Mathematics Ext 1 Sample Examination Materials 2020

• Reading time – 10 minutes• Working time – 2 hours• Write using black pen• Calculators approved by NESA may be used• A reference sheet is provided at the back of this paper• In Questions 11–14, show relevant mathematical reasoning

and/ or calculations

Section I – 10 marks (pages 2–6)• Attempt Questions 1–10• Allow about 15 minutes for this section

Section II – 60 marks (pages 7–12)• Attempt Questions 11–14• Allow about 1 hour and 45 minutes for this section

General Instructions

Total marks: 70

Mathematics Extension 1

NSW Education Standards Authority

Sample HIGHER SCHOOL CERTIFICATE EXAMINATION

– 2 –

Section I

10 marksAttempt Questions 1–10Allow about 15 minutes for this section

Use the multiple-choice answer sheet for Questions 1–10.

1 What is the angle between the vectors 7

1⎛⎝⎜

⎞⎠⎟

and –1

1⎛⎝⎜

⎞⎠⎟

?

A. cos–1(0.6)

B. cos–1(0.06)

C. cos–1(– 0.06)

D. cos–1(– 0.6)

2 The diagram shows a grid of equally spaced lines. The vector OH = h and the vector

OA = a . The point P is halfway between B and C.

A

D

O G H

E F

B C

P

Which expression represents the vector OP?

A. − 12

a − 14

h

B.14

a –12

h

C. a + 14

h

D. a + 34

h

– 3 –

3 Given that cos q – 2 sin q + 2 = 0, which of the following shows the two possible values

for tanθ2

?

A. – 3 or –1

B. – 3 or 1

C. –1 or 3

D. 1 or 3

4 What is the derivative of tan−1 x2

?

A. 1

2( 4 + x 2 )

B. 1

4 + x 2

C. 2

4 + x 2

D. 4

4 + x 2

– 4 –

5 The slope field for a first order differential equation is shown.

3

2

1

–1

–4 4 x

y

–3 3–2 2–1 1O

–2

–3

Which of the following could be the differential equation represented?

A. dydx

= x3y

B. dydx

= – x3y

C. dydx

= xy3

D. dydx

= – xy3

– 5 –

6 Let P(x) = qx3 + rx2 + rx + q where q and r are constants, q ≠ 0. One of the zeros of P(x) is −1.

Given that a is a zero of P(x), a ≠ −1, which of the following is also a zero?

A. − 1a

B. − qa

C. 1a

D. qa

7 Each of the students in an athletics team is randomly allocated their own locker from a row of 100 lockers.

What is the smallest number of students in the team that guarantees that two students are allocated consecutive lockers?

A. 26

B. 34

C. 50

D. 51

8 A team of 11 students is to be chosen from a group of 18 students. Among the 18 students are 3 students who are left-handed.

What is the number of possible teams containing at least 1 student who is left-handed?

A. 19 448

B. 30 459

C. 31 824

D. 58 344

– 6 –

9 A stone drops into a pond, creating a circular ripple. The radius of the ripple increases from 0 cm at a constant rate of 5 cm s−1.

At what rate is the area enclosed within the ripple increasing when the radius is 15 cm?

A. 25p cm2 s−1

B. 30p cm2 s−1

C. 150p cm2 s−1

D. 225p cm2 s−1

10 The graph of the function y = sin–1(x – 4) is transformed by being dilated horizontally with a scale factor of 2 and then translated to the right by 1.

What is the equation of the transformed graph?

A. y = sin−1 x − 92

⎛⎝

⎞⎠

B. y = sin−1 x −102

⎛⎝

⎞⎠

C. y = sin−1 2x − 6( )

D. y = sin−1 2x − 5( )

– 7 –

Section II

60 marksAttempt Questions 11–14Allow about 1 hour and 45 minutes for this section

Answer each question in the appropriate writing booklet. Extra writing booklets are available.

