Grade 12 Memo Topic Exemplar

Grade WZ eclucators Guide ZAODOE?

https://ultradeep.co.za Grade 12 Assessment Exemplars 1 Learning Outcomes 1 and 2 1.1 Assignment : Functions - Memo 3 1.2 Investigation: Sequences and Series – Memo/Rubric 5 1.3 Control Test: Number Patterns, Finance and Functions - Memo 7 1.4 Project: Finance - Guideline 9 1.5 Exam A: Paper 1 - Memo 10 1.6 Exam B: Paper 1 - Memo 18 2 Learning Outcomes 3 and 4 2.1 Assignment: Recap of Grade 11 Data Handling - Memo 26 2.2 Investigation: Polygons - Guideline 28 2.3 Control Test: Geometry and Trigonometry - Memo 31 2.4 Project: Transformation Geometry - Guidelines 33 2.5 Exam A: Paper 2 - Memo 35 2.6 Exam B: Paper 2 - Memo 40 Grade 12 -2- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Assignment: Functions Marks: 1 In each case for each value of x there is only one associated output value 2 a) 2 b) 1 c) 3 3 Those points where the function values are zero which means they are on the x- axis/ The points where the function cuts the x-axis 4 The possible number of x-intercepts are determined by the degree of x in the function 1st degree: one x-intercept 2nd degree: two possible x-intercepts 3rd degree: three possible x-intercepts In the case of 2nd and 3rd degree functions two of the roots may be the same or non-real. 5 y is of the first degree in each case hence only one y-intercept 6 a) min. value b) no min. or max. value c) local min. and local max. values. The points where a function has a min or max value are also called stationary points. These are determined where the gradient of a tangent to the function (derivative) equals zero). To determine whether the point is a min. or max. one has investigate the derivative/gradient on either side of the stationary points. Min or max. can also be determined by the value of the 2nd derivative. 7 7.1 and 7.2 7.1 and 7.2 10 6 9 5 8 4 7 6 3 5 2 4 1 3 2 -4 -3 -2 -1 1 2 3 4 5 6 1 -1 -1 1 2 3 4 5 6 7 8 9 10 -2 -1 -3 -4 −1 7.3 f ( x) not a function – one-to-many mapping 7.3 g −1 ( x) is a function – one-to-one mapping 7.4 domain needs to be restricted to either 3 3 x ≤ or x ≥ 4 4 Grade 12 -3- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 7.1 and 7.2 7.3 f −1 ( x) not a function – one-to-many mapping 7.4 domain needs to be restricted to x ≤ −2 or − 2 ≤ x ≤ 0 or x ≥ 0 8 h(2) and g(2) gives the function value when x = 2. h ′(2) and g ′ (2) gives value of the gradient (derivative) when x = 2. 9 0 10 11 If f ( x) = 2 x 2 − 3 x − 2 then f / ( x) = 4 x − 3 and f // ( x) = 4 ≠ 0 thus no points of inflection. If g ( x) = −5 x + 10 then g / ( x) = −5 thus constant gradient so no points of inflection. If h( x) = x 3 − 12 x − 16 then h / ( x) = 3 x 2 − 12 and h // ( x) = 6 x . h // ( x) = 0 when x = 0 thus h(x) has a point of inflection when x = 0. 12 12.1 k(x) is the reflection of k(x) in the x-axis 12.2 k(– x) is the reflection of k(x) in the y-axis 12.3 k(y) is the reflection of k(x) in the line x = y Grade 12 -4- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Investigation: The Koch Snowflake Marks: 100 Section A 4th stage snowflake on dotty paper (20) Section B 1. 20 marks for first 5 rows (in any form) for each error or omission (4 X 5) Snowflake A:Length of 1 side B: Number of sides C:Perimeter 1 1 3 3 2 3-1 4x3 3 3-2 42x3 4 3-3 43x3 5 3-4 44x3 n 3 1-n 2. Last row of table above (5X3) 3. Column A: geometric sequence: , Column B: geometric sequence: Column C: geometric sequence any valid observations (10) 4. The perimeter at each stage is a term of a diverging geometric sequence since r = . Hence the perimeter increases without limit. Correct answer and valid explanation (5) Grade 12 -5- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Section 3 1. 20 marks for first 5 rows (20) Stage A: Area of each B: number of C: Increase in area D: Total area added triangle triangles added 1 1 2 3 3 4 5 n 2. ie. where and (5) 3. As , and the total area (2) 4. As the number of stages gets very large, the perimeter of the snowflake increases without limit, but the area approaches a fixed limit. (3) Grade 12 -6- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Test: Number Patterns, Finance and Functions Time: 1 hour Marks: 50 1. 1.1. f ( x) = a x substitute (2; 161 ) ∴ 161 = a 2
∴ ( 14 ) 2 = a 2 (2) ∴ a = 14
1.2. ( 161 ;2)
(0;1)

x=0 (3)

1.3. p = −2; q = −1 (2) [7] 2.

2.1. The area of the rectangle depends on the value of x. For each value of x there is only one value for the area. 2.2. 35 + 2 x − x 2 = 20
∴ x 2 − 2 x − 15 = 0 (3) ∴ ( x − 5)( x + 3) = 0
∴ x = 5 ( x = − 3 invalid )
y 2.3. x-intercepts: (-5;0) and (7;0) y-intercept: (0;35)
(1;36)
turning point: (1;36) (0;35) axis of symmetry: x = 1 (5) (-5;0)
Note: full marks if Quad I only. (7;0)
x 2.4. x=1
2.4.1. 0 < x < 7

(2) 2.4.2. x > 1

(2) 2.4.3. 36 square units

(2) Grade 12 -7- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za [17] 3. 3.1. a = 6; r=3 (2) 3.2. 6.(3) n−1 = 1458 3n−1 = 243 (4) n −1 3 =3 5 [6] n= 6 4. Stage 1 Stage 2 Stage 3 Stage 4 Stage n Number of patterned tiles 3 5 7 9 2n +1 Number of black tiles 1 4 9 16 n2 Number of white tiles 2 6 12 20 n2 + n Total number of tiles 6 15 28 45 2n 2 + 3n + 1 Patterned tiles : Tn = 3 + (n − 1)2
Tn = 2n + 1
Black tiles : Tn = n 2 (by inspection)
(5) Total number of tiles : Tn = 2n + 1 + n + n + n
2 2 = 2n 2 + 3n + 1
4.1. 2n + 1 = 21
2n = 20
(3) n = 10
th 10 stage will require 21 patterned tiles. 4.2. Area = 0.25(2n 2 + 3n + 1) m2

