QUESTION 1

1.1

1.1.1

x(x–3)
x = 0 or x = 3

🗸 x = 0
🗸 x = 3

(2)

1.1.2

3x²– 2 x – 1 0  =  0    (-1 Mark for incorrect rounding)
x = – b ± √b² - 4ac
               2a
x =  – (- 2) ± (- 2 )² - 4(3)( - 10)
                         2(3)
x = 2 ± ( - 2 )² - 4(3)(-10)
                  2(3)
x = 2,2 or x = -1,5

🗸Formula
🗸Substitution
🗸 x = 2 , 2
🗸 x = - 1 , 5

1.1.3

2x² – 7 x + 3 ≥ 0
(2x–1) (x– 3) <0
Critical values : 
x = ½ or x ≥3

x = ≤ ½ or x ≥ 3

🗸Critical values
🗸𝑥 ≤ ½ 
🗸 x ≥ 3

(3)

1.2

x² - 4
x + 2
= (x + 2) (x - 2)
          x + 2
= x – 2
= 2 000 000 000 002 – 2
= 2 000 000 000 000
= 2 x 10¹²

2y +  x  =  3........................................ (1)
y  =  x 2   -  x........................................ (2)
Substitute (2) into (1):
2(x² - x ) + x = 3
2x²  - 2x + x - 3 = 0
2x² - x - 3 = 0
x = - (- 1) ± √(-1)² - 4(2) (- 3)
                       2(2)
x = - 1 o r x = ³/₂
y = (1)² - (1) o r y = (³/₂) ² - (³/₂)
y = 0   or   y =³/₄

🗸Substitution
🗸STD form
🗸Substitution
🗸x = –1
🗸 x = ³/₂
🗸 y = 0 o r y = 3

OR
2 y +  x  =  3........................................ (1)
y  =  x 2   -  x........................................ (2)
Substitute (2) into (1):
2(x² - x) + x = 3
2x² -  2x + x - 3 =  0
2x² -  x  - 3 =  0
(x + 1) (2x - 3 ) = 0
x = - 1 o r  x = ³/₂
y = (1)² - (1) o r y = (³/₂) ² - (³/₂)
y = 0   or   y =³/₄

OR
🗸Substitution
🗸STD form
🗸Factors
🗸x = –1
🗸 x = ³/₂
🗸 y = 0 o r y = ³/₄

b²– 4ac <0
(–3)²– 4(2) p <0
9 – 8p <0
p >⁹/₈

🗸 b²–4ac < 0
🗸Substitution
🗸 p >⁹/₈

QUESTION 2

2.1.1

√– 18.√– 12  = √– 3².2.√– 2².3
       √– 6                 √– 2.3
= 3 i.√2 2√3i
     √2 .√3i
= = 6i

🗸Prime factors
🗸 i =            1
🗸 6 i

OR
√– 18.√– 12  = √– 3².2.√– 2².3
       √– 6                 √– 2.3
= 3.2i.√3 
= 6i

2.1.2

log 6 + 2 log 2 0  –  log 3 – 3 log 2
log 6 + log ( 2 0 ) 2 – log 3 – log 2²
= log 6 + log 400 – log 3 – log 8
= log (6 × 400)
            3 x 8
= log 100
= log 10² OR = log (1 0 × 1 0 ) = lo g 10 + log 10

= 2

2.2.1

🗸Power rule
🗸Factors
🗸Simplification
🗸 25

OR

🗸Substitute 5
🗸Zero exponent rule
🗸x = 1

4 log₂x – 1 = log₂8
4 log₂x  = log₂8 + log₂2
log₂x⁴ = log₂16
x⁴  = 16
x⁴ = 24
x = 2

OR

4 log₂x – 1 = log₂3
log₂x⁴ – 1 = 3log₂2
log₂x⁴ = 3 + 1 = 4
x⁴ = 2⁴
x = 2

OR
🗸Power rule
🗸1 = log ₂2
🗸Exponential form
🗸x = 2

QUESTION 3

3.1

3.1.1

z₅  =  z₁ + z₂
= 2+3i+(–3 – 2i)
= –1 + i

🗸-1 Real part
🗸 i Imaginary part

(2)

