QUESTION 1

1.1.1

(3 - x)(x +1) = 0
x = 3  or  x =1
OR
x² - 2x - 3= 0
x = -b ± b ² - 4ac
              2a
= -(- 2) ± (- 2) 2 - 4(1)(- 3)
                 2(1)
x = 3  or   x =-1

x = 3 
x = - 1    

OR

x = 3 
x = - 1
(2)

1.1.2

2x² = 3x + 7
2x² - 3x - 7 = 0 OR - 2x ² + 3x + 7 = 0
x = -  b ± b² - 4ac
              2a
= - (- 3) ± (- 3)2 - 4(2)(-7) = 3 ±√ 65
                  2(2)                       4
x ≈ 2,77 or x ≈ - 1, 27

standard form
SF 
CA
positive x value CA
negative x value CA
AO: Full marks NPR
(4)

x ( x - 5 ) ≤ 0
Critical values: 0 and 5
0 ≤ x ≤ 5 OR  x ∈ [0;5]  OR
x ≥ 0  and x ≤ 5

both critical values A
notation  A
(2)

y + x = 3       and      x² + y² = 89
y = 3 - x
x ² + ( 3 - x ) ² = 89
x ² + 9 - 6x + x ² - 89 = 0
2x ² - 6x -80 = 0
x ² - 3x - 40 = 0
( x - 8)( x + 5) = 0   OR x = -(- 3) ±    (- 3) 2 - 4 (1)(- 40)
                                                          2 (1)
x = 8    or  x = - 5
y = 3 - 8 = -5    or  y = 3 - ( - 5) = 8

OR
y + x = 3   and  x² + y² = 89
x = 3 - y
(3 - y )² + y² = 89
9 - 6 y + y² + y² - 89 = 0
2y²- 6 y - 80 = 0
y² - 3 y - 40 = 0
( y - 8)( y + 5) = 0 OR  y = - (-3)±  (-3) 2 - 4(1)(-40)
                                                       2(1)
y = 8      or   y = - 5
x = 3 - 8 = - 5    or x = 3 - ( -5) = 8

y subject of formula  A
SF
CA
correct standard form  CA
factors/formula
CA
x-values CA
y-values  CA 

OR
x subject of formula A
substitution CA 
correct standard form CA
factors/formula    CA
y-values    CA
x-values/-waardes    CA
(6)

F = K Q₁ Q₂
           r²
F × r² = K Q₁Q₂
r² = K Q₁Q₂
           F
r  = √K Q₁Q₂
            F

transposing r ² A
r subject CA
AO: Full marks
(2)

r  = K Q₁Q₂
           F
r = √(9 x10⁹)(0,5 x 10^-6)(0,2 x 10^-6)
                     2,25 x 10^-4
r = 2 m 

OR
F = K Q₁Q₂
          r²
2,25 x 10^-4 = (9 x 10⁹ )(0,5 x 10^-6 )(0,2 x 10^-6)
                                           r²

r = √(9 x 10⁹ )(0,5 x 10^-6 )(0,2 x 10^-6 )
                       2,25 x 10^-4
r = 2 m

OR
SF   A
value of r   CA
AO: Full marks
NPU
(2)

1101₂ +111₂
1101₂
+ 111₂
10100₂

OR
13 + 7 = 20
20 = 10100₂

10100₂        A

OR
20          A
10100₂    CA
AO: Full marks     (2)
[20]

QUESTION 2

2.1

x = - 3  or    x = 0   
OR          D = 9
Roots are rational

roots or discriminant  A
rational  A
(2)

2.2

x² + p x - 2 p ² = 0
Δ = b² - 4 ac
= (p)²  - 4(1)(- 2 p²)
= p ² + 8p²
= 9p ²
Roots are rational

F   A
SF     A
S      CA
perfect square CA
(4)

[6]

QUESTION 3

3.1.1

√16a⁶
= √(4²) ×(a ⁶) OR (2⁴a⁶)^½
= 4a ³

4 A
a ³    A
(2)

3.1.2

√log ₂ 32 + log100 + 9
= √5log ₂ 2 + log100 + 9
= √5 + 2 + 9
= 4

5log ₂ 2       A
A
S        CA
AO: 0 marks (3)

