maths p1 learner guide july 2011

8/6/2019 Maths P1 Learner Guide July 2011

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MathematicsPaper 1

Winter School 20114 July – 15 July

Learner’s Guide



Mathematics P1 Exam Revision Learner’s 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 andcontextually relevant educational resources for use in the schooling, health andvocational sectors. We distribute our materials through various technology platformslike TV broadcasts, the Internet (www.mindset.co.za/learn ) and on DVDs. Thematerials are made available in video, print and in computer-based multimediaformats.

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. We’ve expanded the broadcast to support youin seven subjects - Mathematics, Physical Sciences, Life Sciences, MathematicalLiteracy, English 1 st 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 topic

and try to answer them without looking up the solutions. If you get stuck and can’tcomplete the answer don’t panic. Make a note of any questions you have. W henwatching the Topic session, compare the approach you took to what the teacherdoes. Don’t 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 ora small study group, as long as everyone is as committed as you are. Mindsetbelieves that the 2011 Winter School programme will help you achieve the resultsyou 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 3 rd to 7 th October andExam School from 19 th October to 22 nd November as well.

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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 OverviewThis 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 TipsIn 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 & AAfter every topic you will get the chance to test yourself.

Live Phone-inThis 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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Question 6Mpho starts a five year savings plan. At the beginning of the month he depositsR2000 into the account and makes a further deposit of R2000 at the end of thatmonth. He then continues to make month end payments of R2000 into the accountfor the five year period (starting from his first deposit). The interest rate is 6% perannum compounded monthly.

6.1 Calculate the future value of his investment at the end of the five yearperiod. (4)

6.2 Due to financial difficulty, Mpho misses the last two payments of R2000. Whatwill the value of his investment now be at the end of the five year period?

(4)Question 7Simphiwe takes out a twenty year loan of R100 000. She repays the loan by meansof equal monthly payments starting three months after the granting of the loan. Theinterest rate is 18% per annum compounded monthly.

7.1 Calculate the amount owing two months after the loan was taken out bySimphiwe. (2)

7.2 Calculate the monthly repayments. (4)

Question 8A clothing company manufactures white shirts and grey trousers for schools.

A minimum of 200 shirts must be manufactured daily.

In total, not more than 600 pieces of clothing can be manufactured daily.It takes 50 machine minutes to manufacture a shirt and 100 machine minutes tomanufacture a pair of trousers.There are at most 45 000 machine minutes available per day.

Let the number of white shirts manufactured in a day be x.Let the number of pairs of grey trousers manufactured in a day be y.

8.1 Write down the constraints, in terms of x and y, to represent the aboveinformation. (You may assume: 0 ,0 y x ) (3)

8.2 Use the attached graph paper to represent the constraints graphically. (3)

8.3 Clearly indicate the feasible region by shading it. (1)

8.4 If the profit is R30 for a shirt and R40 for a pair of trousers, write down theequation indicating the profit in terms of x and y . (2)

8.5 Using a search line and your graph, determine the number of shirts and pairsof trousers that will yield a maximum daily profit. (2)




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Topic 2: Calculus 4 July

Question 1Differentiate f from first principles :

1.1 f(x) = (4)

1.2 f(x) = -2x + 3 (5)

Question 2Use the rules of differentiation to determine

2.1 if y = (2- 5x)² (2)

2.2 if y = x2 – (2)

Question 3The graph below represents the functions f and g with 3 + c x – 2 andg(x) = x – 2. A and ( – 1; 0) are the x -intercepts of f. The graphs of f and g intersectat A and C.

3.1 Determine the coordinates of A. (1)

3.2 Show by calculation that a = 1 and c = – 3. (4)

3.3 Determine the coordinates of B, a turning point of f . (3)




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3.4 Show that the line BC is parallel to the x- axis. (7)

3.5 Find the x -coordinate of the point of inflection of f. (2)

3.6 Write down the values of k for which

will have only ONE root. (3)

3.7 Write down the values of x for which

(1)

Question 4A tourist travels in a car over a mountainous pass during his trip. The height abovesea level of the car, after t minutes, is given as s (t) = metres.The journey lasts 8 minutes.

