Maths P2 Learner Guide July 2011

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MathematicsPaper 2Winter School 20114 July 15 JulyLearners Guide


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Maths Paper 2 Exam Revision Learners GuideWinter School July 2011

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Introduction:

Have you heard about Mindset? Mindset Network, a South African non-profitorganisation, was founded in 2002. We develop and distribute quality and contextuallyrelevant educational resources for use in the schooling, health and vocational sectors.

We distribute our materials through various technology platforms like TV broadcasts,the Internet (www.mindset.co.za/learn) and on DVDs. The materials are madeavailable in video, print and in computer-based multimedia formats.

At Mindset we are committed to innovation. In the last two years, we successfully rana series of broadcast events leading up to and in support of the NSC examinations

Now we are proud to announce our 2011 edition of Matric Exam Revision, which willbegin with our Winter School in July. Weve expanded the broadcast to support you inseven subjects - Mathematics, Physical Sciences, Life Sciences, MathematicalLiteracy, English 1st Additional Language, Accounting and Geography.

During our Winter School, you will get Exam overviews, study tips on each of thetopics we cover, detailed solutions to selected questions from previous examinationpapers, short question and answer sessions so you can check you are on track andlive phone in programmes so you can work through more exam questions with anexperienced teacher.

Getting the most from Winter SchoolBefore you watch the broadcast of a topic, read through the questions for the topicand try to answer them without looking up the solutions. If you get stuck and cant

complete the answer dont panic. Make a note of any questions you have. Whenwatching the Topic session, compare the approach you took to what the teacher does.Dont just copy the answers down but take note of the method used.

Make sure you keep this booklet for after Winter School. You can re-do the examquestions you did not get totally correct and mark your own work by looking up thesolutions at the back of the booklet.

Remember that exam preparation also requires motivation and discipline, so try tostay positive, even when the work appears to be difficult. Every little bit of studying,revision and exam practice will pay off. You may benefit from working with a friend or a

small study group, as long as everyone is as committed as you are. Mindset believesthat the 2011 Winter School programme will help you achieve the results you want.

If you find Winter School a useful way to revise and prepare for your exams,remember that we will be running Spring School from the 3rd to 7th October and ExamSchool from 19th October to 22nd November as well.
http://www.mindset.co.za/learnhttp://www.mindset.co.za/learnhttp://www.mindset.co.za/learnhttp://www.mindset.co.za/learn


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Programme Outline

The Mindset Winter School is designed to focus on two subjects each day. For eachsubject you will find the following sessions:

Examination Overview

This is a 15 minute session that gives details of what you can expect in eachexamination paper. Practical guidelines are also given on how to prepare for theday of the exam.

Topics Tips

In this session you will be given a 15 minutes summary of the key ideas you needto know, common errors and study hints to help you prepare for your exams.

Topic Session

An expert teacher will work through specially selected questions from previousexam papers.

Interactive Q & A

After every topic you will get the chance to test yourself.

Live Phone-in

This is your chance to ask your own questions. So submit your question to theHelp Desk and we might call you back to help you live on TV. All questions yousubmit will be answered within 48 hours as normal.

Winter Broadcast School Schedule


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Topic 1: Transformational and Co-Ordinate Geometry

Question 1

The diagram below shows the points

and

Point A is the midpoint of

PQ. The line AB is perpendicular to PQ and intersects the x-axis at G and the y-axisat B.

1.1 Show that the gradient of PQ is . (1)1.2 Determine the coordinates of A. (2)

1.3 Determine the equation of the line AB. (5)

1.4 Calculate the length of BQ. (3)

1.5 Show that is isosceles. (2)1.6 If is a rhombus, determine the coordinates of R. (3)

[16]


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Question 2

The straight line AB has the equation . Another straight line CD sdrawn to intersect AB at such that the acute angle between AB and CD is .

2.1 Determine the gradient of the line CD. (5)

2.2 Hence, or otherwise, determine the equation of the line CD. (2)[7]

Question 3

3.1 Determine the centre and radius of the circle with the equation . (4)

3.2 A second circle has the equation . Calculate thedistance between the centres of the two circles. (2)

3.3 Hence, show that the circles described in Question 3.1 and Question 3.2intersect each other. (3)

3.4 Show the two circles intersect along the line . (4)[13]


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Question 4

and are vertices of a triangle that lies on the circumferenceof a circle with diameter BD and centre M, as shown in the figure below.

