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GAUTENG DEPARTMENT OF EDUCATION

DIRECTORATE:EXAMINATIONS AND ASSESSMENT

GUIDELINE DOCUMENT FORSCHOOL BASED ASSESSMENT

MATHEMATICS

GRADE 12

2017

1

All assessment activities should be done during contact time and under controlled examination circumstances, unless stated otherwise. Home assignments should include an accountability test or observation sheet.

All schools must comply with the requirements of the assessment activities as prescribed by the Curriculum and Assessment Policy Statements (CAPS). Schools have the following options:

EITHER GDE-set School Based Assessment activities. The Programme of Assessment

will make up 25% of the total mark for Mathematics and comprises seven tasks that are internally assessed.

OR Own School Based Assessment activities (with the proviso that a 70% pass

rate in Mathematics in the previous year was attained).

The conditions if setting own School Based Assessment activities are:

The School Based Assessment activities must cover at least 75% of the year’s learning programme per subject.

The School Based Assessment activities must replicate the GDE-set School Based Assessment activities in length of time, standard, mark allocation and format.

In the event that a school is setting their own tasks the HOD/Subject Head must moderate the School Based Assessment activities and the Subject Advisor must do quality assurance. The moderator must use Annexure A to record their comments on the standard of the School Based Assessment activities. This annexure should also be included in the educator’s file.

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ANNEXURE A

PRE-MODERATION OF SBA ACTIVITIES SET AT SCHOOL LEVEL

MATHEMATICS

DISTRICTSUBJECTGRADENAME OF SCHOOLNAME OF EDUCATOR (S)NAME OF HODNAME OF MODERATORNAME OF SUBJECT ADVISORDATE

MODERATIONFRONT PAGE YES NO COMMENTName of schoolNames of moderator and examinerTime allocationTotal markSubject, e.g. Mathematics or Mathematical LiteracyGrade, e.g. Grade 10 or Grade 11 or Grade 12Assessment activity, e.g. Assignment or Investigation or ProjectDate, e.g. June 2017Are the instructions to candidates clearly specified and unambiguous?REST OF THE ACTIVITY YES NO COMMENTSAll pages numberedMark totals indicated correctly per subsectionsMark totals indicated correctly per questionCorrelation between mark allocation, level of difficulty and time allocationIs “please turn over” indicated?Is the assessment activity complete with grid, memorandum, and diagram sheets?LAYOUT OF THE ACTIVITY YES NO COMMENTSIs the appearance and typing consistent? e.g. font type and sizeAre sketches clear?Are sketches labelled and/or numbered?STANDARD OF ASSESSMENT TASK YES NO COMMENTSDoes the task correspond with the programme of assessment?Is there a verbatim reproduction of questions from previous SBA activities?Are questions ordered from easy to difficult, e.g. Level 1 to Level 4 (different cognitive levels)?Are the subsections grouped by topics?Are questions concise and to the point (not ambiguous)?Are the assessment standards appropriately linked and integrated?Are the questions compliant with CAPS?Is the mark allocation/weighting for the task in accordance with CAPS?ASSESSMENT TOOLS YES NO COMMENTSAre the assessment tools e.g. rubric, memoranda, checklists, etc. for the assessment task included?Are the tools on standard?

3

AREAS OF GOOD PRACTICE

CHALLENGES

RECOMMENDATIONS/FOLLOW-UP

YES NOThe SBA activity is approved.The SBA activity is provisionally approved and requires some adjustments.The SBA activity is not approved and must be resubmitted on the following date:

____________________ ___________________ ___________EDUCATOR SIGNATURE DATE

____________________ ___________________ ___________HOD/ SUBJECT HEAD SIGNATURE DATE(MODERATOR)

____________________ ___________________ ___________DISTRICT FACILITATOR SIGNATURE DATE

Content

4

s1. ASSIGNMENT

2. PROJECT

3. INVESTIGATIONS

4.EXEMPLAR TESTS

5

ENGLISH VERSION

6

ASSIGNMENT

QUESTION 1

1.1. A quadratic pattern has a second term equal to 1, a third term equal to −6 and a fifth term equal to −14 .

1.1.1. Calculate the second difference of this quadratic pattern. (5)

1.1.2. Hence, or otherwise, calculate the first term of the pattern. (2)

1.2. 1 ;3;5 are the first three terms of the first differences of a quadratic sequence. The 7th term of the quadratic sequence is35 .

1.2.1. Determine the 6th and 5th terms of the quadratic sequence. (4)

1.2.2. Determine the nth

term of the quadratic sequence. (5)

[16]

QUESTION 2

2.1. Given the sequence: 2 ;5; 8; . . .

2.1.1. If the pattern continues, then write down the next TWO terms. (1)

2.1.2. Prove that none of the terms of this sequence are perfect squares. (5)

2.2. Consider the sequence of numbers: 2 ;5;2 ;9 ;2 ;13 ;2;17 ; . ..

Calculate the sum of the first 100 terms of the sequence. (5)

2.3. A large quantity of waste material contains metal. At the successive passes through a recovery process the mass of the metal recovered is:

32kg at the first pass; 24kg at the second pass; 18kg at the third pass; 13, 5kg at the fourth pass, and so on, to form a geometric sequence.

2.3.1. How much metal would be recovered at the eighth pass? (Round off the answer to TWO decimal places) (3)

2.3.2. Find the total mass of metal recovered in the first eight passes.(Round off the answer to TWO decimal places) (4)

2.3.3. In order to be economical at least 2, 4kg of metal must be recovered in any pass. Show that the eleventh pass would be uneconomical. (3)

7

[21]

QUESTION 3

3.1. Determine the time taken, in years, for a sum of money to double if the interest rate is 12, 64% p.a. compounded half-yearly. (4)

3.2. Angela wants to buy a house for R700 000. She has a deposit of R50 000 and takes out a loan for the balance at a rate of 18% p.a. compounded monthly.

3.2.1. How much money must Angela borrow from the bank? (1)

3.2.2. Calculate the monthly payment if she wishes to settle the loan in 15 years. (4)

3.2.3. Angela later won the lottery and wished to settle the loan after the 50th payment. What is the outstanding balance? (4)

[13]

TOTAL: [50]

8

ASSIGNMENT MEMORANDUM

QUESTION 1

1.1.1. The 2nd , 3rd , 4th and 5th terms are: 1 ;−6 ;T 4 ;and−14

First differences are: −7 ;T 4+6 ;and−14−T 4

∴d=T3−T 2=T 2−T 1∴ (−14−T 4)−(T 4+6 )=(T 4+6 )−(−7 )

∴−20−2T 4=T 4+13∴−33=3T 4∴T 4=−11d=−11+13=2or−20−2 (−11 )=2

OR

T 5−T2=(T 5−T 4 )+(T 4−T 3)+ (T 3−T 2 )∴−15=(−7+2d )+(−7+d )+ (−7 )∴−15=−21+3 d∴6=3 d∴ d=2

√ −7

√ T 4+6

√ −14−T 4

√ settingup equation

√ answer (5)

√ −7√ −7+d√ −7+2d


√ answer (5)

1.1.2.

