Mathematics: Specialist
Units 3C and 3D
Assessment item bank
Response
2011/15613[v2]
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
1
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2
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Unit 3C/3D—Response items with marking keys
Note: These questions come from the sample examination written in 2010 by the examination
panel. As such the questions appear in the order of the examination paper. The table below sets
the questions out in order of the units and then dot points for reference when choosing individual
items from this item bank. At this stage the item number is the same as the question number in the
sample examination. It is intended that the item bank will be added to from past WACE
examinations.
Item
Section
Unit
8
(a)(b)
calculator
assumed
3C
7
(a)(b)
calculator
assumed
3C
14
(a)(b)
calculator
assumed
3C
4
(a)-(c)
calculator
free
3C
9
(a)(b)
calculator
assumed
3C
5
(a)-(c)
calculator
free
3C
3
(a)-(b)
calculator
free
3C
12
(a)-(c)
calculator
assumed
3C
Syllabus reference
Angle between vectors
1.8 develop the concept of the dot product of vectors in a plane, using
projections, and the formula a  b  a1b1  a2b2  a3b3 and establish the
a  b  a b cos
formula
where a  a1 , a 2 , a3  and b  b1 , b2 , b3  and 
is the angle between the vectors
Division of a line segment
1.9 Calculate the angle between vectors and identify parallelism and
perpendicularity
1.6 add vectors in space using the parallelogram rule and addition of
components
1.7 multiply vectors by scalars and extend this to subdividing line
segments internally
Vectors in 3D
1.10 establish and use the vector equation of a plane
1.11 establish and use the vector equation of a line in space in the
form r  r1  l together with its parametric equivalent
Collision of moving objects
1.12 solve practical problems in three-dimensional geometry using
vector concepts and formulas, and graphical methods where
appropriate
Integration methods of trigonometry and exponentials
2.4 differentiate and integrate the sine, cosine and tangent functions
kf x 
f x 

3.4 integrate functions of the form f x  and kf x e
using the
change of variable (or substitution technique) either by observation,
or provided
Complex numbers
6.2 multiply and divide complex numbers expressed in polar form
Exponentials and logarithms
3.1 review the inverse relationship between exponentials and
logarithms
Exponentials and logarithms
3.2 investigate the logarithmic properties of the function
x1
1
t
dt , define
this as the natural logarithm ln x and review its basic properties
Functions
4.4 determine the equation(s) of the tangent(s) to a function
4.3 find the area under and between curves
Proof by induction
5.4 develop geometric proofs by deduction using vector methods
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
3
Item
Section
Unit
6
(a)-(c)
calculato
r free
1
calculato
r free
3C
2
(a)-(e)
calculato
r free
3D
11
(a)-(f)
calculato
r
assumed
3D
13
(a)-(c)
calculato
r
assumed
3D
17
(a)-(c)
calculato
r
assumed
16
(a)-(c)
calculato
r
assumed
15
(a)-(b)
calculato
r
assumed
3C
3D
3D
3D
3D
3C
3D
Syllabus reference
Regions in the complex plane
6.6 describe regions in the complex plane and Argand diagrams
defined by means of simple systems of equalities and inequalities
Functions
4.1 investigate graphical, geometric and algebraic properties of
absolute value functions (in the complex and Cartesian planes)
Polar coordinates
7.2 draw and interpret polar graphs (including inequalities) of
r = constant, θ = constant and r =k θ
Matrices
1.6 examine the geometric properties of 2 x 2 matrices as linear
transformations in the plane including general rotations and
reflections, and dilations and shears parallel to the coordinate axes
1.7 use matrix multiplication to determine the combined effect of two
linear transformations in the plane
1.8 establish and apply the relationship between the determinant and
areas of shapes before and after transformation
1.4 calculate the determinant and inverse of a 2 x 2 matrix and
recognise a singular 2 x 2 matrix
Leslie matrices
1.9 solve practical problems involving the use of Leslie matrices and
other examples of transition matrices
Simple harmonic motion
2
2.1 investigate the differential equation d y  k 2 y  0 and its
dt 2
solutions y(t )  C cos kt  D sin kt  A cos  kt  1   A sin  kt   2 
as models of simple harmonic motion
Related rates
4.3 solve related rates problems (including those involving
trigonometric functions)
Simple harmonic motion
2
2.1 investigate the differential equation d y  k 2 y  0 and its solutions
dt 2
y(t )  C cos kt  D sin kt  A cos  kt  1   A sin  kt   2 
as models of simple harmonic motion
Rectilinear motion
4.4 solve practical problems by applying calculus techniques to
problems from various branches of the sciences including rectilinear
motion and marginal cost
Solving differential equations
4.6 solve practical problems involving parametric and differential
equations (variables separable)
Growth and decay
3.2 solve practical problems involving models of growth and decay of
the form
10
(a)-(b)
4
calculato
r
assumed
3D
dP
 kP
dt
Complex number roots
6.1 establish properties of sums, products, division and
exponentiation (including combinations of these) of complex numbers
and their conjugates (using real and ‘imaginary’ components)
6.3 find and locate in the complex plane, solutions of z n  C
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item 1
Calculator free
1
3C
Polar coordinates
7.2 draw and interpret polar graphs (including inequalities) of r = constant,
θ = constant and r =k θ
(3 marks)
The diagram below shows the three graphs r  a,   b and r  c where a , b and c are
constants. State the values of a, b and c .
Solution