In Questions 11–14, your responses should include relevant mathematical reasoning and/or calculations.

Question 11 (15 marks) Use the Question 11 Writing Booklet.

(a) A particle is fired from the origin O with initial velocity 18 m s–1 at an angle 60° to the horizontal.

The equations of motion are d2x

dt2= 0 and

d2y

dt2= –10 .

(i) Show that x = 9t. (ii) Show that y = 9 3t − 5t2 . (iii) Hence find the Cartesian equation for the trajectory of the particle.

(b) A function â ( x ) is given by x2 + 4x + 7 .

(i) Explain why the domain of the function â ( x ) must be restricted if â ( x ) is to have an inverse function.

(ii) Give the equation for â –1( x ) if the domain of â ( x ) is restricted to x ≥ –2.

(iii) State the domain and range of â –1( x ), given the restriction in part (ii).

(iv) Sketch the curve y = â –1( x ).

Question 11 continues on page 8

1

2

1

1

2

2

2

– 8 –

Question 11 (continued)

(c) The trajectories of particles in a fluid are described by the differential equation

dydx

= 14

(y – 2)(y – x).

The slope field for the differential equation is sketched below.

y

x

1

O

2

3

–5 –4 –3 –2 –1

–1

1 2 3 4 5

(i) Identify any solutions of the form y = k, where k is a constant.

(ii) Draw a sketch of the trajectory of a particle in the fluid which passes through the point (–3,1) and describe the trajectory as x ± ∞.

End of Question 11

1

3

– 9 –


(a) A recent census showed that 20% of the adults in a city eat out regularly.

(i) A survey of 100 adults in this city is to be conducted to find the proportion who eat out regularly. Show that the mean and standard deviation for the distribution of sample proportions of such surveys are 0.2 and 0.04 respectively.

(ii) Use the extract shown from a table giving values of P (Z < z), where z has a standard normal distribution, to estimate the probability that a survey of 100 adults will find that at most 15 of those surveyed eat out regularly.

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

(b) A force described by the vector F =2

1⎛⎝⎜

⎞⎠⎟

newtons is applied to an object lying

on a line which is parallel to the vector 3

4⎛⎝⎜

⎞⎠⎟

.

(i) Find the component of the force F in the direction of the line .

(ii) What is the component of the force F in the direction perpendicular to the line?

(c) The points A and B are fixed points in a plane and have position vectors a and b respectively.

The point P with position vector also lies in the plane and is chosen so that APB = 90°.

(i) Explain why (a – ) . (b – ) = 0.

(ii) Let m = 12

(a + b) denote the position vector of M, the midpoint of A and B.

Using the properties of vectors, show that p − m 2 is independent of and find its value.

(iii) What does the result in part (ii) prove about the point P?

(d) Use mathematical induction to prove that 23n – 3n is divisible by 5 for n ≥ 1.

2

2

2

1

1

3

1

3

– 10 –


(a) Using the substitution x = sin2 q, or otherwise, evaluate x

1− xdx

0

12⌠

⌡⎮ .

(b) A device playing a signal given by x = 2sin t + cos t produces distortion whenever x ≥ 1.5.

For what fraction of the time will the device produce distortion if the signal is played continuously?

(c) (i) Prove the trigonometric identity cos 3 q = 4 cos3q – 3 cos q.

(ii) Hence find expressions for the exact values of the solutions to the equation 8x3 – 6x = 1.

3

4

3

4

– 11 –


(a) (i) Sketch the graph of y = x cos x for –p ≤ x ≤ p and hence explain why

x cos x dx = 0.−π

2

π2⌠

⌡⎮

(ii) Consider the volume of the solid of revolution produced by rotating about the x-axis the shaded region between the graph of y = x – cos x, the

x-axis and the lines x = – p

2 and x =

p

2 .

p

2p

2–

–1

1

2 y = x – cosx

y

xO

Using the results of part (i), or otherwise, find the volume of the solid.

