QUESTION 1

1.1

111010
- 10101
100001₂

🗸🗸Accurate value
(2)

1.2

1.2.1

x(x - 3) = 0
x = 0 or x = 3

🗸🗸Each correct x-value

(2)

1.2.2

x² + 3x +1 = 0 (correct to ONE decimal)
x = - b ± √b² - 4ac
               2a

x = -3 ± √(3)² - 4(1)(1)  (-1 Mark for incorrect rounding)
                  2(1)  
x ≈ - 0, 4 or x ≈ - 2,6

OR

🗸 Formula

🗸Substitution

🗸 x ≈ -0,4

🗸 x ≈ -2,6

🗸 Each x value

🗸 Expansion

🗸Quadratic factors

🗸 x ≈ -0,4

🗸 x ≈ -2,6

(4)

x² + 3x + 2 < 0
(x+2) (x+1) < 0
Critical values : x = - 1 or x = -2

=>
-2 < x < -1

🗸Standard form

🗸Critical values

🗸Notation

🗸Both values

OR

🗸 🗸 🗸

-2 < x < -1

Accurate answer

🗸 -2< x

x < -1

(4)

y = x²–1 …………. (1) and   y = x+1.................... (2)
x²–1 = x+1 x² – x – 2 =0
(x +1) (x–2) = 0
x= 2 or x = –1
y = 3 or y = 0

🗸Equating (1) and (2)

🗸Standard form

🗸Factors

🗸Both x-values

🗸Both y-values

(5)

b²– 4ac = 0
b²– 4.1.4 = 0
b = 4 or b = – 4

🗸Discriminant = 0

🗸Substitution

🗸🗸Each value of b

(4)

QUESTION 2

2.1

2^x.2¹– 2^x.2^–1
       3.2^x

= 2^x(2 – 2^–1) 
         3.2^x
= 3 
   6
= ½ 🗸

🗸Prime bases

🗸 Factor 2^x

🗸 Factor 2 – 2^–1

🗸 ½

(4)

2.2

🗸Log Rule(numerator)

🗸Log Rule (denominator)

🗸Simplification

  log_a 5   ^–1 
 2log_a 5^–2

🗸Power rule

🗸Prime factors of 25

🗸Prime factors of 125

🗸Power rule (numerator)

🗸Power rule (denominator)

🗸Simplification

(5)

Rabbits = 1000 × 2^0,05(30)
Rabbits = 2828

🗸Substitution

🗸Answer

(2)

8 = 2^0,05t
0, 05t = log₂8
t = 60 days

🗸Substitution

🗸log form

🗸 t = 60 days

(3)

QUESTION 3

3.1

3.1.1

| Z |= √(–2)²  + (1)²
| Z |= √5

🗸Substitution

🗸Answer

(2)

3.1.2

🗸Quadrant

🗸Point/Coordinates

(2)

3.1.3

tanθ = –½
θ = –26, 57^o
θ = 180º – 26, 57º = 153, 43º

Accept angles in radians

🗸tan ratio

🗸Ref Angle

🗸Argument

(3)

3.1.4

 | Z |= √5
θ = 153, 43^o
z = √5 [cos (153, 43^o ) + isin (153, 43^o )]

Accept angles in radians
OR

z = √3cis (153, 43^o )

🗸🗸Accurate polar form

(2)

(x - yi) = –2+ i
              1+ i
x – yi = –2 + i  × 1 – i 
              1 + i     1 – i
x – yi = –2 + 2i +i – i²
                 1 – i²
x – yi = -1 + 3 i
             2     2
∴ x = -1 and y = - 3
          2                2

OR

1(x - yi) + i(x - yi) = –2 + i
x – yi + ix - y (i )²  = –2 + i x – yi + ix+ y = –2 + i
x+ y + ( x – y )i = –2+ i

x+ y = -2.............. (1)
x – y = 1................... (2)

(1)+(2) :

x = -½
and y = - 3
              2

🗸Simplification

🗸Conjugate product

🗸Simplification

🗸 x-value

🗸 y-value

🗸 Multiplication

🗸Simplification

🗸Comparing real values and imaginary values

🗸 x-value

🗸 y-value

(5)

QUESTION 4

4.1

4.1.1

i_norm  = 14% = 0, 035
              4
= 3, 5%quarterly

🗸 Answer

(1)

4.1.2

🗸Formula

🗸Substitution

🗸Interest

(3)

4.1.3

A = 2500(1 + 0, 035)^7×4
A = R6550, 43

🗸Substitution

🗸Correct i = 0,035 and n = 21

🗸 Value of A

(3)

4.2

In A1

i = 0, 08
       12
🗸n = 24
🗸 i = 0,1 = 0,025
         4
🗸n = 12
🗸 A₁ = R687572,9508
In A₂
🗸n = 8
🗸 A₂ =R97 472, 2318
Final Amount R785045,18

(8)

[15]

QUESTION 5

5.1

5.1.1

0 = – ( x – 3)²  + 4
( x – 3)²  = 4
x – 3 = ±2
x = 5 or x = 1
A(1; 0) or B(5; 0)

OR
0 = -²+ 6x - 5
0 = (-x +1)(x - 5)
x = 5 or x = 1
A(1; 0) or B(5; 0)

🗸h(x) =0

🗸Transposition

🗸A co-ordinates

🗸B co-ordinates

🗸h(x) =0

🗸Factors

🗸A coordinates

🗸B coordinates

(4)

5.1.2

h(x) = -x²+ 6x - 5
dy = -2x + 5
dx
0 = -2x + 6
x = 3
h(3) = (3)² + 6 (3) - 5
= 4
∴ D(3; 4)

dy
dx🗸

🗸Coordinates

(2)

5.1.3

x ∈ [0; 6] OR 0 ≤ x ≤ 6

🗸0

🗸6

🗸Correct notation

(3)

5.1.4

Maximum height = 4 units

🗸Answer

5.1.5

y-intercept of h = –5
Beams have height of 5 units

🗸y-intercept

🗸 5 units

(2)

5.1.6

y ≤ 4 OR y Î (-∞; 4] OR - ∞ < y ≤ 4

🗸 Notation

🗸 Value(s)

(2)

5.1.7

x ∈ [3;5] OR 3 ≤ x ≤ 5

🗸3

🗸5

🗸Correct notation

(3)

5.2

No. The truck height (4,5) is greater than the
bridge height (4 units) and the bridge has cross bars on top.