4.1 Calculate the coordinates of M (2)

4.2 Show that lies on the line . (1)4.3 What is the relationship between the line and the circle

centred at M? Motivate your answer. (5)

4.4 Calculate the lengths of AD and AB. (4)

4.5 Prove . (3)4.6 Write down the size of the angle . (1)


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4.7 A circle, centered at a pont Z inside , is drawn to touch sidesAB, BD and DA at N, M and T respectively. Given that BMZN is a kite,calculate the radius of this circle. The diagram is shown below.

(6)

[22]

Question 5

The point lies in a Cartesian plane. Determine the coordinates of P, theimage of P, if:

5.1 P is reflected across the line (1)5.2 P has been rotated about the origin thru in a clockwise direction. (2)


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Question 6 has undergone two transformations to obtain . and are the coordinates of thevertices of

6.1 Describe in words, two transformations of (in the order whichthey occurred), to obtain . (4)

6.2 Write down TWO possible sets of coordinates, the image of H after thefirst transformation. (2)

6.3 Determine: the area of : area of (2)[8]

Question 7

A quadrilateral is transformed to its image in the following way: First, reflect about the line Then, rotate this image thru in a clockwise direction about the origin The second image has a translation od 2 units to the left and three units down

to obtain

Write down the general rule of the transformation of into

. [6]


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Question 8You may NOT use a calculator to answer this question.

8.1 The point is rotated about the origin through an angle of inan anticlockwise direction. Determine and , the coordinates of . (6)8.2 The same rotation sends a point into . Determine the

coordinates of . (4)[10]


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Topic 2 Trigonometry

Question 1If sin 360 cos 120 = p and cos 360 sin 120 = q, determine in terms of pand qthe valueof:

1.1 sin 480 (3)

1.2 sin 240 (3)

1.3 cos 240 (3)

Question 2

Show that : sin2 200 + sin2 400 + sin2 800 = (7)

(Hint : 40 = 60 20 and 80 = 60+ 20)

Question 3

Given: sin =

where 900 2700

With the aid of a sketch and without the use of a calculator, calculate:

3.1 tan (3)

3.2 sin (900+) (2)

3.3 cos 2 (3)

Question 4

Given : f(x) = tan(x -300)

4.1 Sketch the graph of y = f(x) for -900 < x < 900 (3)

4.2 Write down the equation of an asymptote of f (1)

4.3 Describe in words the transformation of f to gif g(x) = tan(300

- x) (2)

Question 5

Given g(x) = 2cos(x 300)

5.1 Sketch the graph of gforx [900 ; 2700]

5.2 Use the symbols A and B to plot the two points on the graph of gfor which cos(x - 300) = 0,5. (2)

5.3 Calculate the x-coordinates of the points A and B. (3)

5.4 Write down the values of x, wherex [900

; 2700

] and g(x) = 0. (2)5.5 Use the graph to solve for x,x [900 ; 2700] and g(x) < 0. (3)


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Topic 3 Trigonometry Applications

Question 1

1.1 Prove that1

sin(45 ).sin(45 ) cos 2

2

(5)

1.2 Hence determine the value of sin75 .sin15 (3)

Question 2Determine the general solution of the following equation.

2sin 2 2sin cos cos 0 x x x x

Round off your answers to one decimal place where appropriate. (7)

Question 3

3.1 Show that the coordinates of P/

, the image of P(x; y) rotated about the originthrough an angle of 135, in the anti-clockwise direction, is given by

(4)

3.2 M/

is the image of M(2 ; 4) under a rotation about the origin through 135, in the

anti-clockwise direction. Determine the coordinates of M/

, using the results(2)


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Question 4A piece of land has the form of a quadrilateral ABCD with AB 20m,

BC 12m, CD 7m and AD 28m . B 110 . The owner decides todivide the land into two plots by erecting a fence from A to C.

4.1 Calculate the length of the fence AC correct to one decimal place. (2)

4.2 Calculate the size of BAC correct to the nearest degree. (2)

4.3 Calculate the size of D , correct to the nearest degree. (3)

4.4 Calculate the area of the entire piece of land ABCD, correct to one decimalplace. (3)

Question 5

Thandi is standing at point P on the horizontal ground and observes two poles, ACand BD, of different heights. P, C and D are in the same horizontal plane. From P theangles of inclination to the top of the poles A and B are 23 and 18 respectively.Thandi is 18 m from the base of pole AC. The height of pole BD is 7 m.

