Mathematics Paper 2 Memorandum

MATHEMATICS PAPER 2 MEMORANDUM

VCAA 2016

Marks: 150 Time: 3 hours

INSTRUCTIONS

1. This question paper consists of 28 pages and an Information Sheet of 2 pages. Please check that your paper is complete.

2. Read the questions carefully.

3. Answer all questions on the question paper and hand this in at the end of the examination.

4. Do not work on loose sheets of paper.

5. Extra space is provided at the end of the paper, should this be necessary.

6. Diagrams are not necessarily drawn to scale.

7. You may use an approved non-programmable and non-graphical calculator, unless otherwise stated.

8. Round off to one decimal place, unless specified otherwise.

9. Ensure that your calculator is in DEGREE mode.

10. All necessary working details must be clearly shown.

QUESTION / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / TOTAL
MAXIMUM / 9 / 12 / 20 / 12 / 10 / 12 / 13 / 10 / 14 / 16 / 15 / 7 / 150
MARK

SECTION A

QUESTION 1

(a) Simplify the following expression:

cos-θtan(180°-θ)cos(θ-90°) (5)

=cosθ.-tanθsinθ
=cosθ.- sinθcosθsinθ
=-1

(b) Evaluate the following expression without using a calculator:

cos69°.cos9°+cos81°.cos21° (4)

=cos69°.cos9°+sin9°.sin69°
or any other correct expansion
=cos60°
=12
9 marks

QUESTION 2

The given data values reflect the masses (in kg) of 20 athletes in the school team:

40 / 47 / 52 / 53 / 55 / 57 / 57 / 58 / 60 / x
63 / 64 / 64 / 65 / 66 / 67 / 67 / 68 / 69 / 73

(a) Determine the value of x if the median of the data is 61,5 kg . (2)

x+632=61,5
x=60

(b) Draw a box-and-whisker diagram of the given data, indicating the necessary values clearly. Use a scale of 2 cm=5 kg. (5)

Q1=56 Q3=66,

less than Q1-(1,5×IQR) or greater than Q3+1,5×IQR.

Determine whether 40kg is an outlier of this data set. (3)

IQR=10,5
Q1-1,5×IQR=56-15,75=40,25
∴ outlier

(d) If 4 kg were added to each data value in the given set, what effect would it have on the resulting (1) mean and (2) standard deviation? (2)

(1) The mean would increase by 4 kg
(2) The standard deviation would remain the same
12 marks

QUESTION 3

(a) A(4;5), B(2:-1) and C(10;3) are given.

BA is produced to D and DC⊥BC.

The circle with centre A passes through E, the midpoint of BA.

Determine:

(1) the coordinates of E. (2)

E(3;2)

(2) the equation of the circle with centre A. (4)

(x-4)2+(y-5)2=r2
(3-4)2+(2-5)2=r2
r2=10
(x-4)2+(y-5)2=10

(3) the angle of inclination of the line BC. (3)

mBC=12
∴tanθ=0,5
θ=26,6°

(4) the equation of the line CD. (3)

mCD=-2
y-3=-2x-10
y=-2x+23

(b) The equation of the circle with centre K is given.

(1) Calculate the coordinates of K and the radius of the circle. (4)

(x+2)2+(y-1)2=10
K-2;1
radius= 10

(2) Hence, determine the equation of the tangent at the point P(1;2) on the circle. (4)

mKP=13
mtan=-3
y-2=-3x-1
y=-3x+5
20 marks

QUESTION 4

(a) Refer to the sketch: Q, V, T, S and R lie on the circle. RV and ST are produced to meet at P. PQ is a tangent to the circle at Q. QS is a diameter.

Complete the table giving the reason for each of the statements. (6)

STATEMENT / REASON
V3=S1 / ∠'s in same segment
Q1=R1 / Tan-chord theorem
T1=R2 / Ext ∠ cyclic quad
Q1+Q2=90° / Tan ⊥ rad
R1+R2=90° / ∠in semi-circle OR tan-chord theorem
T2+R2=180° / Opp ∠'s cyclic quad

(b) AB is a tangent to the circle with centre O. BCOD is a straight line and AE∥BD.