4.1 How high is the car above sea level when it starts its journey on themountainous pass? (2)

4.2 Calculate the car's rate of change of height above sea level with respect totime, 4 minutes after starting the journey on the mountainous pass. (3)

4.3 Interpret your answer to QUESTION 4.2. (2)

4.4 How many minutes after the journey has started will the rate of change ofheight with respect to time be a minimum? (3)

Topic 3: Functions & Graphs 11 July

Question 1 Function notation If and , determine . (3)

Question 2 Hyperbola

Given: 2.1 Write down the equations of the asymptotes of . (2)

2.2 Calculate the coordinates of the x- and y-intercepts of . (3)

2.3 Sketch .Show all intercepts with the axes and the asymptotes. (3)[8]




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Question 3 Quadratic and linear function – (DoE P1 March 2010)

The graphs of and are sketched below. and intersect in D and B. A and B are the x -intercepts of .

3.1 Determine the coordinates of A and B. (4)

3.2 Calculate a , the x -coordinate of D. (4)

3.3 is a point on the graph of , where . ST is drawn parallelto the y -axis with T on the graph of . Determine ST in terms of x . (2)

3.4 Calculate the maximum length of ST. (2)[12]




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Question 4 Exponential function The graph of is drawn below.

4.1 Write in the form (1)

4.2 Sketch the graphs of and (4)

4.3 Use your graphs to solve for x if . (2)[7]

0

y

x

1

0

y

x




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Question 5 Quadratic and Hyperbolic function

Sketched below are the graphs of and .P and Q are the points of intersection of and .

5.1 Show that the coordinates of P and Q are and respectively.(6)

5.2 An axis of symmetry of the graph of is a straight line defined as ,where . Write down the equation of this straight line in theform (2)

5.3 Determine the equation of in the form (2)

5.4 Show algebraically that (3)

[13]




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Question 6 Trigonometric Graphs

Given 6.1 Sketch the graph of for (3)

6.2 Write down the equation of an asymptote of . (1)

6.3 Describe in words the transformation of to if (2)[6]


The graphs of and are sketched below for

7.1 Write down the range of . (1)

7.2 If , write down the coordinates of B. (3)

7.3 Write down the period of . (2)

7.4 Write down a value of for which is a maximum. (2)[8]




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Solutions Topic1Question 1

Question 22.1 1 1 2 3 78

2

7

7

7

1 1 1 1 1.......

2 2 2 2 2

1 112 2 1 1 127

S 1 11 2 128 12812

k

k

expansioncorrect formulacorrect substitution

127128

(4)

2.2 1 1 2 3

2

1 1 1 1............

2 2 2 2

12S 1

1

1 2

k

k

correct formulaanswer

(2)

Question 33 The layers are stacked as follows:

30 ; 29 ; 28 ; 27 ; ..............;1 This forms an arithmetic sequence.

30

30

30S 2(30) (30 1)( 1)

2S 465

There are a maximum of 465 cans that can bestacked in this way.

30 ; 29 ; 28; 27 ;..... ;1 correct formulacorrect substitution465 cans

(4)

1.1 11 13;

32 64

answers(2)

1.2 The numerators represent an arithmeticsequence:T 3 ( 1)(2)

T 3 2 2

T 2 1

n

n

n

n

n

n

The denominators represent a geometricsequence:

1T (2)(2)

T 2

nn

n

n

The general term for the given sequence is:2 1

T2

n nn

3 and 2a d T 2 1n n 2 and 2a r T 2 n

n

2 1

T2

n nn

(5)




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Question 44.1 3; ;10 ; ; 21; ........

(10 ) ( 3) 1

10 3 1

2 126

(21 ) ( 10) 1

21 10 1

2 30

15

a b

a a

a a

aa

b b

b b

b

b

(10 ) ( 3) 1a a 6a

(21 ) ( 10) 1b b

15b (4)

4.2

2

3 ; 6 ;10 ;15 ; 21;........

2 1

12

3 3

13 3

2

3

23

1 33

2 21

1 3T 1

2 2n

a

a

a b

b

b

a b c

c

c

n n

3; 6 ;10;15; 21; ........ 1

2a

3

2b

1c 21 3

T 12 2n n n

(5)

4.3 2 2

2 2

2 2

2

2

1 3 1 31 ( 1) ( 1) 1

2 2 2 2

1 3 1 3 31 ( 2 1) 1

2 2 2 2 21 3 1 1 3 3

1 12 2 2 2 2 2

4 4

( 2)

n n n n

n n n n n

n n n n n

n n

n

1T Tn n expanding

2 4 4n n 2( 2)n

(4)




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Question 5

0,92

A P(1 )

90 000 200 000(1 0,08)

90 000 200 000(0,92)90 000

(0,92)200 000

0, 45 (0,92)

log 0, 45

9,576544593

n

n

n

n

n

i

n

n

It will take approximately 9 years 7 months todepreciate to R90 000.

correct formulacorrect substitutionuse of logsvalue of n

(4)

