MARKING CODES

A

Accuracy

AO

Answer only

CA

Consistent accuracy

M

Method

R

Rounding

NPR

No penalty for rounding

NPU

No penalty for units omitted

S

Simplification

SF

Substitution in the correct formula

 NOTE:

• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed-out version.
• Consistent accuracy (CA) applies as indicated on the marking guidelines.
• Assuming answers/values to solve a problem is NOT acceptable.
  

QUESTION 1

1.1.1

21x² +13x = 0
x (21x +13) = 0
x = 0 or x = - 13
                     21

OR

🗸 Factors

🗸 Both x-values 

OR 

🗸 Substitution 

🗸 Both x-values

A

CA

A

CA

(2)

1.1.2

x + 5 = 7
           x
x² + 5x = 7
x² + 5x - 7 = 0

x = 1 , 14 or x = -6 , 14

🗸 x² + 5x = 7

🗸 Standard form

🗸 Substitution

🗸 Both values of x           R

A

CA

CA

CA

(4)

1.1.3

OR

🗸 Factors  M

🗸 Critical Values        NPR

🗸 Correct notation

OR

A

CA

CA

🗸 Substitution   M

🗸 Critical values      NPR

🗸 Correct notation

A

CA

CA

(3)

1.2

2x + y = 1............................... (1)
x² + y² + 2x - 6 y = 9................. (2)
y = 1- 2x................................ (3)
x² + (1- 2x)2  + 2x - 6 (1- 2x) = 9
x² +1- 4x + 4x² + 2x - 6 +12x = 9
5x² +10x -14 = 0

x = 0 , 95 or x = -2 , 95
∴ y = 1- 2 (0 , 95) or  x = 1- 2 (-2 , 95)
y = -0 , 9 or y = 6 , 9

🗸 Subject of the formula 

🗸 Substitution

🗸 Simplification

🗸 Standard form 

🗸 Substitution into the formula

🗸 Both x-values 

🗸 Both y-values     NPR

A

CA

S

CA

SF

CA

CA

OR 
2x + y = 1................................. (1)
x² + y² + 2x - 6 y = 9................. (2)
x = 1- y  .................................. (3)
        2

OR

🗸 Subject of the formule

🗸 Substitution

🗸 Simplification

🗸 Standard form

🗸 SF

🗸 Both y-values

🗸 Both x-values

A

CA

S

CA

SF

CA

CA

NPR

(7)

1.3.1

110 km/h = 110×1000
                  1×60×60
= 30,55 m/s

AO: 1 Mark

🗸 30,55 m/s                       NPU

A

(1)

1.3.2

30,55 m/s = 3,055 × 10 ¹ m

🗸 3,055 × 10 ¹

A

(1)

1.4

2

315

Remainder

🗸 Method

🗸 1001110112 

AO: 1 Mark

Ignore: No base 

A

A

(2)

2

157

1

2

78

1

2

39

0

2

18

1

2

9

1

2

4

1

2

2

0

2

1

0

1

0

1

 So 315 = 100111011₂

OR

2⁸ + 2⁵ + 2⁴ + 2³ +  2¹ + 2⁰ = 315
1 0 0 1 1 1 0 1 1₂                    = 315

[20]

QUESTION 2

2.1.1

It has one root

🗸 one

A

(1)

2.1.2

Real, Rational and Equal

🗸 Real
🗸 Rational
🗸 Equal

A
A
A

(3)

2.2

(a) P = 1 and b = 3
(d) S = 1 and b = – 3

🗸 a = 1 and/en b = 3
🗸 a = 1 and/en b = – 3

A
A

(2)

[6]

QUESTION 3

3.1

OR

🗸 Prime factors

🗸 Simplification

🗸 Simplification

🗸 3

OR

🗸 Simplification

🗸 Simplification

🗸 Simplification

🗸 3

A

CA

CA

CA

(4)

3.2

🗸 Log Property

🗸 Same Base Rule

🗸 Log Property

🗸 Log Property

S

CA

CA

S

CA

OR

OR

🗸 Log Property 

🗸 Log Property 

🗸 Log Property 

🗸 Log Property 

🗸 Log Property

S

CA

CA

S

CA

(5)