(2) [10] 5. 5.1. A = P (1 + i )n = 42 310(1 + .07 ) 5

(4) = R59 340



2 5.2. 59 340 = x (1 + 0 ,12095 ) + x (1 + 0 ,12095 ) + x (1 + 0 ,12095 ) + ....to 60 terms
3 ( )([ x 1 + 0 ,12095 1 + 0 ,12095 ) − 1]
60 (6) 0 , 095 12 x = R 770 .38

Grade 12 -8- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za PHRPre jest Grade 12 Project: Finance Marks: 50 Notes to Educator 1. The data sheet for this project should be updated to reflect current vehicle prices, interest rates and rate of inflation. Current and historical data on the fixed and linked lending rates of banks can be obtained from the South African Reserve Bank internet site: http:/Awww.reservebank.co.za/economics/netqb2/NetQbsSearch.asp The required document codes are: KBP1181M AND KBP1182M Before learners commence the project, the following concepts and terminology should be discussed in class: loan, interest, compound interest, annuity, present value, future value, outstanding balance, depreciation, sinking fund, inflation and rate of inflation. Assessment Guidelines 1. repayment calculation for fixed- and linked rate options: substitution’ V simplification vv“ paymentYY total’ (8) 2. balance calculation: formulaY substitution’ simplificationY answer ¥ repayment calculation: substitution’ simplification payment total VW comment¥ (9) 3. calculation of balloon payment¥ calculation of loan amountY repayment calculation: substitution’ simplification’ paymentv total” (6) 4. calculation of difference between paymentsv v annuity calculation: substitutionW simplification’ value of nv’ v (6) 5. 5.1. substitution’ answer” (2) 5.2. substitution’ answer” (2) 5.3. Fy of sinking fund substitution” —simplificationw’ answer” (4) 6. Method used: (5) Deducts Fy of T12, T24, Tse, Tasw YY simplification” answers~ Any other method that has mathematical merit and correct calculations, but does not consider the Fy of the holiday payments that are missedV VV VW 7. Valid comment” — substantiation using calculated figures from abovev’ (2 x 4) (8) in respect of each of the following: fixed rate versus linked rate balloon payment payment holiday creation of a sinking fund Grade 12 -9- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Mathematics Exam Paper 1 Time: 3 hours Marks: 150 No. Solutions Comments 1.1.1 ( x − 4)( x − 3) = 2
multiplying x 2 − 7 x + 12 − 2 = 0 x 2 − 7 x + 10 = 0
standard form ( x − 2)( x − 5) = 0
factors ∴ x = 2 or 5
x values (4) 1.1.2 3 x 2 + 2 x + 6 = 10 3x 2 + 2 x − 4 = 0
standard form
formula − 2 ± 4 − 4(3)(−4)
subst x= 6
simplification − 2 ± 52 = 6
x values x = 0,87 or – 1, 54 (5) 1.1.3 (3 x − 2) 2 > 3 x 9 x 2 − 12 x + 4 − 3 x > 0
simplification
standard form 9 x 2 − 15 x + 4 > 0
factors (3 x − 1)(3 x − 4) > 0
(graph or any other method 1 4 used) 1 4 3 3 x< or x> 3 3

critical values/inequality (6) 4 y + 3 x = 50 and x 2 + y 2 = 100 1.2 50 − 3 x
making y the subject y= 4
subst  50 − 3 x  2 ∴ x + 2  = 100  4 
simplification 16 x 2 + 2500 − 300 x + 9 x 2 ∴ = 100
standard form 16 ∴ 25 x 2 − 300 x + 900 = 0
factors ∴ x − 12 x + 36 = 0 2 ( x − 6) 2 = 0
x value x=6
y value y=8 (7) [22] Grade 12 - 10 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 2.1 A = P(1 + i ) n 72  r 

for 320470 = 1500001 +   12  calculating r = 12,72 % nominal interest rate 12  12,72  1 + i = 1 +   12  i = 13,49 %


for 2.2. Loan amount = R114 800 calculating 1 Let monthly instalment be x effective 60 interest rate  0,13   0,13   1148001 +  1 +  −1 [5]  12   12   x=  0,13  60
1 +  −1  12  = R 2612 , 05


for numerator OR

for Equating the current value of the loan with the current value of repayments we denominato get: r 0,13 11 480 × x= 12 = R 2 612,05 (to nearest cent) − 60  0,13  1 − 1 + 

answer  12  [8] After 2 years balance: 2.2. 2  0,13  24  2612 . 05 1 +    12   24  0,13  1148001 +  − = R 77522 , 97  12   0,13  1 +  −1  12  OR Calculating the current value of the sum of the outstanding payments:

for −1 −2 −36  0,13   0,13   0,13  balance 2 612.051 +  + 2 612,021 +  + ... + 2 612,051 +   12   12   12  Grade 12 - 11 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za   0,13  −36  2 612.051 − 1 +     12   = = R77 522,81 0,13 12  0,13  n  3500 1 +  − 1  0,13  n   12   R 77522,971 +  =  12  0,13 12  0,13  n  = 323076,92311 +  − 1

 12   n  0,13  1 +   12  = 4,167499298 n  0,13  1 +  −1  12  n n  0,13   0,13  1 +  = 4,1674992981 +  − 4,167499298

 12   12  n  0,13  1 +  = 1,315706463  12  log 1,315706463 n=  0,13  log1 +   12  n ≈ 25,5 months OR

−n   0,13   [8] 3 500 1 − 1 +     12   77 522,81 = 0,13 12  77 522,81 0,13  log 1 − ×  3 500 12  Then n = = 25,46…  0,13  − log1 +   12  As above there are 25 payments of R3 500 and a final lesser payment. 3.1. The number of HIV positive people increase by 1,2 mill people per year.

1 H = 1,2 t + 2,7

3.1. H = 1,2 (16) + 2,7 = 21,9 million 2

Grade 12 - 12 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 3.1. (6) 3 1 For a series to converge – 1< r < – 1 ; r = thus series converges. 3

