0()cos( AbAbc −+−

a2 = (c – b cos A)2 + (0 – b sin A)2 = c2 – 2bc cos A + b2 cos2A + b2 sin2A = c2 – 2bc cos A + b2(cos2A + sin2A) ... cos2A + sin2A = 1 = c2 – 2bc cos A + b2(1) = b2 + c2 – 2bc cos A

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. draws correct diagram and stops. – One step correct.

Middle partial credit: (6 marks) – Proof substantially correct, i.e. reaches a2 = (c – b cos A)2 + (0 – b sin A)2 or equivalent, but fails to finish or finishes incorrectly.

High partial credit: (8 marks) – Proof complete with one critical step omitted or incorrect.

(ii) In the case that angle A is obtuse, show that a2 > b2 + c2. (5B)

Consider angle A is obtuse 90° < A < 180° cos A < 0 a2 = b2 + c2 − 2bc cos A = b2 + c2 − 2(+)(+)(–) = b2 + c2 + 2x ... where x > 0 a2 > b2 + c2

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down ‘angle A is obtuse cos A < 0’ or similar.

A

a

bx

y

c B c( , 0)A(0, 0)

)sin,cos( AbAbC


Question 3 (25 marks) 3(a) Write down the centre of the circle c: x2 + y2 – 10x – 4y + k = 0 and find the radius-length

in terms of k. (5B)

c: x2 + y2 – 10x – 4y + k = 0

centre = (2

10

−−

, 2

4

−−

)

= (5, 2)

radius-length = cfg −+ 22

= k−−+− 22 )2()5(

= k−+ 425

= k−29

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct centre or radius-length.

3(b) The midpoint of a chord of c is (4, 1) and its length is 24 .

(i) Find the radius-length of c. (5C)

By Pythagoras’ theorem:

R2 = d2 +

2

2

24

R2 = d2 + 8

Points: centre (5, 2), midpoint of chord (4, 1)

d = 22 )21()54( −+−

= 22 )1()1( −+−

d2 = (–1)2 + (–1)2 = 1 + 1 = 2

R2 = 2 + 8 = 10 R = 10

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. shows chord being bisected on diagram. – Correct use of Pythagoras’ theorem, i.e. R2 = d2 + 8, and stops. – Finds correct value for d or d2 and stops (without considering Pythagoras’ theorem).

High partial credit: (4 marks) – Finds correct value for d or d2 and substitutes into Pythagoras’ theorem, but fails to finish or finishes incorrectly.

(ii) Hence, find the value of k. (5B)

k−29 = 10

29 – k = 10 k = 29 – 10 = 19

R

(5, 2)d 22



3(b) (ii) (cont’d.)

** Accept students’ answers from parts (a) and (b)(i) if not oversimplified.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Equates answer from part (a) [ k−29 ]

to radius-length from part (b)(i) [ 10 ], but fails to finish or finishes incorrectly.

3(c) Find the slopes of the tangents to c from the point (–2, 1). (10D)

Equations of tangent

Point (–2, 1), slope m y – y1 = m(x – x1) y – 1 = m(x – (–2)) y – 1 = mx + 2m mx – y + (2m + 1) = 0

⊥distance from centre of circle to tangent

Centre of circle = (5, 2) Radius-length = 10

⊥distance = 22

11

ba

cbyax

+

++

= 22 )1()(

)12()2(1)5(

−+

++−

m

mm

= 1

12252 +

++−

m

mm

= 1

172 +

−

m

m

1

172 +

−

m

m = 10

(7m – 1)2 = ( 10 )2( 12 +m )2

(7m – 1)2 = 10(m2 + 1) 49m2 – 14m + 1 = 10m2 + 10 39m2 – 14m – 9 = 0 (3m + 1)(13m – 9) = 0 3m + 1 = 0 or 13m – 9 = 0 3m = –1 13m = 9

m = –3

1 m =

13

9

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down perpendicular distance formula and stops.– Substitutes (–2, 1) correctly into equation of a line formula.

Middle partial credit: (6 marks) – Substitutes correctly into perpendicular distance formula, but fails to continue or continues incorrectly.

High partial credit: (8 marks) – Finds correct quadratic equation, but fails to finish or finishes incorrectly. – Substantially correct work with one error and both slopes found.



A soft toys manufacturer continuously carries out quality control testing to uncover defects in its products. After testing a large sample of its products, the manufacturer found the probability of a soft toy having poor stitching is 0·03 and that a soft toy with poor stitching has a probability of 0·7 of splitting open. A soft toy without poor stitching has a probability 0·02 of splitting open.

