TECHNICAL MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021 
 

Marking Codes
A Accuracy
AO Answer Only
CA Consistent accuracy
I Identity
F Correct Formula
M Method
NPR No penalty for rounding
NPU No penalty for units
R Rounding
RE Reason
S Simplification
SF Substitution in correct formula
ST Statement
ST/RE Statement with Reason
  • NOTE:
  • If a candidate answers a question TWICE, only mark the FIRST attempt
  • Consistent accuracy applies in all aspects of the marking guidelines where
  • Indicates the questions where tolerance range will be applied: 1 , Q4.2 , Q5.1 , Q8.5 

NOTE: if the candidate used (-3 ; 1) follow the marking guidelines in the addendum 

QUESTION 1

13
NOTE: if the candidate used (–3 ; 1) follow the marking guidelines in the addendum
1.1 a = 1
b = –1
value of A
1.2 KL =√(xK - xL)2 + (yK - yL)2
= √(1-(-3))+ (7-(-1))2
=√80 OR 4√5 OR ≈ 8,94
(2)
SF 
Length 
1.3

M(x+ xL) ; (yK + yL)
         2              2
M(1 +(-3) 7 + (-1) 
         2            2 
M(-1 ; 3)

OR
xM =x1 + x2 , yM y1 + y2 
            2                 2
xM =1 + (3)  ,7 + (-1) 
          2            2 
M(-1;3)

x value
y value
[Penalty of one mark if not simplified]
1.4 mKL = yL - yK     
            x - xL
= - 1 - 7 
   - 3 - 1
= 2
SF           A
gradient
 CA AO Full marks (2)
1.5 tanθ= m = 2
θ ≈ 63,4°
Penalty for rounding
CA from Q 1.4
gradient   CA
value of θ(rounded)CA
AO Full marks(2)
1.6 y = 2x + c
1= 2(-5) + c
c
= 11
y = 2x +11
OR
y - y1 = m ( x - x1 )
y - 1 = 2 ( x - (-5))
y = 2x +10 + 1
y = 2x + 11
ü gradien    CA 
SF (–5; 1)       A
equation    CA
OR
gradient  CA
SF (–5; 1)     A
equation
CA(3)
1.7

= 3/2x + 17/2
LHS = - 2
RHS= 3/2 (-4) + 17/2 = 5/2
LHS ≠ RHS
the point does NOT lie on the line

OR
mKL = 2
nNew = y1 - y2-2 - 1 = - 3    
            x1 - x2    -4 + 5
mNew  mKL
the point does NOT lie on the line

OR
y + 2 = 2( x + 4)
y = 2x + 6
(- 4 ; - 2)
does NOT lie ony = 2x +11

M LHS ≠ RHS         CA
conclusion  CA
OR
 
M mNew    A
conclusion  CA
OR
M equation       CA
conclusion g CA(2)
[15]

 

QUESTION 2

   14
2.1.1 r2 = x2 + y2
= (-5)2 + (12)2
= 169\ x2 + y2 =169  OR    x2 + y2 =132
x = ± √169 - y2 OR y = ± √169 - x2
SF            A  
equation  CA
AO Full marks 
(2)
2.1.2 t = √169 =13 value of t    CA(1)
2.1.3 mOB = -12/5
mtang = 5/12
y = mx + c  OR  y - y1 = m(x - x1 )
12 = 5/12 (-5) + OR y - 12 = ( x 5/12 (x -(-5))
c = 169/12
y =  5/12x  169/12 
OR
x.x1 + y.y1 = r2
x (-5) + y (12) = 169
12y = 5x +169
y 5/12x  169/12
gradient  A 
gradient  CA  
substitution (–5; 1)A   
equation CA  
OR
substitution(–5; 1)
169  CA
S  CA
equation CA
(4)
2.2 15 both x-intercepts A 
both y-intercepts  A 
elliptical shape  CA 
(3)[10]

QUESTION 3

3.1.1 16
(√13)2  = (3) 2 + (m ) 2
13 = 9 + m 2
m 2  = 4      
OR
m = √(13)2  - 32
m = 2
value of  m   A
AO Full marks (
1)
3.1.2 sec2 β + tan 2 β
= (√13) + (2/3)
      3
= 13/94/9
17/9 
OR
sec2 β + tan 2 β
= 1 + tan 2 β + tan β
= 1 + 2 tan 2 β
= 1 + (2/3)
= 1 + 8/9
17/9
CA from/ vanaf Q/V3.1.1
ratio of b    A
ratio of  tan b  CA
simplification   CA  
value of  b + tan 2CA
OR
ratio of tan b CA  
value of sec2 β + tan 2 β CA(4)
3.2.1 cosθ = ½
θ = 60°
value of θ  A(1)
3.2.2 tan a = - 1
ref ∠ = 45°
a = 180° - 45°
a = 135°
ref.
2nd quadrant value
AO Full marks
AA CA (3)
3.2.3 cos ( a - θ)  
= cos(135° - 60° )  
= cos 75°  ≈ 0 , 26        
OR  √6 - √2 
            4 
substitution  CA  
value of (a - θ ) CA
NPR      (2) 
AO Full marks
3.3 2 tan x + 0 , 924 = 0
2 tan x = - 0 , 924
tan x = - 0 , 462
ref∠ ≈ 24 ,8°
x ≈ 180° - 24 ,8° or x ≈ 360° - 24 ,8°
x » 155, 2° or x ≈ 335, 2°
 S
ref∠
x ≈  155, 2°  CA
x ≈  335, 2°  NPR 