3.2. (2) 1 a 5(3) 4 S∞ = = = 607,5 1− r 1 1−
3 formula   1 9 
subst 3.2. 5(3) 1 −    4   3  
answer 2 S 9 == = 607,47 1 (3) 1− 3 607,5 – 607,47 = 0,03
formula 46 ; 60

subst 3.3. 3 Second common difference ∴ Tn = an 2 + bn + c
answer (4) Second common difference = 2 ∴a=1

answer (2) ∴ Tn = n 2 + bn + c

n = 1: T1 = 1 + b + c = 6 ………….. (i) 3.3. (2) 4 n = 2: T1 = 4 + 2b + c = 10 ………….. (ii) (ii) – (i) 3+ b = 4
3.3. b =1 1 c = 4 3.3.
∴ Tn = n 2 + n + 4
2
n + n + 4 = 1264 2 n + n + 1260 = 0 2 (n − 35)(n + 36) = 0 n = 35
(5)
equation
standard form
factors
value of n Grade 12 - 13 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 3.3. (4) 3 Grade 12 - 14 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 41 q= Vv value of q S5=k+1 k=4 v¥v value of k 42 f(x) =a(x-3)? vf) 2=a(0-3)’ v subst a _2 Vv value of a 9 43 g is a one-to-many relation Vv 44 x €[3;%) v xE(- 20:3] Y 45 1 ff: x=k’ +1 x-l=k’ vv log. (x-D=y 46 2 h(x) = -2(x-3)" vv 9 [14] 5.1.1 )x=2 v v y=-4 5 13 /c: =—— -4=-— 51.2 | ¥ ¥-9-2 2 v v x/c: 5 -4=0 x-2 x=34 v v 5.2.1 | 120° 5.2.2 | f(x)=sinxor f(x)= cos(x -90°) vv 5.2.3 |x=-30°: 60° vv [10] Grade 12 -15- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 6.1 f ( x + h) − f ( x ) f / ( x) = lim
h →0 h − 5( x + h) − (−5 x 2 ) = lim
h →0 h − 5 x − 10 xh − 5h 2 + 5 x 2 2
= lim h →0 h
− 10 xh − 5h 2 = lim h≠0 h →0 h = lim [−10 x − 5h] h →0
= −10 x (5) 6.2
power form d  2 −  1 1 1 3 − − 2 4 +

derivatives 4 x − 8 x  = 2 x + 4 x 2 2 2 or dx   x x3 (3) 6.3 Tangent: y=x+4 (– 1 ; 3) is on f : 3 = a (– 1)3 + b (– 1)
3= – a–b
f ( x) = 3ax + b / 2
f / (−1) = 3a + b
1= 3a +b
3=–a–b 2a = 4
a=2 b =–5
(7) Grade 12 - 16 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 7.1 4 xh = 540
540
h= (2) 4x 7.2 A = 8 x + 8h + 2 xh
 540   540  = 8 x + 8  + 2 x 

 4x   4x  (3) = 8 x + 1080 x −1 + 270 7.3 For min A: A/ = 0
A/ = 0 8 − 1080 x −2 = 0
−2 1080 x =8 x = 135 2

x = 3 15 (4) x = 12cm (to nearest cm) 8.1 f ( x) = 2 x 3 − x 2 − 4 x + 3
f (1) f (1) = 2 − 1 − 4 + 3 = 0
f (1) = 0 ∴ x -1 is a factor of f (x) 8.2 2x 3 − x 2 − 4x + 3
quadratic factor = ( x − 1)(2 x 2 + x − 3)
linear factors = ( x − 1)( x − 1)(2 x + 3) y 8.3
derivative (− 2 ; 125 ) 3 27 f / ( x) = 6 x 2 − 2 x − 2 = 0
factors 6x − 2x − 2 = 0 2
x-values
y-values (3 x + 2)( x − 1) = 0 2 x=− or x = 1 0 3 − 32 1 x Graph:
y-intercept

turning points
shape 8.5 f / /( x) = 12 x − 2 = 0

f // 1
f // = 0 x= 6
answer [16] Grade 12 - 17 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 9.1 | 6x+4y < 240 “. 3x+2y <120 v (one mark for each constraint) 5x+5y s 250 x+y <50 v 210 xe Vv y210 v x 9.2 9.2 VV VY (one for mark for each graph) v (for search line) 9.3 vv (for feasible region) Wy Leh GED ty YOY: MME. YY YY YY YY Yyp#u_“yv » ME EEL MELEE LI MME MOLLE dhl bl Ch bbbbthblbilbllh lt llbgt \ MULE i ME EEE WY LL EE UML EL YYYyjyyyyy iE i \N fi 9.4 | P= 3000x + 4000y vv 9.5 | 10 Plasma’s and 40 LCD’s vv {15] Grade 12 - 18 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za exanm [83 Grade 12 Mathematics Exam Paper 1 Time: 3 hours Marks: 150 QUESTION 1 11 . Multiply & 28 Y ply Il. (x=4)? = 49 ¥ both sides simplify ¥” (3) x-4=47 S, hut v Factorise “~~ x=llor x=-3 olution Solution ~“ 112 (x? = 2)(x +3) = x? + 6x x? 43x? -2x-6=x°+6x Vv (5) 3x’ -8x-6=0 v 82136 vv 2 x=3,28 or x=061 ¥ 1.2 Vv v 1.2.1 x=-lor x=5 (2) 1.2.2 \ / Shape ~ 2) = = Intercepts ~~ 1.2.3 -Isxs5 V¥ (1) 13 sub y=x-3 in x’-x=6+y (5) X’-x=64+x-3 Vv x’ -x-6-x+3=0 x°-2x-3=0 ¥ (x-3)(x4+1)=0 % x=3o0r x=-l “ and y=0 or y=-4 v Grade 12 -19- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 1.4 2 Values
1.4.1 x = − and y = 4 (2) 3 And
2 1.4.2 x=− or y = 4 or
(1) 3 [19] QUESTION 2 2.1 T6 = 1500(1,015)5 Formula
2.1.1 (3) = R1615,92 Subst
Answer
2.1.2 1500(1,0156 − 1) (5) S6 =

1,015 − 1 = R9344,33
Ear nings for year : R9344,33 × 2 = R18688,65

2.2 2.2.1 Tn = 162000(1 − 0,22) n −1
(4) T6 = 162000(0,78) 6

= R36482.34
2.2.2 70% of book value = .7 × 36482 .34
(3) = R 25537 ,64
Less than book value
Will not be written of f 2.2.3 x (1 + 012, 75 ) 84 − 1 (4) F= 0 , 75 Formula
12 Subst
80000 = (110,0318...)( x ) Calculation
x = R 727,06 [19] Grade 12 - 20 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za QUESTION 3 3.1


3.1.1 17; 19; 20 (3)

3.1.2 Numbers from 1 to 150 not divisible by 3 (2) 3.1.3 Subtract number of terms in following two sequences: 1; 2; 3; … 150 and 3; 6; 9; … 150 150 = 3 + (n − 1)3
3n = 150
n = 50
(5) Number of terms in sequence = 150 − 50
= 100
3.2 4 2k − 1 1 3 5 7 (4) ∑ 2 k =1 k = + + + 2 4 8 16

37 =

16 3.3 3.3.1


(3) T1 T2 T3 T4 Tn Side 8 4 2 1 8( 12 ) n −1 Perimeter 24 12 6 3 3 × 8( 12 ) n−1 3.3.2 Sum of perimeter of inner triangles = 12 + 6 + 3......
12 =
1 − 12 = 24cm
Thus the limit of the sum of the inner triangles is 24cm and the sum will not exceed 24cm, which is the perimeter of the outer triangle.
(4) [21] Grade 12 - 21 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za QUESTION 4 4.1 Because the given function is a many-to-one function, its inverse is one-to-many and hence is not a function.