∴T 1=1−(−9 )∴T 1=10

OR

2 a=2∴a=1

√ method

√ T 1=10 (2)

9

5 a+b=T 3−T2

∴5 (1 )+b=−7∴b=−7−5=−12

4 a+2 b+c=T2 or 9a+3 b+c=T3

∴4 (1 )+2 (−12 )+c=1∴ c=21

⇒T n=n2−12 n+21⇒T 1=(1 )2−12 (1 )+21=10

√ method

√ T 1=10 (2)

1.2.1. T n=2n−1 for the first difference sequenceT 6=2 (6 )−1=11∴T 6 of original seq .=35−11=24

T 5=2 (5 )−1=9∴T 5 of original seq .=24−9=15

√ 6th termof 1st diff seq .

√ 6th termof quadr . seq .

√ 5th term of 1st diff seq .

√ 5th term of quadr . seq .

(4)

1.2.2. 2 a=2∴a=1c=0

T n=an2+bn+c35=1 (7 )2+b (7 )+0−14=7 b∴b=−2⇒T n=n2−2n

OR

36 a+6b+c=24→ (1 )25 a+5 b+c=15→ (2 )49 a+7 b+c=35→ (3 )(1)−(2) :11a+b=9→ (4 )(3 )−(1 ):13 a+b=911→ (5 )(5 )−( 4 ):2 a=2

∴a=1⇒b=−2⇒T n=n2−2 n

√ value of a

√ value of c

√ T 7=35

√ value of b√ answer (5)

√ 11a+b√ 13a+b

√ value of a√ value of b

10

√ answer (5)

QUESTION 2

2.1.1. 11;14 √ answer (both terms) (1)

2.1.2. T n=3 n−1Nowif 3 n−1=n2

Then n2−3 n+1=0

∴n=3±√52 ⇒ Δ≠a perfect square

⇒nis irrationalBut nmust be natural∴No termis a perfect square. .. .. . .. .. . ..(sin ce Δ≠a perfect square )

√√ for nth term

√ equating n th termton2

√ value of n√ for concluding n is

irrational and fordeducing n isnatural

(5)2.2. (2+2+. ..+2 )

for 50 terms+(5+9+13+.. . )

for 50 terms

=∑i=1

50

(2 )+∑i=1

50

(4 i+1 )

¿2 (50 )+502

[2 (5 )+49 (4 ) ]¿100+25 (10+196 )¿100+5 150¿5 250

√ (2+2+. ..+2 )

for 50 terms

√ (5+9+13+.. . )

for 50 terms

√ settingup∑ exp ression

√ substitution

√ answer (5)

2.3.1. a=32 ;r=34

;n=8

T 8=32( 34 )

7

=2187

512=4 ,27

√ value of r

√ substitution

√ answer (3)

11

2.3.2.Sn=

a (1−r n)1−r

S8=32[1−( 3

4 )8]

1− 34

=128 [1−(34 )8 ]

¿115 , 19

√ formula

√ substitution

√ answer

√ correct rounding−off (4)

2.3.3.T 11=32( 3

4 )10

=1 ,80¿2,4

Hence, the 11th pass is uneconomical.

√ substitution

√ value of T 11

√ conclusion (3)

QUESTION 3

3.1. A=P (1+i )n

∴2 x=x (1+0 ,126 42 )

2n

∴2=(1 ,063 2 )2 n

∴ log2=2n log (1 ,063 2 )

∴2 n=log 2log (1,063 2 )

∴2 n=11 ,310 552 63∴n=5 ,66 years .

√ formula

√ substitution

√ log form

√ answer (4)

3.2.1. Loan Amount=R 700 000−R 50 000=R 650 000 √ answer (1)

3.2.2.Pv=

x [1−(1+i )−n ]i

∴650 000=x [1−(1+0 ,18

12 )−180]

0 , 1812

∴9750=x [1−(1+0 ,1812 )

−180]∴ x=R 10 467 ,74

√ formula

√ substitution

√ simplification

√ answer

12

(4)3.2.3. ∴ A=P (1+i )n

=650000 (1+0 , 1812 )

50

¿650 000 (2 ,105 242 421 )¿R 1368 407 ,57

∴Fv=x [ (1+i )n−1 ]i

¿10 467 , 74 [(1+0 ,18

12 )50−1]

0 , 1812

¿ R 771292 , 69

∴Balance=R 1368 407 ,57−R 771292 ,69=R 597 114 , 88

√ formulae

√ substitution

√ simplification

√ answer (4)

PROJECT: A PRACTICAL APPLICATION OF DIFFERENTIAL CALCULUS

INSTRUCTIONS TOTAL: 501. Answer all the questions. 2. Clearly show all calculations you have used in determining your answers. 3. Round answers off to TWO decimal places, unless stated otherwise. 4. Number your answers correctly according to the numbering system used in this

question paper. 5. Write neatly and legibly, 6. Sketch the containers according to the given specifications. 7. Mathematical methods and formulae need to be used to plan and sketch the containers. 8. All calculations and planning of the side lengths and surface areas must be neatly and

clearly presented in writing and sketches.

CONTAINERS A: A container with a rectangular base B: A container with a circular base C: A container with a triangular base

SPECIFICATIONS Each container must hold exactly one litre of liquid. Each container must have a minimum surface area. The surface area of each container must include the

lid.

13

The length of the rectangular base must be twice the breadth.

The triangular container must have an equilateral base.

FURTHER COMPARISON Apart from your conclusion based on the three options, what other shape of soft drink container would you use in the manufacturing of soft drink cans? Give a reason for your answer.

HINT: The shape in question would be the most economical to manufacture but may not be the most practical choice.

RUBRIC

CRITERIA MAXIMUM MARK

MARKS AWARDED A B C

Correct mathematical formulae 3 x 3 Correct calculations: Measurements of bases 4 x 3 Height of the containers 2 x 3 Logical reasoning and presentation 3 x 3 Submitting on time 2 Conclusion of the least material needed 1 x 3 Final, further comparison 1 x 3

PROJECT MEMORANDUM: A PRACTICAL APPLICATION OF DIFFERENTIAL CALCULUS

14

INVESTIGATION 1: FUNCTIONS AND INVERSES

INSTRUCTIONS 1. Answer all the questions. 2. Clearly show all calculations you have used in determining your answers. 3. Round answers off to TWO decimal places, unless stated otherwise. 4. Number your answers correctly according to the numbering system used in this

question paper. 5. Write neatly and legibly. PART 1: WHICH RELATIONS CONSTITUTE FUNCTIONS? One-to-one relation: A relation is one-to-one if for every input value there is only one

output value. Many-to-one relation: A relation is many-to-one if for more than one input value

there is one output value. One-to-many relation: A relation is one-to-many if for one input value there is more

than one output value. 1.1 Determine the type of relation in each case and give a reason. 1.1.1

______________________________________________________________________

______________________________________________________________________ 1.1.2 {(1 ; 3), (2 ; 5), (6 ; 13), (7 ; 15)}

______________________________________________________________________

18

1.1.3

______________________________________________________________________

______________________________________________________________________

(1)

A function is a set of ordered number pairs where no two ordered pairs have the

same x-coordinate, or put differently: a function is a set of ordered pairs where, for every value of x there is one and only one value for y. However, for the same value of y there may be different values for x.

1.2 Which of the relations (in QUESTIONS 1.1.1 to 1.1.3) are functions? Why? (a) _______________________________________________________________ (2) (b) _______________________________________________________________ (1) (c)__________________________________________________________ (1) The vertical-line test is used to determine whether or not a given graph is a function. To determine whether a graph is a function, do the vertical-line test. If any vertical line intersects the graph of f only once, then f is a function; and if any vertical line intersects the graph of f more than once, then f is not a function.