Solution
 

circle radius  2 
3
ray with  

4
a2
3
b
4
2
c

r  2 when     2  c.
Mark
allocation
1
Item
classification
simple
identifies b  3 /4
1
simple

identifies c  2 / 
1
complex
Specific behaviours
identifies a  2

Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
5
Item 2
Calculator
free
3D
Matrices
1.6 examine the geometric properties of 2 x 2 matrices as linear transformations in
the plane including general rotations and reflections, and dilations and shears
parallel to the coordinate axes
1.7 use matrix multiplication to determine the combined effect of two linear
transformations in the plane
1.8 establish and apply the relationship between the determinant and areas of
shapes before and after transformation
1.4 calculate the determinant and inverse of a 2 x 2 matrix and recognise a singular
2 x 2 matrix
2(a)
2(b)
2(c)
2(d)
2(e)
(8 marks)
The diagram below shows the triangle OPQ, where O is the point (0, 0).
(a)
Triangle OPQ is transformed to OP Q by a rotation of 90º clockwise about the origin. State
the matrix that will do this.
(2 marks)
Solution

0
 0 1  1 2   1
 0 1



  R90  


1
0
1
0

1

2

 
 

 1 0 
Q
O
P
Q'
P'
Specific behaviours
defines the correct rows of the 2 x 2 matrix
(b)
Mark
allocation
2
Item
classification
simple
The object O′P′Q′ is transformed to O′′P′′Q′′ by a dilation of factor 3 about the origin. State
the matrix that will do this.
(1 mark)
Solution
0
0
 3 0  1
 3
 3 0


 
 E 

 0 3   1 2 
 3 6 
 0 3
Specific behaviours
defines the correct 2 x 2 matrix
6
Mark
allocation
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
(c)
Find the single 2 x 2 matrix that will transform OPQ directly on to O′′P′′Q′′.
(2 marks)
Solution
 3 0  0 1  0 3
 0 3



 C 

 0 3   1 0   3 0 
 3 0 
Specific behaviours
combines the matrices in the correct order
defines the correct 2 x 2 matrix
(d)
Mark
allocation
1
1
Given that the area of ΔOPQ is k square units, what is the area of ΔO′′P′′Q′′?
Item
classification
complex
complex
(1 mark)
Solution
Determinant of M is 9, so area of OP Q is 9 k square units.
Specific behaviours
Mark
allocation
1
uses determinant to find area

(e)
Find the single 2 x 2 matrix that will transform O′′P′′Q′′ directly back to OPQ.
Item
classification
simple
(2 marks)
Solution
 0 3
1  0 3 
1
C 
C  

93 0 
 3 0 
Specific behaviours
uses correct determinant
defines the elements correctly
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
7
Item 3
Calculator
free
3C
Functions
4.4 determine the equation(s) of the tangent(s) to a function
4.3 find the area under and between curves
3(a)
3(b)
(5 marks)
The graph of y  e is shown below. A is the point (0, 1), B is the point (1,e) and O is the origin.
x
(a)
x
Find the equation of the tangent to y  e at B.
(2 marks)
Solution
dy
dy
 e x so at B, where x  1, we have
e
dx
dx

Hence the equation of tangent is: y  e  e(x 1)  y  ex

Specific
behaviours
Mark
allocation
1
1
uses derivative to give gradient
 for the linear equation
uses a given point to solve
(b)
Item
classification
simple
simple
Determine the exact area of the region bounded by the straight-line segments OA and OB
and the arc AB.
(3 marks)
Solution
or
Required area 
Required area 
area under y1  e x – area under y2  e.x  triangle 
area under y1  e x – area under y2  e.x  triangle 
Area
1
 
  e x dx 
0
1
 1 e
2
1
1 

   e x  e  dx
2 
0
1
1 

  e x  e.x 
2 0

1
1
 e x   e
 0 2
1
  e  1  e
2
1
 e 1
2
1 

  e  e   1
2 

1
 e 1
2
Specific behaviours
defines the difference of the areas correctly
accurately calculates area under both graphs
accurately evaluates required area
8
Area
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
simple
Item 4
Trigonometry
Integration methods
Complex numbers
2.4 differentiate and integrate the sine, cosine and tangent functions
kf x 
3.4 integrate functions of the form
and kf x e f  x  using the change of
f x 
variable (or substitution technique) either by observation, or provided
Calculator free
3C
4(a)
4(b)
4(c)
6.2 multiply and divide complex numbers expressed in polar form
(7 marks)
2
Determine 
(a)
3tan x
cos 2 x
(2 marks)
dx
Solution
Substitute u  tan x and then
du
1
 sec 2 x 
so
dx
cos 2 x

