MARKING CODES

A

Accuracy

AO

Answer only

CA

Consistent accuracy

M

Method

R

Rounding

NPR

No penalty for rounding

NPU

No penalty for omitting unit

S

Simplification

F

Correct formula

SF

Substitution in correct formula

1.1.1(a)

(12 + 2 x )  OR (12 + x + x )

🗸 length  A (1)

1.1.1(b)

(3 + 2 x )   OR  (3 + x + x )

🗸breadth A (1)

1.1.2

Area = length x breadth
= (12 + 2x )(3 + 2x )
= 36 + 24x + 6x + 4x²
= 4x² + 30x + 36

🗸🗸 SF 
CA (2)

1.1.3

4x² + 30x + 36 = 52
4x² + 30x - 16 = 0 OR    2x² + 15x - 8 = 0
2(2x -1)( x + 8) = 0

OR
x = ½  or  x ≠ -8 
Outsidelength = 12 m + 2(½)m = 13m

🗸equation CA
🗸factors/formula CA
🗸both x values CA
🗸length   CA
NPU 
(4)

³/_x = 7x - 5
7x² - 5x - 3= 0
x = -b ±  b² - 4ac = -(-5) ± √(-5)² - 4(7)(-3)
               2a                        2(7)
x = 5 ± √109
          14
x ≈ 1,10  or x ≈ -0,39

🗸standard form A
🗸 SF    CA
🗸both values of x
NPR
(3)

x ∈ {Real numbers}
OR x ∈ (-∞; ∞)  OR x ∈

🗸 x ∈ Real Numbers A (1)

y - x = 3 and 3x² + xy - y² = - 3
y = x + 3
3x² + x (x + 3) - (x + 3)² = -3
3x² + x² + 3x -(x² + 6x + 9) + 3 = 0
3x² - 3x - 6 = 0    OR   x² - x - 2 = 0

3(x - 2)(x +1) = 0    OR x = -(-1) ± √(-1)² - 4(1)(-2)
                                                        2(1)
x = 2 or  x = -1
y = 2 + 3=5  or y = -1+ 3= 2

OR
y - x =3    and  3x² + xy - y² = - 3
x = y - 3
3( y - 3)² + y ( y - 3) - y² = -3
3y² -18y + 27 + y² -3y - y² + 3 = 0
3y² - 21y + 30 = 0   OR    y² - 7 y +10 = 0

3( y - 2)( y - 5) = 0  OR  y = -(-7) ±    (-7)² - 4(1)(10)
                                                       2(1)
y = 2 or  y = 5
x = 2 - 3=-1 or x = 5 - 3= 2

🗸 subject A
🗸substitution CA
🗸 S     CA 
🗸 factors or formula CA
🗸 both x-values CA
🗸 both y-values CA

OR
🗸subject   A
🗸substitution CA
🗸S     CA
🗸factors or formula CA
🗸both y-values   CA
🗸both x-values  CA
(6)

X_C =    1    
        2πfC
f =    1      
    2πX_CC   OR   f = (2πX_CC)^-1

🗸making f the subject A (1)

f =    1      
    2πX_CC
=                 1                       OR    (2π x 63, 66 x 50 x 10^-6)^-1
  2π x 63, 66 x 50 x 10^-6
≈ 50 hertz

OR
X_C =    1    
        2πfC
63, 66 =          1             
               2πf x 50 x 10^-6

f =                   1                  
     2π x 63, 66 x 50 x 10^-6
≈ 50 hertz

🗸substitution  CA
🗸value of f   CA

OR
🗸substitution CA
🗸value of CA
NPR     NPU
(2)

110011₂ + 111101₂ = 1110000₂
OR
32 + 16 + 2 +1 + 32+16+ 8 +1 = 112 = 1110000₂

= 64 + 32 + 16 =  112

OR

51 + 61= 112

🗸M CA
🗸decimal       CA

OR
🗸M      CA
🗸decimal  CA
AO: Full marks (2)

[24]

2.1.1

3 x ² + 2x + 2 = 0
Δ = b² - 4ac
= (2)² - 4 x 3 x 2
= - 20

🗸 substitution A

🗸value of Δ  CA
AO: Full marks  (2)

