MATHEMATICS
PAPER 2
GRADE 12 
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2018

NOTE: 

  • If a candidate answers a question TWICE, only mark the FIRST attempt.
  • If a candidate has crossed out an attempt of a question and not redone the question, mark the  crossed out version.
  • Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the  second calculation error. 
  • Assuming answers/values in order to solve a problem is NOT acceptable. 
GEOMETRY  
 S  A mark for a correct statement
(A statement mark is independent of a reason.)
 R

A mark for a correct reason  
(A reason mark may only be awarded if the statement is correct.)

 S/R Award a mark if the statement AND reason are both correct.

MEMORANDUM 

QUESTION 1

Days

10

Units of blood

45 

59 

65 

73 

79 

82 

91 

99 

101 

106

1.1.1

x =  800 
       10
= 80
Answer only: full marks 

✔ 800 (addition of units) 
✔ answer (CA if ÷ 10) (2)

1.1.2

σ =18,83
No penalty for rounding

✔✔ answer (A)  (2) 

1.1.3

(61,17 ; 98,83) 
Days 1, 2, 8, 9 and 10 lie outside 1 standard deviation from the  mean 
∴5 days
Correct answer only: full marks provided  that 1.1.1. & 1.1.2 both correct

✔ mean – 1 SD 
✔ mean + 1 SD 
✔ answer   (3) 

1.2.1 

Skewed to the left or negatively skewed

✔ answer  (1) 

1.2.2 

A = 65 
B = 99
Answers without labelling: 1/2

✔ answer 
✔ answer (2) 

1.3 

New total = 95 × 10 = 950 
∴Units not counted = 950 – 800 = 150 

✔ answer (CA from 1.1.1)  (1) 

 [11]

QUESTION  2 
2

2.1 

Outlier  (100 ; 100)
accept: 100 as answer

✔ answer (1)

2.2 

a = 94,50273… 
b = 2,913729…  
yˆ= 94,50 + 2,91x 

  • Integral values: max 2/3 
  • Swopped a and b: 2/3

✔ value of a 
✔ value of b 
✔ equation  (3)

2.3

yˆ= 2,91(240) + 94,50 (CA from 2.1) 
 = 792,90 
Value = R793 000 

OR

yˆ=793,7978142 (calculator) 
Value = R794 000

  • Penalise 1 mark if answer  not in thousands of Rands

✔ substitution 
✔ answer in   thousands  of Rands (2) 

✔✔ answer in  thousands  of Rands  (2)

2.4 

b = 2,913729…  
∴ R2 914 OR/OF R2 910 (calculator)

  • Answer only: full marks

✔ value of b 
✔ answer  (2) 

[8]

QUESTION 3 
3

3.1

x = 3

✔ answer  (1)

3.2

mQP = tan71,57° 
 = 3

  • Answer only: full marks

✔ mQP= tan71,57°   
✔ answer  (2)

3.3

y = mx+c                                    y − y1 = m (x − x1
− 2 = 3(−7) + c        or                  y + 2 = 3(x + 7) 
y = 3x +19 

(m CA from 3.2 if > 0) 
✔ substitution of m & Q
✔ equation  (2)

3.4 

3.4

(wrong R: CA if x > 0) 
✔ substitution 
✔ answer (in surd form)  (2)

3.5

tan(90° − θ) = mQR
= 0 - (-2)
   3 - (-7)
= 1/5

  • Answer only: full  
  • tanθ =  1/5 ; 1/3 

(wrong R: CA if x > 0)
✔ gradient of QR/RN/QN 
✔ substitution of Q & R 
✔ answer    (3) 

3.6 

3.6

OR

3.6 A

✔ RN 
✔ SR 
✔ diagram (5 & √26) 
✔ use of correct area rule 
✔ substitution of sinθ 
✔ answer  (6) 

✔ RN 
✔ SR 
✔ diagram 
✔ use of correct area rule 
✔ substitution of sinθ 
✔ answer  (6)

✔ SR 
✔✔ ⊥ height 
✔ use of correct area   formula 
✔ substitution of sinθ 
✔ answer  (6)  

  

 [16]

 QUESTION 4 
4
 

4.1

OK = √180 or 6√5

✔ answer  (1)