3tan 2 x
2
cos x
So

3tan 2 x
cos 2 x
dx   3u 2 du  u 3  C
dx  tan 3 x  C
Specific behaviours
uses the chain rule
carries through to correct solution
1
6 x2
0
x3  1
Evaluate 
(b)
Mark
allocation
1
1
Item
classification
simple
simple
(2 marks)
dx
Solution
d 3
( x  1)  3 x 2
dx
we see that
Since
1

0
6 x2
x3  1
dx
1
3x 2
0
x3  1
 2
dx
1
 2  ln( x3  1) 

0
 2(ln 2  ln1)  2ln 2
Specific behaviours
recognises form
f ( x )
f ( x)
carries through to correct solution
Mark
allocation
Item
classification
1
simple
1
simple
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
9
(c)
 3 
 
3cis 
  8cis  3 
4


 
Simplify
 
 5 
2cis    6cis 

6
 12 
(3 marks)
Solution
 3 
 
 3  
3cis 
24cis 
 
  8cis  3 
4
3


  
 4
 
 5 
  5 
2cis    6cis  
 12cis  

6
 12 
 6 12 
 13 
2cis 

 12   2cis  16 

 12 
 


cis   
 4
 4 
 2 
 2cis 
 2cis  


 3 
 3 
Specific behaviours
simplifies ‘r’ component
simplies ‘  ’ component correctly
uses the Principal Domain
10
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
simple
Item 5
Calculator free
5(a)
5(b)
5(c)
3C
Exponentials and logarithms
3.2 investigate the logarithmic properties of the function
x1
1
t
dt , define this as the
natural logarithm ln x and review its basic properties
(10 marks)
x1
The natural logarithm can be expressed as ln x   1 dt
t
ab 1
a1
ab 1
Explain why  dt   dt   dt where 1  a  b and a and b are constants.
a t
1t
a t
(a)
Solution
(2 marks)
y
Graphically it is clear that the area under the
curve y 
1
between 1  t  a plus that
t
between a  t  ab equals that
t  ab.
between 1

1
a
Mark
allocation
1
1
Specific behaviours
diagram
describes the two areas
(b)
t
Use the substitution u  in the integral to deduce that,
a
ab 1
  ln  b   ln  a  .
a
t
t
ab
Item
classification
complex
complex
(3 marks)
Solution
If u 
t
then
a
1
dt and when t  ab  u  b and when t  a  u  1
a
ab 1
b 1
b1
Then  dt   (adu )   du
a t
1 au
1u
du 
ab 1
a1
ab 1
From part  a   dt   dt   dt
1 t
1t
a t
ab 1
a1
b1
t
1t
1t
 dt   dt   dt
1
ln  ab   ln  a   ln  b 
Specific behaviours
substitutes for t with the integrand and the limits
substitutes for t with the limits
rearranges part (a) to prove the result
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
11
(c)
Use your result from part (b) and proof by induction to prove that ln a n  n ln a for all n  1.
(5 marks)
Solution
Clearly holds for n  1 since 1  ln a  ln a1
Let us assume the result is true for some positive integer n  k
k ln a  ln a k .................................(1)
Required to prove true for n  k  1
If
k ln a
i.e. ln a ( k 1)  (k  1) ln a
 ln a k .......... from (1)
ln a  k ln a
 ln a  ln a k
 k  1 ln a
 ln a  ln b where a k  b
 ln ab
 ln a.a k
 ln a k 1
Since statement is true for n  1 and assuming it is true for n  k
implies it is true for n  k  1 then it is true for all n.
Specific behaviours
proves proposition is true for n = 1
makes the assumption that proposition is true for n = k
uses assumption to prove proposition for n = k + 1
makes the inductive statement
Mark
allocation
1
1
2
Item
classification
complex
complex
complex
1
complex
"If n  1 is true and assuming n  k implies n  k  1
then the statement is true for all n"
12
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item 6
Calculator free
6
(a)-(c)
3C &
Regions in the complex plane
3D
Functions
6.6 describe regions in the complex plane and Argand diagrams defined by
means of simple systems of equalities and inequalities
4.1 investigate graphical, geometric and algebraic properties of absolute value
functions (in the complex and Cartesian planes)
(7 marks)
Sketch the following regions in the complex plane.
(a)
Im  z   2 Re  z   1
(2 marks)
Solution
Im  z   2 Re  z   1
Specific behaviours
shows correct upper boundary line
correctly shades below the line
(b)
Mark
allocation
1
1
1  z  3  4i  6
Item
classification
simple
simple
(3 marks)
Solution
1  z  3  4i  6
Specific behaviours
shows correct centre – either scale shown or centre labelled
correctly identifies length of radius
shows correct shading including dotted inner circle
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
simple
13
(c)
For the region in (b) above, state the maximum value of z .
(2 marks)
Solution
Distance from the origin to the centre of the circle;
32  42  5 units
Thus, the maximum value of z  5  Radius
56
 11 units
 See diagram above 
Specific behaviours
determines the distance from the origin to the centre of circle
states maximum value of z
Mark
allocation
1
1