2.1.2

non-real

🗸description   CA
Accept imaginary (1)

2.2.1

x² - 2 p x = 3 p²
x² - 2 p x - 3p² = 0

OR
-x² + 2 p x + 3p² = 0

🗸standard form A (1)

2.2.2

Δ = (-2 p)² - 4(1)(-3p² )
Δ = 4 p² +12 p²
= 16 p²

Δ isa perfect square \ roots will be rational

🗸subst. in discriminant  CA
🗸 S         CA
🗸perfect square  CA (3)

[7]

3.1.1 #

log 3 + log 27
log 81 - log 9

= log 3 + log 3³
   log 3⁴ -log 3²

=  log 3+ 3log 3 
  4 log 3 - 2 log 3

= 4 log 3 
   2 log 3
= 2

OR
log 3 + log 27
log 81 - log 9

= log (3 x 27)
     log(⁸¹/₉)
= log 81
    log 9

= log 3⁴    OR log 9² OR  log₉81
   log 3²            log 9

= 4 log 3  OR  2 log 9 OR   log₉9² = 2 log₉9
   2 log 3           log 9
= 2

🗸prime bases  A
🗸log property CA
🗸S      CA
🗸S       CA 

OR
🗸log property A
🗸 prime bases or log prop     CA
🗸S         CA

🗸S         CA
AO: 1 mark (4)

2ⁿ √32 + 2ⁿ √2    = 2ⁿ √2 x 16 + 2ⁿ √2
    2ⁿ √50                    2ⁿ√2 x 25

= 2ⁿ4 √2 + 2ⁿ √2
       2ⁿ× 5√2
= 2ⁿ √2 (4 + 1)
     2ⁿ× 5√2
= 1

OR
2ⁿ√32 + 2ⁿ √2 = 2ⁿ √32  +  2ⁿ√2  
  2ⁿ√50              2ⁿ√50      2ⁿ√ 50
= 4√2 + √2 
   5√2   5√2
= 4  + 1 
   5     5
= 1

OR
2ⁿ√32 + 2ⁿ√2  = 2n (2⁵)^½ + 2ⁿ2^½
     2n√50              2ⁿ (5² × 2)^½
= 2ⁿ 2^½  + 2ⁿ 2^½
       2ⁿ5 × 2^½
= 2ⁿ 2^½ ( 2² + 1)
      2ⁿ5 × √2^½
= 1

OR
2ⁿ√32 + 2ⁿ√2  = 2n (√32 +√2)
     2ⁿ√50                2n(50)
= 4√2 +√2
       5√2
= 5√2
   5√2
= 1

🗸simplified surds      A
🗸S       CA
🗸common factor or like terms CA
🗸 S        CA

OR
🗸 Separating terms      A
🗸S         CA
🗸 S        CA
🗸 S       CA 

OR
🗸exponent form     A
🗸S       CA
🗸common factor    CA
🗸 S              CA

OR
🗸Common factor     A
🗸S        CA
🗸common factor      CA
🗸 S        CA 
AO: 1 mark (4)

log_x32 + log_x4 - log_x16 = log₅125
log_x32 x 4 = log₅5³ 
         16
log_x 8= 3
x³ = 8 = 2³
x = 2

OR
log_x32 + log_x4 - log_x16 = log₅125
5log_x2  + 2log_x2 - 4log_x2 = 3log₅5
3log_x2 = 3
log_x2 = 1
x = 2

OR
log_x32 + log_x4 - log_x 16 =log₅125
log_x32 x 4 = log₅5³
          16
log_x 8= 3
x³ = 2³  OR   log_x2³ =log_xx³
x = 2

🗸log property      A
🗸power form A
🗸S     CA
🗸exp form CA
🗸value of x  CA

OR
log property  A
🗸log identity    A
🗸S     A
🗸S    CA
🗸 value of x    CA

OR
🗸log property     A
🗸power form A
🗸S        CA
🗸exp form  CA
🗸 value of x CA (5)

Z_T = 4 + 5i - 4 - 4i
= i

🗸total impedance  A (1)

z_T = i
r = 1
tanθ =¹/₀
θ = 90°   OR θ = ½π
z_T = 1(cos 90°+i sin 90°)  OR   zT = 1 (cos½π + isin½π)