🗸 No

🗸 Bridge less than truck height

OR Truck height greater than the bridge height

🗸Cross bar

(3)

At F, x = 3

y = – 3+5 =2

FD = D – F
FD = 4 – 2
FD = 2 units

🗸y-value at F

🗸Subtracting y-values

🗸 FD

(3)

QUESTION 6

6.1

6.1.1

0 = –2 + 1
       x
x = 2
(2; 0)

🗸y =0

🗸 Coordinates

(2)

6.1.2

f(x) = 2⁰

y =1

🗸 Value of y

(1)

6.1.3

y=0 for f(x)

x = 0 and y =1 for g(x)

–1 Mark for 1 omitted asymptote

🗸y =0

🗸x =0 and y=1

(2)

6.2

🗸Shape of f

🗸 y-intercept of f

🗸 y=1 Asymptote of g

🗸 1 more point on f

🗸Shape of g

🗸 1 more point on g

🗸 x- intercept of g

(7)

6.3

6.3.1

x ∈ R, x ≠ 0

🗸Restriction

🗸Domain value

(2)

6.3.2

x ∈ (0; +∞ ) OR x > 0

🗸Correct inequality

(1)

QUESTION 7

7.1

f (' x)= lim  f ( x+ h) –  f ( x)
         h→0          h

f '(x)= lim –2 ( x+ h)2  – (–2x² )
        x→0            h

f (x)= lim  –2x² – 4xh – 2h² + 2x²
       x→0                 h
f'(x)= lim h (–4x – 2h)
       x→0        h
f '(x)= –4x

🗸Formula

🗸Substitution

🗸Expansion

🗸Factors

🗸 f '(x)= –4x

(5)

-1 Mark for incorrect notation in 7.1 or 7.2

7.2

y = 2√x – 1
                x
y = 2x^½ – x^-1

dy = x^½ + x^-2
dx

OR

dy = 1  + 1 
dx   x^½   x²

🗸2x^½ 

🗸x^-1

🗸 x^½ 

🗸x^-2

7.3

g'(x) = 2x - 2
m_tangent = 2(2) – 2 = 2
y = 2²– 2.2 = 0
(2; 0)
y = mx+c
0 = 2.2 + c
c = – 4
y = 2x – 4

🗸 g'(x)

🗸 m_tangent

🗸(2;0)

🗸 c = – 4

🗸 y = 2x – 4

(5)

[14]

QUESTION 8

8.1

f (–1) = (–1)³ + 4(–1)² + (–1) – 6
f (–1) = – 4 ≠ 0
So x+1 is not a factor of f(x) because f (–1) is not equal to 0.

🗸f (–1) = – 4 ≠ 0

8.2

f (1) = (1)³ + 4(1)² + (1) - 6 = 0
(x -1) is a factor of f

       1        4         1           -6
1     0        1         5            7
       1        5         6            0

f(x) = (x–1) (x²+5x+6)
f(x) = (x–1) (x+3) (x+2)
x =1 or x = –3 or x = –2
(1;0), (–2;0), (–3;0)

🗸 f(x) =0

🗸 First linear factor

🗸Quadratic factor

🗸Factors of x²+5x+6

🗸All coordinates

(5)

8.3

y-intercept = –6

🗸Answer

(1)

8.4

f '(x)= 3x²+ 8x+1
0 = 3x² + 8x+1
x= – 8 ± √64 – 4.3.1
               2.3
x = – 0,13 or x = –2, 54
(– 0,13; – 6, 06) or (– 2, 54; 0,89)

🗸 f '(x) = 0

🗸x- values

(-0,13; -6,06)

(-2,54; 0,89)

OR

Y – coordinates of TP

🗸 y= -6, 06

🗸 y= 0,89

🗸Shape

🗸x-intercepts

🗸Max. Turning point

🗸Min. Turning point

🗸 y-intercepts

(5)

[16]

QUESTION 9

9.1

9.1.1

Surface Area = 2(2x.x +2x.h + x.h)
4x² + 4xh + 2xh = 120
6x.h = 120 – 4x²2

∴ h = 120 - 4x²
            6x
h = 20 - 2x
       x     3

🗸 Formula

🗸Substitution

🗸 Simplification

🗸 h

(3)

9.1.2

V = l.b.h

🗸V = l.b.h

🗸Substitution

(2)

9.1.3

dV = 40 – 4x²
dx
0 = 10 – x²
x = √10 or x ≠ - √10

🗸 dV
   dx
🗸 dV = 0
    dx

🗸 x = 10 ≈ 3,16 cm³

(3)

T = t³ - 9t ²+ 50t - 66
dT = 3t ² -18t + 50
dt
dT = 3(5)² -18(5) + 50
dt
dT = 35º C.s^-1
dt

🗸 3t ² -18t + 50

🗸Substitution by 5

dT =35 C.s^-1
dt🗸 

(3)

QUESTION 10

10.1

🗸 x³

🗸 - x ²
      2

🗸 c

10.2

🗸Integration expression

🗸Simplification

🗸Substitution by 1 and 0

🗸 1 square units
   6

(4)