QUESTION 4

4.1 cosθ (tanθ + cotθ )
=cosθ (sinθcosθ)
            cosθ   sinθ
= cos θ (sin2 θ+ cos 2θ)
               cosθ × sinθ  
= cosθ(       1       )
           cosθ× sinθ    
=     1      OR cosecθ
    sinθ 
OR
cosθ (tanθ + cotθ )
= cosθ × tanθ+ cosθ ×cotθ
= cosθ × sinθ + cosθ× cosθ
              cosθ             sinθ
= sinθ + cos2θ 
              sinθ
sin2θ + cos2θ
          sinθ
=    1        OR cosecθ
   sinθ 
OR
cosθ( tanθ+    1     )
                     tanθ  
cosθ x(tan2 θ +1)
               tanθ 
cosθ x (sec2θ)
              tanθ
= cosθ × (    1     . cosθ)
                 cos2θ   sinθ
=   1     OR cosecθ
   sinθ
(5)
4.2 sin 2 (180° + B) ×cosec(π- B)
 sec( 2π - B) ×cos (180° - B)
sin 2 B×cosecB
   sec B×(- cos B)
= sin2B × 1=   1   
                    sinB
-    1      ×cos B
  cosB
= - sin B  
OR 
sin 2 (180° + B) ×       1        
                            sin (π - B)
          1          ×cos (180° - B)
cos ( 2π - B)
sin 2 B× 1=     1     
                   sin B
     1    ×( - cos B)
cos B
sin 2 B ×   1   
             sinB  
-    1     ×cos B
  cosB
= - sin B

sin 2 B                        A
cosec B                      
secB                          
- cosB                        
A                          
AsinB
I                  
AcosB
S                               
CA
OR        
Asin (p - B)     
Acos( 2p - B)
sin 2 B                       
sin B                           
cos B                          
- cosB                         
A  
                               
CA (7)


QUESTION 5 

 5.1   17
5.2.1 x = 90° and x = 270° 90°
270°(2)
5.2.2 x ∈ (90ο ; 135°] or 
x =180°  
OR
90° < x ≤ 135° or x =180°

x ∈ (90ο ; 135°]
x =180° 
OR
90° < x  135° CA
x =180°  CA(2)
[12]

 

QUESTION 6

   18
6.1 PR 2 = QR 2 + PQ 2 - 2QR×PQcos Q
= ( 750 ) 2 + (1200 ) 2 - 2 (750 ) (1200 ) cos 60°
= 1 102 50
PR = 1 050 m
ü cosine rule SF  
value/PR  A (3)
6.2 S = 120° size of S A(1)
6.3    PS     =   PR  
 sin R1     sin S
    PS       = 1 050 
sin 40,5°    sin120°
PS = 1 050 sin 40 , 5°
               sin120°
PS ≈ 787, 41m
sine rule
SF  
value of PS  NPR 
(3)
6.4 Area ΔQPR =½QR ×QPsin Q
= 1 (750)(1 200)sin 60°
≈389711, 43 m 2
area rule   A
SF      A
value of  CA(3)
[10]

Related Items

 

QUESTION 7

7.1 are equal answer    A(1)
7.2 19
7.2.1(a) PTS = R1 = 56° (∠s in thesame segment)
OSR =R1 =56°(∠s opp. = sides) 
TPS = PTS = 56º
∠s opp. = sides =
OR= chords subtend
ST
RE    A
ST  RE   A
ST 
A
(5)
7.2.1(b) PSR =90° (∠s in semicircle)
P1 + 90°+ 56° =180°  (sum of ∠s of Δ) 
P1 = 34°
OR
O1 =112°  ∠ at centre = 2 x ∠at circum 
P1 = S1 =34° (∠s opp. = sides)
ST
RE        A
value of P1                                       CA
OR
ST RE     A
value of P1                                       CA(3)
7.2.1(c) 34°+ P2  =56°
P2  = 22°
S3 = P2 = 22° (∠s in same segment)
OR
S1 + S2 + S3 = 90°  (∠ in the semi-circle)
S1  + S2    = 180°- 112°   (sum of ∠s of Δ)
= 68°
S3 = 90° -68° =  22°
OR
O2 + O3 =112°
∠at centre = 2 x ∠at circum.
S2 = T2 = 34° [∠s opp. = sides
S3 = 90° - 68° = 22° 
∠ in the semi-circle
ST     CA
ST       CA
RE          A
OR
ST   CA
ST        A
ST         CA
OR
ST
ST
ST    CA
A   CA(3)
7.2.2