14
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
Item 7
Calculator
assumed
7(a)
3C
Vectors in 3D
1.10 establish and use the vector equation of a plane
1.11 establish and use the vector equation of a line in space in the form
r  r1  l together with its parametric equivalent
7(b)
(4 marks)
P is the point with coordinates  2, 1, 1 and
Q is the plane with equation; 3x  2 y  5 z  2  0
(a)
Give a vector equation for the plane that contains P and is parallel to Q.
(2 marks)
Solution
Required plane has the same normal as Q so is of the form;
3x  2 y  5 z  c
As P  2,1,1 lies on the plane then c  3(2)  2(1)  5(1)  6  2  5  9.
Thus vector equation of the required plane is;
r (3, 2,5)  9.
Specific behaviours
defines parallel plane with unknown constant c and solves for c
using point P
correctly defines the vector form of the plane
(b)
Mark
allocation
1
Item
classification
simple
1
simple
Give a vector equation for the line through P and perpendicular to Q.
(2 marks)
Solution
Normal to the plane, n is parallel to  3, 2,5  so required vector equation is
r  a tn
r  (2,1,1)  t (3,  2,5)
 (2  3t ,1  2t ,1  5t )
Specific behaviours
defines normal vector n correctly
uses normal n and point P to correctly define vector equation
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
15
Item 8
Calculator
assumed
Angle between vectors
Division of a line segment
1.8 develop the concept of the dot product of vectors in a plane, using
projections, and the formula a  b  a1b1  a2b2  a3b3 and establish the formula
3C
a  b  a b cos where a  a1 , a 2 , a3  and b  b1 , b2 , b3  and 
8
(a)-(b)
between the vectors
1.9 calculate the angle between vectors and identify parallelism and
perpendicularity
1.6 add vectors in space using the parallelogram rule and addition of
components
1.7 multiply vectors by scalars and extend this to subdividing line segments
internally
Point M has position vector 8i  24 j  k and point N has position vector 22i  3j  50k.
(a)
is the angle
Find, to the nearest degree, the angle between the vectors OM and ON .
(5 marks)
(2 marks)
Solution
Using the dot product:
OM ON  (8, 24,1) (22,3,50)  176  72  50  298.
Now if  is the required angle then
298
298
cos 

OM ON
64  576  1 484  9  2500
cos  0.2151
  cos 1 (0.2151)  77.58o
  78o
Using a CAS Calculator;
or
OM  (8, 24,1) and ON  (22,3,50)
sMON  77.58o  78o
Specific behaviours
uses appropriate technology to calculate correct angle
gives the answer to the desired degree of accuracy and includes
units
16
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
(b)
Find the position vector of the point P that divides MN internally in the ratio 2:5.
(3 marks)
Solution
M (22,3,50)
The vector MN  MO  ON
MN  14i – 21 j  49 k 2
OP  OM  MN
7
2
 8i  24 j  k  (14i – 21j  49 k ) 7
 12i  18 j  15k
Specific behaviours
defines MN in terms of i, j, k components
defines position vector for P in terms of MN
accurately carries through calculation in terms of i, j, k
O
P
N (8,24,1)
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
simple
17
Item 9
Calculator
assumed
9(a)
9(b)
3C
Exponentials and logarithms
3.1 review the inverse relationship between exponentials and logarithms

(5 marks)