4.2

a2 + b2 = 180
b = 1/2a
a2 + (1/2a)2 = 180
a2 + 1/4a2 = 180
a2 = 144     ∴a = -12
b = 1/2 (-12)
K (–12 ; – 6) (given) 

  • No penalty if x and y are not  converted to a and b
  • Error in simplification:  max 2/4

OR 

a2 + b2 = 180
a = 2b 
(2b)2 + b2 = 180  
5b2 = 180
b2 = 36       ∴ b = –6 
a = 2(–6) 
K (–12 ; – 6) (given) 

✔ b in terms of a 
✔ substitution 
✔ a2=  144
✔ substitution   (4) 

✔ a in terms of b 
✔ substitution 
✔ b2 =36 
✔ substitution  (4)

4.3.1  4.3.1

✔mPT= −2 
✔ substitution of m & K(–12 ; – 6)
✔ equation  (3)

4.3.2

4.3.2

OR

4.3as
3MK = OK 
9MK2 = OK2 = 180
∴MK2 = 20
Let M(x ; y), then y = ½x:
(x + 12)2 + (y + 6)2 = 20
(x + 12)2 + (  ½x + 6)2 = 20
4(x + 12)2 + (x + 12)2 = 80
(x + 12)±4
x = -16     x#-8 [since M is outside the large circle]
y = -8
M(-16; -8)

✔ 3MK = OK 
✔ OM = 4/3 OK 
✔✔ M = 4/3 (-12 ; -6) 
✔ x-coordinate 
✔ y-coordinate (6) 

✔ 3MK = OK 
✔ MK2= 20 
✔ equation 
✔ substitution 
✔ x-coordinate 
✔ y-coordinate (6) 

✔ 3MK = OK 
✔ ✔✔ diagram with  values OR  valid explanation
✔ x-coordinate 
✔ y-coordinate  (6)

✔ 3MK = OK 
✔ MK2= 20 
✔ equation  
✔ substitution 
✔ x-coordinate 
✔ y-coordinate  (6)

4.3.3 (x - (-16))2 + (y - (-8))2 = (1/3√180)2
(x + 16)2 + (y + 8)2 = 20

✔ LHS (CA from 4.3.2)
✔ RHS (CA from 4.1)  (2)

4.4 maths 4.4
Answer only: full marks 
(No need to simplify)

✔✔ values 
✔ inequality  (3) 

4.5

x2 + 32x + (16)2 + y2 + 16y + (8)2 = 256 +64 - 240
(x + 16)2 + (y + 8)2 = 80
New circle:
Centre(−16 ;−8)&  r = 4√5 
Original circle: M(−16 ; −8)& r = 2√5 
This circle will never cut the circle with centre M as they  have the same centre (concentric circles) but unequal  radii

✔ equation in centre,   radius form 
✔ Centre:(−16;−8) 
✔ r = 4 5(new) 
✔ r = 2 5(original) 
✔ conclusion  (“concentric” must be  stated)  (5)  [24]

QUESTION  5

5.1.1 

[ no calculator in 5.1] 

cos2θ = - 5/6, where 2θ ∈[180°;270°]
maths 5.1.1 jsah
y2 = 62 − (−5)2 [Pythagoras] 
y = ±√11 
(5 ; y) is in 3rd quadrant: 
∴y = −√11 
sin 2θ = − √11/

  • Getting to sin 2θ = − √11/ =: 3/4

OR 

sin22θ = 1 - cos22θ 
= 1 − (-5/6)2
= 1 - 25 /36
= 11 / 36
sin 2θ = − √11/

✔ diagram  (3rd quadrant only) 
✔ using Pythagoras 
✔ y – value 
✔ answer   (4) 

 

 

 

 

 

 

 

 

✔sin22θ = 1 - cos22θ 
✔ substitution 
✔ value of sin2
✔ answer  (4)

5.1.2

cos2θ =1− 2sin2θ 
2sin2θ = 1 - cos 2θ 
maths 5.1.2

✔cos2θ =1− 2sin2θ 
✔ substitution 
✔ answer   (3)