3.1.2

z₆  = z₅ x z₃
= (–1 + i )( –4 + i )
= 4 – i – 4i + i²
=3– 5i

🗸Expansion
🗸3– 5i

(2)

3.1.3

Output =   z ₆ = 3 - 5i
                 z₄     2 + 1
= 3 - 5i  x  2 - i
    2 + i       2 - i

= 6 - 3i - 10i + 5i ²
         4 - i²
= 1 - 13i
       5
=  1  -  13i 
    5       5 

🗸Conjugate product

🗸Expansion

🗸Simplification

1  -  13i 
5       5 

(4)

3.2

3.2.1

🗸Substitution
🗸|Output|

(2)

🗸Correct quadrant
🗸Coordinates

No, the learner did not manage to cut out a circular piece, because the modulus of the output is less than 5 and the quadrant of the argument of the output is in the fourth quadrant; so a square was cut.

🗸Conclusion
🗸Modulus and Argument

QUESTION 4

4.1

4.1.1

A = P( 1 - i )ⁿ
4000 = 12000 (1 - i) ³⁰ 

Rate = 3,6 %

🗸Formula
🗸Substitution
🗸 1 - (¹/₃)^1/₃₀
🗸r = 3,6%

(4)

4.1.2

A = P (1 - i) ⁿ
A = 12000 (1 - 0, 036) ⁶⁰ 
A = 1329.85 Bacteria

🗸Substitute A and i
🗸Substitute n
🗸A = 1329.85

(3)

4.2.1

18% of R600 000 = R108 000
Loan amount = R600 000 – R108 000
Loan amount = R492 000

OR

Percentage loaned = 100% - 18%
Percentage loaned = 82%

Loan amount = 82% of R600 000
Loan amount = R492 600

🗸11,9% of R600 000
🗸Loan amount

OR
🗸Percentage loaned
🗸Loan amount

(2)

4.2.2

A = P (1 + i )ⁿ
1204860, 32 = 492000 (1 + ^0,15/₁₂) ^{12 x n} 
12n = log_1,0125 (1204860,32)
                             492000
n = 6years

🗸Formula
🗸Substitution
🗸Logarithm
🗸n = 6 years

[13]

QUESTION 5

5.1

q =2

🗸Accurate answer

(1)

5.2

0 = a  + 2
    - 2
a = 4

🗸Substitution
🗸a =4

(2)

5.3

x = 0 and y =2

🗸Horizontal asymptote
🗸Vertical asymptote

(2)

5.4

x ∈  R , but x ≠ 0 o r x ∈ ( - ∞ ; 0 ) ∪ ( 0 ; ∞)

🗸Excluding x =0
🗸x values

(2)

[7]

QUESTION 6

6.1

A (–2; 0) and B(2; 0)

🗸 A (–2; 0)
🗸 B (2; 0)

(2)

6.2

2 = 2⁰ + q
q = 1

🗸q =1

(1)

6.3

m = 0 - 2 = -1
       2 - 0
c = 2
y = -x + 2

🗸m = - x
🗸c =2

(2)

6.4

- 73 < x < 0  OR  x ∈  (-73 ;0)
  50                               50

🗸Notation
🗸Values

(2)

[7]

f (x) = - (x - 2) 2 + 40 = - (x -  2) 2 + 4
0 = [- ( x - 2) + 2 ] [ (x -  2) + 2]
- (x - 2) + 2 = 0 o r (x - 2) + 2 = 0

OR

0 = - ( x - 2)   + 4
( x - 2)²   = 4
x - 2 = ± 2
x = 4 o r x = 0

OR

0 = - ( x - 2)² + 4
0 = - ( x² - 4 x + 4 ) + 4
0 = - x² + 4 x
0 = - x ( x - 4 )
x = 4 o r x = 0

