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
?? ≤ 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