(1) By using the given reason in each case, write a true statement that includes D. (3)

STATEMENT / REASON
D=A3 Pa / Tan-chord theorem
D=A1 Pa / Alt ∠’s, AE∥BD
D=E1 Pa / ∠’s in the same segment

(2) Giving reasons, show that D+C1+C2=90°. (3)

STATEMENT / REASON
A2=90° / ∠in semi-circle
D+C1+C2=90° / Int ∠'s of ∆
12 marks

QUESTION 5

(a) Without using a calculator, determine the value of sin2x if it is given that cosx=-45

and 180°<x<360°. (5)

Sketch or Pythag
y=-3
sin2x=2sinx.cosx
=2-35-45
=2425

(b) If cos62°=p express the following in terms of p without using a calculator:

(1) sin208° (2)

(2) cos31° (3)

(1) sin208°= -sin28°
=-cos62°
=-p
(2) cos62°=2cos231°-1
p+1=2cos231°
p+12=cos31°
10 marks

QUESTION 6

(a) A, B, C and D are points on a circle with centre O and diameter BD.

C1=10° and A2=53°

Calculate, with reasons, the sizes of the following angles:

(1) O1 (4)

(2) C2 (3)

STATEMENT / REASON
(1)
O2=106° / ∠at centre = 2 × ∠at circle
O1=74° / adj ∠'s on a str line
Alternate
A1=37° / ∠in semi-circle
O1=74° / ∠at centre = 2 × ∠at circle
(2)
D2=53° / ∠’s in same segment
C2=53° / ∠’s opp equal sides
Alternate
D2+C2=106° / Int ∠’s of ∆
D2=C2=53° / ∠’s opp equal sides

(b) In ∆AEG it is given that

BD∥FE and CD∥AE

Calculate the values of m and k by referring to the lengths on the diagram. (5)

STATEMENT / REASON
36=m8 / Line ∥ one side;BD∥FE
m=4
12=72+k / Line ∥ one side;CD∥AE
k=12
12 marks

TOTAL SECTION A: 75

SECTION B

QUESTION 7

An athlete’s ability to take and use oxygen effectively is called his / her VO2-max.

Twelve athletes with pre-recorded VO2-max readings ran for one hour. The distances (in kilometres) that they each covered are represented in the table:

VO2-max reading / 20 / 55 / 30 / 25 / 40 / 30 / 50 / 40 / 35 / 30 / 50 / 40
Distance run (km) / 8 / 18 / 13 / 10 / 11 / 12 / 16 / 14 / 13 / 9 / 15 / 12

(a) Draw a scatter plot of the data. (Place VO2-max on the horizontal axis.) (4)

(b) Use the correlation coefficient to describe the correlation between the two sets of data. (2)

r≈0,9
Strong positive linear

A=3,7 B=0,2
y=0,2x+3,7
x;y=(37,1;12,6)
On graph

(d) Use the method of interpolation to predict the distance run if an athlete has a VO2-max value

of 26. (2)

Distance=10.5 km
On graph Type equation here.
OR
13 marks

QUESTION 8

(a) Without using a calculator, determine the general solution of:

cos40°=sinx.cos20°- cosx.sin20° (6)

cos40°=sin(x-20°)
sin50°=sin(x-20°)
x-20°=50°+k.360° / x-20°=130°+k.360°
x=70°+k.360° / x=150°+k.360° k∈Z
Alternate
cos40°=sin(x-20°)
cos40°=cos(110°-x)
110°-x=40°+k.360° / 110°-x=-40°+k.360°
x=70°+k.360° / x=150°+k.360° k∈Z

(b) Prove that = 2 (4)