y = f −1 ( x ) = − − x 4.2


2 4.3 (− ∞;0]
x 4.4 At points of intersection: − − = −2 x 2 2 x ∴− = 4x 4 2 ( ∴ x 8x 3 + 1 = 0 ) 1 Hence x = 0 or x = − 2 ^y O >x (−0,5; −0,5)



− 18 4.5 f −1 [ f (− 3)] = f −1 (− 18) = − − = − 9 = −3

2 2 [ ]  And f f (− 3) = f  − −1 3   = −2 −  2  3  = −3 2 

  [14] Grade 12 - 22 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za QUESTION 5 5.1 4 (1) m=− and c = 400 5 Therefore equation is : 4 h ( x ) = − x + 400
5 5.2 400 4 (5) = − x + 400
x +1 5 2000 = −4 x ( x + 1) + 400(5)( x + 1)
2000 + 4 x 2 + 4 x − 2000 x − 2000 = 0 4 x 2 − 1996 x = 0
4 x( x − 499) = 0
x = 0 or x = 499 Cable touches ground again at 499 m
[6] QUESTION 6 6.1 (1) 1
-180o -135o -90o -45o 45o 90o 135o 180o -1 6.2 The graph repeats itself every 180 degrees.

(2) 6.3 Each x-intercept would be 30 degrees more than it currently is.
(1) 6.4 0
(1) [5] Grade 12 - 23 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za QUESTION 7 7.1  f ( x + h) − f ( x)  (5) f ′( x ) = lim   h→0  h  − 2 − −2  = lim  x +h x 
h→0  h   − 2 x + 2 x + 2h 1  = lim  × 
h→0  ( x + h) x h  2h  = lim  
h→0  xh ( x + h)   2  = lim  
h→0  x ( x + h )  2 = 2
x 7.2 7.2.1 f ( x) = 7 x3 − 4 x + 6 (3) f ′( x ) = 21x 2 − 4

7.2.2 1 f ( x) = 3 x − 3x 1 f ( x ) = 3 x 2 − x −1 1
3
3 −3 1 f ′( x ) = x 2 + x − 2 2 3

[12] QUESTION 8 8.1 f ( x) = x 3 + x 2 − 5 x + 3 (2) f (1) = 1 + 1 − 5 + 3
=0
∴ x − 1 is a factor of f ( x )

x 3 + x 2 − 5 x + 3 = ( x − 1)( x 2 + 2 x − 3) 8.2 (3) = ( x − 1)( x − 1)( x + 3)
8.3 x-intercepts: x = 1 or x = -3
y-intercept: y = 3
(2) Grade 12 - 24 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 8.4 Stationery points at : − 53 f(x) 1 f ′( x ) = 0 f ′′( x ) = 6 x + 2 -8 8
3x + 2 x − 5 = 0 2 Local Local (3 x + 5)( x − 1) = 0
max min 5 x = − or x = 1
3 Turning points: (− 1;6; 9,5) or (1; 0)
(4) 8.5 Point of inflection at 6x+2=0 C: (-1.6, 9.5) 10 (1) 1 x=−
3 8 8.6 Sketch

(2)
6 8.7 x < −1,6 or x > 1
(2)

4 8.8 Shift graph more than 3 units to right. (2) D: (0.0, 3.0) Or: Shift graph down more than 1,6 units and 2 right more than one unit. − g ( x) = x 3 + x 2 − 5 x + 3

B: (-3.0, 0.0) A: (1.0, 0.0) 8.9 (3) g ( x) = − x − x + 5 x − 3
3 2 -2 8.10 ∆f ( x ) Avg rate of change = ∆x (2) 3−0 =
0 −1 = −3
8.11 f ' ( −2) = 3(−2) 2 + (−2) + 5 =9
(4) f ( − 2) = 9
Eqn of line : y − 9 = 3( x − ( −2))
y = 3 x + 15
8.12 x 3 + x 2 − 5 x + 3 = 3 x + 15
x 3 + x 2 − 8 x − 12 = 0 (4) ( x + 2)( x − x − 6) = 0 2
[31] ( x + 2)( x + 2)( x − 3) = 0
∴ Tangent cuts curve at : x = −2 or x = 3
Grade 12 - 25 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 2 QUESTION 9 (dy) b 91 2x+2b=72 b |? , b= 72-2x 2 b=36-x Vv 2 92 length=x-4 V breadth = 36-x-4=32-x “ (2) 93 A(x) = (x-4)(32-x) Q) = -x’ +36x-128 cm? Vv 94 A(x) = -2x+ 36 ¥ x x<9 9 x9 Max at A'(x) = 0V A(x) + 0 - (4) -2x+36=0% Local max x=9 ¥ [9] QUESTION 10 10.1 A:ys4” A B: xs3 v ; v C: 14x+10y255 D: x+ys5 i (4) ee 10.2 ondiagram Y ———_ (1) j Cc 10.3 A: ys4 Vv N (1) 10.4 Must be integer solutions, therefore | feasible solutions are (2; 3) and B (3:2) vv a (2) 10.5 More ball skills coaches. Therefore : choose (2; 3) v dd) 10.6 Cost = 14x2x15+10x3x1l0 v (3) = R720 (12) Grade 12 -26- Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Assignment: Recap of Grade 11 Data Handling Marks: 50 Question 1 1.1.1 x = 14736,19

1.2 Cumulative 1.1.2 σ = 2089,60

Class Interval Tally Frequency Frequency less than 10 999 11 000 - 11 499 l 1 1 1.3 14 000 - 14 499
11 500 - 11 999 l 1 2 12 000 - 12 499 ll 2 4 12 500 - 12 999 l 1 5 13 000 - 13 499 lllll llll 9 14 13 500 - 13 999 llll 4 18 14 000 - 14 499 llll 4 22 14 500 - 14 999 lll 3 25 15 000 - 15 499 llll 4 29 15 500 - 15 999 lll 3 32 16 000 - 16 499 l 1 33 16 500 - 16 999 l 1 34 17 000 - 17 499 ll 2 36 17 500 - 17 999 ll 2 38 18 000 - 18 499 l 1 39 18 500 - 18 999 l 1 40 19 000 - 19 499 l 1 41 19 500 - 19 999 l 1 42



University Fees Ogive 1.4 45


40


35 Cumulative Frequency 30 25 20 15 10 5 0 1099 1199 1299 1399 1499 1599 1699 1799 1899 1999 9 9 9 9 9 9 9 9 9 9 Fees in Rand Grade 12 - 27 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za 1.5




University Fees 10999 11999 12999 13999 14999 15999 16999 17999 18999 19999 1.6 Skewed to the right, more clustered to left of median than to the right, large range, smaller interquartile range, a few high figures skewing the data (any reasonable explanation)