Question 66.1

61

(1 ) 1F

0,062000 1 1

12F

0,0612

F R142 237,76

n x i

i

correct formula

0,0612

value of n answer

(4)

0T 1T 2T 60T3T0,06

0,00512

0T 1T 2T 58T3T0,06

0,00512

59T 60T




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6.2

59

2

(1 ) 1F

0,062000 1 1

12 0,06F . 1

0,06 1212

F R138 227,76

n x i

i

correct annuityformula

59n

2

0,061

12

answer (4)

Question 7

7.1

2

2

At T :

A 100 000 1,015

A R103 022,50

correct formulaanswer

(2)

7.2

238

1 (1 )P

1 (1,015)103 022,50

0,015

R1591,35

n x i

i

x

x

correct annuityformula

238n P 103 022,50 answer

(4)

0T 1T 2T 240T3T

0,180,015

12




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(300;300)

Question 88.1 200 x 8.2 See diagram

600y x

4500010050 y x 8.3 See diagram

8.4 30 40 P x y

40 30 P

3 P4 40

y x

y x

3 cuts on y -axis

4 cuts on x -axis

8.5 Search line

Maximum at (300;300)




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Solutions Topic 2: Calculus

1.1 f(x) = –

= –

=

= x

= -

= -

1.2 f / (x) = –

=

=

=

=

=

2.1 y = (2- 5x)²y = 4 – 20x + 25x 2

= -20 + 50x

2.2

3.1 For A g(x) = x-2 = 0 (x-intercept of g)0 = x – 2x = 2A (2 ; 0)

3.2 f(-1) = 0 : -a + c = 2

f(2) = 0 : 8a – 2c = 2a = 1 and c = -3




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3.3 f’(x) = 0 3x 2 - 3 = 0x 2 – 1 = 0x = 1 , x = -1

B (1;-4)

3.4 x -2 = x 3 – 3x -20 = x 3 – 4x0 = x (x 2 – 4)0 = x (x -2)(x +2)x = 0 , x = 2 or x = -2

x C = -2 , y C = (-2) 3 – 3(-2) – 2 = -4C(-2;-4)

MBC = —

= 0BC is parallel to the x - axis.

orFollowing from C( – 2 ; – 4), B and C have the same y – coordinate, y= – 4.BC is parallel to the x -axis.

3.5 f’’(x) = 0

6x = 0x = 0

3.6 k < -4 or k > 0

3.7 f’’(x) < 0-1 < x < 1or

3(x 2 – 1) < 0if (x – 1)(x + 1) < 0

-1 < x < 1

4.1 S(0) = 5(0) 3- 65(0) 3 + 200(0) 100= 100 metres

4.2 S(t) = 5t 3 – 65t 2 + 200t + 100

S / (t) = 5t 2 – 65t + 200

S / (t) = 5(4) 2 – 65(4) + 200

= -80 metres per minute




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4.3 The height of the car above sea level is decreasing at 80 metres per minuteand the car is travelling downwards hence it is a negative rate of change

Or The vertical of the car at t = 4 is 80 metres per minute velocity

4.4 S / (t) = 15t 2 – 130t + 200S // (t) = 30t – 1300 = 30t – 130130 = 30tt = 4,33

Solutions Topic 3: Functions & Graphs

Question 1 Function notation (DoE P1 March 2010 Q 1.3)

Question 2 Hyperbola (DoE P1 March 2010 Q 5)

2.1 and

2.2 For y-intercept; let

For x-intercept; let

2.3

1

31 1

y




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Question 3 Quadratic and linear function (DoE P1 March 2010 Q 6)

3.1 x-intercepts:

or and

3.2 At D:

or

3.3

3.4 Maximum length at

Max length is: ORMaximum occurs at

Max length is:

Question 4 Exponential function – (DoE P1 March 2010 Q7)

4.1 Function: Inverse:

4.2

4.3

1 3

0

y

x

1




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Question 5 Quadratic and Hyperbolic function – (DoE P1 Nov 2009 Q6)

5.1 By substituting the points into both and to see ifthey are common.

the point P lies on and .

the point P lies on and .OrAt the points of intersection:

or or

Substituting into

5.2 For , . For straight line .Passes thru point

5.3 Inverse:

5.4

RTP:

LHS: RHS:




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6.1

6.2

6.3 Reflection about the x-axis

Question 7 Trigonometric Graphs – (DoE P1 Nov 2009 Q7)

7.1 Various correct forms of notation are acceptable

7.2 x-value is to the left of .

7.3 Period of

7.4

maths p1 learner guide july 2011

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