4(a) (i) Represent this information on a tree diagram, showing clearly the probability associated with each branch. (5C)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. draws branch with any new correct probability.

High partial credit: (4 marks) – Draws tree diagram with all relevant branches shown with no more than one probability omitted or incorrect.

(ii) Use your tree diagram, or otherwise, to find the probability that a soft toy, chosen at random, has exactly one of these defects. (5C)

P(exactly one defect) = (0·97)(0·02) + (0·03)(0·3) = 0·0194 + 0·009 = 0·0284

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. finds correct probability along one branch, i.e. (0·97)(0·02) or (0·03)(0·3). – Correct probabilities chosen, but operator(s) incorrect.

High partial credit: (4 marks) – Correct probabilities and operator(s) chosen, i.e. (0·97)(0·02) + (0·03)(0·3), but fails to evaluate or evaluates incorrectly.

Not splitopen

PoorStitching

Toy

GoodStitching

Not splitopen

Splitopen

Splitopen

0 97·

0 02·

0 98·

0·7

0·3

0 03·



4(b) The manufacturer also found that the colours used in the soft toys can fade. The probability of this defect occurring is 0·05 and it is independent of poor stitching or splitting open. A soft toy is chosen at random.

(i) Find the probability that the soft toy has none of these three defects. (5C)

P(can fade) = 0·05 P(does not fade) = 1 – 0·05 = 0·95

P(none of three defects) = (0·97)(0·98)(0·95) = 0·90307

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. finds correct P(does not fade) = 0·95. – Correct probabilities chosen, but operator(s) incorrect.

High partial credit: (4 marks) – Correct probabilities and operator(s) chosen, i.e. (0·97)(0·98)(0·95), but fails to evaluate or evaluates incorrectly.

(ii) Find the probability that the soft toy has exactly one of these three defects. (10C)

P(exactly one defect) = P(poor stiching only) + P(split only) + P(fade only) P(one defect) = (0·03)(0·3)(0·95) + (0·97)(0·02)(0·95) + (0·97)(0·98)(0·05) = 0·00855 + 0·01843 + 0·04753 = 0·07451

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. one case correct, (0·03)(0·3)(0·95) or (0·97)(0·02)(0·95) or (0·97)(0·98)(0·05).

High partial credit: (7 marks) – All three cases correct, but wrong operator chosen. – All cases correct and operator correct, but fails to evaluate or evaluates incorrectly.



Two events A and B are such that P(A) = 0·33, P(B) = 0·45 and P(A ∩ B) = 0·13.

5(a) (i) Find P(A ∪ B), the probability that either A or B occurs. (5B)

[ 0 ]·35

[0 2 ]· [0 13 ]· [0 32 ]·

A B

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0·33 + 0·45 – 0·13 = 0·65

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. draws Venn diagram with some substitution. – Writes down P(A) + P(B) – P(A ∩ B) and stops. – Correct figures (no expression given), but not executed accurately (provided answer lies in [0, 1]).

(ii) Find the conditional probability P(A′| B′). (5C)

P(A′| B′) = )(

)(

'BP

'B'AP ∩

= 35020

350

⋅+⋅⋅

= 550

350

⋅⋅

= 11

7

= 0·636363...

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. defines or explains conditional probability.

– States P(A′| B′) = )(

)(

'BP

'B'AP ∩ and stops.

High partial credit: (4 marks) – Correct expression, but not executed accurately (provided answer lies in [0, 1]).– Finds correct value for P(B′| A′).



5(a) (cont’d.)

(iii) State whether events A and B are independent and justify your answer. (5C)

Condition of independence P(A ∩ B) = P(A) × P(B)

P(A ∩ B) = 0·13 P(A) × P(B) = (0·33)(0·45) = 0·1485

P(A ∩ B) ≠ P(A) × P(B) A and B are not independent

or

P(A | B) = P(A)

P(A) = 0⋅33

P(A | B) = )(

)(

BP

BAP ∩

= 450

130

⋅⋅

= 0·288888...

P(A | B) ≠ P(A) A and B are not independent

or P(B | A) = P(B)

P(B) = 0⋅45

P(A | B) = )(

)(

AP

BAP ∩

= 330

130

⋅⋅

= 0·393939...