Question 2 2.1 2.2 σ = 11,3

Values x−x (x − x) 2 40 -23 529 2.3 How far from the mean most 56 -7 49 figures are, the average deviation from the norm 59 -4 16

60 -3 9 60 -3 9 2.4 large range, most learners in 62 -1 1 excess of 10% away from the 65 2 4 mean, couple of high marks 69 6 36 skewing data 75 12 144

84 21 441 x = 63 ∑ (x − x)2 1238

Var 123,8

Question 3 Question 4 3.1 Boys: x = 1249 ÷ 20 = 62,45

4.1


m = 61
Girls: x = 1361 ÷ 22 = 61,86

Learners attending University m = 60
45 40 3.2 Boys have large range, some 35 high and some low, very little No of Learners 30 bundling in the middle. Girls 25 marks don’t start as low 20 but have fewer girls scoring 15 as high as the boys. Mean and 10 median very similar but marks 5 very different. Low scores of 0 2000 2001 2002 2003 2004 2005 2006 2007 2008 boys are offset by high scores. Year (any reasonable explanation)




4.2 Exponential

Grade 12 - 28 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Grade 12 Investigation: Polygons with 12 Matches Marks: 100 Some answers and Marking Rubric Triangles 1 Isosceles triangle with sides 2, 5, 5 Area = 0,5 × base × ht = 0,5 × 2 × 24 =2 6 2 Right angled triangle with sides 3, 4, 5 Area = 0,5 × 3 × 4 =6 3 Equilateral triangle with sides 4, 4, 4 Area = 0,5 × 4 × 4 × sin 60 0 3 = 8× 2 =4 3 4 3 = 48 > 6 = 36 > 2 6 = 24 So the equilateral triangle has the greatest area. There are no other triangles because the longest side must be shorter than the sum of the other two sides, otherwise there is no triangle! So longest side < 6 If longest side = 5, then the sum of the other sides is 7. They can be 5 and 2 or 3 and 4. If the ‘longest’ side is 4, then the sum of the other sides is 8 and the only possibility is the equilateral triangle 4, 4, 4. Quadrilaterals with equal angles If all angles are equal, each is 900 and the quadrilaterals will be rectangles The possible rectangles are 1, 1, 5, 5 2, 2, 4, 4 3, 3, 3, 3 The 3 by 3 square has the greatest area. 9 > 8 > 5 Grade 12 - 29 - Educators Guide 2008 https://tvetpapers.co.za

https://ultradeep.co.za Pentagons Without breaking the matches, it is not possible to create a regular pentagon (or a regular polygon with n sides where n is not a factor of 12). It is also not possible (as can be proved using the sine and cosine rules) to create a pentagon in which all angles are 1080 and the perimeter 12 matches. A 3 matches 3 matches B 90° 90° E 2 matches 2 matches F C 2 matches D This pentagon has an area of 6 + 2 3 which can be calculated as follows: Let AF be the altitude from the vertex A to the opposite side, CD. In ∆ABC : AC = 2 2 + 3 2 = 13 (Th. of Pythagoras) And in ∆ACF AF 2 = ( 13 ) − 1 (Th. of Pythagoras) 2 2 Hence altitude AF = 12 = 2 3 Area of pentagon = ∆ABC + ∆ACD + ∆ADE = 3 + 2 3 + 3 ≈ 9,464... (bigger than the square). Hexagons A regular hexagon has sides each 2 matches long and angles of 1200. This can be broken down into 6 equilateral triangles with all sides 2 matches long. 12 3 Area = 6 × 0,5 × 2 × 2 × sin 60 0 = = 6 3 ≈ 10,392... getting bigger! 2 Other hexagons have areas less than this. There are endless possibilities. Twelve sides 10 × 180 0 Each angle in the regular 12 sided polygon is = 150 0 12 1 This polygon can be broken into twelve isosceles triangles, each having equal sides of r = 2− 3 (calculated using the cosine rule) with the included angle being 300. Hence the area is 3 12 × 0,5 × r 2 sin 30 0 = ≈ 11,196... 2− 3 Again other 12 sided polygons can be constructed but none has an area as large as this. 2 12 6 6 36 The circle with a perimeter of 12 has a radius r = = and the area is π   = ≈ 11,459... 2π π π  π It should be clear that the biggest possible area that can be enclosed with n matches is a regular n sided polygon: as close to a circle as possible. Grade 12 - 30 - Educators Guide 2008 https://tvetpapers.co.za

Rubric Communication Special cases Generalisation and Presentation Total justification 32-40: Clear, coherent, logical 28 to 35: Interesting, comprehensive 16 to 20: Quality 5: Neat, striking explanations, supported by set of special cases. All calculations arguments to support all visual impact appropriate sketches and tables accurate. claims. of results.All main ideas covered: the more sides, the greater the possible area; regular polygons larger than others with the same number of sides; the limiting area is that of the circle with a perimeter of 12 sides. Some logical extension. 28 to 31: Clear, logical 25 to 27: Areas and sketches of the 14 to15: Quality 4: Neat pleasing explanations, supported by three triangles, the square, the regular arguments to support impact. appropriate sketches and hexagon and the regular 12 sided almost all claims. tables. One main idea might be polygon correctly shown and some excluded or no extension ‘other’ examples to support the ‘regular is biggest’ theory. 24 to 27: Clear logical 21 to 24: Areas and sketches of the 12 to 13: Quality 3: Neat explanations, supported by three triangles, the square, the regular arguments to support appropriate sketches and hexagon and the regular 12 sided most claims. tables. One main idea missing polygon correctly shown. and no extension. 20 to 23: Satisfactory 18 to 20: Areas and sketches 10 to 11 Some valid 3: Neat explanations with appropriate provided of the three triangles, the attempts at explanations sketches and tables. More than square, the regular hexagon and the and justifications. one main idea missing. regular 12 sided polygon. Accurate calculations with at most one minor error. 16 to 19: Adequate explanations 14 to 17: Areas and sketches 8 to 9: Attempts at 2: Rather untidy. with some sketches and tables. provided of the three triangles, the justifications but flawed At least one main idea square, the regular hexagon and the or inadequate. satisfactorily covered. regular 12 sided polygon. Accurate calculations with at most two minor errors. 12 to 15 Some logical 11 to 13: Areas and sketches 6 to 7: no valid 1: Very untidy. discussion with at least some provided of the three triangles, the explanations or sketches and tables. square, the regular hexagon and the justifications for claims. regular 12 sided polygon. Accurate calculations with at most three minor errors or one major error or omission. 0 to 11: No main ideas 0 to 10: Major errors and/or 0 to 5 0: A mess: clearly satisfactorily covered. Sketches, omissions. no effort made. tables omitted, flawed or inadequate. Grade 12 - 31 - Educators Guide 2008