3.1.3

(4√5 + 2)( 2 - 4√5)
= 4√10 -80 + 2 - 4√10
= - 78

OR
= - (4√5 - 2)(4√5 + 2)
= -80 + 2
= – 78

product      A
S           CA

OR
product    A
S     
AO: 0 marks/punte CA (2)

3.2 #

log₃x = 3 - log₃(x + 6)
log₃x + log₃(x + 6) = 3
log₃ (x² + 6x )  = 3
x² + 6x = 3³OR  log₃( x² + 6x ) = 3log₃
x² + 6x = 27
x² + 6x - 27 = 0
(x + 9 )(x - 3) = 0
x ≠ - 9  or  x = 3

OR
log₃x = 3 - log₃(x + 6)
log₃ x = 3log₃3 - log₃ (x + 6)
log₃ x = log₃ 27 - log₃(x + 6)
log₃ x = log₃    27   
                    (x + 6)
x =   27   
     x + 6
x² + 6x - 27 = 0
(x + 9 )(x - 3) = 0
x ≠- 9    or / of     x = 3

log property  A
exponential form   A
standard form  CA
factors/formula  CA
value of x  CA
validity   CA

OR
log property A
log property     A
standard form CA
factors/formula  CA
value of x  CA
restriction   CA
(6)

z = 2(½ + 3i) - 7i = 1 + 6i - 7i
= 1 -1i

substitution w
S
AO: Full marks
A A
(2)

z = r = √x² + y² = √(1)² + (-1)² = √2
tanθ =- ¹/₁ = -1
ref. angle = 45°
θ = 360° - 45° = 315°
z = √2 cis 315°

modulus CA from
value of  tan θ
CA from Q3.3.1
ref angle
angle in the correct quadrant
polar form
AO: Full marks

a +b + ia - bi = 5 -3i
a + b + ( a - b )i = 5 - 3i
a + b = 5............................... 1
a - b = - 3............................. 2
Equat.1+ Equat. 2
a + b = 5
a - b = - 3
   2a = 2
a = 1
b = 4

OR
a = 5
a - b = -3
(5 - b) - b = -3
-2b = -8
b = 4
a = 5 - 4
a = 1

OR
a + b - 5 = bi - ai - 3i
a + b - 5 = 0
- a + b - 3 = 0
(1) -(2) :
2b
- 8 = 0
2b = 8
b = 4
a + 4 - 5 = 0
a = 1

equation
equation
value of a
value of b

OR 
equation
equation
value of b
value of a

QUESTION 4

4.1.1

x ∈ x ≠ 0  OR  x ∈ (-∞; 0)  ∪  (0;∞)
OR
-∞ < x < 0  ∪  0 < x <∞

domain    A
(1)

4.1.2

P (-4 ; 0)

coordinates of P          A

(1)

4.1.3(a) #

P( - 4 ; 0) , S( 2 ; 0) and U(1 ; 10)
y = a (x - x₁) (x - x₂ )
y = a (x + 4) (x - 2) 10 = a (1+ 4)(1- 2)
a =- 2
f(x) = -2(x + 4)(x - 2) OR f (x) = -2x² - 4x + 16

OR
y = a(x + 1) ² + q
10 = a(1 + 1) ² + q
10 = 4a + q
0 = a(2 + 1)² + q
q = - 9a
10 = 4a - 9a
a = - 2
q = - 9(- 2) = 18
f (x) = - 2(x + 1) ² + 18

subst. roots  A
subst  U     CA
a = – 2     CA
eq. of f      CA

OR
subt. (1 ;10)   A
subt. (2 ; 0)       A
values of a and q CA
eq. of f   CA
(4)

4.1.3(b)

h ( x ) = ^k/_x + q
= ^k/_x + 9
10 = ^k/₁  + 9
k = 1
h( x ) = ¹/_x + 9

F (asymptote)
A
subst. (1; 10)   A
eq. of h   CA
(3)

At / by R: y = - 2(-1)² - 4(-1) + 16 = 18
At / by V: y =  1  + 9 = 8
                    (-1)
RV = 18 - 8 = 10 units

