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

? ≠ 0 ??, ? ∈ ? ?? − ∞ <
? < ∞ ? 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