5.2

sin(180° − x).cos(−x) + cos(90° + x).cos(x −180°)
= sin x.cos x − sin x(−cos x) 
= 2sin x.cos x 
= sin 2x 

  • Second line written as: sinx cos x + sinx cos x: max 5/6

✔sin x
✔cos x 
✔ − sin x
✔− cos x 
✔ simplification 
✔ answer  (6)

5.3

sin3x.cos y + cos3x.sin y 
sin(3x + y) 
= sin 270° 
= –1

✔ compound angle 
✔ answer  (2)

5.4.1

2cos x = 3tan x 
2cos x = 3sin x
               cos x
2cos2x = 3sin x 
2(1 - sin2x) = 3sin x 
2 - 2sin2x = 3sin x 
2sin2x =  3sinx - 2 = 0 

✔ tan x  = sin x
                cos x
✔ multiplying by cosθ 
✔cos2x  =1− sin2x  (3)

5.4.2

2sin2x +  3sin x - 2 = 0 
(2sin x −1)(sin x + 2) = 0 
sin x = 1/2  or     sin x = −2 (no solution) 
x = 30° + k.360°    or       x =150° + k.360° ; k ∈Z

✔ factors 
✔ both values of sin x
✔ no solution 
✔30°+ k.360° 
✔150°+ k.360° ; k ∈Z   (5)

5.4.3

5y = 30º + k.360º        or         5y = 150º + k.360º
y = 6°+ k.72°               or                y = 30°+ k.72°
∴ y = 144° + 6°           or              y = 144° + 30° 
 y = 150°                     or                    y = 174° 

OR

144° ≤ y ≤ 216° 
720° ≤ 5y ≤ 1080° 
5y = 750° or 5y = 870° 
y = 150° or y = 174°

✔y = 6°+ k.72° 
✔y = 30°+ k.72° 
✔ 150°
✔ 174°    (4) 

✔ 5y = 750°
✔ 5y = 870°
✔ 150°
✔ 174°    (4)

5.5.1

g(x) = −4cos(x + 30°) 
maximum value = 4 

✔ answer   (1)

5.5.2 

range of g(x):  
− 4 ≤ y ≤ 4            OR                y ∈[−4 ; 4] 
∴range of  g(x) + 1:  − 3 ≤ y ≤ 5 OR     y ∈[−3 ; 5] 
Answer only: full marks

✔ range of g(x) 
✔ answer   (2)

5.5.3

y = −4cos(x + 30°) 
shifted to the left: 
y = -4cos( x + 30º +  60º ) 
= -4cos(x + 90º)
= 4sinx
∴h(x) = −4sin x 
Answer only: full marks

✔ shift of 60° to the left 
✔ reduction 
✔ equation of h  (3)  [33]

QUESTION  6 
maths 1 jhgh

6.1.1

tan θ = PQ  =  PQ 
            QR       x 
∴PQ = xtanθ 
Answer only: full marks

OR   =   PQ  
sinP     sinPRQ 
∴PQ  = x. sinθ
          sin(90º - θ ) 

✔ trig ratio 
✔ answer  (2) 

✔ trig ratio 
✔ answer (2)

6.1.2

    AR        =     QR       
sinAQR         sinQAR 
AR = xsin(90º + θ) 
             sinθ 
Answer only: full marks  

✔ use of sine rule  
✔ substitution into sine  rule correctly (2)

6.2 sin2θ  = AB
              AR
AB = ARsin2θ 
= xsin(90º + θ ).sin2θ 
               sinθ 
= xcosθ .sin2θ 
          sinθ 
= xcosθ .2sinθ cosθ 
             sinθ 
= 2x cos2θ 

✔ substitution into trig   ratio and AB as subject
✔ substitution of AR 
✔ co-ratio 
✔sin 2θ = 2sinθ cosθ   (4)

6.3 AB = 2cos212º
QP     xtan12º
= 9

✔ substitution  CA from 6.1.1) 
✔ answer (2)   [10]

QUESTION  7 
maths 2 ohjhag 

7.1.1

T1 = 70° [ext ∠ of cyclic quad] T 70  

✔ S ✔ R   (2)

7.1.2

Q1 = Q2 = 35º [equal chords;equal∠s/] 

✔ S ✔ R  (2)

