MATHEMATICS PAPER 2
NATIONAL SENIOR CERTIFICATE
GRADE 12
MEMORANDUM

SEPTEMBER 2018

QUESTION 1

1

1.1

a = - 6 ,54
b = 1, 05
yˆ = 1, 05 x - 6 ,54

? value of
? value of
? equation 

(3)

1.2

y =  - 6 ,54  + 1,05 ( 41 )

= 36 ,51 » 37

? substitution

? answer 

(2)

1.3

On the scatter plot

?? x-intercept 

6 < x < 8  and

(45;41) both correct

OR

?? (69,87;66,73) and (45;41) both correct 

 (2)

1.4

r =  0 ,88

? ? answer 

(2)

1.5

Yes. The strong positive correlation

 

? Yes 

? strong positive

(2)

   

[11]

 

QUESTION 2

2.1

Range = 29 - 10
= 19

?answer 
(1)

2.2

x = 15 + 23  + 17  + 24  + 26  + 18  + 28  + 13  + 10  + 28  + 29
                                                    11

 231    Answer ONLY full marks
     11
= 21

?? answer
(2)

2.3

σ  = 6 ,37

? min  ? max
? answer
(3)

2.4

(21 - 6 , 37 ; 21 + 6 , 37 ) = (14 , 63 ; 27 , 37 )
5 weeks

? 231
? answer 
(2)

    [8]

 

QUESTION 3
2
3.1 (√89)2 = (-3- t)2 + (-3 - 5)2
89 = 9 + 6 t + t2 + 64
t2 + 6t - 16 = 0
( t - 2 ) ( t + 8 ) = 0
t = 2 or  t = 8
?substitution
? simplification
? standard form
? factors
? value of t
(5)

3.2

mAB     = - 3
3 + 3 = - 3
p + 3
- 3p - 9 = 6
p = - 5

? mAB
? mAB in terms of p
? equating
? value of

(4)

3.3

- 3 x - 12 = 0
x = - 4
E ( - 4 ; 0 )

? y = 0
? x = 4

(2)

3.4

M = (- 5 + 2 ; 3 + 5)
             2         2
= (-3/2 ; 4)

? x – coordinate
? y – coordinate

(2)

3.5

mE = 0 - 4  
           -4 + 3/2 
=8/5
mBC = 5 + 3 
           2 + 3
=8/5
EM BC [ = gradients]

? correct substitution
? mEM
? mBC
? = gradients 

 (4)

3.6

tan θ = - 3
θ  = 108 , 4349488º
tan β = 8/5
β = 57 ,99461679º
ABC = 50 ,44º

? size of θ
? size of α
? size of  A Bˆ C 

 (4)
 

 

   [21]

 

QUESTION 4

3

4.1

PR = √(0 - 8 )2 + (- 3 + 9 )2
= 10

?correct substitution
?answer

(2)

4.2

M = (  0+8  3+9 )
            2         2
=(4;- 6)

?x-coordinate
?y-coordinate

(2)

4.3

mPQ = - 3 + 6
              0-9
= -1/3
mQR  =  - 6+9  
               9 - 8
= 3
PQR = 90º [ mPQ x mQR = -1/3 x 3 = -1]

?correct substitution
?mPQ
?mQR
?mPQ x mQR = -½ x 2

(4)

4.4

(x- 4)2 + (y + 6)2 = 25

?r2  =  25
?equation

(2)

4.5 mrad -6+3  
              4-0
=-3/4
mtan4/3 
y = 4/3 x - 3

?correct subst.
?mrad  = -3/4
?mtan = 4/3 
?Subst . (0 ; -3) & m
?equation

 
4.6

(√146)2 = (cosθ - 8)2 +  (sinθ + 9)2
146 = cos2θ- 16cosθ + 64 + sin2 θ + 18 sin θ+ 81
0 = - 16  cosθ + 18 sinθ
0 = - 16  cosθ + 18 sinθ
sinθ = 16
cosθ   18
tanθ = 8/9

?correct substitution
?simplification
?sin2 θ + cos2θ = 1
?equation
?tanθ = 8/9
(5)
      [20]

 

QUESTION 5

5.1

4

sin ( A + 30º ) = sinA cos 30º  + cosA sin 30º
= (-3/7)(√3/2) + (-2√10)(½)
                            7
= -  3√3 - 2√10 
            14

?x  = - 2√10 
?Expansion
?Both √3  & ½
             2
?(-2√10)
       7

(4)

5.2

- sin2 (90º - x ) - tan cos (- x). sin (- x - 360º)
=  - (cos x )2 - sin x (cos x )(-  sin x )
                      cos x
= - cos2 x + sin 2x
= - (cos2 x - sin 2x)
= - cos2 x

?(cos x)2
?sin  x
   cos x
?cos x
?(- sin x )
?- (cos 2 x - sin 2 x )
- cos 2 x

(6)

5.3

x2 - 2 x sin   A = cos 2   A
cos 2 A +  2 x sin   A - x  = 0
Δ =  (2 sin A )2 -  4 (- 1 )(cos 2 A )
= 4 (sin 2   A + cos 2   A )
= 4

?standard form
?correct substitution
?Δ = 4

(3)
5.4

LHS / LK = cos 3xsin 3x  
                    sin x     cos x
= cos 3x cos x + sin 3x sin x
               sin x cos x
= cos (3x x)
   sin x cos x
= cos 2
  ½sin 2x
=   2   
  tan 2x