1
2
1
Find f ( x) . What is its range?
Let f ( x)  ln 4 x 2  1 , x 
(a)
(3 marks)
Solution


y  ln 4 x 2  1 , x 
1
2
 e y  4 x2  1 , x 
 x2 
1
2
ey 1
1
,x 
4
2
ey 1
1 y

e 1
4
2
Need to select the correct branch :
 x
as the original function has domain x 
1
2
we need the positive sign so that f 1 ( x) 
1 x
e 1
2
Range of f 1 is the domain of f so
1
Range of f 1 is y 
2
Specific behaviours
uses inverse relationship between log and exponential
rearranges the exponential expression correctly
states the correct range
18
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
(b)
Determine the coordinates of the point(s) of intersection of f ( x) and f 1 ( x) correct to 2
decimal places. What do you notice?
(2 marks)
Solution
Point of intersection given by
1 x
ln 4 x2  1 
e  1 and x  0
2


By calculator, solutions are (0.95, 0.95) and
(4.28, 4.28). They lie on the line y = x
Specific behaviours
gives both points correct to 2 d.p
recognises line of reflection y = x
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
19
Item 10
Calculator
assumed
10(a)
3D
6.1 establish properties of sums, products, division and exponentiation
(including combinations of these) of complex numbers and their conjugates
(using real and ‘imaginary’ components)
6.3 find and locate in the complex plane, solutions of z n  C
10(b)
(a)
Complex number roots

State the exact value of 2  2 3i

(9 marks)
4
in Cartesian form.
(1 mark)
Solution
2  2 3i  128(1
4
3i).

Specific behaviours
expands the bracket correctly in Cartesian form
(b)
Mark
allocation
1
Determine exact values for all the roots of z 4  8  8 3i
Item
classification
simple
(5 marks)
Solution
1
(2  2 3i) 4
16
then one root of the equation is clearly;
1
z0  (2  2 3i)  1  3i
2
Other roots of the equation are equally spaced in the Argand diagram
z 4  8  8 3i 
so are rotated by

from this first root.
2
Thus other roots are z1   3  i, z2  1  3i and z3  3  i
Specific behaviours
takes a factor of
1
outside bracket
16
identifies the first root
states that the rotational symmetry =
correctly gives the other solutions
20

2
Mark
allocation
1
Item
classification
complex
1
complex
1
complex
2
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
(c)
Sketch all the roots from (b) on the Argand diagram below. Identify all the important features.
(3 marks)
Solution
Specific behaviours
shows 4 point symmetry – each root makes an angle of 90º with
its neighbouring roots
shows scale or correct points
Mark
allocation
2
Item
classification
complex
1
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
21
Item 11
Calculator
assumed
11 (a) – (f)
CA
Leslie matrices
1.9 solve practical problems involving the use of Leslie matrices and other
examples of transition matrices
(10 marks)
The following table describes the size of a herd of beef cattle and its breeding and survival rate for
the year 2002. Throughout this question, only the female members of the herd are to be
considered.
Age (years)
0–1
1–2
2–3
3–4
4–5
5–6
Female population
800
720
620
500
310
140
0
0
1.6
1.8
1.2
0.4
0.6
0.7
0.8
0.5
0.3
0
Annual breeding rate of
females
Annual female survival
rate
(a)
How many females were in the herd in the year 2002?
(1 mark)
Solution
800 + 720 + 620 + 500 + 310 + 140 = 3090
Specific behaviours
carries out correct addition
(b)
Mark
allocation
1
Complete the Leslie matrix generated by this information.
Item
classification
simple
(1 mark)
Solution
0 1.6 1.8 1.2 0.4
0
0.6 0
0
0
0
0 

 0 0.7 0
0
0
0


0 0.8 0
0
0
0
0
0
0 0.5 0
0


0
0
0 0.3 0 
 0
Specific behaviours
sets up matrix from information in table
22
Mark
allocation
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
(c)
If none of the cows was sold for meat, how many females would there be in
(i)
2012?
(ii)
2017?
(2 marks)
Solution
(i)
H 2002  800 720 620 500 310 140 
T
and if L is the Leslie matrix in b  , then
H 2012  L10 H 2002
and number of females is
(1 1 1 1 1 1) H 2012  10012
(ii)
Similarly
H 2017  L15 H 2002 and
the number of females is;
(1 1 1 1 1 1) H 2017  17484
Specific behaviours
accurately calculates the total of females for 2012
accurately calculates the total of females for 2017
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
23
Some cows aged between 2 and 4 years are regularly harvested with the intention of maintaining a
stable population. The harvesting rate, h, affects both the breeding and the survival rates.
(d)
If h = 0.24, what would the breeding rate for cows be that are aged
(i)
2–3 years?
(ii) 3–4 years?
(2 marks)
Solution
(i) New breeding rate = 1.6  0.76  1.216
(ii) New breeding rate = 1.8  0.76  1.368

