Grade 12 June Paper II 2014

ST TERESA’S SCHOOL

MID YEAR EXAMINATION

JUNE 2014

GRADE 12 MATHEMATICS: PAPER 2

(ANALYTICAL GEOMETRY, EUCLIDEAN GEOMETRY, TRIGONOMETRY AND STATISTICS)

Time: 3 hours 150 Marks

Examiner: K MOFFAT Moderator: K MARSH

PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY

  1. This question paper consists of 26 pages including a two page formula sheet. Please check that your paper is complete.
  2. Answer all questions on the QUESTION PAPER.
  3. You may use an approved, non-programmable, and non-graphics calculator, unless otherwise stated.
  4. Answers must be rounded off to two decimal places, unless otherwise stated.
  5. It is in your own interest to show all your working details, to write legibly and to present your work neatly.
  6. Diagrams are not drawn to scale.

“There are no secrets to success. It is the result of preparation, hard work and learning from failure.”

General Colin Powell

SECTION A: [75 Marks]

QUESTION 1: [10]

The ogive curve (cumulative frequency) below represents the finishing times of the 590 runners who completed the RAC Old Parks 10km race on 1 June 2008.

(a)  Estimate in how many minutes a runner would have had to complete the race in order to place in the 15th percentile or better. (2)
From the graph 37 – 40 minutes acceptable answer aAaA

(b)  If a silver medal is awarded to all runners completing the race in under 40 minutes, estimate the number of runners who would receive a silver medal. (3)
Under 40 minutes, from the graph approximately 15 % of the runners. aM
15% of 590=88,5 aM
Answers of between 87 and 89 runners is acceptable. aA

(c)  Draw a box and whisker plot to summarise the data represented on the graph. (5)
aA aA aA aA aA

QUESTION 2: [15]

(a)  If sin2θ-cos2θ=1 and θ∈0 ; 90o, find the value of the following, without the use of a calculator:
(1) cos2θ (2)
cos2θ=cos2θ-sin2θ aM
cos2θ=-sin2θ-cos2θ=-1 aA
(2) θ (1)
2θ=±cos-1-1+k360
2θ=±180o+k360o
θ=90o aA
(3) the value of φ if φ∈0 ; 90o and θ=90O: (5)
sinθ-φ-cosθ-φ=0
sinθ-φ-cosθ-φ=0
sin90o-φ-cos90o-φ=0 aM
−=0 aA
= aM
=1 aM
=45o aA

(b)  Evaluate the following, without the use of a calculator: (7)
sin108otan126o.cos234o.sin324o
sin108otan126o.cos234o.sin324o
=sin72-tan54.-cos54.-sin36 aA aA aA aA
=2sin36cos36-sin54cos54.cos54.sin36 aM
=2cos36-sin54 aM
=-2cos36cos36
=-2 aA

QUESTION 3: [6]

(a)  Sketch the following trigonometric graphs for x∈0 ; 360o on the same set of axes provided below. (4)
y=sinx2 and y=12cosx

(b)  From your graph, write down the value(s) of x for which: (2)
sinx2=12cosx+112
x=180o a A aA

QUESTION 4: [13]

In the diagram below, PQRS is a quadrilateral on the Cartesian Plane with Q(2;10) and S(6;2). The diagonals of PQRS bisect each other at right angles, at M. T is the point of intersection of line PR with the y-axis and P is the x-intercept of line PR.

(a)  Determine the gradient of PR. (3)
mQS=10-22-6 aM
mQS=8-4=-2 aA
∴mPR=12 lines ⊥ aCA

(b)  Show that the equation of PR is given by 2y=x+8. (3)
M2+62 ; 10+22 aM
=M4;6 aA
y-6=12x-4 aM
2y-12=x-4
2y=x+8

(c)  Determine the co-ordinates of R. (4)
20=x+8
x=-8 aM
∴P-8;0 aA
-8+x2=4 0+y2=6
-8+x=8 y=12
x=16
R(16;12) aA aA

(d)  Calculate the size of RTO. (3)
mPR=tanθ
tanθ=12 aM
θ=26,57o
RTO=26,57+90 ext ∠ of a ⊿ aM
RTO=116,57o aA

QUESTION 5: [11]

P(4;-3), Q(-5;0) and R(-3;k) are three points in the Cartesian Plane.