(b) The population of foxes on an island is modelled by the logistic equation dydt

= y(1 – y), where y is the fraction of the island’s carrying capacity reached

after t years.

At time t = 0, the population of foxes is estimated to be one-quarter of the island’s carrying capacity.

(i) Use the substitution y = 1

1 – w to transform the logistic equation to

dwdt

= –w.

(ii) Using the solution of dwdt

= –w, find the solution of the logistic equation

for y satisfying the initial conditions.

(iii) How long will it take for the fox population to reach three-quarters of the island’s carrying capacity?

Question 14 continues on page 12

3

3

2

2

2

– 12 –

Question 14 (continued)

(c) The diagram below is a sketch of the graph of the function y = â ( x ).

y ƒ x

(0,7.5)

(0.5,2.5)

8

6

4

2

–1 1 2 3 4 5

–2 (2,–1.75)

y

xO

(i) Sketch the graph of y = 1ƒ x( )

.

Your sketch should show any asymptotes and intercepts, together with the location of the points corresponding to the labelled points on the original sketch.

(ii) How many solutions does the equation 1

ƒ x( )= x have?

End of paper

3

1

© 2020 NSW Education Standards Authority

– 1 –XXXX


2020 HIGHER SCHOOL CERTIFICATE EXAMINATION

Mathematics AdvancedMathematics Extension 1Mathematics Extension 2

– 2 –

– 3 –

– 4 – © 2020 NSW Education Standards Authority


Page 1 of 17

Mathematics Extension 1 Sample HSC Marking Guidelines

Section I

Multiple-choice Answer Key

Question Answer

1 D 2 D 3 D 4 C 5 D 6 C 7 D 8 B 9 C

10 A

NESA Mathematics Extension 1 Sample HSC Marking Guidelines

Page 2 of 17

Section II

Question 11 (a) (i) Criteria Marks

• Provides correct solution 1

Sample answer:

xi = 18cosπ3

= 9 ms−1

yi = 18sinπ3

= 9 3 ms−1

x = d2x

dt2 .dt⌠

⌡⎮

= 0.dt⌠

⌡⎮

∴ x = C xi = 9 ms–1

∴ x = 9 ms–1

x = x dt⌠

⌡⎮

= 9.dt⌠

⌡⎮

x = 9t + ′C when t = 0, x = 0 ∴ ′C = 0

∴ x = 9t


Page 3 of 17

Question 11 (a) (ii) Criteria Marks

• Provides correct solution 2 • Determines y or equivalent merit 1

Sample answer:

y = d2y

dt2 .dt⌠

⌡⎮

y = −10.dt⌠

⌡⎮

y = −10t + C when t = 0, y = 9 3 ms–1

∴ y = 9 3 – 10t

y = y dt⌠

⌡⎮

y = 9 3 −10t( )dt⌠

⌡⎮

y = 9 3 t − 5t2 + ′C when t = 0, y = 0 ∴ ′C = 0

∴ y = 9 3 t − 5t2

Question 11 (a) (iii) Criteria Marks

• Provides correct solution 1

Sample answer:

x = 9t → t = x

9

y = 9 3 t − 5t2 substitute t = x

9

= 9 3x

9⎛⎝

⎞⎠ − 5

x

9⎛⎝

⎞⎠

2

∴ y = 3 x − 5x2

81


Page 4 of 17

Question 11 (b) (i) Criteria Marks

• Refers to horizontal line test, or equivalent merit 1

Sample answer:

ƒ x( ) = x2 + 4x + 7 is a parabola. Therefore, for each value of ƒ x( ) in the range (except at the turning point), there are two x-values. (A horizontal line will cut the graph twice.) ∴ If x and y are swapped, each x in the domain will have two y-values, and so the inverse will

not be a function. Question 11 (b) (ii) Criteria Marks • Provides correct solution 2 • Swaps x and y or equivalent merit 1