O3 = 44°
∠at centre = 2 x ∠at circum
O3 ≠ R1
OT is not parallel to SR (alt. ∠s are not equal)

OR
O3 = 44°
∠at centre = 2 x ∠at circum
O2 = 68° 
∠at centre = 2 x ∠at circum
SOT + OSR = 44°+ 68°+ 56°=168°¹180°
OT is not parallel to SR (co-int ∠s ≠180°)

OR
S2 = T2 = 34° [∠s opp. = sides]∠ in the semi-circle/
S3 = 90° - 68° = 22
T2≠ S3
OT is not parallel to SR  (alt. ∠s are not equal)

ST
RE A 

OR
ST    A
ST  A
RE  A 

OR
ST  A
ST  A
RE  A  (3)

 

QUESTION 8
20

8.1

M = 98° ≠ 90°
LN is not a diameter (∠ subtended by LN ≠ 90°)

OR
P2 + 98°=180° (Opp. ∠s of cyclic quad.)
P2 =82°¹ 90°
LN is not a diameter (∠ subtended by LN ≠ 90°)

ST M = 98° ≠ 90° A
RE              A 

OR
ST P2 =82° ≠ 90° A 
RE      A(2)

8.2.1 P2 + 98°=180° (Opp. ∠s of cyclic quad.)
P2 =82°
ST / RE         A 
P2  =82°     A(2)
8.2.2

P1 +82°=180°  ( ∠s on straight line)
P1 =98° 

OR
P1 =98°   (Ext. ∠ of a cyclic quad.)

ST / RE    A
P1 = 98°    CA

OR
ST / RE        A
P1 = 98°    CA(2)

8.2.3 L1 = 27° (tan-chord theorem) ST        A
RE        A(2)
8.3.1

K is common
L1 = N1   (both = 27° / tan- chord)
P1 = KLN ( 3rd ∠ of D )
ΔKLP|||ΔKNL (∠∠∠)

OR Equiangular

ST          A
ST        A 
ST/RE     A(3)
8.3.2 KL = KP    (|||∠s)
KN   KL
KL2 =KN.KP
ST                A
RE            A(2)
8.4 KL2 =KN.KP
(6)2  = 13.KP
KP ≈ 2,77 units
subst            A 
value of  KP     (2)
8.5

K + 27°+ 98°=180°  (∠s of/van D)
K = 55°
K + M = 55°+ 98°¹ 180°
KLMN is not a cyclic quad.

OR
K + L1 + = 86° ext∠ = sum of opp.in t ∠s 
K = 55°
K + M = 55°+ 98°¹ 180°
KLMN is not a cyclic quad.

ST/RE        CA
value of  K   A 
Conclusion    A 

OR
ST/RE          CA
value of  K    A 
Conclusion

    [18]

 
QUESTION 9
 
21

9.1.1 AB = AC   (Prop. theorem; DE || BC)
DB   EC
1,8 =
DB    2
DB = 2/x 1 , 8 m
DB=1,2 m
ST/RE   A    
length of  DB    A
(2)
9.1.2 AD= 1,8/3 = 0,6 m  or AD =1,8 -1, 2 = 0, 6 m
DF = 3/2 (0,6 m) = 0,9m
M       CA 
length of DF   CA(2)
9.2

CF ==1  (BF = FC; F is the midpoint of BC)
FB    1
CE == 2
EA    1
CF CE
FB    EA
EF is NOT parallel to AB (sides are not prop)

OR
BF = FC; F is the midpoint of BC
AE≠ EC; ; E is NOT the midpoint of AC
EF is NOT parallel to AB(FE not joining midpoints of two sides of a triangle)

ST             A 
ST            A 
Conclusion      CA

OR
F is the midpoint of BC          A
E is NOT the midpoint of  AC       A
Conclusion   CA(3)

    [7]

 

QUESTION 10

 22
10.1.1(a) BC = 6,95 - 4 = 2,95m

height of segment A
NPU           (1)
10.1.1(b)

h = 2, 95 m and d =13, 9 m
4h2 - 4dh + x2 = 0
4(2, 95)2 - 4(13, 9) (2, 95) + x2 = 0
-129, 21+ x2 = 0
x2 = 129, 21
x = 11, 36ED ≈ 11, 37 