3.3

3× 2²ⁿ⁺¹ - 8^n-1 = 4ⁿ
-2^3n-3 + 3.2²ⁿ⁺¹ = 2²ⁿ
2²ⁿ (-2ⁿ.2^-3 + 3.2) = 2²ⁿ
-2ⁿ.2^-3 + 6 = 1
- 2ⁿ = 1- 6
 8
2ⁿ= 40
n = log₂ 40
n = 5, 32

🗸 Simplification 
🗸 Log form
🗸n = 5,32

CA
CA
CA

3.4.1

z = 5 – 3i

🗸 5 – 3i

A

(1)

3.4.2

🗸Substitution
🗸Modulus

A 

CA

(2)

3.4.3

tanθ = 3
           5

Reference ∠ = 30,96°
θ = 360° - 30,96° = 329,04°
z = √34 cis (329,04°)

🗸 Tan ratio
🗸 Reference angle
🗸 329,04°
🗸 √34cis (329,04°)

A
CA
CA
CA

(4)

(2 - 3i)i + 7 y + 9 = 11+13ix
2i - 3i² + 7 y + 9 = 11+13ix
2i + 3 + 7 y + 9 = 11+13ix
12 + 7 y + 2i = 11+13ix
\12 + 7 y = 11 and 2 = 13x
y = - 1 and x = 2
        7             13

🗸 Simplification
🗸 i² = – 1
🗸 Simplification
🗸 y =- ¹/₇
🗸 x = ²/₁₃

S A
S
CA
CA

QUESTION  4

4.1.1

0 = 2^x– 1
1 = 2^x
2⁰ = 2^x
x = 0

🗸 y = 0
🗸 2⁰ = 2^x/ Logarithm
🗸 x = 0

A
CA
 

CA

(3)

4.1.2

y = –1

🗸 y = –1

A

(1)

4.1.3

x = 0 and y = –1

🗸 x = 0
🗸 y = – 1

A
A

(2)

4.1.4

k(x) = 0
⸫ x = 1

🗸 k(x) =0
🗸 x = 1

A
CA

(2)

4.1.5

🗸 x-Intercept
k:
🗸 Shape
🗸 x-intercept

CA
CA
CA
CA

(5)

4.1.6

𝑥 ≠ 0 𝑶𝑹, 𝑥 ∈ 𝑅 𝑶𝑹 − ∞ <
𝑥 < ∞ 𝑥 ¹ 0

🗸

CA

(1)

4.1.7

f (x) > – 1

🗸 Critical value
🗸 Notation

CA
CA

(2)

4.2.1

q = – 4

🗸

– 4

A

(1)

4.2.2

0 =  √ 9 - x²
x = 3

🗸 h (x) = 0
🗸 x = 3

A
CA

(2)

4.2.3

p = -1+ 3
         2
∴ p = 1

🗸 Mid-point
🗸 Value of

A
CA

(2)

4.2.4

g(x) = a (x – 1)² – 4
0 = a (– 1– 1)² – 4
4 = 4a
⸫ a = 1

OR

g(x) = a (x – 1)² – 4
0 = a (3 – 1)² – 4
4 = 4a
⸫ a = 1

🗸 Substitute p and q
🗸 Substitute point (−1; 0)
🗸 a = 1

OR 

🗸 Substitute p and q
🗸 Substitute point  (3; 0)
🗸 a = 1

CA
CA
CA
CA
CA
CA

(3)

4.2.5

y = (0 – 1)² – 4 = – 3
D (3; – 1)

🗸 D (3; – 1)

CA

(1)

4.2.6

0 ≤ x < 3

🗸 Critical values
🗸 Correct notation

CA
CA

(2)

[27]

QUESTION 5

5.1

Effective Interest Rate
= 4,59% » 4,6%

🗸 Formula
🗸 Substitution
🗸 4,6%
Accept  4,59%

A
A
CA

(3)

5.2.1

y = 350 kPa

🗸 350

A

(1)

5.2.2

(c) Compound depreciation

🗸 Compound depreciation

A

(1)

5.2.3

27,216 = 350 (1- i)⁵

i = 0,40
Rate of Depreciation
= 40%

🗸 Substitution SF
🗸 Simplification  S
🗸 i
🗸 40%

A
CA
CA
CA

(4)

5.3

OR

🗸 Formula
🗸 Substitute R50 000
🗸 Substitute i and n
🗸 Substitute i and n
🗸 R80 165, 96