1.3 Determine whether or not the following graphs are functions. Give a reason for your answer.

19

(a)____________________________________________________________ (1)

(b)___________________________________________________________ (1)

(c)___________________________________________________________ (1)

(d)___________________________________________________________ (1)

(e)___________________________________________________________ (1)

(f)___________________________________________________________ (1)

(g)___________________________________________________________ (1)

(h)___________________________________________________________ (1)

20

PART 2: THE INVERSE OF AN EXPONENTIAL FUNCTION 2.1 Consider the equation g(x) = 2x. Now complete the following table: (1)

2.2 Sketch the graph of g.

2.3

Sketch the graph of f (x) = x as a dotted line on the same set of axes as g. (1)

2.4

Complete the table below for h, if h is g when the x and y values are interchanged.

x

y

21

x – 3 – 2 – 1 0 1 2 y

Draw h on the same set of axes as g. (4) 2.5 Hence, write down the x-intercept of each of the following graphs below.

y = 2x x = 2y

2.5.1___________________________________________________________ (2)

2.5.2___________________________________________________________

2.6 Write down the domain and range of:

2.6.1 y = 2x

Domain: ___________________ Range: ____________________ (2)

2.6.2 x = 2y

Domain:____________________ (2) Range :_____________________

2.6.3 What is the relationship between the domain and the range of the two graphs in 2.6.1 and 2.6.2

________________________________________________________ ________________________________________________________ (1) 2.6.4 Are both graphs functions? Give a reason for your answer. __________________________________________________________ (2)

2.6.5 Write the equation of x = 2y in the form y = (1) __________________________________________________________

2.6.6 Do you notice any line of symmetry in your sketch? What is the equation of this line?

__________________________________________________________ (1)

2.6.7 In mathematics we call h the inverse of g. Make a conjecture about the graph and its inverse.

__________________________________________________________ (3)

22

PART 3: WHEN IS THE INVERSE OF A QUADRATIC FUNCTION ALSO A FUNCTION? 3.1 Given: f (x) = 2x2, for x∈

3.1.1 Write down the equation of the inverse of f.

________________________________________________________ (1)

3.1.2 Write down the turning points of both f and its inverse. ___________________________________________________________ (2)

3.1.3 Sketch the graphs of f and its inverse on the same set of axes.

3.1.4 Decide whether or not the inverse of f is a function, and give a reason for your answer.

___________________________________________________________ (2)

3.1.5 Explain how you would restrict the domain of f such that its inverse is a function.

___________________________________________________________

3.1.6 Hence, write down the corresponding range of the inverse of f if:

23

(a) x≤0 __________________________________________________ (1)

(b) x≥0 __________________________________________________ (1)

3.1.7 On separate sets of axes, sketch the graphs of the inverse of f

with restricted domains as in QUESTION 3.1.6. Indicate the domain and range of each.

3.1.8 Are the two graphs in QUESTION 3.1.7 functions? Give a reason or reasons for your answer.

24

___________________________________________________________

TOTAL: 50

25

INVESTIGATION 1 MEMORANDUM: FUNCTIONS AND INVERSES

26

INVESTIGATION 2: APPLICATIONS OF DIFFERENTIAL CALCULUS

33

INVESTIGATION 2 MEMORANDUM: APPLICATIONS OF DIFFERENTIAL CALCULUS

35

EXEMPLAR TEST 1

Sequences and Series, Functions, Inverse of Functions and Financial Mathematics

Time: 1 hour Total: 50 marks

INSTRUCTIONS

This paper consists of 5 questions.

Answer ALL the questions.

Show ALL your calculations.

All working must be clearly shown.

Write neatly and legibly.

Calculators may be used where necessary.

Number your answers according to the system used in this question paper.

Start each question on a new page.

GOOD LUCK.

Question 1

The first two terms of an arithmetic sequence are 7 and 3.

1.1 Determine the eighteenth term. [3]

1.2 Which term of the sequence will be equal to – 157? [3]

[6 marks]

Question 2

2.1 Determine the value of n if ∑k=1

n

(5 k+2 ) = 6 475 [5]

2.2 For which values of x will the geometric series (5x + 2) + (- 3x) + (x + 1) + …

converge? [4]

[9 marks]

46

Question 3

The diagram represents the graphs of y = f(x) = - 1(x + 1) + 4 and y = g(x) = 4

( x+1 ) + 2.

3.1 Write down the coordinates of point C, the turning point of f. [1]

3.2 Write down the equations of the asymptotes of g. [2]

3.3 Calculate the lengths of :

3.3.1 AB [3]

3.3.2 DE [1]

3.4 Calculate the length of PQ if OR = 2 units. [3]

[10 marks]

Question 4

In the sketch below, f(x) = bx + 2, and T (- 1; 5) is a point on f.

4.1 Calculate the value of b. [3]

47

4.2 Determine the equation of h, which is a reflection of f about the y – axis. [2]

4.3 Determine the domain off−1, the inverse of f. [1]

4.4 If ( x ; 8 ) is a point on the graph f calculate the value of x. [3]

[9 marks]

Question 5

5.1 At what annual percentage interest rate, compounded quarterly, should a lump sum be

invested in order for it to double in 6 years? [5]

5.2 Joe uses his inheritance money to start a small trucking business. He spends R900 000

on a truck which depreciates by 25% per annum on a reducing balance. The cost of a

new truck is expected to appreciate by 12%. Joe’s truck will have to be replaced in

five years’ time.

5.2.1 Calculate the re-sale value of the existing truck in five years’ time. [3]

5.2.2 Joe sets up a sinking fund to pay for a new truck in five years’ time. How

much he accumulates if he uses the old truck as a trade-in? [3]

5.2.3 At the end of each month, after the purchase of the truck, payments are made

into the sinking fund account which pays interest at the rate of 9.5% p.a.

Calculate the monthly payments into the fund. [5]

[16 marks]

TOTAL: 50 marks

48

EXEMPLAR TEST 1 MEMORANDUM

1.1 a = 7; d = 3 – 7 = - 4; n = 18; T 18 = ?T 18=a+17 d = 7 + 17(- 4) = - 61

substitute√d = -4√

T 18=−61 √1.2 a = 7; d = - 4; n = ?; T n = - 157

T n=a+ (n−1 ) d(−157 ) = 7 + (n−1 ) (−4 ) (−157 )=7−4n+4

4 n = 168 n = 42

T n = - 157√-157=7-4n+4√n = 42√

2.1 ∑k=1

n

(5 k+2 ) = 7 + 12 + 17 + … = 6 475

a = 7; d = 5; n = ?; Sn=6475

Sn=n2[2 a+(n−1 ) d]

6 475 = n2[14+(n−1 )5]

12 950 = n(14 + 5n – 5) 0 = 5n2+9n−12950 0 = (5n+259 ) (n−50 )

n = −259

5 (which is not applicable) or n = 50

expansion√substitution√standard form√factors√n = 50√

2.2 T3

T2=

T2

T1 = r

x+1−3 x

= −3 x5 x+2

( x+1 ) (5 x+2 )=9 x2

5x2+7 x+2=9 x2

0 = 4x2−7 x−20 = (4 x+1 ) ( x−2 )

x = −14∨x=2

If x = −14

r=1∧if x=2 r=−12 , so the series will converge if x = 2

since r < 1

x+1−3 x

= −3 x5 x+2

√

0=

4 x2−7 x−2√Values of x√Conclusion√

3.1 (−1 ;4 ) (−1 ;4 ) √3.2 Vertical asymptote: x = - 1

Horizontal asymptote: y = 2x = - 1√y = 2√

3.3.1 0 = - 1( x+1 )2+4( x+1 )2 = 4x + 1 = - 2 or x + 1 = 2

Equating to zero√x = - 3

49

x = - 3 or x = 1So, AB = 4 units

or x = 1√answer√

3.3.2 D (0; 6), E (0; 3) then DE = 6 – 3 = 3 units. 3 units√3.4 At P, x = 2 then y P=

42+1

+2=3 13 .