Specific behaviours
calculates the correct result for (i)

calculates the correct
result for (ii)
24
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
(e)
With h = 0.24, how many females would there be in
(i)
(2 marks)
2012?
(ii) 2017?
Solution
If the new Leslie matrix is denoted by K then
10
(i) H 2012  K H 2002 and the number of females is (1 1 1 1 1 1) H 2012  3107
Similarly,
15
(ii) H 2017  K H 2002 and thenumber of females is (1 1 1 1 1 1) H 2017  3083
Using CAS:
Specific behaviours
modifies the Leslie matrix
calculates the total females in both cases
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
25
(f)
Comment on whether or not this is a reasonable value for h in order to maintain a stable
population.
(2 marks)
Solution
A few trial values suggest that in the years 2012, 2017, 2022, 2027 and 2032 the number of
females will be 3107, 3083, 3030, 2980 and 2931 respectively.
Thus the decreases in female population over each five year period is 24, 53, 50 and 49. There
appears to be a near-constant decrease over each interval suggesting that the population will not
be stable but, rather, steadily reduce.
Mark
Item
Specific behaviours
allocation
classification
gives additional data as part of the comment
1
complex
comment is appropriate and consistent with data
1
complex
26
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item 12
Calculator
assumed
12
(a)-(c)
3C
Proof by induction
5.1-5.4 develop geometric proofs by deduction using vector methods
(6 marks)
In the diagram below , O is the centre of the circle and AB is a chord with midpoint M .
The vectors OA and OB are denoted by a and b respectively.
B
O
M
A
(a)
Express AB in terms of a and b .
(1 mark)
Solution
AB  AO  OB  b  a
Specific behaviours
correctly writes the expression for AB in terms of a and b
(b)
Mark
allocation
1
Expressin OM terms of a and b .
Item
classification
simple
(2 marks)
Solution
OM  OA  AM
1
OM  OA  AB
2
1
OM  a   b  a 
2
1
OM   a  b 
2
Specific behaviours
correctly writes a vector expression for OM
substitutes correctly for a and b into the expression and
simplifies
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
simple
simple
27
(c)
Use vector methods to prove that OM is perpendicular to AB.
(3 marks)
Solution
Required to show OM  AB
OM  AB  OM AB  0
1
a  b  b  a 
2
1 2
2
OM AB 
a  b since a  b  radius
2
OM AB  0  OM  AB
OM AB 


Specific behaviours
uses dot product OM AB to test for perpendicularity
correctly carries out dot product in terms of a and b
recognises a  b  radius
makes logical statement of proof
28
Mark
allocation
1
1
1
Item
classification
complex
complex
complex
1
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item 13
Calculator
assumed
3D
Simple harmonic motion
d2y
 k2y  0
2
dt
solutions y(t )  C cos kt  D sin kt  A cos  kt  1   A sin  kt   2 
2.1
13(a)
13(b)
13(c)
investigate
the
differential
equation
and
its
as models of simple harmonic motion
(6 marks)
Tom is exercising on a rowing machine and realises that the motion of his seat is describing simple
harmonic motion.
(a)
If the seat completes one full cycle every 2 seconds, write down the differential equation
which describes the relationship between the acceleration of the seat and its displacement,
cm, from its central position.
(2 marks)
Solution
The SHM equation is
d 2x
dt
2
 k 2 x for some constant k
Since the period of motion T 
Thus our equation is
d 2x
dt
2
2
 2sec s then k  
k
  2 x
Specific behaviours
uses correct acceleration equation
correctly evaluates k
(b)
Mark
allocation
1
1
Item
classification
complex
complex
If Tom starts rowing in the furthest forward position and the seat can move a distance of 25
cm in either direction from the central position, find the displacement of the seat at time, t
seconds, after Tom begins.
(2 marks)
Solution
d 2x
  2 x for some constants A and B
dt 2
Then x  A cos  t  B sin  t....................(1)
Since t  0 x  25  A  25
getting the derivative from (1) above
dx
  B cos  t   A sin  t.......................(2)
dt
dx
Also when t  0
0 B0
dt
Hence
x  25cos  t...........................(3)
Specific behaviours
defines constant A from the displacement equation 1
defines constant B from the derivative in equation 2
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
29
(c)
Calculate the maximum speed of the seat (in cm/sec) and state the time that this first
occurs.
(2 marks)
Solution
From equation  2  A  25 B  0 
maximum magnitude
dx
 25 sin  t
dt
dx
1
 25 when sin  t  1  t 
dt
2
Hence maximum speed is
dx
1
 25 cms 1 and first occurs when t  second
dt
2
Thus greatest speed is 25 cm/s and this first happens when
sin t 1 t 1/2 second

Specific behaviours
defines correct maximum speed using appropriate equation
solves for t using velocity/time equation
30
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
Item 14
Calculator
assumed
14(a)
14(b)
3C
Collision of moving objects
1.12 solve practical problems in three-dimensional geometry using vector
concepts and formulas, and graphical methods where appropriate
(9 marks)
2
 