OR
RV = f (x)- h (x)
= - 2(-1)² - 4(-1) + 16 - (  1   + 9)
                                     (-1)
= 18 - 8
= 10

y valueCA from Q 4.1.3(a)
y value  CA from Q 4.1.3(b)
length       CA

OR
f(x) value CA from Q 4.1.3(b)
h(x) value CA from Q 4.1.3(b)
length   CA
(3)

x = - 4
or x = 2
or x = 0

x = - 4
from Q 4.1.2
x = 2
x = 0
(3)

4.2.1 (a)

y = (1,495)⁰ - 5 = - 4

OR
(0 ; – 4)

4.2.1 (b)

0 = (1, 495) ^x - 5
x = log_{1, 495}5
x ≈ 4

y = 0
log form
(3)

g:
all intercepts
shape
asymptote
p:
all intercepts
shape

- 4 ≤ y ≤ 0
y ∈ [- 4;0]
- 4 ≤ y and y ≤ 0

both endpoints
CA
from Q 4.2.1(b)
notation  A(2)

4.2.3 (b)

m =  0 - 4   = 1
       - 4 - 0
y = mx + c  OR y - y₁ = m(x - x₁ )
y = 1x + ( - 4)       y - 0 = 1(x - (-4))
y = x - 4 

OR
 x  +  y  = 1
 4     -4
x - y = - 4
y = x - 4

gradient CA
equation
CA

OR
M
equation  CA

4.2.3 (c)

0 < x < 4

OR
x ∈ (0 ; 4)

OR
0 < x and x < 4

both endpoints  CA
ünotation    CA
(2)

- 4 < x ≤ 0
x ∈ (- 4 ; 0]
- 4 < x  and x < 0

both endpoints
CA
notation
CA
(2)

QUESTION 5

5.1

formula    CA
m = 4     CA
rate   CA

OR
formula     CA
m = 4   CA
rate       CA
(3)

5.2

A = P (1 - i)ⁿ
152 523 = P (1 - 0,11)³
152 523 = 0,704969 P
P ≈ R216 354, 19

🗸 SF   A
🗸 152 523 = 0,704969P   CA
🗸 value of CA
AO: Full marks (3)

A_Martin = P (1 + in)
= 13 000 (1 + 5 x 0,058)
= R16 770, 00

OR
SI = P x i x n
= 13 000 x 5,8% x 5 = R 3 770
AMartin = 13 000 + 3 770 = R16 77

🗸SF      A
🗸 answer  CA

OR
🗸SF    A
🗸 answer (2)
AO: Full marks

A_Nosizwe = P (1 + i) ⁿ
= 8 000(1 + 0, 0764)²⁴(1 + 0, 0812)³+ 5 000 (1 + 0, 0812)³
                       12
= R 18 094, 50
A_Martin + A_Nosizwe = R16 770, 00 + R18 094, 50
= R 34 864, 50
R 34 864, 50 < R 35 000
They will NOT have enough

OR
A_Nosizwe = P(1 + i)ⁿ
= 8 000(1 +0,0764)²⁴
                     12
= R 9 316,222013
New P = R 9 316,222013 + R 5 000 = R14 316 , 22
A_Nosizwe= R14 316 , 222013(1 + 0 , 0812)³
= R18 094 , 50
A_Martin + A_Nosizwe = R16 770, 00 + R18 094, 50
= R 34 864 , 50
R 34 864 , 50 < R 35 000
They will NOT have enough

🗸(1 +0,0764)²⁴
          12
🗸 R18 094,50  A
🗸 calculating interest after adding R5000 at 8,12%
🗸 R 34 864,50      CA
🗸conclusion  CA

OR
🗸 (1 +0,0764)²⁴
            12
🗸 calculating interest after adding R5000 at 8,12%
🗸 R18 094,50   CA
🗸 R 34 864,50 CA
🗸conclusion
CA NPR

[13]

QUESTION 6

Penalty (1 mark) for incorrect notation only in QUESTION 6.1

6.1

f (x) = 2x + 3
f ' (x) = lim  f( x + h) - f ( x)
           h→0        h
= lim 2(x + h) + 3 - (2x + 3)
   h→0             h
= lim 2x + 2h + 3 - 2x - 3
  h→0            h
= lim    2h 
  h→0   h
= lim (2)
   h→0
= 2

definition
SF
S   CA
S   CA
A   CA 
2   CA
AO: 0 marks/ Volpunte

6.2.1

y = - x ^-5 + 3x ^4
dy/_dx = 5 x ^-6 + 12 x ³

5x ^-6  A
12x³ A
(2)