At Q, x = 2 then yQ=−(−1 (2+1 )2+4 ) = - 5

So, PQ = 313−(−5 )=10

3+5=25

3

y P=3 13

√

yQ=−5√answer√

4.1 f ( x )=bx+25 = b−1+2b−1=3

So, b=13

substitution√b−1=3 √

answer√

4.2f( x )=(13 )

− x

+2

h( x )=(3 )x+2

( 13 )

−x

√

(3 ) x √4.3 x ∈ (2 ;∞ ) OR x ¿2 answer√4.4

8 = ( 13 )

x

+2

6 = ( 13 )

x

x = log 13

6 = - 1.63

substitution√

x = log 13

6 √

answer√

5.1 A = P(1+i )n

2P = P(1+ r4 )

24

2 = (1+ r4 )

24

(1+ r4 )=2

124

r = 4(2124−1)=0.1172=11.72% p .a . compounded

quarterly.

OR the values 2P, r

400 and 24 can be used where r

400 is already in

% form. In this case finally, r = 400 (21

24−1)=11.72% p .a .

compounded quarterly.

2P √r4∧24√

(1+ r4 )=2

124 √

r = 4(2124−1)√

answer√

5.2.1 Depreciated value of the truck after five years is given by: F v=Pv (1−i )n=900 000 (1−0.25 )5=R 213 574.22

n = 5√substitute in correct formula√

50

answer √5.2.2 Price of new truck = Pv (1+i )n = 900 000(1+0.12 )5

= R1 586 107.52Sinking fund = price of new truck – trade-in on old truck = 1 586 107.52 – 213 574.22 = R1 372 533.30

substitute√answer√

answer√

5.2.3 Let x be the monthly payment into the sinking fund. The timeline looks like this:

The geometric series:R1 372 533.30 = x + x(1+i )−1+x ( x+i )−2+…+x (x+i )59 1 372 533.30 = x¿¿

So, x = 1372 533.30i¿¿

n = 60√substitute into correct formula√

0.09512

√

Making x subject√answer√

51

EXEMPLAR TEST 2 QUESTION 1

Sketched below is the graph of f−1, the derivative of f ( x )=−2 x3−3 x2+12 x+20. A, B and C are the intercepts of f−1 with the axes.

1.1 Determine the coordinates of A. (2)

1.2 Determine the coordinates of B and C. (3)1.3 Which points on the graph of f will have exactly the same x –values as B

and C? (1)

1.4 For which values of x will f ( x ) be increasing? (2)1.5 Determine the y-coordinate of the point of inflection of f . (4)

[12]

QUESTION 2

Refer to the figure below showing the parabola given by f ( x )=4− x2

4 with

0≤ x≤ 4. D is the point (x ;0) and DB is parallel to the y –axis, with B on the graph of f .

2.1 Write the coordinates of B in terms of x. (2)

2.2 Show that the area, A, of ∆OBD is given by: A¿2 x− x3

8.

(3)

2.3 Determine how far D should be from O in order to make the area of ∆OBD as large as possible.

(5)

52

2.4 Hence, calculate the area of ∆OBD when D is at the point determined in (2.3) above.

(2)

[12]

QUESTION 3

The figure below represents a triangular right prism with BA=BC=5 units, ABC=50° and FÂC = 25°.

3.1 Determine the area of ∆ABC. (2)

3.2 Calculate the length of AC. (3)

3.3 Hence determine the height FC, of the prism. (3)

[8]

QUESTION 4

In the diagram below, A.B and C are points in the same horizontal plane. P is a point vertically above A. The angle of elevation from B to P is α . AĈB =β, ABC=θ and BC = 20 units.

4.1 Write AP in terms of AB and α . (2)

4.2Prove that AP=

20 sin β tan αsin(α+β)

(3)

53

4.3 Given that AB=AC, determine AP in terms of α and β in its simplest form. (3)

[8]

QUESTION 5

In the diagram below, A.B and C are points in the same horizontal plane. P is a point vertically above A. The angle of elevation from B to P is α . AĈB =β, ABC=θ and BC = 20 units.

5.1 Determine with reasons the value of EM. (3)

5.2 Prove that SM¿ JY. (3)

5.3 If M is the midpoint of JY, calculate Area ∆ AEMArea ∆ AEY . (4)

[10]

TOTAL: 50

54


55

EXEMPLAR TEST 3

QUESTION 1

The data in the table below represents the score in percentage of 12 Mathematics learners in their Grade 12 trial examination and their corresponding final examination.

Trial Exam 76 64 90 68 70 79 52 64 61 71 84 70

Final Exam 82 69 94 75 80 88 56 81 76 78 90 76

1.1 Determine the equation of the least squares regression line for this set of data. (3)

1.2 Hence, predict the final percentage for a learner obtaining 73% in the trial examination. Give your answer to the nearest percentage. (2)

1.3 Calculate the correlation coefficient for the above data. (2)

1.4 Do you think that by using the least squares regression line one can accurately predict a learner’s final percentage? Provide Mathematical justification for your answer. (2)

[9]

QUESTION 2

2. In the diagram below, BD is a tangent to the circle with centre C (3;-2) at point B (0;2).

61

x

y

C(3;-2)

4D

B

2.1 Show that the equation of the circle is given by (4)

2.2 Determine the equations of the vertical tangents to the circle. Give the coordinates of the respective points of contact. (4)2.3 Find the length of BD and hence show that (4;5) are the coordinates of D. (6)

2.4 Determine the equation of the new circle that will be formed if the circle in question 2.1 is translated four units to the left and two units upwards. (3)

2.5 Determine the inclination of line DC. (3) [20]

QUESTION 2

3.1 In ABC, the angles are acute. Prove that

asin A

= bsin B . (4)

3.2 A lamp post TC is placed vertically in the corner, C, of a horizontal triangular field. The distance between A and B is 25 m and the angle of elevation of T from A is . Furthermore, C A B=αand B=β

3.2.1 Express A C B in terms of α andβ. (1)

62

BA

C

3.2.1 Prove that the height of the lamp post, h, is given by the formula:

h = 25 sin β . tan φsin ( α + β ) (5)

[10]

QUESTION 4

4.1 If P ( A )=38 and P (B )=1

4 , find:

4.1.1 P(A∨B) if A and B are mutually exclusive events. (1)

4.1.2P(A∨B)

if A

and B

are independent events. (3)

4.2 A car park has 14 VOLKSWAGEN cars and 18 BMW’s. There are no other cars. During the afternoon two cars are stolen – one early afternoon, the other later. Determine the probability that:

4.2.1 Both cars were BMW’s. (4)

4.2.2The first one stolen was a BMW and the second one a Volkswagen.