P(B | A) ≠ P(B) A and B are not independent

or

P(A′| B′) = P(A′)

P(A′) = 0·32 + 0·35 = 0·67 From part (ii)

P(A′| B′) = )(

)(

'BP

'B'AP ∩

= 0·636363...

P(A′| B′) ≠ P(A′) A and B are not independent

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. defines or explains independence. – Writes down any condition of independence i.e. P(A ∩ B) = P(A) × P(B) or P(A | B) = P(A) or similar. – Correct figures (no expression given), but not executed accurately (provided answer lies in [0, 1]).

High partial credit: (4 marks) – Correct expression, but not executed accurately (provided answer lies in [0, 1]).– Finds correct answer, but no justification or incorrect justification given.



5(b) Event C is such that P(C) = 0·2. Events A and C are mutually exclusive, while events B and C are independent.

(i) Represent the probability of each event on a Venn diagram. (10D)

[ 0 ]·24

[ 0·23 ][ 0·2 ] [ 0·09 ][ 0·13 ] [ 0·11 ]

A [ 0·33 ] B [ 0·45 ] C [ 0·20 ]

or

[ 0 ]·24

[ 0·2 ]

[0·11 ]

[0·09 ]

[0·13 ] [0·23 ]

A [ 0·33 ] B [ 0·45 ]

C [ 0·2 ]

As B and C are independent events P(B ∩ C) = P(B) × P(C) = (0·45)(0·2) = 0·09

(ii) Hence, find P((B ∪ C)′), the probability that neither B nor C occurs.

P((B ∪ C)′) = 0·2 + 0·24 = 0·44

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any first step correct, e.g. draws suitable Venn diagram showing P(A ∩ C) = ∅. – Writes down A ∩ C = ∅. – Calculates correct value for P(B ∩ C).

Middle partial credit: (6 marks) – Draws suitable Venn diagram with three or more entries correct.

High partial credit: (8 marks) – Venn diagram correct, but fails to find or finds incorrect value for P((B ∪ C)′).


Question 6 (25 marks) 6(a) ABC is a triangle such that | AB | = 8 cm, | AC | = 6 cm and | BC | = 12 cm. (10D)

(i) Given the line segment [ AB ] below, construct the triangle ABC.

(ii) On the same diagram, construct the circumcentre and the circumcircle of the triangle ABC, using only a compass and a straight edge. Show all construction lines clearly.

6cm

12 cm

BA

C

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. swings two arcs from A of lengths 6 cm and 12 cm to find point C with or without C indicated. – Constructs triangle correctly (labels not necessary) and stops.

Middle partial credit: (6 marks) – Constructs triangle correctly with further work of merit, e.g. one perpendicular bisector constructed properly. – Constructs triangle correctly and draws incircle or centroid of triangle correctly.

High partial credit: (8 marks) – Constructs triangle correctly with the circumcentre established correctly, but no circle drawn. – Constructs triangle correctly with both mediators and circle drawn correctly, but no evidence of construction.



6(a) (cont’d.)

(iii) Under what condition(s) does the circumcentre of a triangle lie inside the triangle? Justify your answers. (5B)

Condition Any 1: – all the angles in the triangle are acute // – angle A is acute // – the angle opposite the longest side is acute

Justification – if the angle | ∠BAC | is acute, then the angle on the same arc is less than 180° - the centre lies on the near side of BC from A //

– the angle at the centre of a circle standing on a given arc is twice the angle at the point of the circle standing on the same arc - therefore, if angle | ∠BAC | is acute, the angle at the centre will be less than 180°, i.e. the centre of the circle will be inside the triangle // etc.

[if angle | ∠BAC | is equal to 90°, then CB is a diameter]

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Condition correct, but justification incorrect or incomplete.

6(b) In the diagram, PQ is parallel to RS. PS and QR intersect at O, which lies on the circle.

| PO | = 3 cm, | OS | = 5 cm and | ∠SOR | = 90°. Let | RS | = x.



6(b) (cont’d.)

(i) Prove that | PO |.| RS | = | OS |.| PQ |. (5C)

ΔPOQ and ΔSOR are similar / equiangular as | ∠SOR | = | ∠POQ | ... vertically opposite angles

are equal | ∠RSO | = | ∠QPO | ... alternate angles are equal | ∠ORS | = | ∠OQP | ... alternate angles are equal

or ... the third angle in two triangles

are equal if the other two angles in the triangles are also equal

As the two triangles are similar / equiangular corresponding sides are in proportion

i.e. ||

||

RS

PQ =

||

||

OS

PO

| PQ |.| OS | = | PO |.| RS |

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. identifies one pair of corresponding angles (with brief explanation) or two pairs of corresponding angles (with no explanation).– Assumes that triangles are similar and finishes correctly.