🗸value of modulus CA
🗸tan ratio     CA
🗸correct angle  CA
🗸z in polar vorm    CA
AO: 1 mark

k = 6 + 4( i - 9)+ 2mi
k - 2mi = 6 + 4i -36
k - 2mi = - 30 + 4i
k = - 30     and - 2m = 4
k = - 30 and  m = - 2

OR
k = 6 + 4( i - 9)+ 2mi
k - 6 = 4i -36 + 2mi
k = - 30 + (2m + 4)i
k = - 30   and - 2m = 4
k = - 30 and  m = - 2

OR
k - 6 - 2mi =- 36 + 4i
k - 6 = - 36 and  - 2mi = 4i
k = - 30 and   m = - 2

🗸product      A
🗸S        CA
🗸value of k       CA
🗸value of m     CA

OR
🗸product     A
🗸S          CA
🗸value of k    CA
🗸value of m    CA

OR
🗸product  A
🗸S          CA 
🗸value of k CA
🗸value of  m CA
(4)

[22]

4.1.1

radius= 4,5 units

🗸length of radius       A (1)

4.1.2 #

CA from Q 4.1.1f:
🗸both intercepts    CA
🗸negative straight line  CA
🗸x-intercepts  CA
🗸y-intercept  CA
🗸semi-circle CA (5)

4.1.3

x ∈ [ - 4,5;4,5]  OR- 4,5 ≤ x ≤ 4,5

🗸end points CA
🗸correct notation   A (2)

4.2

🗸Horizontal asymptote  A
🗸x-intercept   A
🗸shape (both sections) A (3)

g(x) = a ( x - x₁ )( x - x₂ )
g(x) = a ( x + 4)( x - 2)
16 = a ( 0 + 4)( 0 - 2)
a = - 2
g(x) =- 2 ( x + 4 ) ( x - 2 ) OR - ^b/_2a = -1
g(x) =- 2x² - 4x + 16         - b   = -1
                                        2(-2)           
g ^/ (x) = 2ax + b = 0
OR   2(-2)(-1) + b = 0
b = - 4

OR
subst. U (2 ; 0):
0= a(2)² + b(2) + 16
4a + 2b = - 16......... (1)
subst./ verv. S ( 1 ; 10):
10 = a(1)² + b(1) + 16
a + b = - 6  ⇒ 2a + 2b =-12.......... (2)
(1) -(2) : 2a = - 4
a =- 2
2(- 2) + 2b = -12
b = - 4

OR
y = a(x + p)² + q y = a(x +1)² + q
Subst./verv. (2;0) : 0 = a (2 +1)2 + q
0 = 9 a + q............. (1)
Subst./verv. (1;10) : 10 = a (1+1)2 + q
10 = 4 a + q........... (2)
(1) - (2)          -10 = 5 a
a = - 2
10 = 4 a + q
10 = 4 (-2) + q
q = 18
y = -2(x +1)² +18
= - 2x² - 4x -16
b = - 4

🗸substitution in intercept form      CA

🗸value of a CA

🗸substitution CA

🗸value of b CA

OR

🗸substitution A

🗸substitution A 

🗸value of a CA

🗸value of b

OR/OF

🗸substitution A

🗸substitution A

🗸value of a CA

🗸value of b 

(4)

g(x) =- 2x² - 4x + 16
subst. x =-1
g(-1) =- 2(-1)² - 4(-1) + 16
y = 18

OR 
( R(-1; 18)

🗸substitution
CA ( Q4.3.2)
🗸 y-coordinate of R CA

OR
🗸🗸 y-coordinate of  R  CA (2)

h ( x ) = k ^x + 8
10 = k ¹ + 8
k = 2
h ( x ) = 2^x + 8

subst. x =-1
At W : y = 2^-1 + 8 = ¹⁷/₂ = 8,5
VW = ¹⁷/₂- 8   OR   VW = 8,5 - 8
= 0,5 units

OR
At W : y = 2^-1 + 8 =¹⁷/₂ = 8, 5
VW = √(1-1)2  + (8,5 - 8)²
=√0, 25
= 0,5 units