(2)

4.3 Eight boys and seven girls are to be seated randomly in a row. What is the probability that:

4.3.1 The row has a girl at each end? (3)

4.3.2The row has girls and boys sitting in alternate positions?

(3)

[16] TOTAL MARKS: 55

63


Question 1

Question 2

2.1 ( x−3 )2+¿ x2−6 x+9+ y2+4 y+4−25=0x2−6 x+ y2+4 y−12=0

(x-3)2 + (y-2)2

25expansion/ uitbreiding (4)

2.2 x = -2 with point of contact (-2;-2) and/ en x = 8 met raakpunt (8;-2)

x values/ x waardes points of contact/ raakpte (4)

2.3 In ∆ BCD:B=90° … tan−chord / raaklyn- koordBC = BD = 5 … isosc/ gelykb

Eqn BD: y−2=¿ 34(x−0)

y=34

x+2……………… (1)

BD2 = x2+( y−2)2=25…… (2)

Subst./verv (1): x2+( 34

x+2−2)2

=25

( 54

x)2

=25

x=4 or/of x≠−4y=5

D(4;5)

B=90 BC = BD = 5eqn/ vgl (1)eqn/ vgl (2)

subst.

( 54

x)2

=25 (6)

2.4 ( x−3+4 )2+¿( x+1 )2+¿

OR/OF

( x+1 )2

¿25 (3)

64

x2+2 x+1+ y2=25 x2+2 x+ y2−24=0

2.5 mCD=5+24−3

=7

tanθ=7 = 81,87°

mCD

tan = 7answer/ antwoord (3) [20]

Question 3

3.1 sin A=CDAC en/and sin B=CD

BC

AC sin A=CD en/and BC sin B=CD

AC sin A=BC sin B

ACsin B

= BCsin A

asin A

= bsin B

Construction/ konstruksie

sin A=CDAC

sin B=CDBC

AC sin A=BC sin Ba

sin A= b

sin B

(4)

3.2

3.2.1 A C B=180 °− (α+β ) A C B (1)

3.2.2 ∴ ACsin β

= 25sin (180°−(α+ β ))

∴ AC= 25 sin βsin (α+ β)

∴ tan ¿ hAC

h=AC tan

sin formula

sin(α+β )AC ¿

tan¿ hAC

h=AC tan

65

∴h=25 sin β tan❑sin (α+β )

h=25 sin β tan❑sin(α+β )

(5) [9]

Question 4

4.1.1 P ( A∨B )=38+ 1

4=5

8 58

(1)

4.1.2 P ( A∧B )=38

× 14= 3

32

P ( A∨B )=P ( A )+P ( B )−P ( A∧B )=38+ 1

4− 3

32=17

32

P ( A ∩B )or

P ( A∧B )= 332

P ( A∪B )=P ( A )+P ( B )−P ( A ∩ B )

1732

(3)

4.2.1 VW VW;VW

1331

1432 VW

1831 BMW VW;BMW

VW BMW;VW

1431

1832 BMW

1731 BMW BMW;BMW

P (both BMW )=1832

× 1731

=153496

≈ 0,31

Boomdiagram/Tree diagram (takke/branches)

1832

× 1731

153496

≈ 0,31(4)

4.2.2 P (BMW …VW )=1832

× 1431

= 63248

≈ 0,25 1832

× 1431

63248

≈ 0,25(2)

4.3.1 n ( E )=7.13! .6n ( S )=15 !

n ( E )=7.13! .6n ( S )=15 !

66

P ( E )=n(E)n(S)

=7.13 ! .615 !

=15

15

(3)

4.3.2 n ( E )=8 !7 ! n ( S )=15 !

P ( E )= n(E)n(S)

=8 !7 !15 !

= 16435

n ( E )=8 !7 !

P ( E )= n(E)n(S)

=8 !7 !15 !

1

6435

(3)

[16]

67

AFRIKAANS VERSION

68

ASSIGNMENT

VRAAG 1

1.3. ŉ Kwadratiese patron het ŉ tweede term gelyk aan 1, ŉ derde term gelyk aan −6 en ŉ vyfde term gelyk aan −14 .

1.3.1. Bereken die tweede verskil as die patroon kwadraties is. (5)

1.3.2. Vervolgens, of andersins bereken die eerste term van die patroon. (2)

1.4. 1 ;3;5 is die eerste drie terme van die eerste verskil van ŉ kwadratiese ry. Die 7de term van die kwadratiese ry is 35 .

1.4.1. Bepaal die 6de en 5de term van die kwadratiese ry. (4)

1.4.2. Bepaal die nde term van die kwadratiese ry. (5)

[16]VRAAG 2

2.4. Gegee die ry: 2 ;5; 8; . . .

2.4.1. Sou die patron aanhou, skryf die volgende TWEE terme neer. (1)

2.4.2. Bewys dat geen terme in hierdie ry ŉ volkome vierkant sal wees nie. (5)

2.5. Ondersoek die volgende getalle: 2 ;5;2 ;9 ;2 ;13 ;2;17 ; . ..

Bereken die som van die eerste 100 terme van die ry. (5)

2.6. ŉ Groot hoeveelheid afval material bevat metaal. Teen ŉ opeenvolgende suksesvolle herwinnings proses word die volgende hoeveelheid metaal herwin:

32kg tydens die eerste herwinning; 24kg tydens die tweede herwinning; 18kg tydens die derde herwinning; 13, 5kg tydens die vierde proses, en so aan, om ŉ meetkudige ry te vorm.

2.6.1. Hoeveel metaal sal teen die agste herwinning herwin word? (Rond jou antwoord af tot TWEE desimale plekke) (3)

2.6.2. Vind die totale hoeveelheid metaal herwin tydens die eerste agt prosesse.(Rond jou antwoord af tot TWEE desimale plekke) (4)

2.6.2.1.1. Vir hierdie proses om ekonomies te wees moet ten minste 2, 4kg metaal herwin wees na enige proses. Toon aan dat die elfde proses onekeonomies sal wees.

(3)

69

[21]

VRAAG 3

3.3. Bepaal die tydperk, in jare, vir ŉ bedrag geld om te verdubbel as die rentekoers 12, 64% p.a. saamgesteld, halfjaarliks gehef word. (4)

3.4. Angela wil ŉ huis vir R700 000 koop. Sy het ŉ deposito van R50 000 en gaan ŉ lening aan vir die balans teen ŉ koers van 18% p.a. maandeliks saamgestel.

3.4.1. Hoeveel moet Angela by die bank leen? (1)

3.4.2. Breken die maandelikse paaiement as sy die lening in 15 jaar wil afbetaal. (4)

3.4.3. Angela wen die Lotto en wens sy het eerder die lening na die 50ste paaiement afgelos. Wat is die uitstaande bedrag? (4)

[13] TOTAL: [50]

70

ASSIGNMENT MEMORANDUM

QUESTION 1

1.1.3. The 2nd , 3rd , 4th and 5th terms are: 1 ;−6 ;T 4 ;and−14

First differences are: −7 ;T 4+6 ;and−14−T 4

∴d=T3−T 2=T 2−T 1∴ (−14−T 4)−(T 4+6 )=(T 4+6 )−(−7 )

∴−20−2T 4=T 4+13∴−33=3T 4∴T 4=−11d=−11+13=2or−20−2 (−11 )=2

OR

T 5−T2=(T 5−T 4 )+(T 4−T 3)+ (T 3−T 2 )∴−15=(−7+2d )+(−7+d )+ (−7 )∴−15=−21+3 d∴6=3 d∴ d=2

√ −7

√ T 4+6

√ −14−T 4


√ answer (5)

√ −7√ −7+d√ −7+2d


√ answer (5)

1.1.4.