High partial credit: (4 marks) – Identifies three pairs of corresponding angles or sides correctly (with brief explanations), thereby showing ΔPOQ and ΔSOR are similar (must be stated), but fails to show or shows incorrect sides in proportion.

(ii) Hence, find the area of the circle, in terms of x. (5C)

||

||

RS

PQ =

||

||

OS

PO

| PQ |.| OS | = | PO |.| RS | | PQ |(5) = (3)(x) | PQ | = diameter of circle = 2r (2r)(5) = 3x 10r = 3x

r = 10

3x

area of circle C = πr2

= π2

10

3

x

= 100

9 2xπ

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Substitutes correctly into ratio, but fails to find or finds incorrect value for r or d (diameter).

High partial credit: (4 marks) – Finds correct value of r or d, but fails to find or finds incorrect area of circle.


Section B Contexts and Applications 150 marks

Answer all three questions from this section.


Intelligence Quotient (IQ) is a score derived from standardised tests designed to assess intelligence.

7(a) The following table shows the countries with the highest recorded mean IQ scores.

(i) Using a calculator, or otherwise, calculate the mean and standard deviation for the above set of data, correct to one decimal place. (5B, 5B)

Mean, x = n

x

= 27

711,2

= 100·4074 ≅ 100·4 ... calculator work

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down the relevant formula for the mean. – Finds Σx = 2,711 and stops or continues.– Uses n = 25 (last number in table) and finds x correctly [ans. 108·44 ≅ 108·4].

standard deviation, = n

xx − 2)(

= 2·738 ≅ 2·7 ... calculator work

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Any relevant first step, e.g. writes down the relevant formula for standard deviation. – Correct substitution into formula with or without calculation.



7(a) (cont’d.)

(ii) Explain why standard deviation is the most appropriate measure of variability by which to analyse the above data. (5B)

Explanation Any 1: – standard deviation uses all of the data within the dataset

- whereas the range and interquartile range uses only two values from the data //

– range and interquartile range are less appropriate due to the narrow band of the data // etc.

** Accept other appropriate material.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Partial answer given, but incomplete or insufficient.

(iii) Suggest one limitation of using standard deviation in this case. Give a reason for your answer. (5B)

Limitation – data is ranked //

Reason Any 1: – only data from the countries with the highest mean

scores is considered // – as all the data provided is ranked, the standard deviation

is skewed, as only a narrow band of data (27 countries out of 200+) is considered // etc.

or

Limitation – the standard deviation may be skewed due to the accuracy of the data collected within each country // etc.

Reason Any 1: – data collected within each country may not be

representative of the entire population of that country (small sample size, small population size, only certain segments of the population may be tested) //

– the population tested within each country may not be large enough to draw reasonable comparisons with other countries (Hong Kong, Singapore vs. China) // etc.

** Accept other possible limitations and appropriate reasons.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Limitation correct, but reason incorrect or incomplete.



7(b) The mean IQ score of people who have been tested in Ireland is 96 and the standard deviation is 15. Assume that IQ scores are normally distributed and the number of people who have been tested is so large that it can be assumed to represent a population.

(i) Find the probability that an IQ score chosen at random from this population is greater than 100. (5C)

Let x be an IQ score chosen at random x = 100 = 96 = 15

z = σ

μ−x

P(x > 100) = P

z >

15

96100 −

= P(z > 0·266) = P(z > 0·27) = 1 – 0·6064 ... from z-tables = 0·3936

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down relevant formula. – Substitutes correctly into relevant formula and stops and continues.

High partial credit: (4 marks) – Finds correct P(z < 0·27) [ans. 0·6064], but fails to finish or finishes incorrectly.

Random samples of size 50 are repeatedly selected from this population and the mean of each sample is calculated and recorded. 500 such sample means are recorded.

(ii) Describe the expected distribution of all possible sample means from this population. You should refer to the shape of the distribution and to its mean and standard deviation. (5C)

Answer Two points: – the Law of Large Numbers says that these sample means

should be normally distributed – with a mean of 96 and standard deviation,

of 50

15 = 2·12 (Central Limit Theorem)

** Phrases underlined (or similar wording) must be included for any mark to be awarded.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One point complete.

High partial credit: (4 marks) – One point complete, but other point only partially complete.

x –



7(b) (cont’d.)