A small rocket is fired at noon, from position  3  kilometres,
7
 
 3
 
with a constant velocity of v   1  km per minute.
 5
 
 20 
 
A stationary weather balloon is at position  4  kilometres.
 38 
 
It is known that the rocket just misses the balloon.
(a)
Find
(i)
(6 marks)
at what time the rocket is closest to the balloon, to the nearest minute.
Solution
2
Since the rocket is fired from the point  3 ; at time t minutes later it is at P
 7 
 2  3t 
and its position is OP   3  t 
 7  5t 
The relative position of the rocket from the Balloon B is;
 2  3t   20  3t  18 
 3






BP   3  t    4    t  7  and v  1
 7  5t   38  5t  31
5
At time of closest approach BP is perpendicular to the velocity vector of the rocket.
3t  18  3
Hence  t  7   . 1  0
5t  31 5
 9t  54  t  7  25t  155  0
 35t  216  0
t 
216
min
35
Hence closest when t 
216
min  approx 6.17 minutes  i.e. 12 : 06 pm
35
Specific behaviours
defines vector equation for path of the rocket
determines BP in terms of t
uses dot product to check for perpendicularity
Mark
allocation
1
Item
classification
complex
1
1
complex
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
31
uses dot product to solve for t
(ii)
1
complex
the distance the rocket and the balloon are apart at that time (to the nearest m).
Solution
 0.5143 
BP   0.8286 
 0.1429 
BP  0.51432  0.82862  0.14292  0.986
Hence closest distance of approach is 0.986 km approximately
Using CAS:
Specific behaviours
substitutes for t into BP (component form)
evaluates magnitude of BP
32
Mark
allocation
1
Item
classification
complex
1
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
A second rocket is also launched at noon, also with a constant velocity, but is fired from position
3
1  kilometres and aimed so as to collide with the first rocket at exactly 12.07 pm.
 
 4 
(b)
Determine the velocity of the second rocket that will ensure collision takes place at the
required time.
(3 marks)
Solution
 
Let the velocity of the second rocket v    
 
3  7 
At 12.07pm this rocket is at position r  1  7  
 4  7 
 23 
At this time the position of the first rocket   4 
 42 
3  7   23  3  7  23
Hence 1  7     4   1  7   4 
 4  7   42   4  7  42 
i.e.  
20
3
38
,   and  
7
7
7
 20   2.8571
1  
Hence velocity of second rocket is v   3    0.4286  km/minute.
7
 38  5.4286 
Specific behaviours
defines velocity of second rocket in component form
defines vector equation for second rocket
equates both vectors and solves
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
33
Item 15
Calculator
assumed
15(a)
15(b)
3C &
Solving differential equations
3D
Growth and decay
3.2 solve practical problems involving models of growth and decay of the
form
dP
 kP
dt
4.6 solve practical problems involving parametric and differential equations
(variables separable)
(7 marks)
Medical authorities have been notified that 40 people in a town with a population of 10 000 have
fallen ill with swine flu. The disease is known to spread at the daily rate of 0.3% of the remaining
healthy inhabitants. If P is the number of people who have already been infected, then
dP
 0.003(10000  P)
dt
where t denotes the number of days that have elapsed since the initial 40 cases were noted.
(a)
Find P as a function of t.
(4 marks)
Solution
dP
 0.003(10000  P)
dt
dP

  0.003dt

10000  P

ln(10000  P )  0.003t  c

10000  P  P0 e( 0.003t )
t  0 P  40  C  10000  40  9960
Hence P  10000  9960e( 0.003t )
or
Specific behaviours
defines integral by separation of variables
integrates correctly
defines the expression in terms of e
carries through correct solution
34
Mark
allocation
1
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
complex
(b)
If no preventative measures are put in place to limit the spread of the disease, determine on
which day the number of people infected reaches 10% of the population.
(3 marks)
Solution
When 10% of the population is infected P  1000
then 9960e( 0.003t )  9000
 0.003t  ln(
900
) t  33.78 days.
996
Hence 10% of the population becomes infected on the 34thday.
Specific behaviours
substitutes correctly into the equation
carries through to solve correctly
rounds correctly
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
35
Item 16
Calculator
assumed
16
(a)-(c)
3D
Rectilinear motion
4.4 solve practical problems by applying calculus techniques to problems
from various branches of the sciences including rectilinear motion and
marginal cost
(8 marks)
A train moves along a straight track and at time, t hours, is x km from a fixed reference point.
The velocity, v km/h, of the train is given by,
v  5 x  2, where x is a function of t and when t  0 its position is x  1.
(a)
Determine the initial acceleration.
(2 marks)
Solution
Acceleration a  t  
dv
and v  5 x  2
dt
d
5x  2
dt
dx
 5  5v
dt
Given when t  0, x  1  v  7 km/h.
a t 