∴T 1=1−(−9 )∴T 1=10

OR

2 a=2∴a=1

√ method

√ T 1=10 (2)

71

5 a+b=T 3−T2

∴5 (1 )+b=−7∴b=−7−5=−12

4 a+2 b+c=T2 or 9a+3 b+c=T3

∴4 (1 )+2 (−12 )+c=1∴ c=21

⇒T n=n2−12 n+21⇒T 1=(1 )2−12 (1 )+21=10

√ method

√ T 1=10 (2)

1.2.1. T n=2n−1 for the first difference sequenceT 6=2 (6 )−1=11∴T 6 of original seq .=35−11=24

T 5=2 (5 )−1=9∴T 5 of original seq .=24−9=15

√ 6th termof 1st diff seq .

√ 6th termof quadr . seq .

√ 5th term of 1st diff seq .

√ 5th term of quadr . seq .

(4)

1.2.2. 2 a=2∴a=1c=0

T n=an2+bn+c35=1 (7 )2+b (7 )+0−14=7 b∴b=−2⇒T n=n2−2n

OR

36 a+6b+c=24→ (1 )25 a+5 b+c=15→ (2 )49 a+7 b+c=35→ (3 )(1)−(2) :11a+b=9→ (4 )(3 )−(1 ):13 a+b=911→ (5 )(5 )−( 4 ):2 a=2

∴a=1⇒b=−2⇒T n=n2−2 n

√ value of a

√ value of c

√ T 7=35

√ value of b√ answer (5)

√ 11a+b√ 13a+b

√ value of a√ value of b

72

√ answer (5)

QUESTION 2

2.1.3. 11;14 √ answer (both terms) (1)

2.1.4. T n=3 n−1Nowif 3 n−1=n2

Then n2−3 n+1=0

∴n=3±√52 ⇒ Δ≠a perfect square

⇒nis irrationalBut nmust be natural∴No termis a perfect square. .. .. . .. .. . ..(sin ce Δ≠a perfect square )

√√ for nth term

√ equating n th termton2

√ value of n√ for concluding n is

irrational and fordeducing n isnatural

(5)2.2. (2+2+. ..+2 )

for 50 terms+(5+9+13+.. . )

for 50 terms

=∑i=1

50

(2 )+∑i=1

50

(4 i+1 )

¿2 (50 )+502

[2 (5 )+49 (4 ) ]¿100+25 (10+196 )¿100+5 150¿5 250

√ (2+2+. ..+2 )

for 50 terms

√ (5+9+13+.. . )

for 50 terms

√ settingup∑ exp ression

√ substitution

√ answer (5)

2.3.1. a=32 ;r=34

;n=8

T 8=32( 34 )

7

=2187

512=4 ,27

√ value of r

√ substitution

√ answer (3)

73

2.3.2.Sn=

a (1−r n)1−r

S8=32[1−( 3

4 )8]

1− 34

=128 [1−(34 )8 ]

¿115 , 19

√ formula

√ substitution

√ answer

√ correct rounding−off (4)

2.3.3.T 11=32( 3

4 )10

=1 ,80¿2,4

Hence, the 11th pass is uneconomical.

√ substitution

√ value of T 11

√ conclusion (3)

QUESTION 3

3.2. A=P (1+i )n

∴2 x=x (1+0 ,126 42 )

2n

∴2=(1 ,063 2 )2 n

∴ log2=2n log (1 ,063 2 )

∴2 n=log 2log (1,063 2 )

∴2 n=11 ,310 552 63∴n=5 ,66 years .

√ formula

√ substitution

√ log form

√ answer (4)

3.2.2. Loan Amount=R 700 000−R 50 000=R 650 000 √ answer (1)

3.2.2.Pv=

x [1−(1+i )−n ]i

∴650 000=x [1−(1+0 ,18

12 )−180]

0 , 1812

∴9750=x [1−(1+0 ,1812 )

−180]∴ x=R 10 467 ,74

√ formula

√ substitution

√ simplification

√ answer

74

(4)3.2.3. ∴ A=P (1+i )n

=650000 (1+0 , 1812 )

50

¿650 000 (2 ,105 242 421 )¿R 1368 407 ,57

∴Fv=x [ (1+i )n−1 ]i

¿10 467 , 74 [(1+0 ,18

12 )50−1]

0 , 1812

¿ R 771292 , 69

∴Balance=R 1368 407 ,57−R 771292 ,69=R 597 114 , 88

√ formulae

√ substitution

√ simplification

√ answer (4)

PROJEK: 'N PRAKTIESE TOEPASSING VAN DIFFERENSIAALREKENE

INSTRUKSIES TOTAAL: 50

1. Beantwoord al die vrae.2. Toon alle berekeninge wat jy in die bepaling van jou antwoorde gebruik het, duidelik

aan.3. Rond antwoorde tot TWEE desimale plekke af, tensy anders vermeld.4. Nommer die antwoorde korrek volgens die nommeringstelsel wat in hierdie vraestel

gebruik word. 5. Skryf netjies en leesbaar.6. Skets die houers volgens die gegewe spesifikasies.7. Wiskundige metodes en formules moet gebruik word om die houers te beplan en te

skets.8. Alle berekeninge en die beplanning van die sylengtes en oppervlaktes moet netjies en

duidelik in geskrewe dele en sketse aangebied word.

HOUERS

A: 'n Houer met 'n reghoekige basisB: 'n Houer met 'n sirkelvormige basisC: 'n Houer met 'n driehoekige basis

SPESIFIKASIES

• Elke houer moet presies een liter vloeistof kan bevat.• Elke houer moet 'n minimum oppervlakte hê.• Die oppervlakte van elke houer moet die deksel insluit.

75

• Die lengte van die reghoekige basis moet twee maal die breedte wees.• Die driehoekige houer moet 'n gelyksydige basis hê.

VERDERE VERGELYKINGAfgesien van jou gevolgtrekking op grond van die drie opsies, watter ander vorm sou jy gebruik vir 'n koeldrankhouer in die vervaardiging van koeldrankblikkies? Gee 'n rede vir jou antwoord.

WENK: Die betrokke vorm kan die mees ekonomiese een wees om te vervaardig maar nie die mees praktiese keuse nie.

RUBRIEKCRITERIA MAXIMUM

MARK MARKS AWARDED

A B C Korrekte wiskundige formules 3 x 3 Korrekte berekeninge:

Afmetings van basisse 4 x 3 Hoogte van die houers 2 x 3

Logiese redenering en aanbieding 3 x 3 Tydige inlewering 2 Gevolgtrekking oor die minste materiaal gebruik

1 x 3

Laaste verdere vergelyking 1 x 3

PROJEK MEMORANDUM: 'N PRAKTIESE TOEPASSING VAN DIFFERENSIAALREKENE

76

ONDERSOEK 1: FUNKSIES EN INVERSE

80

ONDERSOEK 1 MEMORANDUM: FUNKSIES EN INVERSE

90

ONDERSOEK 2: TOEPASSING VAN DIFFERENSIAALREKENE

97

ONDERSOEK 2 MEMORANDUM: TOEPASSING VAN DIFFERENSIAALREKENE

100

EXEMPLAR TOETS 1

Rye en reekse, Funksies, Inverses van Funksies en Finansiële Wiskunde.

