MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATEΒ 
MEMORANDUM

JUNE 2018

NOTE:

  • If a candidate answered a question TWICE, mark the FIRST attempt ONLY.
  • Consistent accuracy(CA) applies in ALL aspects of the memorandum.
  • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.
  • The mark for substitution is awarded for substitution into the correct formula.

QUESTION 1

1.1.1

(π‘₯ βˆ’ 2)(3π‘₯ βˆ’ 1) = 0
π‘₯ βˆ’ 2 = 0 or 3π‘₯ βˆ’ 1 = 0
π‘₯ = 2 or π‘₯ =Β 1/3

βœ“βœ“ π‘₯-values
(2)

1.1.2

2π‘₯2Β + 3π‘₯ βˆ’ 7 = 0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β  Penalise 1 mark for incorrect rounding off
x = -b ± √b2 - 4ac    
Β  Β  Β  Β  Β  Β  Β 2a
x = -(3) ± √(3)2 - 4(2)(-7)
Β  Β  Β  Β  Β  Β  Β  Β  Β  2(2)

x = -3 ± √65
Β  Β  Β  Β  Β  Β  4
∴ x = 1, 27 or  x = -2,77

βœ“ substitution

βœ“βœ“ π‘₯-valuesΒ (3)

1.1.3

-x2 - 2x +15 < 0
x2 + 2x -15 > 0
1

βœ“factorsΒ 

βœ“critical values with method

βœ“βœ“ answer (accuracy)

(4)

1.1.4 2

βœ“ simplifying LHS

βœ“ simplifying RHSΒ 

βœ“ same baseΒ 

βœ“ equating exponents

βœ“ answer

(5)

1.2 x - 3y = 1................... (1)
y2 + 2xy - x2 = -7...... (2)
fromΒ (1) : x = 3y +1..... (3)
(3) in (2) :Β Β Β Β Β Β Β Β Β Β  y2 + 2 y(3y +1) - (3y +1)2 = -7
y2 + 6 y2 + 2 y - 9 y2 - 6 y -1+ 7 = 0
- 2 y2 - 4 y + 6 = 0
y2 + 2 y - 3 = 0
( y +Β 3)( y -1) =Β 0
y = -3Β  or / ofΒ Β Β  y = 1
x = -8 or / of x = 4

βœ“ x subject of the formula Β 

βœ“ substitutionΒ 

βœ“ removing brackets

βœ“ standard form

βœ“ 𝑦-values

βœ“ x-values

(6)

1.3

For equal roots Β :
n2 + 4mn = 0
n(n + 4m) = 0
∴ n = 0 or / of n = -4m

3

βœ“Ξ”Β  = 0

βœ“ both n-values

βœ“ substitution

βœ“ x = 0

βœ“ x = -2

(5)

Β  Β  [25}

Β 

QUESTION 2

2.1.1

15Β  ; 10Β  ; 7Β  Β  Β  Β ;Β  Β  Β π‘₯Β  Β ;Β  Β  Β 7
Β  Β βˆ’5Β  Β βˆ’ 3Β  Β (π‘₯ βˆ’ 7)Β  Β (7 βˆ’ π‘₯)Β  - 1st differences
Β  Β  Β  Β  Β  2Β Β Β Β Β  (π‘₯ βˆ’ 4) (βˆ’2π‘₯ + 14) -Β 2nd differences
x - 4 = 2Β  orΒ  Β - 2x +14Β = x - 4
x = 6Β  orΒ  Β - 3xΒ = -18
x= 6

βœ“ 2nd differencesΒ 

βœ“ equatingΒ 

βœ“ answerΒ 

(3)

2.1.2

2π‘Ž = 2Β  Β  Β  3π‘Ž + 𝑏 = βˆ’5Β  Β  Β a + b + c = 1
π‘Ž = 1Β  Β  Β  Β  3(1) + 𝑏 = βˆ’5Β  Β 1 - 8 + c = 15
Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  𝑏 = βˆ’8Β  Β  Β  Β  Β  Β  Β  Β c = 22
∴ 𝑇𝑛 = 𝑛2 βˆ’ 8𝑛 + 22