OR
h(x) = 2^x+ 8 eq. of the asympt y = 8
VW= 2^x+ 8 - 8 = 2^x
x = -1
VW = 2^-1 = 0, 5 units

value of y at W  A
M           CA
length of VW CA

OR
value of at W       A
M    CA
length of VW CA

OR
value of y at W   A
M         CA
length of VW CA
AO: Full marks (3)

[27]

5.1.1

90% of R 250 000 = R 225 000

OR
10 % of  R 250 000 = R 25 000
Loan value:
R 250 000 -R25 000 = R 225 000

Loan value  A

OR
Loan value   A (1)

5.1.2

OR
A = P( 1 + i )n
Let P = R100
A = 100 (1 + 6,3%)¹² 
                      12
= R106, 49
int erest = R106, 49 - R100
= 6, 49
i ≈ 6, 49 ≈ 6, 5

F             A
SF          A
value of ieff greater than 6,3%    CA

OR
F        A
SF          A
value of ieff greater than 6,3% CA
AO: Full marks NPR (3)

A = P (1 - i)ⁿ
60 =P(1 - 5, 43%)⁴
      60      = P
(1 - 5, 43%)⁴   
P ≈75,01
There were 75 unskilled workers during April 2019
Incorrect formula: one mark for value of n

F         A
n = 4     A
SF       A
Number of unskilled Workers  CA
Accept 76

Value of the investment at the end of the first 2 years
A = P (1+ i)ⁿ
=R 85000(1 + 5,4 % )^2x2
                         2
≈ R 94558,53

SF     A
R 94558,53     CA
PR
Incorrect formula: no marks (2)

Value of the investment after change in interest rate for 2 years
A = R94558,53(1 +  6% )^2x12
                                12
≈ R106582,57
Value of the investment after withdrawing:
P=R106582,57 -R20 000 =R86582,57
YES it will be more.

OR 
A =  R94 558,53(1 + 6 % )^4x12  - 20 000(1 + 6 %)^2x12
                                 12                                  12
≈R 97592,39
YES, it will be more. 

OR
A =[R94 558,53(1 + 6 %)^2x12 - 20 000 (1 + 6 %)^2x12
                                 12                                12
≈ R 97592,39
YES, it will be more.

CA from Q 5.3.1
SF         CA
R106582, 57        CA
M subtracting 20000   A
difference      CA
conclusion CA

OR
M        A
SF                                 CA
value of A_final CA
conclusion   CA

OR
M     A
SF       CA
value of i and n     A
value of A_final  CA
conclusion CA
NPR                                 (6)

[16]

6.1

definition     A
SF            A
S            CA
Penalty of 1 mark if incorrect notation used(4)

6.2.1

dA = 2πr
dr

derivative  A (1)

6.2.2

g(x) = ax² - x sub( -1; -1)
-1 = a(-1)² - (-1)
-1 = a +1
a = -2 

OR
3x - y + 2 = 0
y = 3x + 2
m_tan = 3
g ( x ) = ax² - x
g ^/ ( x ) = 2ax - 1

2a ( -1) - 1 = 3
- 2a = 3 + 1
a = - 2 

OR
ax² - x = 3x + 2
ax² - 4x - 2 = 0
b² - 4ac = 0(equal roots) tangent touches)
(-4)² - 4(a)(-2) = 0
16 + 8a = 0
a = -2

subst  A
S            CA 
value of A      CA

OR
gradient of tan.      A
derivative A
g ^/ ( x ) = 3         CA
substitution  CA
value of a  CA 

OR
equating A
std form     A
equal roots     CA
S   CA
value of a CA  (5)

[15]

7.1

y = f (0) = -(0 - 1)² (0 + 3) = -3
OR
(0 ;  –3)

y-intercept  A (1)

7.2

f ( x ) = -( x - 1)² ( x + 3)
x = 1 or x =- 3

OR
(1 ; 0 )   or  (-3 ;0)

x = 1           A
x =- 3         A (2)