Hence initial acceleration a  t   5  7  35 km/h2 .
Specific behaviours
uses implicit differentiation to get an expression for acceleration
substitutes correctly into acceleration expression
carries through to correct acceleration
36
Mark
allocation
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
(b)
Obtain an expression for x in terms of t.
(4 marks)
Solution
dx
dx
 5x  2 
 dt
dt
5x  2
1
1
Hence ln(5 x  2)  t  C and t  0, x  1  C  ln 7
5
5
ln(5 x  2)  ln 7  5t
v t  


 5x  2 
 ln 
  5t
 7 
 5 x  2  7 e 5t
x
7 e 5t  2 1 5t
 (7e  2).
5
5
so
Using CAS:
or
Specific behaviours
defines the integral from the differential equation
solves integral correctly
calculates constant of integration
gives exponential form correctly
Mark
allocation
1
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
complex
37
(c)
After how many minutes is the train travelling at 100 km/h? Give your answer correct to the
nearest minute.
(2 marks)
Solution
Solve for t when v  100 km/h
1
98
v  100  x  (100  2)  km
5
5
1 5t
98
(7e  2) 
 7e5t  100
5
5
1  100 
t  ln 
0.532 hours
5  7 
Hence, required time is 0.532 hours; i.e. approximately 32 minutes.
Specific behaviours
chooses velocity equation to obtain x when v = 100
uses the x(t) equation to solve for t
38
Mark
allocation
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
Item 17
Calculator
assumed
Related rates
Simple harmonic motion
4.3 solve related rates problems (including those involving trigonometric
functions)
2
2.1 investigate the differential equation d y  k 2 y  0 and its solutions
dt 2
y(t )  C cos kt  D sin kt  A cos  kt  1   A sin  kt   2 
as models of simple harmonic motion
3D
17(a)
17(b)
17(c)
(11 marks)
Laura is travelling on a Ferris wheel of radius 40 metres, that is that is turning at a constant angular
speed of one revolution every 5 minutes. Initially, Laura’s cabin is at ground level, at point S.
4

0
m
Cabin
N
S
(a)
Shadow
Find the rate as an exact value, in metres per minute, at which the height of Laura’s cabin is
increasing when it is 20 metres above the ground.
(6 marks)
Solution
Let the radius joining the cabin to the centre of the wheel make an angle  with the
d 2
Wheel rotates every 5 minutes 

radians per minute
dt
5
The distance of the cabin below the centre of the wheel is 40cos ,
Hence the distance above the ground z  40(1  cos  )
dz dz d
d


.
 40sin 
dt d dt
dt
1
3
z  20  cos    sin  
2
2
dz
3 2
 40. .
 8 3  43.53 metres per minute
dt
2 5
Specific behaviours
defines appropriate variables
identifies rate of change of θ
defines height above ground in terms of θ
calculates sinθ for z = 20
uses chain rule to calculate appropriate rate of change
accurately carries through calculation

Mark
allocation
1
1
1
1
1
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
complex
complex
complex
complex
39
(b)
If x metres is the displacement of the shadow from point S on the ground, obtain an
expression for x in terms of t, where t is time in minutes. By finding as expression for the
acceleration in terms of x, show that the shadow moves with simple harmonic motion.
(3 marks)
Solution
d 2
2

 d 
dt
5
5
2

t  c and  = 0 and t  0  c  0
5
2
x  40sin   x  40sin
t
5
dx 2
2

 40cos
t
dt
5
5


d2x
2
 2 
 
  40sin 5 t
2
5
dt


2
d2x
 k 2 x which implies simple harmonic motion.
dt 2
Using CAS:
or
Specific behaviours
uses
d
to express θ in terms of t
dt
uses equation in (b) to define x(t)
identifies acceleration equation using 2nd derivative
40
Mark
allocation
1
Item
classification
complex
1
1
complex
complex
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
(c)
The sun is directly overhead and casts a shadow on the ground directly below Laura’s
cabin. Find the speed at which the shadow is moving horizontally, when the cabin is
20 metres above the ground.
(2 marks)
Solution
If metres denotes the distance of the shadow from S then;
x  40sin 
dx dx d
d
1
d 2

.
 40cos 
, when z  20  cos   also

dt d dt
dt
2
dt
5
dx
1 2
 40  
 8 m/min.
dt
2 5
Thus the shadow moves along the ground at a speed of 8 metres/minute.
Mark
Specific behaviours
allocation
defines horizontal measurement in terms of θ
1

calculates horizontal rate of change
1
Mathematics: Specialist—Assessment item bank, Units 3C and 3D Response
Item
classification
complex
complex
41