(iii) Find the number of sample means you would expect to be greater than 100. (10D)

Let sample mean be a random variable which is normally distributed: x = 100 = 96 = 15 n = 50

z =

n

xσ

μ−

Seeking number of sample means > 100

P(x > 100) = P

z >

50

1596100 −

= P

z >

...12132

96100

⋅−

= P(z > 1·8856...) = P(z > 1·89) = 1 – P(z < 1·89) = 1 – 0·9706 ... from z-tables = 0·0294

# sample means expected greater than 100 = 500 × 0·0294 = 14·7 ≅ 15

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down relevant formula for z. – Substitutes correctly into relevant formula for z and stops or continues. – Correct value for z.

Middle partial credit: (6 marks) – Finds correct P(z < 1·89) [ans. 0·9706], but fails to finish or finishes incorrectly.

High partial credit: (8 marks) – Finds correct probability, i.e. P(x > 100) = 0·0294, but fails to find or finds incorrect # sample means expected greater than 100.

– .



7(b) (cont’d.)

The sample mean is used to estimate the population mean.

(iv) Find a 95% confidence interval for the population mean and interpret this interval in the context of the question. (5C)

95% confidence interval for the population mean is x ± 1·96n

σ

z-value = 1·96 95% confidence interval

. =

96 – 1·96

50

15, 96 + 1·96

50

15

= [ 96 – 1·96(2·1213...), 96 + 1·96(2·1213...) ] = [ 96 – 4·1578, 96 + 4·1578 ] = [ 91·8422, 100·1578 ]

Interpretation – 95% confident that the interval [ 91·8422, 100·1578 ] will contain the true population mean

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down relevant formula for confidence interval or 95% confidence interval implies z-value of ±1·96. – Substitutes correctly into relevant formula, but fails to find or finds incorrect interval.

High partial credit: (4 marks) – Finds correct confidence interval, but fails to contextualise answer correctly. – Incorrect confidence interval (no more than one error), but contextualises answer correctly.

.



7(c) The principal in a large school in Ireland claims that the students in her school are above average intelligence. A sample of 50 students from the school are chosen at random and their IQ scores are determined. The mean IQ score of the sample is calculated to be 102.

(i) Conduct a hypothesis test at the 5% level of significance to decide whether there is sufficient evidence to conclude that the principal’s claim is legitimate. Write the null hypothesis and the alternative hypothesis and state your conclusion clearly. (10D)

H0 : = 96 – average intelligence of students in the school lies within the population mean which is normally distributed

H1 : ≠ 96 – average intelligence of students in the school lies outside the population mean which is normally distributed

Method 1

x = 102 = 96 = 15 n = 50

z =

n

xσ

μ−

=

50

1596102 −

= 2·8284... ≅ 2·83 2·83 > 1·96

Conclusion – reject the null hypothesis and accept H1, i.e. we can conclude that there is sufficient evidence to suggest that the principal’s claim is valid in relation to her pupils

or

Method 2

102 is outside the 95% confidence interval for the mean IQ of the population i.e. 91·8422 < < 100·1578

Conclusion – reject the null hypothesis and accept H1, i.e. we can conclude that there is sufficient evidence to suggest that the principal’s claim is valid in relation to her pupils

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down null hypothesis or alternative hypothesis.– Some work towards finding z-value. – Mention of 5% level of significance and therefore comparing to z-value of ±1·96.

Middle partial credit: (6 marks) – Finds correct z-value. – Either null or alternative hypothesis stated and relevant work to find z-value. – Confidence interval from part (b)(iv) and either null or alternative hypothesis stated.

High partial credit: (8 marks) – Finds correct z-value calculated and compared to ±1·96 but: - Not stating null hypothesis and/or alternative hypothesis correctly. - Not accepting or rejecting hypothesis. - Incorrect conclusion for hypothesis. – Incorrect use of 102 and confidence interval, but otherwise correct.

– .



7(c) (cont’d.)

(ii) Find the p-value of the test you performed in part (ii) above and explain what this value represents in the context of the question. (5C)

p-value

x = 102 = 96 = 15 n = 50

z =

n

xσ

μ−

=

50

1596102 −

= 2·8284... ≅ 2·83

P(z > 2·83) = 1 – P(z < 2·83) = 1 – 0·9977 ... from z-tables = 0·0023

p-value = 2 × 0·0023 = 0·0046 < 0·05

Explanation Any 1: – the p-value is the probability that the test statistic or

a more extreme value could occur if the null hypothesis is true //

– if the mean IQ of the Irish population is 96, then the probability that the mean IQ of the students in the school would be 102 by chance is 0·46% - it is because this is less than a 5% chance that we reject the null hypothesis

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Finds 1 – P(z < 2·83) or 1 – 0·9977 correctly in an effort to find P(z > 2·83) [ans. 0·0046].