7.2.1

T2 =Q1 = 35º [alt ∠s ; PQ || TR] 

✔ S ✔ R  (2)

7.2.2

PT= QR  [prop theorem ; PQ || TR] 
TS   RS 
TR = QR  [PT = TR]
   TS    RS 

✔ S ✔ R (2)  

[8]

QUESTION 8 
MATHS 3 KJHGAJG

8      

PTR = 90° [∠ in semi-circlel] 
x = 90º +  P  [ext ∠ of Δ] 
∴ R = x − 90°  
STP = x − 90° [tan chord theorem] 
x + x − 90° + y =180° [sum of  ∠s in ∆]
∴ y = 270° – 2x

✔ S/R 
✔ S/R 
✔ S ✔R 
✔ S 
✔ answer   [6]

QUESTION  9 
maths 4 jhghj

9.1 

equiangular ∆s OR (∠∠∠) 

✔ answer (1)

9.2

∴ GE  = DE       [||| Δs] 
   GF     GE 
GE2 = 45 ×  80 
GE = 60

✔ proportion 
✔ substitution
✔ answer  (3)

9.3 

In ΔDEH and ΔFGH: 
DHE = FHG [vert opp ∠s =]
DEH = FGH    [||| Δs]  
EDH = GFH    [sum of ∠s  in ∆]
∴ΔDEH ||| ΔFGH  

OR

In ΔDEH and ΔFGH: 
DHE = FHG [vert opp ∠s =]
DEH = FGH            [||| Δs]  
∴ΔDEH ||| ΔFGH    [∠∠∠]

✔ S/R 
✔ S/R 
✔ S  (3) 

✔ S/R 
✔ S/R 
✔ R (3)

9.4

GHFG    [||| Δs] 
EH     DE 
      GH        =  80 
  60 -  GH        45
45GH = 80(60 – GH) 
45GH = 4800 – 80GH
125GH = 4800 
GH = 38,4

✔S  
✔ substitution 
✔ answer (3) 

[10]

QUESTION 10 
maths 5 nbjh

10.1 

Construction: 
AO is drawn and produced to M 
O1 = A1 + B [ext ∠ of Δ] 
But A1 = B [∠s opp = radii]  
∴ O1 = 2A1 
Similarly : O2 = 2 A2 
∴ O1 + O2  = 2A1 +  2A
 = 2 (A1 + A2
 BOC = 2BAC

✔ Constr 

✔ S/R 

✔ S/R 

✔ S 

✔ S  (5)

10.2 
maths 6 kjhj

10.2.1(a)

F1 = 2x     [∠ centre = 2∠ at circum ]  

✔ S ✔ R  (2)

10.2.1(b)

C = x [∠s in the same seg]

OR

C = x [∠ centre = 2∠ at circum]

✔ S ✔ R  (2) 

✔ S ✔ R (2)

10.2.2

D3 = x      [∠s opp equal sides] 
E3 = 2x    [ext ∠ of Δ] 
∴ F1 = E3 = 2x
∴ AFED is a cyclic quadrilateral [converse ∠s in the same seg] 

✔ S/R  
✔ S/R 
✔ S  
✔ R (4)

10.2.3

A2 +  A3  +  D1 + F1= 180°      [sum of ∠s in ∆]
A2 + A3 = D1                            [∠s opp = sides]
∴ A2 + A3 = 90° − x 
E1 =  A2 + A3                          [ext∠of cyclic quad] 
= 90° − x 
FKE = 90°                              [line from centre bisects chord]
F3 = x                                    [sum of ∠s in ∆]

✔ S 
✔ S 
✔ R 
✔ S 
✔ S ✔ R  (6)

10.2.4

BAC = D3          [∠s in the same seg]
AE = BE            [sides opp equal ∠s] 
area ΔAEB1/2(BE)(AE).sinAEB 
area ΔDEC      1/2(EC)(ED).sinDEC 
6,25 =  AE2
           ED2  
AE=  2,5 
   ED 

✔ S 
✔ S 
✔ substitution   into area rule 
✔ simplification  of RHS 
✔ answer  (5) 

[24]

TOTAL: 150

Last modified on Tuesday, 17 August 2021 07:36