OR
Using the half chord of PQ 
½ ED = √(6,95)2  - (4)2
½ED = √32,3025
½ED = 5,68
ED ≈ 11,37

formula            A 
SF         CA
S           CA  
length    CA 

OR
Pythagoras       A
SF         CA
S               CA  
length   CA 
(4)
NPR

 23
10.1.2(a)

angle of sector, FOG = 20% x 2π
= 2/5π = 1, 26 rad

OR
angle of sector, FOG = 360°x 20/100 = 72°
72° = 72° x π/180°
= 2/5π

OR 1, 26 rad

OR
Circmf. = 2π
= 2π(6, 95)
= 43, 67m
20% x 43, 67m = 8,73
s =
8, 73 = 6, 95θ
θ = 1, 26 rad

A
radian         CA

OR
angle size   A
M                A
radian   A

OR
Circumf.   A
M           CA
radian  CA
NPR       NPU         (3)

10.1.2(b)

A = r 2θ 
        2
= (6,95)2(1,26)
             2
≈30,43 m 2

OR
A = rs 
      2
(6,95)(8,73)
          2
= 30,43 m 2

Formula  A
SF              A
area    CA
NPU

OR
Formula    A
SF            A
area   CA
NPU
NPR 
 
(3)

10.2.1

n =   18   
      3600
n (in rev/sec) = 0, 005 rev / sec 
n
(in rad/sec) = 0, 005 rev / sec x 2π
= 0, 01p rad / sec or
0, 03141..rad 

M    n (in rev)     A
value of n   CA
NPU
NPR  
AO Full marks(2)
10.2.2

D = 2 x 10 m = 20 m
v = π D n                           

OR
v = 2πr n
=π x 20 x (  18  )   = 2π x 10 x (  18  )
                 3600                         3600 
= 0,1p m/s   

OR
 0, 31 m/s

Formula  A
SF         CA
circum.velocity     CA
NPU
NPR      (3)
10.2.3

w = 2πn
= 2π (  18  )
          3600 
= 0, 01π rad

OR
3,14 x 10 -2 rad

Formula
SF            CA
ang.velocity      CA
NPU
NPR          (3)
    [19]

 

QUESTION 11

24
11.1.1

Area = length´breadth
187,5 =length x 7,5

OR
length = A/b = 187,5/7.5
length = 25m

M           A 
length    A
AO Full marks
(2)
11.1.2 p = 15m values of p CA(1)
11.1.3

AT = a(o1 + on+ o2 + o3 + ....+ on-1)
                2
= 1, 5 (15 + 7 + 12 +18 +15 +11) m2
                2
=1, 5 (11+ 12 +18 +15 +11) m2
=100, 50 m2
Damaged area =187, 5 -100, 50 = 87m2
It will take 87 x 0, 25 hours = 21, 75 hours to repair the damaged area

OR
AT = a(m1 + m2 + m3 +....+ mn)
=1.5(15 + 1212 + 1818 + 11515 + 1111 + 7)m2
              2              2              2               2            2
= 1, 5 (13, 5 + 15 +16, 5 +13 + 9) m2
=100, 50 m2
Damaged area = 187, 5 -100, 50 = 87m2
It will take 87 x 0, 25 hours = 21, 75 hours to repair the damaged area

OR
AT = a(o1 + on+ o2 + o3 + ....+ on-1)
                2
= 1, 5 (10 + 18 + 13 +7 +10 +14) m2
                2
= 1.5(14 + 13 + 7 + 10 + 14)m2
= 87m 
It will take 87 x 0, 25 hours = 21, 75 hours to repair the damaged area

formula  A
value of a   A
SF    CA 
value of ATCA
87m2   CA 
time  CA

OR
F         A
value of a  
A
S     CA
value of AT  CA
87m2    CA
Time    CA 

OR
F         A
value of a  
NEW ordinates 
S     CA
value of AT  CA
87m2    CA
Time    CA 

11.2     25
11.2.1 1 l = 1 000 cm3
1,5l = 1500 cm3
value of  A
11.2.2 TSA/TBO = 4( ½ side length of base x slant height) + (sidelength )2 
=4(½ x 3 x 3.81) + (3 x 3)
= 22.86 + 9
= 31.86cm2
F        A  
SF           A 
value of TSA   CA
NPR/NPU
AO Full marks
11.2.3 Volume of pyramid = 1/3 (length x breadth) x ⊥ Height
1/3 (3 x 3) x 3,5
= 10, 5 cm3
number of small pyramids = 1500
                                              10,5
≈142,86
142Remaining milk= 1500 - (142 x 10, 5)
OR 0,86 x 10,5
= 9cm3OR 9 ml
SF             A
value of V pyramid  CA
M           CA 
value of  CA  
NPU/NPR(4)[17]

TOTAL: 150 

Last modified on Thursday, 24 February 2022 06:34