βœ“π‘Ž = 1

βœ“ 𝑏 = βˆ’8

βœ“ 𝑐 = 22

βœ“ answer
(4)

2.1.3

T50Β =Β (50)2Β Β -Β 8(50) +Β 22
= 2122

βœ“ substitution

βœ“ answer
(2)

2.2.1

a +14d = 74
a + 6d = 34
8d = 40
d = 5

βœ“ setting up 2 equationsΒ 

βœ“ method

βœ“ answerΒ 

(3)

2.2.2

a + 6(5) = 34
a = 4

S40Β =Β 40Β [2Β (4) +Β (40 -1)(5)]
Β  Β  Β  Β  Β  2
= 4060

βœ“ value of a

βœ“ substitution into correctΒ formula

βœ“ answer

(3)

2.2.3

Β 4

βœ“βœ“ answerΒ  (accuracy as one unit)
(2)

2.3.1

TkΒ =Β 3Β k
Β  Β  Β  Β 15
=Β  3Β kΒ Β 
Β  Β 5 x 3

=Β 1Β Β x 3k -1
Β  Β 5

βœ“ factors of 15Β 

βœ“ answerΒ 

(2)

2.3.2 Sn =Β  a(rn - 1)Β 
Β  Β  Β  Β  Β  Β  r - 1
241/5 =1/5(3n - 1)
Β  Β  Β  Β  Β  Β  Β  Β  Β 3-1
482/5 =Β 1/5(3n - 1)
242 = 3n -1
243 = 3n
35 = 3n
n = 5

βœ“ a and r

βœ“ substitution into correct formula

βœ“ exponential equationΒ Β 

βœ“ answerΒ 

(4)

2.3.3

NO
r = 3 > 1
r not in the interval -1 < r < 1, rΒ β‰  0

βœ“ NOΒ 

βœ“ reason

(2)

2.4

P = 91/3 x 91/9 x 91/27 x ....to infinity
=Β 91/3Β + 91/9Β + 91/27 ... to infinity

S∞ =    a    
Β  Β  Β  Β  Β  1 - r

5

βœ“ adding exponentsΒ 

βœ“ sum to infinity

βœ“ answerΒ 

βœ“ P = 9Β½Β  / √9

(4)

Β  Β  Β [29]

Β 

QUESTION 3

3.1.1

P (2; 4)
Q(0;1)

βœ“ coordinates of PΒ 

βœ“ coordinates of Q

(2)

3.1.2

y = a(x - 2)2 + 4
1 = a(0 - 2)2 + 4
-3 = 4a
a = - 3
y = bx
4 = b2
b = 2

βœ“ substitution

βœ“ answer

βœ“ substitution

βœ“ answerΒ 

(4)

3.1.3

xΒ β‰₯ 2 / x ≀ 2

βœ“βœ“ answer / antwoord

3.1.4

h(x) = 2 f(x) is a maximum when f (x) is a maximum
max valueof f (x) = 4Β 
max of h(x) = 24 = 16Β 

βœ“ max. value of f(x)Β 

βœ“answer

(2)

3.2.1

𝑦 β‰₯ 1, 𝑦 ∈ 𝑅 OR 𝑦 ∈ (1; ∞)

βœ“ answer

(1)

3.2.2

p(x) = x2 +1Β  Β /Β  Β r(x) = x2 + 2x
p(x +1) - 2 = (x +1)2 +1- 2
= x2 + 2x +1+1- 2
= x2 + 2x

Shift 1 unit to the left and 2units downΒ 

OR

Turning Point of p(x) = (0 ; 1)
Turning Point of r(x) = (-1 ; -1)