High partial credit: (4 marks) – Finds correct p-value, but fails to contextualise answer correctly.

– .



8(a) The surface of the oceans are curved. Although we tend to think about water forming large flat sheets, the surface of a large body of water is not actually flat at all – it follows the curvature of the Earth.

The diagram below represents part of a circular cross-section of the Earth. From point A on the top of a mountain, 4⋅83 km above sea level, an observer measures the angle of depression to the ocean horizon, H, as 2⋅23°.

(i) Find | ∠AHO | and give a reason for your answer. (5B)

| ∠AHO | = 90°

Reason Any 1: – AH is tangential to the circular cross-section which

is perpendicular to [ OH ] // – a tangent to a circle is perpendicular to the radius

of the circle that goes to the point of contact // – if H is a point on a circle c, and a line l through H is

perpendicular to the radius of the circle c, then the line l is a tangent to the circle c // etc.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Measure of angle correct, but no reason or incorrect reason given.

r



8(a) (cont’d.)

(ii) Hence, use a suitable trigonometric ratio to show that, correct to the nearest kilometre, the radius of the Earth is 6,373 km. (10D)

| ∠OAH | = 90° – 2·23° = 87·77° sin 87·77° = 0·999242...

sin | ∠ | = |Hyp|

|Opp|

sin | ∠OAH | = ||

||

OA

OH

= sin 87·77° = 0·999242...

= 834⋅+r

r

834⋅+r

r = 0·999242...

r = (0·999242...)(r + 4·83) = 0·999242...r + 4·826342... r – 0·999242...r = 4·826342... r – 0·999242...r = 4·826342... 0·000757...r = 4·826342...

r = ...0007570

...8263424

⋅⋅

= 6,372·919970... ≅ 6,373 km

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down | OA | = r + 4·83, | ∠OAH | = 87·77° or

sin | ∠OAH | =

||

||

OA

OH and stops.

– Finds sin | ∠OAH | = 834⋅+r

r and stops.

Middle partial credit: (6 marks) – Finds sin | ∠OAH | = 834⋅+r

r or

sin 87·77° = 834⋅+r

r and stops.

High partial credit: (8 marks) – Finds 834⋅+r

r = 0·999242..., but fails

to find or finds incorrect radius of the Earth.



8(b) In astronomy, the same concept can be used to determine the distance between the Earth and objects in space.

Let O be the centre of the Earth, E be a point on the equator and S represent the nearest point of the object in space, as shown (diagram not to scale).

Given the Earth is positioned in such a way that | ∠OES | = 90°, then = | ∠OSE | is called the equatorial parallax of the object in space. In the case of the sun, has been observed as 0·00244°.

Using the result from part (a) (ii) above, or otherwise, find, correct to the nearest kilometre, the minimum distance between the surface of the Earth and the surface of the sun. (10C*)

sin = ||

||

OS

OE

sin 0·00244° = ||

373,6

OS

| OS | = °⋅002440sin

373,6

= ...00004258.0

373,6

= 149,650,001.207886... ≅ 149,650,001 km

Minimum distance between the surface of the Earth and the surface of the sun, d = 149,650,001 – 6,373 = 149,643,628 km

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down

sin = ||OS

r or ||

373,6

OS and stops.

– Finds sin 0·00244° = ||

373,6

OS and stops.

High partial credit: (7 marks) – Finds correct value for | OS |, but fails to finish or finishes incorrectly.


r

r

d



8(c) An observer on the equator measures the angle from one visible edge of the sun to the other opposite visible edge of the sun, as shown (diagram not to scale).

| ∠PEQ | = 32′ 4″.

(i) Prove that | ∠PEC | = | ∠QEC |. (5C)

For ΔECP and ΔEQC | ∠EPC | = | ∠EQC | ... both triangles are right-angled (R) | EC | = | EC | ... common hypotenuse (H) | PC | = | QC | ... both radii of the sun (S) ΔPEC ≡ ΔQEC ... congruent triangles (R.H.S.)

As ΔPEC and ΔQEC are congruent | ∠PEC | = | ∠QEC |

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. identifies one pair of corresponding angles or sides (with brief explanation) or two pairs of corresponding angles / sides (with no explanation).