Tyd: 1 hour Totaal: 50 punte

INSTRUKSIES

Hierdie toets bestaan uit 5 vrae.

Beantwoord ALLE vrae.

Toon AL jou berekenings.

Alle bewerkings moet duidelik getoon word.

Skryf netjies en leesbaar.

Sakrekenaars mag gebruik word waar nodig.

Nommer al jou vrae volgens die vraag nommering op die vraestel.

Begin elke vraag op ŉ nuwe bladsy.

STERKTE.

Vraag 1

Die eerste twee terme van ŉ rekenkundige ry is 7 en 3.

1.1 Bepaal die agtiende term. [3]

1.2 Watter term van die ry sal gelyk wees aan – 157? [3]

[6 punte]

Vraag 2

2.1 Bepaal die waarde van n as ∑k=1

n

(5 k+2 ) = 6 475 [5]

111

2.2 Vir watter waarde van x sal die meetkundige reeks (5x + 2) + (- 3x) + (x + 1) + …

konvergeer? [4]

[9 punte]

Vraag 3

Die diagram hieronder toon die grafieke van y = f(x) = - 1(x + 1) + 4 en

y = g(x) = 4

( x+1 ) + 2.

3.1 Skryf die ko-ordonate van C, die draaipunt van f neer. [1]

3.2 skryf die vergelykings van die asymptote van g neer. [2]

3.3 Bereken die lengtes van:

3.3.1 AB [3]

3.3.2 DE [1]

3.4 Bereken die lengte van PQ as OR = 2 eenhede. [3]

[10 punte]

Vraag 4

In die skets hieronder is f(x) = bx + 2, en T (- 1; 5) is ŉ punt op f.

112

4.1 Bereken die waarde van b. [3]

4.2 Bepaal die vergelyking van h, wat die refleksie van f om die y – as is. [2]

4.3 Bepaal die definisie versameling van f−1, die inverse van f. [1]

4.4 As ( x; 8 ) ŉ punt op die grafiek f is, bereken die waarde van x. [3]

[9 punte]

Vraag 5

5.1 Teen watter jaarlikse rentekoers, kwartaaliks saamgestel, sou ŉ lompsom belê moet

word om te verdubbel in 6 jaar? [5]

5.2 Joe gebruik sy erfgeld om ŉ klein vragmotor besigheid te begin. Hy spandeer

R900 000 op ŉ vragmotor wat 25% per jaar depesieer, volgens die vemindering

balans. Die verwagte koste van ŉ nuwe vragmotor word bereken teen

waardevermeerdering van 12%. Joe’s verwag dat die vragmotor oor vyf jaar vervang

sal moet word.

5.2.1 Bereken die her-verkoop waarde van die huidige vragmotor oor vyf jaar. [3]

5.2.2 Joe belê in ŉ delgingsfonds om die nuwe vragmotor te betaal oor vyf jaar.

Hoeveel sal sy aangegroeide bedrag waaarde moet wees as hy die huidige

vragmotor gebruik as inruiling teen die nuwe vragmotor? [3]

5.2.3 Na die aankoop van die nuwe vragmotor word ŉ maandelikse betaling teen die

delgingsfonds gemaak, wat rente van 9.5% p.a hef. Bereken die maandelikse

paaiement wat in die fonds betaal word. [5]

[16 punte]113

TOTAAL: 50 punte

114

EXEMPLAR TOETS 1 MEMORANDUM

1.1 a = 7; d = 3 – 7 = - 4; n = 18; T 18 = ?T 18=a+17 d = 7 + 17(- 4) = - 61

substitute√d = -4√

T 18=−61 √1.2 a = 7; d = - 4; n = ?; T n = - 157

T n=a+ (n−1 ) d(−157 ) = 7 + (n−1 ) (−4 ) (−157 )=7−4n+4

4 n = 168 n = 42

T n = - 157√-157=7-4n+4√n = 42√

2.1 ∑k=1

n

(5 k+2 ) = 7 + 12 + 17 + … = 6 475

a = 7; d = 5; n = ?; Sn=6475

Sn=n2[2 a+(n−1 ) d]

6 475 = n2[14+(n−1 )5]

12 950 = n(14 + 5n – 5) 0 = 5n2+9n−12950 0 = (5n+259 ) (n−50 )

n = −259

5 (which is not applicable) or n = 50

expansion√substitution√standard form√factors√n = 50√

2.2 T3

T2=

T2

T1 = r

x+1−3 x

= −3 x5 x+2

( x+1 ) (5 x+2 )=9 x2

5x2+7 x+2=9 x2

0 = 4x2−7 x−20 = (4 x+1 ) ( x−2 )

x = −14∨x=2

If x = −14

r=1∧if x=2 r=−12 , so the series will converge if x = 2

since r < 1

x+1−3 x

= −3 x5 x+2

√

0=

4 x2−7 x−2√Values of x√Conclusion√

3.1 (−1 ;4 ) (−1 ;4 ) √3.2 Vertical asymptote: x = - 1

Horizontal asymptote: y = 2x = - 1√y = 2√

3.3.1 0 = - 1( x+1 )2+4( x+1 )2 = 4x + 1 = - 2 or x + 1 = 2

Equating to zero√x = - 3 or x = 1√

115

x = - 3 or x = 1So, AB = 4 units

answer√

3.3.2 D (0; 6), E (0; 3) then DE = 6 – 3 = 3 units. 3 units√3.4 At P, x = 2 then y P=

42+1

+2=3 13 .

At Q, x = 2 then yQ=−(−1 (2+1 )2+4 ) = - 5

So, PQ = 313−(−5 )=10

3+5=25

3

y P=3 13

√

yQ=−5√answer√

4.1 f ( x )=bx+25 = b−1+2b−1=3

So, b=13

substitution√b−1=3 √

answer√

4.2f( x )=(13 )

− x

+2

h( x )=(3 )x+2

( 13 )

−x

√

(3 ) x √4.3 x ∈ (2 ;∞ ) OR x ¿2 answer√4.4

8 = ( 13 )

x

+2

6 = ( 13 )

x

x = log 13

6 = - 1.63

substitution√

x = log 13

6 √

answer√

5.1 A = P(1+i )n

2P = P(1+ r4 )

24

2 = (1+ r4 )

24

(1+ r4 )=2

124

r = 4(2124−1)=0.1172=11.72% p . a . compounded

quarterly.

OR the values 2P, r

400 and 24 can be used where r

400 is already in

% form. In this case finally, r = 400 (21

24−1)=11.72% p . a .

compounded quarterly.

2P √r4∧24√

(1+ r4 )=2

124 √

r = 4(2124−1)√

answer√

5.2.1 Depreciated value of the truck after five years is given by: F v=Pv (1−i )n=900 000 (1−0.25 )5=R 213 574.22

n = 5√substitute in correct formula√answer √

116

5.2.2 Price of new truck = Pv (1+i )n = 900 000(1+0.12 )5

= R1 586 107.52Sinking fund = price of new truck – trade-in on old truck = 1 586 107.52 – 213 574.22 = R1 372 533.30

substitute√answer√

answer√

5.2.3 Let x be the monthly payment into the sinking fund. The timeline looks like this:

The geometric series:R1 372 533.30 = x + x(1+i )−1+x ( x+i )−2+…+x (x+i )59

1 372 533.30 = x [(1+i )60−1]i

So, x = 1372 533.30i¿¿

n = 60√substitute into correct formula√

0.09512

√

Making x subject√answer√

117

EXEMPLAR TOETS 2

118


119

EXEMPLAR TOETS 3

VRAAG 1

Die data in die onderstaande tabel verteenwoordig punte van 12 wiskunde leerlinge, as ’n persentasie, in hulle Graad 12 voorbereidende eksamen en hulle ooreenstemende finale eksamen.