Calculate, correct to TWO decimal places:5.1The distance from Thandi to the top of pole BD (2)

5.2The distance from Thandi to the top of pole AC (2)

5.3The distance between the tops of the poles, that is the length of AB, if ABP =42(4)

110


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sinDE

sin( ) cos

b

2000 metres, 43 and 27b

Question 6In the diagram below A, B and C are three points in the same horizontal plane. D isvertically above B and E is vertically above C. The angle of elevation of E from D is .F is a point on EC such that DF || BC.

BAC , ACB and AC metresb

6.1 Prove that: (6)

6.2 Calculate DE if (3)


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Solutions Topic 1

Question 1

1.1

1.2 1.3

Equation of AB is

Equation of AB is OR

Equation of AB is Equation of AB is

1.4 B is the point

1.5

is isosceles

1.6 If is a rhombus then A is the midpoint of BR.Let the coordinates of R be

OR

By symmetry: BQ x moves from 0 to 4 (4 units up) PR x moves from 0 to 4BP y moves from -3 to 0 (3 units up) PR y moves from 2 to 5


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Question 2

2.1 AB is defined as which can be written as: Let angle of inclination of AB be

.

Let angle of inclination of CD be . Gradient of CD

2.2 Equation of CD is

Equation of CD is Question 3

3.1 The centre is and the radius is

3.2 Centre of second circle is Distance between centres is

3.3 Sum of radii Sum of radii > distance between centres circles overlap

3.4 Equation of second circle:

Let be either of the two points of intersection, then: Solving simultaneously by subtracting:

Both points of intersection lie on this line

is the equation of the common chord.


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Question 4

4.1 Midpoint BD 4.2

A lies on the line4.3

to the lineThe line is a tangent to the circle at A.

4.4

4.5

4.6 4.7 Let the radius of the circle TNM be r

properties of a kite

TZNA is a square

Question 55.1

5.2


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Question 66.1

So swop x and y and halve

Reflection across followed by contraction by

6.2 6.3 Length

Length 1

Question 77.1 For anti-clockwise rotation:

7.2 .equation 1

. Equation 2Substitute equation 2 into equation 1


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Solution Topic 2 Trigonometry

Question 1


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Question 2

Question 3

3.1 sin > 0 is in the second quadrant.

y = 17 and r = 17, using Pythagoras, x =-15,

tan = -

3.2 sin (900+) = cos

= -

3.3 cos 2 = 1 2sin2

= 1-2(

)2

=


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Question 44.1

4.2 x = -600

4.3 tan(300- x) = - tan(x -300)Reflection about the x-axis


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Question 55.1

5.2

5.3


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5.4

5.5


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Solutions Topic 3: Trigonometry Applications

Question 11.1

2 2

sin(45 ) .sin(45 )

sin 45 cos cos45 sin sin 45 cos cos45 sin

2 2 2 2cos sin cos sin

2 2 2 2

2 2(cos sin ) (cos sin )

2 2

2(cos sin )(cos sin )

4

1 (cos sin )2

1cos2

2

1.2sin 75 .sin15

sin(45 30 ).sin(45 30 )

1cos2(30 )

2

1 1 1 1cos602 2 2 4

Question 22

2

sin 2 2sin cos cos 0

2sin cos 2sin cos cos 0

2sin (cos 1) cos (cos 1) 0

(cos 1)(2sin cos ) 0

cos 1 or 2sin cos

sin 1

cos 2

tan 0,5

x x x x

x x x x x

x x x x

x x x

x x x

x

x

x

0 360 153,4 360

180 360 333,4 360

x k x k

x k x k

OR

0 180 153,4 180 x k x k


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OR

180 180 333,4 180

OR

180 360 45 180

x k x k

x k x k

Question 3

3.1 3.2


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Question 44.1

2 2 2

2

AC (12 ) (20 ) 2(12 )(20 )cos110

AC 708,1696688

AC 26,6

m m m m

m

4.2sinBAC sin110

12 26,6

12 sin110sinBAC

26,6

sin BAC 0,4239214831

BAC 25

m m

m

4.32 2 2 (26,6 ) (7 ) (28 ) 2(7 )(28 ) cos D

392cosD 125,44

cosD 0,32

D 71

m m m m m

4.4

2

Area ABCD

1 1(12 )(20 )sin110 (7 )(28 )sin 71

2 2

205,4

m m m m

m

110


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Question 55.1

5.2

5.3

Question 66.1

BC

sin 180 ( ) sin

BC

sin( ) sin

sinBCsin( )

But BC DF

sinDF

sin( )

DFNow cos

DE

DFDE

cos

sinDE

sin( ) cos

b

b

b

b

b

6.2

2000sin 43DE

sin79 cos27

DE 1559,50m

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Maths P2 Learner Guide July 2011