LHS=sin6A.cos2A-cos6A.sin2Asin2A.cos2A
=sin4Asin2A.cos2A
=2sin2A.cos2Asin2A.cos2A
=2=RHS
10 marks

QUESTION 9

(a) The circle (x+3)2+(y-1)2=9 with centre T is given. Calculate the value(s) of m if the length of the tangent from point A(m;7) outside the circle to the point P on the circumference of the circle is 213. (6)

Centre -3;1
AP2+PT2=AT2 Pythagoras
52+9=m+32+(7-1)2
m2+6m-16=0
m+8m-2=0
m=-8 or m=2

(b) Refer to the diagram alongside and determine the equation of the circle, with centre M, that touches the x-axis at and passes through the point

(8)

M2;y
AM2=BM2 distance
y2=2-42+(y+6)2
y2=4+y2+12y+36
y=-103
(x-2)2+(y+103)2=1009
14 marks

QUESTION 10

(a) The curves of fx=a tanbx and gx=csindx are sketched for x∈0°;180°.

(1) Write down the values of a, b, c and d. (4)

a=2 b=1
c=-3 d=2

(2) Write down the new equation of f if it is translated 45° to the right and 3 units down.

(2)

y=2tanx-45°-3

(b) A is the foot of a vertical tower AD. B and C are two points in the same horizontal plane as A. It is given that ABC=x° and AC=BC=d, while the angle of elevation of D from B is y°.

If AD=h prove that:

h=2dtanycosx (6)

cos(x)=AB/2 d
AB=2d cos⁡(x)
tan(y)=h 2d cosxP
∴h=2dtanycosx
Alternate:
hAB=tany
h=AB.tany
AB sin(180-2x)=dsinx
AB=d.2sinxcosxsinx
=2d.cosx
∴h=2dtanycosx

The line OP is rotated about the y-axis to form a cone.

Total surface area= πr2+πrs

where s is the slant height and r the radius.

Determine the shaded outer surface area of the resulting cone. (4)

s2=32+42
s=5
r=3
Area =πrs
=π35
=47,1 u2
16 marks

QUESTION 11

(a) O is the centre of the circle and OM ⊥AC.

The radius of the circle is 5 cm and BC=8 cm.

Calculate the length of AM (7)

STATEMENT / REASON
AM=MC / ⊥from circle centre to chord
AO=OB=5 / radii
OM=12BC=4 / Midpoint Th
AM=3 / Pythag
Alternate
M1=90° / Given
C=90° / ∠in semi-circle
A is commom
∴∆AMO½½½∆ACB / AAA
AC=6 / Pythag
AM=3 / ⊥from circle centre to chord
Alternate
M1=90° / Given
C=90° / ∠in semi-circle
OM∥BC / corr ∠'s equal
AC=6 / Pythag
AM6=510 / Line ∥ one side ∆
AM=3

(b) KLMN is a trapezium with KL∥NM.

MO=ML and O2=K

Prove:

KL×LM=ON×KN (8)

STATEMENT / REASON
In ∆KLN and ∆ONM / similarity
K=O2 / Given
L1=N1 / Alt ∠’s, KL∥NM
∴ ∆KLN½½½∆ONM / AAA
KLON=KNOM
∴KL×OM=ON×KN
but OM=LM / Given
∴KL×LM=ON×KN
15 marks

QUESTION 12

In the figure P, A, R, C, Q and B are points on the circle.

Prove that BPA+BQC+ARC=360°.

(7)

STATEMENT / REASON
P+C2=180° / Opp ∠’s cyclic quad
Q+A2=180° / Opp ∠’s cyclic quad
R+B2=180° / Opp ∠’s cyclic quad
P+C2+Q+A2+R+B2=540°
But C2+A2+B2=180° / Int ∠’s of ∆
P+Q+R=360°
7 marks

TOTAL SECTION B: 75

VCAA 2016: Mathematics Paper 2 Marking Guidelines Page 25 of 25