Sample answer:

ƒ x( ) = x2 + 4x + 7 x ≥ −2

= x + 2( )2 + 3

ƒ −1 x( ): x = y + 2( )2 + 3

x − 3 = y + 2( )2

y + 2 = x − 3 − x − 3 is discarded as y must be ≥ −2( )y = x − 3( ) − 2

∴ ƒ −1 x( ) = x − 3( ) − 2


Page 5 of 17

Question 11 (b) (iii) Criteria Marks

• States correct domain and range 2 • States correct domain or range 1

Sample answer: Domain: x ≥ 3 as x − 3 ≥ 0

Range: y ≥ −2 as x − 3 ≥ 0 Question 11 (b) (iv) Criteria Marks

• Provides correct sketch 2 • Provides graph with correct shape, or equivalent merit 1

Sample answer:

Question 11 (c) (i) Criteria Marks

• Correct answer 1

Sample answer:

y = 2


Page 6 of 17

Question 11 (c) (ii) Criteria Marks

• Provides correct answer 3 • Provides correct sketch 2

• States that y → 2 as x → ±∞ , or equivalent merit 1

Sample answer:

The y-coordinate of the particle approaches 2 from below as x → ± ∞. Question 12 (a) (i) Criteria Marks

• Provides correct mean and standard deviation 2 • Provides correct mean or standard deviation 1

Sample answer:

x = np = 20, σ 2 = np 1− p( ) = 16 so σ = 4

∴ xproportion = 20100

= .2 σ proportion = 4100

= 0.04


• Provides correct answer 2 • Calculates z = 0.15 − 0.20( ) / 0.04 OR uses the table appropriately with

an incorrect value for z 1

Sample answer:

P Z < 0.15 − 0.20( ) / 0.04( ) = P Z < –1.25( ) = 1− P z < 1.25( ) so estimate is

1 – 0.8944 = 0.1056.


Page 7 of 17


• Provides correct answer 2

• Attempts to find the projection of F in the direction of 1

Sample answer:

A unit vector in the direction of is υ̂ = 1

5

3

4⎛⎝⎜

⎞⎠⎟

, so the component of F in the direction of is

F ⋅υ̂( )υ̂ = 2υ̂ =

1.2

1.6⎛⎝⎜

⎞⎠⎟

.

Question 12 (b) (ii) Criteria Marks


Sample answer:

The component of F perpendicular to is F −

1.2

1.6⎛⎝⎜

⎞⎠⎟=

0.8

–0.6⎛⎝⎜

⎞⎠⎟

.


• Provides correct explanation 1

Sample answer:

PA = a − p( ) , while PB = b − p( ) . Since we are given that they are perpendicular, the dot

product of these two vectors is zero.


Page 8 of 17


• Obtains a correct expression for p − m 2

involving only a, b or m 3

• Replaces m by

12

a + b( ) and uses the result of part (i) appropriately 2

• Replaces m by

12

a + b( ) and attempts to simplify 1

Sample answer:

From part (i) a ⋅ b − p ⋅ a + b( ) + p ⋅ p = 0, so

p − m 2 = 14

2p − a + b( )( ) ⋅ 2p − a + b( )( )

= p ⋅ p − p ⋅ a + b( ) ⋅ 14

a + b( ) ⋅ a + b( )

= 14

a + b( ) ⋅ a + b( ) – a ⋅b independent of p

OR

4 p − m 2 = 2p − 2m( ) ⋅ 2p − 2m( )= p − a( ) + p − b( )( ) ⋅ p − a( ) + p − b( ) since 2m = a + b

= p − a( ) ⋅ p − a( ) + 2 p − a( ) ⋅ p − b( ) + p − b( ) ⋅ p − b( ) expanding

= p − a( ) ⋅ p − a( ) − 2 p − a( ) ⋅ p − b( ) + p − b( ) ⋅ p − b( ) by part (i)