?Simplification
?cos 2x
? ½  sin 2x

(3)
5.5.1

sin  68º  = cos  22º 
= p

OR
5
sin  68º = p

?cos 22º 
?p

OR
?y = √1 - p2
?sin  68º = p

(2)
5.5.2

cos 16º  =  cos  (38º  - 22º)
=  cos38º . cos  22º   + sin   38º  . sin   22º
= √1 - q2 . p  +  q . √1 - q2
=  p √1 - q2  + q 1 - q2

?cos (38 0 - 22 0 )
?Expansion 
?cos 38º i.t.o 
?sin   22º i.t.o

(4)
      [22]

 

QUESTION 6 

6.1

a =  - 2
p = 30º

?a =  - 2
?p = 30º

(2)

6.2

6  
 

6.2.1

x = 60º

 

(1)

 

6.2.2

cos( x - 60 ) = sin 3 x
cos(   x - 60 ) = cos( 90º - 3 x )
x - 60 = ± (90  0  - 3 x ) + 360º .k   k ∈ Z
4 x = 150º + 360º .k
OR
- 2 x = - 30º + 360º.k   k Î Z
x = 37 ,50º + 90º.k or   x = 15º  - 180
x = - 52 ,50º and   x = 15º

?co-function
?ref ∠
?4 x  = 150º + 360º .k
?&
- 2 x  =  - 30º + 360º .k
?x = 15º
?x =  - 52 ,50º

(6)

 

6.2.3

- 52 ,50º<  x < 15º

?both critical values
?notation

(2)

       

[11]

 

QUESTION 7

7

7.1

AB = k
AC = √(2k)2 + k 2 - 2 . 2 k .k . cos 2θ
= √ 5 k2 - 4k2 . cos 2θ
= √ k2(5 - 4 (1 - 2 sin 2 θ ))
= √k2 (5 - 4 + 8 sin 2 θ)
√ 1 + 8 sin 2θ

? AB i.t.o 
? cosine rule formula in ∆ ABC
? correct subst.
? cos 2θ = 1 - 2 sin 2θ
? simplification

 (5)

7.2

AC = 139 ,5 √1 + 8 sin2 (42º)
299 m

? correct substitution
? answer

(2)

     

[7]

 

QUESTION 8

8

8.1

D2 = 28º[ ∠s subt by = chords ] 

? S      ? R
(2)

8.2

Alternate ∠s = 

? R
(1)

8.3

EB = EC [ radii ]
but 
EB = CD [given]
EC = CD

? S        ? R
(2)

8.3.1

D1=  28º [ ∠s opp = sides] 
C2 = 124º [∠s of  a  Δ ] 
B2 =  56º  [ opp . ∠s of a cyclic quad]

? S/R
? S
?S       ? R
(4)

8.3.2

E2 = 96º [ ∠s of a Δ]
BAC = ½( 96º + 28º) [ ∠ at centre=  2 x ∠ at circumf]
= 62º

? S

?S          ? R

 
     

 

QUESTION 9

9

9.1

YU = XV
UZ    VZ     [ prop the o , UV || YX or line || to one side of a Δ]

YV [prop the o , UV || ZW or line to one side of a Δ ]
   VW
XV = YV 
V Z   VW

? S ?R
?S
(3)

9.2

Area of ΔXVY½ x 3r x 3s x sin V2
Area of ΔWVZ   ½ x 4r x 4s x sin  V5
but V2 [vert . opp . ∠s ]
= 9/16

?substitution  
?substitution
?S/R
?answer
(4)

9.3

X1 =  V4 [alt . ∠s , XY || WX]
V3 =  V4  [given ] 
V3 =  W2 [corresp  ∠s , WZ || UV ]
X1 =  W1 
WXYZ is a cyclic quad [converse ∠s same segment or line subt = ∠s]

? S/R
? S/R
? R
(3)

9.4 V3 = V4 = X1 
UV is a tangent to circle XYV [converse of tan chord theo.]
?S
?R
(2)

 

QUESTION 10
10.1  10  
  Constr: Mark M on PQ and N on PR such that PM = AB and PN = AC
Proof : In ΔABC and ΔPMN
AB = PM [constr]
A = P [given]
AC = PN [constr]
Δ ABC º ΔPMN [S∠ S]
B =PMN
=Q [given]
MN II QR [corresp ∠s =]
PM = PN  [prop theo, MNQR]
PQ    PR
but AB = PM and AC = PN [const]
AB = AC
PQ    PR
constr
SSS
R
S/R
S/R
S
10.2  11  

10.2.1

N1 = 90º [ ∠subt by diameter /∠ in semi - circle]
LN = NP [line from centre ⊥ to chord

? S ? R
? R
(3)

10.2.2

P4 =  L4 [ tangent chord  theorem] 
LPR = 90º [ ∠subt by diameter]
R2 = 90º - P4  [ ∠s of Δ LPR]
R1 =  90º -  P4  [∠s of ΔRPQ]

? S ? R
? S/R
? S
(4)

10.2.3

N1 =  Q [both = 90º]
P2 =  L [∠s opp. = sides] 
= P4
M2 =  R1 [3rd ∠ ]
ΔPNM ||| ΔPQR [ ∠∠∠ ]

? S
?S ? R
?R
(4)

10.2.4

In ΔPLR and ΔQPR
LPR = Q [both =90º]
R2 =  R1  [proved]
L =  P4 [3rd ∠]
ΔPLR ||| ΔQPR [ ∠∠∠ ]
LR = PR
PR   QR
LR = 30 2
        15
= 60

? SSS
? R
? ratios
?substitution
?LR
(5)
10.2.5 NM || PR [co - int ∠s supp OR corresp ∠s =]

NM = ½ PR[midpoint theorem]
sin x =30√3
            15
x = 60º

? R
? R
? ratio
? value of x
(4)

    [26]
Last modified on Tuesday, 07 September 2021 11:05