High partial credit: (4 marks) – Identifies three pairs of equal angles and sides correctly (with brief explanations), thereby showing ΔPEC and ΔQEC are congruent (must be stated), but fails to conclude that | ∠PEC | = | ∠QEC |. – Identifies three pairs of equal angles and sides correctly and finishes correctly, but no explanations given.



8(c) (cont’d.)

(ii) Hence, using your answer to part (b) above, find the radius of the sun, correct to the nearest kilometre. (10D*)

| ∠PEQ | = 0° 32′ 04″

| ∠PEC | = 2

04320 '''°

= 0° 16′ 02″ = 0·267222...°

sin | ∠PEC | = ||

||

EC

PC

sin 0·267222...° = 628,643,149+R

R

4·663890... × 10–3 = 628,643,149+R

R

R = 4·663890... × 10–3(R + 149,643,628) R(0·995336...) = 697,921.525548...

R(0·995336...) = ...9953360

...525548921,697

⋅⋅

= 701,191.807501... = 701,192 km

** Accept students’ answers from part (b) if not oversimplified.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down | OA | = r + 4·83, | ∠OAH | = 87·77° or

sin | ∠OAH | =

||

||

OA

OH and stops.

– Finds sin | ∠OAH | = 834⋅+r

r and stops.

Middle partial credit: (6 marks) – Finds sin | ∠PEC | or sin 0·267222...°

= 628,643,149+R

R and stops.

High partial credit: (8 marks) – Finds 628,643,149+R

R = 4·663890... × 10–

3, but fails to find or finds incorrect radius of sun.




9(a) The area of an equilateral triangle is 9 3 cm2.

(i) Find the length of each side of the triangle. (10C*)

Area ΔABC = 2

1ab sin C

For an equilateral triangle of side x

2

1(x)(x)sin 60° = 9 3

2

2x ×

2

3 = 9 3

4

2x = 9

x2 = 9 × 4 = 36 x = 6 cm

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any relevant first step, e.g. writes down relevant area formula or angle = 60°. – Some correct substitution into relevant area formula (not stated).

High partial credit: (7 marks) – Finds

2

2x ×

2

3 = 9 3 , but fails to find

or finds incorrect value of x.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘cm’) - apply only once in each section (a), (b), (c), etc. of question.

(ii) Find the area of the largest circle that fits in the triangle, in terms of π. (5C*)

tan 30° = 3

r

r = 3 tan 30°

= 3 × 3

1

= 3 cm

Area of circle = πr2 = π( 3 )2

= 3π cm2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. draws sketch with incentre or angle bisector shown.

– Writes down tan 30° = 3

r or equivalent

and stops.

High partial credit: (4 marks) – Finds correct radius, but fails to find or finds incorrect area.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘cm2’) - apply only once in each section (a), (b), (c), etc. of question.

30° r

3 cm



9(b) A circle, centre O, has a radius of 90 cm, as shown. [ OM ] intersects the chord [ AB ] at M.

| OM | = 30 cm.

(i) Find | AB |, in surd form. (5C*)

Using Pythagoras’ theorem | MB |2 + | OM |2 + = | OB |2 | MB |2 + 302 = 902 | MB |2 + 900 = 8,100 | MB |2 = 8,100 – 900 = 7,200 | MB | = 200,7

= 60 2 cm

| AB | = 2 | MB | = 2(60 2 ) = 120 2 cm

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down Pythagoras’ theorem. – Some correct use of Pythagoras’ theorem.

High partial credit: (4 marks) – Finds correct value of | MB |, but fails to find or finds incorrect value of | AB |.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘cm2’) - apply only once in each section (a), (b), (c), etc. of question.

(ii) Find | ∠AOB |, correct to two decimal places. (5C*)

cos | ∠BOM | = ||

||

OB

OM

= 90

30

= 3

1

| ∠BOM | = cos–1

3

1

= 70·528779...°

| ∠AOB | = 2| ∠BOM | = 2(70⋅528779...) = 141·057558... ≅ 141·06°

or

a2 = b2 + c2 – 2bc cos A | AB |2 = | OA |2 + | OB |2– 2| OA |.| OB | cos | ∠AOB | (120 2 )2 = (90)2 + (90)2 – 2(90)(90) cos | ∠AOB | 28,800 = 8,100 + 8,100 – 16,200 cos | ∠AOB | 16,200 cos | ∠AOB | = 8,100 + 8,100 – 28,800 16,200 cos | ∠AOB | = –12,600

cos | ∠AOB | = –200,16

600,12

= –0·777777... | ∠AOB | = cos–1(–0·777777...) = 141·057558... ≅ 141·06°



9(b) (ii) (cont’d.)