= p − a( ) − p − b( )( ) ⋅ p − a( ) − p − b( )( ) factoring

= b − a( ) ⋅ b − a( ) simplifying

= b − a( ) 2independent of p

so p − m 2 = 14

b − a( ) 2

OR


Page 9 of 17

Let u = 12

b − a( ) so m = a + u = b − u

Then

p − m 2 = p − a( ) − u( ) ⋅ p − b( ) + u( )= p − a( ) ⋅ p − b( ) + u ⋅ p − a( ) − p − b( ) − u( ) expanding

= 0 + u ⋅ b − a − u( ) by part (i)

= u 2 independent of p

OR

p − m 2 − m − a 2 = p − m( ) ⋅ p − m( ) − m − a( ) ⋅ m − a( )

= p − m( ) − m − a( )( ) ⋅ p − m( ) + m − a( )( )= p − 2m − a( )( ) ⋅ p − a( )

= p − b( ) ⋅ p − a( )= 0

Therefore p − m 2 = m − a 2 which is independent of p.

Question 12 (c) (iii) Criteria Marks

• Provides correct statement 1

Sample answer: P lies on the circle whose diameter is AB. OR

P lies on a circle centre M and radius

14

a + b( ) ⋅ a + b( ) − a ⋅b .


Page 10 of 17

Question 12 (d) Criteria Marks

• Provides correct proof 3 • Attempts to do the induction step 2

• Proves cases for n = 1, or equivalent merit 1

Sample answer:

If n = 1, then 23 − 3 = 8 − 3

= 5 which is divisible by 5.

Assume true for n = k

ie 23k − 3k = 5 j for some integer j.

Then if n = k +1

23 k+1( ) − 3k+1 = 23k+3 − 3k × 3

= 8 5 j + 3k( ) − 3 × 3k (using the assumption)

= 8 × 5 j + 5 × 3k

= 5 8 j + 3k( ) which is divisible by 5.

Hence the claim is true for n = k +1. Since shown true for n = 1, so is true for n = 2, 3, …

and so true for all integers n ≥ 1.


Page 11 of 17

Question 13 (a) Criteria Marks

• Provides correct solution 3 • Attempts to use a double angle result, or equivalent merit 2 • Obtains correct integrand in terms of θ, or equivalent merit 1

Sample answer:

x = sin2θ

dx = 2sinθ cosθ dθ

x

1− x× dx = sin2θ

1− sin2θ× 2sinθ cosθ dθ

0

π4⌠

⌡⎮

0

12⌠

⌡

⎮⎮

= sinθcosθ

× 2sinθ cosθ dθ0

π4⌠

⌡⎮

= 2 sin2θ dθ0

π4⌠

⌡⎮

= 1− cos2θ( )dθ0

π4⌠

⌡⎮

= θ − 12

sin2θ⎤

⎦

⎥⎥⎥

0

π4

= π4− 1

2sin

π2

⎛⎝

⎞⎠ − 0( )

= π4− 1

2


Page 12 of 17

Question 13 (b) Criteria Marks


• Considers when cosθ ≥ 32

but obtains incorrect proportion, or

equivalent merit 3

• Correctly deduces that 2 sin t + cost = 3 cos t −α( ) for some α, or equivalent merit

2

• Attempts to write 2 sin t + cost in the form Acos t −α( ) or equivalent merit

1

Sample answer:

x = 2 sin t + cost = 313

cost + 23

sin t⎛⎝⎜

⎞⎠⎟= 3 cos t −α( ), where tanα = 2.

Thus x ≥ 1.5 whenever cos t −α( ) ≥ 32

.

Now the region in the interval 0,π2

⎡⎣⎢

⎤⎦⎥

where cosθ ≥ 32

is 0,π6

⎡⎣⎢

⎤⎦⎥

, and other intervals

between multiples of π2

are similar.

So distortion occurs 13

of the time.