(a)  Determine k if:

(1)  P, Q, and R are collinear. (4)
mPQ=mQR aM
0+3-5-4=k-3+5 aM aA
k=6-9=-23 aA

(2)  PR=7 units. (4)
4+32+k+32=7 aM aA
49+k+32=49 aM
k+32=0
=−3 aA

(b)  If P, Q and R are collinear, determine the equation of the line perpendicular
to PQR at P. (3)
mPQR=-13 aM
m⊥=3 aM
y+3=3x-4
y=3x-15 aA

QUESTION 6: [20]

(a)  Use the diagram below to prove the theorem that states that:
“Opposite angles of a cyclic quadrilateral are supplementary.” (5)
Given: Circle centre O and points A, B, C and D on the circumference of the circle
RTP: B+D=180o
Proof:
Construct AO and OC
O1=2B ∠ at centre=2 × ∠ at circle aA
O2=2D ∠ at centre=2 × ∠ at circle aA
O1+O2=360o ∠s round a point aA
2B+2D=360o aA
B+D=180o aA

(b)  O is the centre of the circle. AE is a diameter. BA is a tangent to the circle at A. BCE is a secant to the circle.
OD=CD=DE

(1)  Prove that A1=A2 (6)
A1=E tanchord theorem aM
E=O1 isos ⊿ aM
C1=90o ∠ in a semi-circle aM
D=90o line from centre to mpt chord aM
∴OD∥AC aA
∴O1=A2 corresp ∠s=OD∥AC aM
∴A1=A2

(2)  Prove that AODB is a cyclic quadrilateral. (4)
A1+A2=90o tan⊥radius aM
D1=90o line from centre to mpt chord aM
∴AODB is cyclic ext ∠=int opp ∠ aA

(3)  Prove that AE.OD=CE.OE. (5)
In ⊿EAC and ⊿EOD
E = E common ∠
OAC = EOD corresp ∠s OD∥AC aM
C1 = D1 3rd ∠ of a ⊿
∴ ⊿EAC ∕∕∕ ⊿EOD AAA aA
∴ AEEO=ECED similar⊿s aA
∴ AE.ED=CE.OE aM
but ED=OD given aM
∴AE.OD=CE.OE

SECTION B: [75]
QUESTION 7: [8]

(a)  Given that cosα+β=cosαcosβ-sinαsinβ, deduce the compound angle formula for sinα+β. (3)
sinα+β=cos90-α+β aA
sinα+β=cos90-α+-β aA
sinα+β=cos90-αcos-β-sin90-αsin-β aA
sinα+β=sinα.cosβ+cosα.sinβ

(b)  Prove that: 2sin5Acos4A-sin9A=sinA. (5)
2sin5Acos4A-sin5A+4A aM
=2sin5Acos4A-sin5Acos4A+cos5Asin4A aM
=2sin5Acos4A-sin5Acos4A-cos5Asin4A aM
=sin5Acos4A-cos5Asin4A aM
=5A-4A aM
=

QUESTION 8: [8]

Three numbers 2, x and y have a mean of 5 and a standard deviation of 6.

Determine x and y.

2+x+y3=5 aM


2+x+y=15


=13− …1 aA


.=2-52+x-52+y-523=6 aM


9+x-52+y-523=6 aA


9+x-52+13-x-52=18 aM


9+x2-10x+25+64-16x+x2=18


2x2-26x+80=0 aA


x2-13x+40=0


x-5x-8=0 aM


=5 =8


=8 =5 aA

QUESTION 9: [10]


In the diagram above, BA and BC are tangents to the base of a right cylindrical silo with A and C the points of contact. D is a point h metres above C and the angle of elevation of D from B is θ. M is the centre of the base that has a radius of r metres. BAC=α.
B, A, C and M are all in the same horizontal plane.