OR

🗸 Formula
🗸 Substitute R50 000
🗸 Substitute i and n
🗸 Substitute i and n
🗸 R80 165, 96

A
A
A
A
CA

A
A
A
A
CA

(5)

[14]

QUESTION 6

NOTE: PENALISE 1 MARK FOR NOTATION IN QUESTION 6

6.1

🗸 Substitution
🗸 Simplification
🗸 Simplification
🗸 f ¢( x) =13a

CA
CA
CA
CA

(5)

6.2.1

x (- 6x + y ) = x³
- 6x + y = x²
y = x² + 6x
dy = 2x + 6
dx

🗸 Divide by x
🗸 y Subject
🗸 2x
🗸 6

A
CA
CA
CA

(4)

6.2.2

🗸 Exponential form
 Simplify fraction
🗸 ²/₃ t ^1/₃
🗸 -x.t^-2

A
CA
CA
CA

(4)

g (-2) = (-2)²  + 3(-2) - 4
g (-2) = - 6
g (1) = (1)²  + 3(1) - 4
g (1) = 0
Average gradient = 0 - (-6)
                               1- (-2)
∴ gradient = 2

🗸 g (-2) = -6
🗸 g (1) = 0
🗸 2

A
A
CA

QUESTION 7

7.1

y = 6

🗸 6

A

(1)

7.2

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

OR

f (2) = 0
x = 2 or – x² + 4x – 3 = 0
x = 2 or (x – 1)(x – 3) = 0
x = 2 or x = 1 or x = 3 

OR 

f (3) = 0
x = 3 or – x² + 3x – 2 = 0
x = 3 or (x – 1)(x – 2) = 0
x = 3 or x = 1 or x = 2

🗸 f (1) = 0
🗸 Quadratic factor
🗸 x = 2
🗸 x = 3 

OR 

🗸 f (2) = 0
🗸 Quadratic factor
🗸 x = 1
🗸 x = 3 

OR 

🗸 f (3) = 0
🗸 Quadratic factor
🗸 x = 1
🗸 x = 2

A
CA
CA
CA

A
CA
CA
CA

A
CA
CA
CA

(4)

7.3

🗸 f ¢( x) = -3x² +12x -11

🗸 f ¢( x) = 0

🗸 Substitution/Vervanging

🗸 (1,42; − 0;39)

🗸 (2,58 ; 0,39)

A

A

CA

CA

CA

7.4

🗸 Shape
🗸 y-intercept
🗸 All 3 x-intercepts
🗸 (1,42; − 0;39)
🗸 (2,58; 0,39)

A
CA
CA
CA
CA

(5)

7.5

f'( x) = m_tangent
-3x² +12x -11 = -11
-3x ( x - 4) = 0
x = 0 or x = 4

🗸 f '(x) = m_tangent
🗸 Factors/substitution
🗸 x = 0
🗸 x = 4

A
CA
CA
CA

(4)

[19]

QUESTION 8

8.1

H (0) = 400⸰C

NPU

🗸 400 ℃

A

(1)

8.2

H (3) = −2 (3)² + 20 (3) + 400
H (3) = 442⸰C

🗸 Substitution
🗸 442 ℃

A
CA

(2)

8.3

dH = - 4t + 20
dt
- 4t + 20 = 0
t = 5s
0 = −2t² + 20t +400
0 = t² − 10t – 200
0 = (t – 20) (t + 10)
⸫ t = 20 or/of t ¹  –10
Time = 20 – 5 = 15 seconds

🗸 dH = - 4t + 20
    dt
🗸 dH = 0
    dt
🗸 5
🗸 H ( x) = 0
🗸 Factors/substitution
🗸 t = 20
🗸 t ¹ –10
🗸 15 seconds   NPU

A
A
CA
A
CA
CA
CA
CA

[11]

QUESTION  9

9.1

9.1.1

A
A

(2)

9.1.2

🗸 Simplification
🗸t ²
   2x
🗸 - x²t² + C
       2

S
A
CA

(3)

9.2

🗸🗸 Integral form
🗸 integral
🗸🗸 Substitution
🗸 Simplification
🗸 Area

A
A
CA
CA
CA
CA

(7)

[12]

TOTAL:

150