• Provides correct proof 3 • Obtains a correct expression for cos3θ involving only cosθ and sinθ 2

• Obtains a correct expression for cos3θ involving only cosθ , sinθ , cos2θ and sin2θ 1

Sample answer:

cos3θ = cosθ cos2θ − sinθ sin2θ angle sum formula

= cosθ cos2θ − sin2θ( ) − sinθ 2sinθ cosθ( ) angle sum formula

= cos3θ − 3cosθ sin2θ

= cos3θ − 3cosθ 1− cos2θ( ) cos2θ + sin2θ = 1

= 4cos3θ − 3cosθ


Page 13 of 17


• Finds correct expressions for the three solutions 4

• Deduces that 3θ = ± π3± 2nπ , where n is an integer 3

• Deduces that cos3θ = 12

2

• Makes the substitution x = cosθ 1

Sample answer:

Writing x = cosθ we get 8cos3θ − 6cosθ = 2cos3θ = 1.

Consequently cos3θ = 12

, and so 3θ = ± π3± 2nπ , where n is an integer.

Thus θ = ± π9± n

2π3

, and so x = cosπ9

⎛⎝

⎞⎠ or cos

7π9

⎛⎝

⎞⎠ or cos

13π9

⎛⎝

⎞⎠ .

NB cos7π9

⎛⎝

⎞⎠ = −cos

2π9

⎛⎝

⎞⎠ and cos

13π9

⎛⎝

⎞⎠ = cos

5π9

⎛⎝

⎞⎠ = −cos

4π9

⎛⎝

⎞⎠ .


Page 14 of 17

Question 14 (a) (i) Criteria Marks

• Provides correct sketch and explanation 3 • Provides correct sketch, or equivalent merit 2 • Provides a sketch that is an odd function or has three zeros, or equivalent

merit 1

Sample answer:

x cos x dx = 0−π

2

π2⌠

⌡⎮ because the function is odd

ie x cos x dx =−π

2

0⌠

⌡⎮ − x cos x dx

0

π2⌠

⌡⎮


Page 15 of 17


• Provides correct answer 3 • Expands integrand and evaluates at least one of the resulting integrals

between − π2

and π2

2

• Provides correct integrand for volume of revolution 1

Sample answer:

V = π x − cos x( )2 dx−π

2

π2⌠

⌡⎮

= π x2 − 2x cos x + cos2 x( )dx−π

2

π2⌠

⌡⎮

= π x2 − 2x cos x + 12

1+ cos2x( )⎛⎝

⎞⎠ dx

−π2

π2⌠

⌡⎮

= π x2 + 12

⎛⎝

⎞⎠ dx

−π2

π2⌠

⌡⎮ by part (i)

= π x3

3+ x

2

⎡⎣⎢

⎤⎦⎥−π

2

π2

= π 4 + 6π 2

12


Page 16 of 17



• Provides correct expression for dy

dt in terms of w and

dw

dt or deduces

that y 1− y( ) = −w

1− w( )2

1

Sample answer:

dy

dt= ′w

1− w( )2and y 1− y( ) = −w

1− w( )2, so

dw

dt= −w.

Question 14 (b) (ii) Criteria Marks


• Deduces that w = −3et or deduces that y = 11− ke−t for some k 1

Sample answer:

w = −3 when t = 0 so w = −3e−t . Thus y = 11+ 3e−t .

Question 14 (b) (iii) Criteria Marks


• Makes some progress towards finding the value of t for which y = 34

1

Sample answer:

1+ 3e−t = 43

so e−t = 19

so t = ln9 years.


Page 17 of 17


• Provides correct sketch including location information 3 • Provides correct shape and asymptotes 2 • Provides a sketch with asymptotes at x = 1, x = 3 and y = 0 1

Sample answer:


• Identifies that the equation has (at least) five solutions 1

Sample answer: The equation has at least five solutions. (It might have more depending on the behaviour outside the region shown in the sketch in the question.)