Control Test Grade 12 Test: Geometry and Trigonometry Time: 1 hour Marks: 50 110 xt +y? =(-5) +(12) =169 vv 1.2 (13 +(0) =169 v 13 Yri2_0-12_ 2 Vv x+5 134+5 3 2. By -36 = -2x-10 w. By +2x = 26 vv 1.4 Line through the origin, which is perpendicular to AB will cut the circle at the points of contact of the required tangents. This line is y = = VW . . . . > (3x ° Points of intersection are at points where x° + 3) 169 ie 13x* =4x 169 ox = 33V13 “ys 42/13 vv One tangent is: y= 2v13 --2 the other is y+2vi3 --2 x-3V13 3 x4+2v13. 3 2 By 42x =12V13 oBy+2x=-12/13 Vv 21 Reflection in the y-axis Vv 2.2 Glide (1 to the right) and reflection about the x-axis vv 2.3 Rotation of Habout the origin vv 2.4 Anti-clockwise rotation of 60° about the origin vv 2.5 Rotation through 180° (because k <0) v Enlargement by a factor of — k ( or reduction by a factor of -*) vv Reflection about the line y = x vv 3.1 sin 2x =cos(x-15°) an cos(90° -2x)=cos(x-15°) vv 190° = 2x = #(x-15°)+k.360°;kKEZ VW 3x =105° +k.360° or x =75° +k.360° v 1x =35°+k.120° or x=75°+k.360°9KEZ 3.2 Yes, Mvuyo is correct: 155° = 35° +1x120° (or substitute) VW Grade 12 - 32- Educators Guide 2008

2 x + tan 45 ( ) 1tan 0 4.1 tan 2 x + 45 0 =

− tan 2 x tan 45 0 2 tan x +1 = 1 − tan 2 x

2 tan x 1− .1 1 − tan 2 x 2 tan x + 1 − tan 2 x 1 − tan 2 x = × 1 − tan 2 x 1 − tan 2 x − 2 tan x 2 tan x + 1 − tan 2 x =

1 − tan 2 x − 2 tan x 4.2 Identity not valid for 2 x + 45 0 = 90 0 + k .180 0 ; k ∈ Z
( ) ie for x = 22,5 0 + k .90 0 as tan 2 x + 45 0 is not defined for these values of θ
Also not defined for x = 90 + k .90 since tan x is not defined for these 0 0 values of θ
5.1 In ∆ABC BC 2 = x 2 + x 2 − 2 x.x. cos 30 0

= 2 x 2 − 3x 2
∴ BC = x 2 − 3
5.2 Area ∆ABC = Area ∆ACD = Area ∆ADB = 0,5 × x 2 sin 30 0
2 x =
4 2 Area ∆BCD = 0,5 x 2 − 3  . sin 60 0
  2 3 = 0,5 ×  x 2 − 3  ×   2 = ( x2 2 3 − 3 )
4 ∴ Total surface area = 3x 2 + ( x2 2 3 − 3 ) 4 4 3 2 = x square units
2 Grade 12 - 33 - Educators Guide 2008

Grade 12 Project: Escher and Transformation Geometry Marks: 100 (All M.C. Escher works (c) 2007 The M.C. Escher Company - the Netherlands. All rights reserved. Used by permission. www.mcescher.com ) Some Guidelines The object of this project is to show that transformation geometry is not just a lot of rules and formulae, but that the concepts are used in the fascinating work of the Dutch artist, M.C. Escher. It is suggested that a rubric is drawn up, with agreement from the learners about the need to show a clear understanding of the transformations learnt in the course of their years at school. It is hoped that the learners will enjoy investigating the various patterns. The solutions below should be helpful to the marker: Task 1 This is Escher’s version of filling the plane with this fish. Learners may come up with other options. Measure each on its merit. 1. Mark the point of the tail (not the extreme point, the one about 15mm from the extreme point), A. 2. Place A at the origin so that the fish faces upwards. Trace the fish and call it 1. 3. Rotate the fish anti-clockwise, about A, through 900. Trace the fish again and call it 2. 4. Rotate again through 900 around A and call this fish 3. 5. Rotate for a third time and call the resulting fish 4. 6. Place your template fish on fish 1 and then slide (translate) it down and slightly to the right to fit between fish 3 and fish 4. Trace the fish again and call it 5. 7. Repeat steps 3 to 5 and call the resulting fish 6, 7 and 8. 8. Place the template on fish 2. Translate it to the right and slightly up until it lies between fishes 1 and 4. Call this fish 9. And so on… Grade 12 - 34 - Educators Guide 2008

Task 2  Three shapes are used: a fish, a bird and a lizard/gecko.  The fish meet at the mouth and each is a rotation of another through 1200 about the point where the mouths meet.  The birds fit between the fish and are also 1200 rotations of each other.  The geckos fit into the tails of the fish and the birds.  The geckos are reflections of each other about a vertical line through the centre of the fish.  There is also an axis of reflection at an angle of 600 to the vertical.  The complete motifs of all three creatures can be translated horizontally and vertically so that the geckos fit on to each other. Task 3  The birds and fish are repeated by a series of glide reflections.  The axis of reflection is a horizontal line through the middle of the frieze.  The glide is a translation about 2cm to the right. Task 4  Start with a 160cm by 160cm square.  Place the first large gecko in the bottom right hand corner as shown in the given art work.  Rotate this motif three times, either clockwise or anti-clockwise through the central point (80cm from the bottom and 80cm from the left) and trace copies in the other three corners.  The other eight boundary geckos are a reduction of the biggest geckos by a factor of three, but they first need to be flipped over.  Place a small gecko horizontally at the nose of the bottom left big gecko and then translate it horizontally to form the other.  Repeat this process (either vertically or horizontally) to complete the outer ring.  25cm from the bottom, to the right of the central vertical line, is a reduction of the largest gecko by a 15 factor of . This gecko is then rotated seven times, about the central point, though an angle of 450 7 each time.  52cm from the bottom is another ring of geckos, this time reductions of the largest gecko by a factor of 5. This gecko is also rotated seven times through 450 to form the ring.  The next ring of black geckos is formed in the same way.  In between the rings of black geckos, the space is filled by a collection of reductions (and in some cases also reflections) of the original. Each one is rotated through 900 three times to create its identical partners. Grade 12 - 35 - Educators Guide 2008