Voorbereidende Eksamen 76 64 90 68 70 79 52 64 61 71 84 70

Finale Eksamen 82 69 94 75 80 88 56 81 76 78 90 76

1.1 Bepaal die vergelyking van die kleinstekwadrate-regressielyn vir hierdie data. (3)

1.2 Voorspel die finale persentasie vir ’n leerling wat 73% in die voorbereidende eksamen behaal het. Gee jou antwoord tot die naaste persentasie. (2)

1.3 Bereken die korrelasie koeffisiënt vir die bovermelde data. (2)

1.4 Dink jy deur die gebuik van die kleinstekwadrate-regressielyn kan ’n person akuraat ’n leerling se finale persentasie voorspel ? Voorsien Wiskundige regverdiging vir jou antwoord. (2)

[9]

VRAAG 2

2. In die onderstaande skets, is BD ’n raaklyn aan die sirkel met middelpunt C (3;-2) by punt B (0;2).

125

x

y

C(3;-2)

45˚D

B(0;2)

2.1 Wys dat die vergelyking van die sirkel gegee is deur: (4)

2.2 Bepaal die vergelyking van die vertikale raaklyn aan die sirkel. Gee die koördinate van die onderskeie kontakpunte . (4)

2.3 Vind die lengte van BD en wys dan dat (4;5) die koördinate van D is. (6)

2.4 Bepaal die vergelyking van die nuwe sirkel wat gevorm sal word, as die sirkel in vraag 2.1, vier eenhede na links en twee eenhede opwaarts beweeg. (3)

2.5 Bepaal die inklinasie van die lyn DC. (3)[20]

VRAAG 3

3.1 In ABC, is die hoeke skerp. Bewys dat

asin A

= bsin B . (4)

3.2 ’n Lamppaal TCword vertikaal in ’n hoek geplaas, C, van’n horisontale driehoekige veld. Die afstand tussen A en B is 25 m en die hoogte hoek T vanaf A is θ. Verder, C A B=αen B=β

3.2.1 Druk A C B in terme van α en β uit. (1)

3.2.1 Bewys dat die hoogte van die lamppaal, h,word gegee deur die formule:

126

BA

C

h = 25 sin β . tan θsin ( α + β ) (5)

[10]VRAAG 4

4.1 As P ( A )=38 en P (B )=1

4 , vind:

4.1.1 P(A of B) as A en B onderling uitsluitende gebeurtenisse is. (1)

4.1.2P(A of B)

as A

en B

onafhanklike gebeurtenisse is. (3)

4.2 ’n Motor parkeerterrein het 14 VOLKSWAGEN motors en 18 BMW’s. Daar is geen ander motors nie. Gedurende die middag word twee motors gesteel – die een vroeër die middag, en die ander een later. Bepaal die waarskynlikheid dat:

4.2.1 Albei motors BMW’s was. (4)

4.2.2Die eerste een wat gesteel is,was

’n BMW en die tweede een ’n Volkswagen. (2)

4.3 Agt seuns en sewe dogters moet willekeurig in’n ry sit . Wat is die waarskynlikheid dat :

4.3.1 Die ry ’n dogter het aan die einde? (3)

4.3.2Die

ry dogters en seuns wat op alternatiewe posisies sit het? (3)

[16] TOTAAL PUNTE: 55

127


Question 1

Question 2

2.1 ( x−3 )2+¿ x2−6 x+9+ y2+4 y+4−25=0x2−6 x+ y2+4 y−12=0

(x-3)2 + (y-2)2

25expansion/ uitbreiding (4)

2.2 x = -2 with point of contact (-2;-2) and/ en x = 8 met raakpunt (8;-2)

x values/ x waardes points of contact/ raakpte (4)

2.3 In ∆ BCD:B=90° … tan−chord / raaklyn- koordBC = BD = 5 … isosc/ gelykb

Eqn BD: y−2=¿ 34(x−0)

y=34

x+2……………… (1)

BD2 = x2+( y−2)2=25…… (2)

Subst./verv (1): x2+( 34

x+2−2)2

=25

( 54

x)2

=25

x=4 or/of x≠−4y=5

D(4;5)

B=90 BC = BD = 5eqn/ vgl (1)eqn/ vgl (2)

subst.

( 54

x)2

=25 (6)

2.4 ( x−3+4 )2+¿( x+1 )2+¿

OR/OF

( x+1 )2

¿25 (3)

128

x2+2 x+1+ y2=25 x2+2 x+ y2−24=0

2.5 mCD=5+24−3

=7

tanθ=7 = 81,87°

mCD

tan = 7answer/ antwoord (3) [20]

Question 3

3.1 sin A=CDAC en/and sin B=CD

BC

AC sin A=CD en/and BC sin B=CD

AC sin A=BC sin B

ACsin B

= BCsin A

asin A

= bsin B

Construction/ konstruksie

sin A=CDAC

sin B=CDBC

AC sin A=BC sin Ba

sin A= b

sin B

(4)

3.2

3.2.1 A C B=180 °− (α+β ) A C B (1)

3.2.2 ∴ ACsin β

= 25sin (180°−(α+ β ))

∴ AC= 25 sin βsin (α+ β)

∴ tan ¿ hAC

h=AC tan

sin formula

sin(α+β )AC ¿

tan¿ hAC

h=AC tan

129

∴h=25 sin β tan❑sin (α+β )

h=25 sin β tan❑sin(α+β )

(5) [9]

Question 4

4.1.1 P ( A∨B )=38+ 1

4=5

8 58

(1)

4.1.2 P ( A∧B )=38

× 14= 3

32

P ( A∨B )=P ( A )+P ( B )−P ( A∧B )=38+ 1

4− 3

32=17

32

P ( A ∩B )or

P ( A∧B )= 332

P ( A∪B )=P ( A )+P ( B )−P ( A ∩ B )

1732

(3)

4.2.1 VW VW;VW

1331

1432 VW

1831 BMW VW;BMW

VW BMW;VW

1431

1832 BMW

1731 BMW BMW;BMW

P (both BMW )=1832

× 1731

=153496

≈ 0,31

Boomdiagram/Tree diagram (takke/branches)

1832

× 1731

153496

≈ 0,31(4)

4.2.2 P (BMW …VW )=1832

× 1431

= 63248

≈ 0,25 1832

× 1431

63248

≈ 0,25(2)

4.3.1 n ( E )=7.13! .6n ( S )=15 !

P ( E )= n(E)n(S)

= 7.13 ! .615 !

= 15

n ( E )=7.13! .6n ( S )=15 !

15

(3)

4.3.2 n ( E )=8 !7 ! n ( E )=8 !7 !

130

n ( S )=15 !

P ( E )=n(E)n(S)

=8 !7 !15 !

= 16435

P ( E )=n(E)n(S)

=8 !7 !15 !

1

6435(3)

[16]

131

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