6.2.2

f ( x ) =  3 -  x  
            x ⁴  √x
= 3x^-4 - x^½
f ^/ (x) = - 12x ^-5 - ½ x^-½

OR
f ^/( x ) =- 12 -  1   
              x ⁵  √2x

3x^-4 A
x^½ A 
 -12x^-5 CA
- ½ x^-½ CA

(4)

6.2.3

factors  
S
1
(3)

6.3

m_ave = f (x2) - f  x1)
                x₂ - x₁
= [- 2( 2)² + 2] - [- 2( 0)² + 2]
                2 - 0
= - 6 - 2
      2
= - 4

OR
y₁ = f ( x₁ ) = - 2(0)² + 2 = 2
y₂ = f ( x₂ ) = - 2( 2)² + 2 = - 6
m_ave = y₂ - y₁
            x₂ - x₁   
= - 6 - 2
     2 - 0
= - 4

SF  A
S    CA
average gradient  CA

OR
both values of y A
SF    CA
average gradient CA
(3)

6.4.1

g ( x ) = 1 - x²
g ^/ ( x ) = - 2x
m_tan = g ^/( -3) = - 2( -3) = 6

derivative A
gradient  CA
(2)

6.4.2

g ( - 3)= 1 - (- 3) 2 = - 8
y = mx + c   OR   y - y₁ = m (x - x₁)
-8 = 6 (-3) + c      y + 8 = 6(x + 3)
- 8 + 18 = c
c = 10
y = 6x + 10

y-value A
SF        CA
equation CA (3)

[22]

QUESTION 7

7.1.1

B(0 ; - 5)

Coordinates of B A (1)

7.1.2

h ( x ) = x³ - 3x² - 9x - 5
h( -1) = ( -1)³ - 3(-1)² - 9(-1) - 5 = 0
x + 1 is a factor

SF       A
0          A
(2)

7.1.3

By inspection
h ( x ) = ( x² + 2x + 1)( x - 5)
h ( x ) = ( x + 1)² ( x - 5)
\ x = - 1 or   x = 5
D(5 ; 0)

OR
h( x ) = ( x + 1)( x² - 4x - 5)
h( x ) = ( x + 1)( x - 5)( x + 1)
x-intercepts ; h ( x ) = 0
x = - 1 or  x = 5
D(5 ; 0)

OR
0 = ( x + 1)( x² - 4x - 5)
x = -1 or x = -(-4) ± √(-4)² - 4 (1)(-5)
                                 2 (1)
x = - 1 or   x = 5
D(5 ; 0)

quadratic factor   A
Other intercepts       A
Coordinates of   D   CA

OR
quadratic factor A
Other factors    A
Coord. of D        CA

OR
quadratic factor  A
Other intercepts   A
Coord. of  D       CA
AO: Full marks
(3)

7.1.4

h^/ ( x ) = 3x² - 6x - 9
3x² - 6x - 9 = 0   
x² - 2x - 3 = 0
( x - 3)( x + 1) = 0

OR
x =- (-2) ± (- 2)2 - 4(1)(- 3)
                  2(1)
x = 3  or x = - 1
h (3) = (3)³ - 3(3)² - 9(3) - 5 = - 32
C( 3 ; -32)

derivative  M
equating derivative to 0   M
factors/formula  CA
x = 3          CA
y = - 32        CA
(5)

7.2

x < - 1 or  x > 3

OR
x ∈ (-∞;-1) ∪ (3;∞)

both crit. values CA
notation        CA

OR
both crit. values CA
notation CA(2)

[13]

30 - 6t    A
value of t CA
NPU
(2)

s = 30(5) - 3(5)²   m
= 75 m

substitution  A
distance  CA
AO: Full marks NP

75 m > 70 m
Therefore car A will collide with the stationary car B

reason    CA
conclusion  CA
(2)
[9]

QUESTION 9

9.1.1

- x ^-1 or -¹/_x 
ln x
C
(3)

9.1.2

(2)

9.2

Area notation using intergrals M
 x⁴
 2      
- 4x  
SF   CA
bounded area CA

OR
Area notation using intergrals
 x ⁴
 2      
- 4x  
- 4,5    CA
bounded area (6)
[11]

TOTAL: 150