7.3

f ( x ) = - x³ - x² + 5x - 3
f ^/ ( x ) = -3x² - 2x + 5
-3x² -2x + 5 = 0
3x² + 2x - 5 = 0
- (2) ±   (2)² - 4(3)(-5)
(3 x + 5)( x - 1) = 0  
x = - ⁵/₃ or  = 1
y = - 256 ≈ - 9,5 or y = 0 27
         27

derivative   A
f ^/ ( x ) = 0     A
factors/formula CA
both values of  CA
both values of y  CA

If derivative is first degree then Max 2 marks (5)

cubic shape  A
y-intercepts  CA from Q7.1
both
x-intercepts
CA from Q 7.2
both turning points
CA from Q7.3 (4)

Using calculator to generate table, maximum 3 marks

8.1.1

D(10) = - 0,5(10)²  + 20(10)
= 150 m

distance      A
NPU   (1)

8.1.2

velocity = D^/ (t ) = - t + 20
D^/ (12) = - (12) + 20
= 8 m/s

derivative   A
substitution in derivative CA
velocity   CA

8.2.1(a)

TSA. = ( 4x )(3x ) + (5x )( y ) + (4x )( y ) +(3x)( y )
= 12x ² + 5xy + 4xy + 3xy 3600 = 12x ² + 12xy
300 = x ² + xy
xy = 300 - x ²
y = 300 - x ²
          x

OR
TSA = ( 3x + 4x + 5x ) y + 2( ½× 3x × 4x)
12 xy + 12 x² = 3600
300 = x ² + xy   OR      xy = 300 - x ²    
y =300 - x ²
         x

area         A
equat. area to 3 600 A
S               CA

OR
area  A
equat. area to 3 600      A
S             CA
(3)

8.2.1(b)

V = ½ (3x)( 4x)(300 - x² )
                             x
= 6x (300 - x²)
= 1800x - 6x³

SF          CA
S     CA (2)

8.2.2

V = 1800x - 6x³
dV = 1800 - 18x²
dx
1800 - 18x ² = 0  OR  18(100 - x ²) = 0
x ² = 100
x = 10

derivative CA
equating derivative to 0 A
value of x  CA (3)

[12]

Penalize for constant C in either Q 9.1.1 or Q 9.1.2 

9.1.1

∫2^x dx
=  2 ^x  + C
  ln 2

  2 ^x  + C
ln 2 (2)

9.1.2

power vorm   A

7 lnx    A
-x^-4

OR
- 1   .....     A
 x⁴   (4)

Unshaded area = 125 - 34
                              6       3
= ¹⁹/₂ square units
the unshaded area is LESS than the shaded area

OR
Unshaded area above the x - axis

The unshaded area is LESS than the shaded area

area notation using integrals    A
integration      A
subst.  A
S       CA

area notation using Integrals
integration   A
subst  CA
subst.     CA
S     CA
M unshaded area CA
conclusion CA
AO (conclusion): 1 mark
(7)

[13]

ITEM

DESCRIPTION

1.1.2

Factors must have a variable x and product should lead to a quadratic equation.

1.1.3

If 4x² + 30x -16 = 0 is used and leading to negative x-values with not valid conclusion, maximum 2 marks

1.2.1

Linear equation, no marks

1.3

If simplification leads to linear equation,maximum 3 marks

1.5.1

If base 2 is omitted,no penalty

2.2.1

Order of terms not necessary

2.2.2.

• If p is omitted,accept Δ = 16 ,maximum 2 marks
• If Δ is irrational based on CA from Q2.2.1, maximum 3 marks

3.3.2 #

• tanθ = ¹/₀ can be implied
• If 1 is omitted, no penalty
• Accept ZT = 1cis 90

If:
Year 1: 60 ¸ (1- 5, 3%) = 63, 45
Year 2: 63,45 ¸ (1- 5, 3%) = 67, 09
Year 3: 67,09 ¸ (1- 5, 3%) = 70, 94
Year 4: 70,94 ¸ (1- 5, 3%) = 75, 01
\75 workers

• Simple interest used , maximum 3 marks
• Depreciation Compound used,maximum 4 marks

• If point by point plotted and Turning Point not shown maximum 3 marks
• If 2 Turning points are correctly plotted, award a mark