Grade 12 Mathematics Exam Time: 3 hours 1100 AC = 87 +47 v =4/5 v yg MOLL _t vv x+3 543 2 Equationis 2y+a+1=0 v 1.3 Atpoint of intersection of AC and BD: 2(2x-8)+x+1=0 vv -.4x-164+x4+1=0 J.3x=15 “x=3 andy=-2 Vv But the mid-point of BD is (353 -(32) vw And gradient of AC * gradient of BD = ~ 3x2 =-1 vv Hence AC is the perpendicular bisector of BD 1.4 Area of kite ABCD = 2x area AACD v =2x05x4V5 x (4-3) +(0+2) vv = 20 square units v 1.5. The inclination of AB = wo( SS 7 v 2+3 = 135° v 1.6 The inclination of AD = w(5] =171,2 (correct to 1 decimal place) BAD =171,9° - 135° = 37° (correct to nearest degree) vv 2.1 The equation of the circle is (x - 2)" + (r+ 3) = (6-2) + (-1+3) =20 or x* -4xt yt +6y =7 2.2. The mid-point of PQ is (3;-4) The gradient of PQ = ; =1 . . _ yt Hence the perpendicular bisector has the equation 3 =-1 x- Substituting the co-ordinates of the centre into this equation: tus -2*4 ~~ 2H -3-3 Hence the perpendicular bisector of PQ passes through the centre of the circle. 2.3 The radius of the given circle is f20 or aaf5 Grade 12 The distance between R (-1:2) and (2;-3) is .f(-1- 2)? +(2+3% -/34 Since ./34 > ./20, the point R lies outside the circle. - 36 - (axann /4\ Paper 2 Marks: 150 VV vv vv vv vv v Educators Guide 2008

2.4 2.5 3.1 3.2 3.3 3.4 41 4.2.1 Grade 12 The co-ordinates of Af‘ are [2-5] v and the co-ordinates of ”" are (°-3} v Area of original circle = a{J20y a Area of enlarged circle centre AZ” is (9 - 37 + [- 2 + :) = 457 Vv Area original circle : area of enlarged circle = 207: 452 = 4:9 vv Dis (2:4) vv 4OAB has been reflected about the line » =x vv F is the point [Fx-3 ——x?2- pee Ba = (2:32) vv AOAB has been rotated anticlockwise about the origin through an angle of 45° Vv His (- 2:32) vv AO#F has been reflected about the y axis ie. AG4.3 has been rotated anticlockwise about the origin through an angle of 45° and then reflected about the y axis. vv Jis the point (= 2:4) vv AOAB is reflected first about 3» = x and then about the y axis. vv sin (90° + x}.cos(-x)- tan{180° - x) cos(180° + x) sin{-x -720°) =cosx.cos x - (-tan x)(—cos x)(- sin x) vvv = cos? x + 2% cosz sin x vv cosx =1 v 7 —_ 15 5 = F VVV 5 “ 13 3 7 -21 W721, 32 8 =-13 24 v 13 -37- Educators Guide 2008

4.2.2 4.2.3 43 5.1 5.2 6.1 Grade 12 sin 2a = 2sin wcosa v Zoos? F cos A+ v 8 =| -—]+1 v [7] cos? fe v 2 3 “cos B= v 7 f34 1+sin 28-4sin? g=0 cos? @ + sin? @+2sin @cos@ -4sin?@ =0 vv (cos 8 +3sin A)(cos 8 -sin 8) = 0 v ‘tan @= => v or tan 8=1 v 18 tan7 “3 +180"; kez or G=45°+E.180°, kez 1, 8e{-184°:459}+%180°, te Z (correct to 1 decimal place) 9 VVY Bearing = 60° v AB? = 3007 +500? - 2.300.500. cos 120° VVv 1 = 90 000 + 250 000-300 oe - 3] vv = 490 000 . AB = 700 metres Race distance = 300 + 500 + 700 = 2 200 metres v CDB = 180° -[30° +2] = 150° -¢ v - 38 - Educators Guide 2008

6.2 stm v BC Bo =? v tan 2 63 Inagcp,— > __ = _8_ Vv sin fi50 -8) smo ar “Re Bsin(150 a) m2 InAace, 2 =tane v BC p= BCtan 8 . oL _ Bsinll50-2) Sim _ 8{sin 150° cos @-cos150°sin 8) sin & V sin 8 “cosé d ose- 3 sin 2 2 2 = Vv cos? cos@ sm 2 =4 +4 v cos@ neers = 4(1 +/3tan 8) v 74 cos 2x = sin {x + 30") "cos 2x = cos(90° - (x = 30°) vv .2x=3(60°-x) +4360", kez vv 3x = 60° +%.360" or x = 60" +%.360" Jom = 20° +4120" or x= -60" +4.360"; kEZ For xe|-180°:180°| xe {-100",-60°;20° 140°] vv 7.2 Function values correct to 1 decimal place where applicable. Cra a0",-0,5) €120°;-1) (90", 1) eos) A (A00°;-0,5) © @neng) B (-6P -0,5) D (140°9),2) 4 marks for each graph 73 ~~ 60° v Grade 12 - 39- Educators Guide 2008

8.1 3 marks for Boys and 3 marks for Girls (need not split the 20s into 20-24 and 25-29) Boys Girls 1/7 9 3.3 210 0/2|0 0011 2 3 88 7 65 5 5 5/2 2|3]0 8.2 Box and whisker plot of Grade 2 Girls’ Masses VW¥VW” 1s 20 22 24 26 28 30 - (x- ny" ; 8.3 Standard deviation = 5 7 3,3 (correct to 1 decimal place) VVVV 8.4 Modal mass=25kg v 8.5 Interquartile range = 25 - 22 = 5kg VVVV 8.6 90% of 15 is 13,5. On the ogive, the weight that corresponds to 13,5 is approximately 27,5 kg vv 9.1 True: 50% of the data items are within the inter-quartile range compared to 68% within one standard deviation. vv 9.2 True: the data is spread more to the left of the median than to the right. vv 9.3 False: the greater the time, the lower the water level, so the correlation is negative. vv 9.4 False: the scatter plot will be a straight line: h = -* is the height. C is the original capacity of the dam, t the number of hours and r the radius of the dam. All values except h and t are constants. h is a linear function of t with a negative gradient. Vv Grade 12 - 40- Educators Guide 2008

Grade 12 Mathematics Exam Paper 2 Time: 3 hours Marks: 150 1.1 AC = (5 − 0)2 + (0 − 3)2 = 34

AB = (− 3 − 0) + (− 2 − 3) = 34 2 2
BC = (5 + 3) + (0 + 2) = 68 2 2
Hence AB=AC and the triangle is isosceles
And BC = AB + AC and hence the triangle is right angled at A 2 2 2
Right angle can also be proved by showing that gradient AB × gradient AC = − 1 1 1.2 Area ∆ABC = × 34 × 34 = 17 square units

2 −3+5 −2+0 1.3 M, the mid-point of BC is  ;  = (1;−1)