🗸 f ( x ) = 0
🗸Factors
x = 4 o r x = 0
🗸 x  =  4 o r x  =  0

OR
🗸 f ( x ) = 0
🗸Square root both sides

🗸 x  =  4 o r x  =  0

OR
🗸 f ( x ) = 0
🗸STD form

🗸 x =  4 o r x  =  0

f ( x ) = - ( x - 2 ) 2   + 4
y = - ( 0 - 2 ) 2 + 4
y = 0

🗸Shape
🗸y-intercept🗸x-intercepts
🗸Turning point

🗸x-coordinate
🗸 y-coordinate

QUESTION 8

8.1

f (1) = 2(1) 2 + (1) - 1 = 2
f (3) = 2(3) 2 + (3) - 1 = 20
Average gradient = f(3) - f (1)
                                   3 - 1
Average gradient = 20 - 2
                                  2
= 9

🗸f (1)
🗸f (3)
🗸Average gradient formula
🗸Substitution
🗸Average gradient = 9

8.2

f (x) = 3x
f ' (x) = lim f(x + h) - f (x) 
        x → 0          h
f ' (x) = lim 3(x + h) - 3x
       x → 0       h
f ' (x) = lim  3h 
        x → 0  h
f ' (x) = 3

🗸Formula
🗸Substitution
🗸Simplification
f ' (x) = 3

(4)

-1 Mark for incorrect notation in 8.2 or 8.3

8.3.1

3x - 2y = √x
y = 3x -√x
         2
y =   3x  -  x  ^½ 
         2      2
dy =3 -x^-½
dx    2   4 

🗸y = 3x -√x
         2
🗸Exponential form
🗸 ³/₂
🗸 x  ^-½
   4 

🗸Exponential form

🗸 4 x^4/₃ 
    3
🗸 - 4x - 5

f (5) = (5 - 5) (5 + 1) = 0
Then, (5; 0) is the x-intercept of f.
(- 1; 0) is the other x-intercept of f.

🗸 f (5) = 0
🗸 ( - 1; 0)

0 = (x - 5) (x + 1)
(x - 5) = 0 or (x + 1) = 0
x = 5 i s the intercept of f
( - 1;0) is the other x-intercept of f.  

🗸 f (x) = 0
🗸 ( - 1; 0)

f (x) = x³ - 3x²   - 9x - 5
f '( x ) = 3x² - 6x - 9
0 = 3x² - 6 x - 9
0 = x²   - 2x - 3
0 = (x - 3) ( x + 1 )
x = 3 or x = - 1
x = –1;
y = –1 – 3 + 9 – 5 = 0
Turning point: (–1; 0)
x = 3;
y = 27 – 27 – 27 – 5 = –32
Turning point : (3; –32)

🗸Factors
🗸Both x values
🗸(–1; 0)
🗸 (3; –32)

🗸Shape
🗸x-intercepts
🗸y-intercept = –5
🗸Turning Point (–1; 0)
🗸Turning Point (3; –32)

QUESTION 10

10.1

A = 6 + 4t - t ²
A = 6 + 4 (0) - (0)
A = 6 cm²

🗸Substitute 0
🗸 A =  6 cm 2

(2)

10.2

dA = 4  -  2 t
dt
Rate of increase in A at t =1: = 4 - 2 (1)
= 2 cm²

🗸 4 - 2 t
🗸Substitution by 1
🗸2 cm²

(3)

10.3

dA = 4 -  2t
 dt
dA = 4 -  2 t  =  0
dt
2t =  4
t = 2 seconds

🗸 dA =0
    dt
🗸t =2 seconds

(2)

10.4

A = 6 + 4 (2) - 2²
= 10 cm²

🗸Substitution
🗸 A = 10 cm²

(2)

[9]

QUESTION 11

11.1

∫( 2 x - 4 ) dx
= x² - 4 x + C

🗸 x²
🗸 - 4x
🗸C

(3)

11.2

=7,75 square units

🗸A1 definite integral formula
🗸Simplify A1 integral
🗸Substitution in A1
🗸A1 value
🗸A2 definite integral formula
🗸Substitution in A2
🗸A2 value
🗸 A₁  + A₂
🗸7,75 square units

(9)

[12]

TOTAL:

150