Page 1 of 2

Mathematics Extension 1 Sample HSC Mapping Grid

Section I

Question Marks Content Syllabus outcomes Targeted performance

bands

1 1 ME-V1 Introduction to Vectors ME12-2 E2-E3

2 1 ME-V1 Introduction to Vectors ME12-2 E2-E3

3 1 ME-T3 Trigonometric Equations ME12-3 E2-E3

4 1 ME-C2 Further Calculus Skills ME12-1 E2-E3

5 1 ME-C3 Applications of Calculus ME12-4 E2-E3

6 1 ME-F2 Polynomials ME11-2 E3-E4

7 1 ME-A1 Working with Combinatorics ME11-5 E3-E4

8 1 ME-A1 Working with Combinatorics ME11-5 E3-E4

9 1 ME-C1 Rates of Change ME11-4 E3-E4

10 1 ME-T1 Inverse Trigonometric Functions ME11-3 E3-E4


Page 2 of 2

Section II

Question Marks Content Syllabus outcomes Targeted performance

bands

11 (a) (i) 1 ME-V1 Introduction to Vectors ME12-2 E2-E3

11 (a) (ii) 2 ME-V1 Introduction to Vectors ME12-2 E2-E3

11 (a) (iii) 1 ME-V1 Introduction to Vectors ME12-2 E2-E3

11 (b) (i) 1 ME-F1 Further Work with Functions ME11-1 E2-E3

11 (b) (ii) 2 ME-F1 Further Work with Functions ME11-1 E2-E3

11 (b) (iii) 2 ME-F1 Further Work with Functions ME11-1 E2-E3

11 (b) (iv) 2 ME-F1 Further Work with Functions ME11-1 E2-E3

11 (c) (i) 1 ME-C3 Applications of Calculus ME12-4 E2-E3

11 (c) (ii) 3 ME-C3 Applications of Calculus ME12-4; ME12-7 E2-E4

12 (a) (i) 2 ME-S1 The Binomial Distribution ME12-5 E2-E3

12 (a) (ii) 2 ME-S1 The Binomial Distribution ME12-5 E2-E3

12 (b) (i) 2 ME-V1 Introduction to Vectors ME12-2 E3-E4

12 (b) (ii) 1 ME-V1 Introduction to Vectors ME12-2 E3-E4

12 (c) (i) 1 ME-V1 Introduction to Vectors ME12-2 E2-E3

12 (c) (ii) 3 ME-V1 Introduction to Vectors ME12-2 E2-E4

12 (c) (iii) 1 ME-V1 Introduction to Vectors ME12-2; ME12-7

E2-E3

12 (d) 3 ME-P1 Proof by Mathematical Induction ME12-1 E2-E4

13 (a) 3 ME-C2 Further Calculus Skills ME12-1; ME12-4 E2-E4

13 (b) 4 ME-T3 Trigonometric Equations ME12-3 E2-E4

13 (c) (i) 3 ME-T3 Trigonometric Equations ME12-3 E2-E4

13 (c) (ii) 4 ME-T3 Trigonometric Equations ME12-3 E2-E4

14 (a) (i) 3 ME-F1 Further Work with Functions ME11-1 E2-E3

14 (a) (ii) 3 ME-C3 Applications of Calculus ME12-4 E2-E4

14 (b) (i) 2 ME-C3 Applications of Calculus ME12-4 E2-E4

14 (b) (ii) 2 ME-C3 Applications of Calculus ME12-4 E2-E4

14 (b) (iii) 2 ME-C3 Applications of Calculus ME12-4 E2-E4

14 (c) (i) 3 ME-F1 Further Work with Functions ME11-2; ME11-7 E2-E4

14 (c) (ii) 1 ME-F1 Further Work with Functions ME11-2; ME11-7 E2-E4

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Mathematics Ext 1 Sample Examination Materials 2020

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