 2 2  Circle is ( x − 1) + ( x + 1) = (1 + 3) + (− 1 + 2 ) = 17 2 2 2 2 1.4


0+2 1 1.5 Gradient BC = =

5=3 4 ∴ gradient of tangent = − 4
y−0 ∴ equation of tangent is = −4

x−5 i.e. y = −4 x + 20
−1− 3 Gradient MA = = −4
1− 0 Gradient of tangent = gradient MA hance these lines are parallel
2.1 Radius = 10
y −1 3 −1 2.2 Equation of AB is =


x + 3 1+ 3 ∴ 4 y − 4 = 2 x + 6 i.e. 2 y = x + 5

At points if intersection: x 2 + (− 2 x ) = 10 2 2.3

∴ 5 x 2 = 10 ∴x = ± 2

and y 2 = 8 ∴ y = ±2 2

( So the points of intersection are − 2 ; 2 2 and) ( 2 ;− 2 2 )

Grade 12 - 41 - Educators Guide 2008

2.4 ACˆ B = inclination AC − inclination BC

0 −1 0−3 = tan −1 − tan −1

10 + 3 10 − 1 = −170,78... − 125,78... 0 0

= 45 0
Hence AOˆ B = 2 × ACˆ B 3.1.1 P ′(5;−1)

3.1.2 P ′(1;−5)

3.2.1 The co-ordinates of the vertices of the images are reduced by a factor of 2: i.e. they are half the distance of the original vertices from the origin.

1 The area of the transformed triangle is the area of the original triangle.

4 3.2.2 The vertices are reflected about the x-axis

and translated 2 units to the right.

This transformation is a glide reflection which is rigid and hence the area of the transformed triangle is equal to the area of the original triangle.

3.3 ^y P' P(a;b) 60° α >x O a = OP cos α and b = OP sin α and hence ( the x co-ordinate of P ′ = OP ′ cos α + 60 0 )
= OP[cos α cos 60 − sin α sin 60 ) 0 0
1 3 a 3b = OP cos α . − OP sin α . = −
2 2 2 2 Grade 12 - 42 - Educators Guide 2008

and the y co-ordinate of P ′ = OP ′[sin α + 60 0 ( )
 1 3 = OP ′sin α . + cos α . 
 2 2  b 3a = +
2 2 ( sin (− α ) cos 90 0 − α = ) (− sin α )(sin α ) 4.1 ( cos α cos 180 + α 0 ) cos α (− cos α )


− sin 2 α =
− cos 2 α = tan 2 α
4.2.1 cos 27 0 = 1 − sin 2 27 0 = 1 − t 2

4.2.2 tan 153 0 = − tan 27 0
t =−
1− t2 4.2.3 cos 243 0 = − cos 630
= − sin 27 0 = −t
4.2.4 ( cos 54 0 = cos 2 × 27 0 )
= 1 − 2 sin 27 0 = 1 − 2t 2 2

4.3 ( ) tan 3 x + 75 0 = −1 ∴ 3 x + 75 0 = −45 0 + k .180 0 ; k ∈ Z

∴ 3 x = −120 + k .180 0 0
∴ x = −40 0 + k .60 0 ; k ∈ Z

4.4 sin 15 0 + 3 cos 195 0 = sin 15 0 + 3 cos 180 0 + 15 0 ( )
2 2 2 2 1 = . sin 15 0 + 2 2 3 − cos 15 0 ( ) = sin 30 0 sin 15 0 − cos 30 0 cos15 0

= − cos 30 0 + 15 0 ( )
2 =−
2 Grade 12 - 43 - Educators Guide 2008

r2 + r2 − k2 5.1 In ∆ADC : cos θ =
2r.r 2r − k 2 2 =
2r 2 r 2 + (2r ) − (2k ) ( ) 2 2 5.2 In ∆ABD : cos 180 0 − θ =
2.r.2r 5r − 4 k 2 2 ∴ − cos θ =
4r 2 4 k 2 − 5r 2 ∴ cos θ =
4r 2 2 r 2 − k 2 4 k 2 − 5r 2 5.3 Hence =
2r 2 4r 2 ∴ 4 r 2 − 2 k 2 = 4 k 2 − 5r 2 ∴ 9r 2 = 6k 2 3 and k 2 = r 2
2 3 2r 2 − r 2 2 1 So cos θ = 2 =
2r 4 6. A B θ α β C x metres D AC x In ∆ACD = 6.1 [ sin β sin 180 − (α + β ) 0 ]

x sin β ∴ AC =

sin (α + β ) AB In ∆ABC : = sin θ
AC x sin β ∴ AB = . sin θ
sin (α + β ) 40. sin 70 0. sin 15 0 6.2 Height =
sin 120 0 = 11,23 metres (correct to 2 decimal places)
or 11 metres (correct to the nearest metre) Grade 12 - 44 - Educators Guide 2008

1 6.3 Area ∆ACD = AC.CD. sin α
2 1 x sin β = × .x. sin α
2 sin (α + β ) = 664,97 m 2 (correct to 2 decimal places
2 2 or 665 m (correct to the nearest m ) 7.1 At A sin 2 x = 0
∴ x = k .90 0 ; k ∈ Z ( A − 90 0 ;0 )
( B is 60 0 ;2 )
At C x = 0 so y = 2 cos − 60 0 = 1 ( )
C is (0;1)
7.2 ( DE = 2 cos 120 0 − 60 0 − sin 2 × 120 0 ) ( )

= 2 cos 60 + sin 60 0 0
1 3 3 =2× + = 1+
2 2 2 7.3 At G: x = −30 0
( At H: y = sin 2 × −30 0 = − sin 60 0 )
3 =−
2 Hence the equation of the new curve is y = 2 cos(x − 60 0 ) − 3

2 8.1 x = 185 g
8 ( xi − 185) 2 σ= ∑ i =1 8 =25,98 g (correct to 2 decimal places)
or 26 g (correct to the nearest g

8.2 Standard deviation is a measure of dispersion: the larger the standard deviation, the wider the spread of data items.

Grade 12 - 45 - Educators Guide 2008

9.2 9.3 9.4 9.1 Ogive: Marks for 245 Students 300 250 3 ae = 200 s 7 E g 150 Z] z 100 = eo ZL 40 VVVS 0 Se VVVY A cS e & & © & e Marks Median ~ 59% (read at A) vv Lower quartile ~51 —_ (read at B) vv Upper quartile ~ 71% (read at C) vv Mean = 2x15+10x35+43x45 + 72x55 +5365 + 37x75 +25x85+3x95 245 VVVY = 60.84% (correct to 2 decimal placesO vv Mean ~ 61% , median ~ 59% and mode 55% . Because all these values are approximate, and the differences are not significant, we can say that the distribution is fairly symmetric (the data is not skewed) VV Grade 12 - 46 - Educators Guide 2008