(a)  Find BAM, giving your answer in degrees. (1)
BAM=90o tan⊥radius aA

(b)  Find MAC in terms of α. (1)
MAC=90o-α given aA

(c)  Find AMC in terms of α. (2)
A1=C1=90o-α isos ∆ aM
M=2α ∠s of a ⊿ aA

(d)  Prove that AC=2hcosαtanθ. (6)

tanθ=CDBC aA
=htanθ
⊥ tan⊥radius aM


B=180-2α ∠s of a quadrilateral aM


ACsinB=BCsinA2 sine rule aM


=BCsin180-2αsinα aM


=hsin2αtanθsinα


=h.2sinαcosαtanθsinα aM


=2hcosαtanθ

QUESTION 10: [15]

In the diagram above A(2;3) is the midpoint of the radius OP and lies on the circumference of the smaller circle with centre M and diameter OA. The equation of the straight line BC is 3y+2x=26 with B and C the x- and y-intercepts respectively.

(a)  Find the co-ordinates of P. (2)
P4;6 aA aA

(b)  Find the equation of the larger circle. (2)
42+62=52 aM
x2+y2=52 aA

(c)  Determine the equation of the smaller circle. (5)
M1;32 aA aA
AM2=2-12+3-322=134 aM aA
x-12+y-322=134 aA

(d)  Calculate the equation of the tangent to the circle AOD at D, the point of intersection of the circle and the x-axis. (6)

x-12+y-322=134


x-12+0-322=134 aM


x-12+94=134


x-12=1 aM


x-1=±1


=0 =2


∴2;0 aA


mMD=32-01-2=-32


∴mt=23 tan⊥radius aM


−0=23x-2 aM


=23x-43 aA

QUESTION 11: [13]

In the diagram above, O is the centre of the circle with radius r. CA is a tangent to the circle. BOD is a diameter that extends to the point C, such that OD=DC.

AOB=2θ.

(a)  Prove that AB=2rsinθ. (5)
A1=180-2θ2 ∠s of an isos ⊿ aM
A1=90-θ aA
ABsin2θ=rsin90-θ sine rule aM
=r.2sinθcosθcosθ aM aA
=2

(b)  Determine AC in terms of r. (4)
OA=r given aM
=2 = aM
AC2=2r2-r2 pythagoras aM
=3r aA

(c)  Express BAC in terms of θ and hence determine the area of ∆ABC in
terms of r and θ. (4)

BAC=180-θ tan⊥radius aA
⊿= 12AB.ACsinBAC area rule aM
⊿=12.2rsinθ.3r.sin180-θ aM
⊿= 3r2sin2θ aA

QUESTION 12: [11]

In the above diagram, KT bisects LKM so that K1=K2. Note that H is not the centre of the circle.

(a)  Prove that TL is a tangent to circle LHK. (3)
L1=K1 ∠s in the same segment aM
K1=K2 given aM
∴L1=K2 aA
∴ tanchord thm converse

(b)  Show that TL2=TK.TH (4)
In⊿ KLT and ⊿LHT
L1 = K2 above aM
T = T common aM
LT = H1 3rd ∠ of a ⊿
⊿ KLT /// ⊿LHT AAA aA
∴TLTH=TKTL similar ⊿s aA
TL2=TK.TH

(c)  Show that KH.KT=KL.KM (4)

In⊿KLT and ⊿KHM
K1 = K2 above aM
T = M ∠s in the same segment aM
LT = H2 3rd ∠ of a ⊿
∴ ⊿ /// ⊿KHM AAA aA
∴ KLKH=KTKM similar ⊿s aA
∴.=.

QUESTION 13: [10]

In the above diagram, the diagonal DB of the rhombus BCDE is produced to A.

A is joined to E.

O is the point of intersection of the two diagonals.

(a)  Prove that AE2=AB2+2AB.BO+BE2 (5)
O2=90o diag of a rhombus bisect at right angles aA
AE2=AO2+OE2 pythag aM
AE2=AB+BO2+OE2 aM
AE2=AB2+2AB.BO+BO2+OE2 aM
BO2+OE2=BE2 pythag aA
∴AE2=AB2+2AB.BO+BE2

(b)  Show that AE2=AB.AD+BE2. (5)
BO=OD diags of a rhombus bisect aA
AE2=AB2+2AB.BO+BE2 above aM
AE2=AB2+2AB.12BD+BE2 aA
AE2=AB2+AB.BD+BE2 aM
AE2=ABAB+BD+BE2 aM
AE2=AB.AD+BE2

MATHEMATICS

INFORMATION SHEET

; ;

y=a+bx b=x-xy-yx-x2

Page 19 of 26