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. ΔOAM or ΔOMB with sides labelled (i.e. hyp., adj., etc.).

– Finds | ∠BOM | = cos–1

3

1 and stops.

– Some correct substitution into formula for cosine rule, but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds | ∠BOM | = 70·528779...° or 70·53°, but fails to find or finds incorrect | ∠AOB |.– Correct substitution into formula for cosine rule, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only if incorrectly rounded or is not rounded or for the omission of or incorrect use of units (‘°’) - apply only once in each section (a), (b), (c), etc. of question.

(iii) Hence, find the area of the shaded region, correct to the nearest cm2. (5C*)

Shaded area = area of sector OAB – area of ΔOAB

= °

θ360

× πr2 – 2

1| Base | × | ⊥Height |

= [360

06141⋅ × π × 902] – [

2

1 × 120 2 × 30]

= [360

06141⋅ × π × 8,100] – 1,800 2

= 9,970·943843... – 2,545·584412... = 7,425·359431... ≅ 7,425 cm2

** Accept students’ answers from parts (b)(i) and (b)(ii) if not oversimplified.

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down formulae for both areas. – Finds correct area of triangle only.

High partial credit: (4 marks) – Finds correct area of sector (with or without area of triangle), but fails to find or finds incorrect area of shaded region.




9(c) The diagram shows a horizontal cylindrical oil tank of length 2·2 m and radius 0·9 m.

(i) Find, in litres, the capacity (volume) of the oil tank. Give your answer correct to two significant figures. (5B*)

Capacity (m3) = Maximum volume of the tank = πr2h = π(0·9)2(2·2) = 5·598318...

Capacity (litres) = 5·598318... × 1,000 = 5,598·318108... ≅ 5,600 litres

Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Correct formula with some substitution.


‘Dipping’ is a method by which the level of the oil remaining in the tank can be checked. A graduated stick is used to measure the level of oil from the bottom of the tank. It was found that the depth of oil in the tank was 60 cm.

(ii) Find, in litres, the volume of oil in the tank. (5C)

Shaded area = 7,425 cm2 ... answer to part (b)(iii) Volume (cm3) = 7,425 × 220 = 1,633,500 cm3

1000 cm3 = 1 litre

Volume (litres) = 000,1

500,663,1

= 1633·5 litres

** Accept students’ answers from part (b)(iii) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any relevant first step, e.g. writes down volume = area shaded × 220.

High partial credit: (4 marks) – Final answer correct, but given in cm3.



9(c) (cont’d.)

After a fill of oil is delivered, the tank was dipped again and the depth of oil in the tank was found to be 160 cm.

(iii) Find, in litres, the volume of oil delivered, correct to the nearest whole number. (10D*)

After delivery, the depth of oil is 160 cm

cos = 90

90160 −

= 90

70

= 9

7

= cos–1

9

7

= 38·942441... 2 = 77·884882...°

Area not filled with oil = Area of sector OPQ – area of ΔOPQ

= °

θ360

× πr2 – 2

1| OP |.| OQ | sin | ∠POQ |

= 360

...88488277 ⋅π(902) –

2

1(90)(90) sin 77·884882...°

= 5,505·357933... – 3,959·797974... = 1,545·559959... cm2

Volume of the empty portion of the tank Volume (cm3) = 1,545·559959... × 220 = 340,023·19098... cm3

Volume (litres) = 000,1

...19098023,340 ⋅

= 340·023190... ≅ 340 litres

Volume of oil delivered = 5,600 – (1,633·5 + 340) = 5,600 – 1,973·5 = 3,626⋅5 ≅ 3,627 litres

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any relevant first step e.g. finds

cos = 90

90160 − or

90

70 or

9

7.

– Finds correct value of or 2.

Middle partial credit: (6 marks) – Finds correct value for the area not filled with oil.

High partial credit: (8 marks) – Finds correct volume of the empty portion of the tank, but fails to finish or finishes incorrectly.


New level

160

P Q

O

90


Notes:


Documents

Mathematics - Microsoft · Mathematics Higher Level – Paper 1 Marking Scheme (300 marks) Section A Concepts and Skills 150 marks Answer all six questions from this section. Question