Shift 1 unit to the left and 2 units down

βœ“ calculation

βœ“ 1 unit to the left

βœ“ 2 units downΒ 

(3)

Β  Β 

[14]

Β 

QUESTION 4Β 

4.1

y =Β  -3Β Β  + 5
Β  Β  Β 0 +1
= 2

βœ“ y-intercept (1)

4.2

Β  -3Β  Β  + 5 = 0
x +1

Β  Β -3Β  Β  = -5
Β  x +1

-5x - 5 = -3
-5x = 2
x =- 2
Β  Β  Β  5

βœ“ simplificationΒ 

βœ“ answerΒ 

(2)

4.3

Β 6

βœ“ asymptotes

βœ“ x- and y-intercepts x

βœ“ shape

(3)

4.4

Β 7

βœ“ substitution

βœ“ simplification

βœ“ reflection

(3)

Β 

[9]

Β 

QUESTION 5

5.1

f (x) = log3 x
x = log3 y
f -1(x) = 3x

βœ“ interchanging x and y

βœ“ answer

(2)

5.2

f -1 is a reflection of f in the line y = x

βœ“ answer

(2)

5.3

y = log3 x
-2 = log3 k
∴ k = 3-2

=Β 1Β 
Β  Β 9

βœ“ substitution

βœ“ answer

(2)

5.4

0 < x < 1Β 
Β  Β  Β  Β  Β  Β  9

ORΒ 

log3 x < -2 and x > 0
x < 1 and x > 0
Β  Β  Β  9
0 < x < 1Β 
Β  Β  Β  Β  Β  Β  9

βœ“ βœ“answer

(2)

5.5

x β‰₯ 1

βœ“βœ“ answer

(2)

Β 

[10]

Related Items

Β 

QUESTION 6

6.1

8
effective rate =25,59% p.a

βœ“formula

βœ“ substitution

βœ“ answer

(3)

6.2

Β 9

βœ“i and/en n

βœ“ substitution

βœ“ answer

(3)

6.3.1

Β 10

βœ“ A and P

βœ“ substitution

βœ“ making n subject of the formulaΒ 

βœ“ answer (4)

6.3.2

Β 11

βœ“ substitution

βœ“ answerΒ 

(2)

6.3.3

Amount = R457 551,55 – R50 710.00
= R406 841,55

βœ“ answer

(1)

Β  Β 

[13]

Β 

QUESTION 7

Β  Penalise 1 mark forΒ incorrect notation in the questionΒ 
7.1 𝑓(π‘₯) = 1 βˆ’ 3π‘₯2
𝑓(π‘₯ + β„Ž)Β = 1 βˆ’ 3(π‘₯ + β„Ž)2

= 1 βˆ’ 3(π‘₯2 + 2π‘₯β„Ž + β„Ž2)
= 1 βˆ’ 3π‘₯2 βˆ’ 6π‘₯β„Ž βˆ’ 3β„Ž2

𝑓'(π‘₯) = limΒ   𝑓(π‘₯ + β„Ž) - 𝑓(π‘₯)
Β  Β  Β  Β  Β β„Žβ†’0Β  Β  Β  Β  Β  Β β„Ž
= limΒ  Β 1 βˆ’ 3π‘₯2 βˆ’ 6π‘₯β„Ž βˆ’ 3β„Ž2 βˆ’ (1 βˆ’ 3π‘₯2)
Β  β„Žβ†’0Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  β„Ž

= limΒ  Β βˆ’6π‘₯β„Ž βˆ’ 3β„Ž2
Β  β„Žβ†’0Β  Β  Β  Β  Β β„Ž
Β = limΒ  Β  β„Ž(βˆ’6π‘₯ βˆ’ 3β„Ž)
Β  Β β„Žβ†’0Β  Β  Β  Β  Β  β„Ž
= lim(βˆ’6π‘₯ βˆ’ 3β„Ž)
Β  β„Žβ†’0Β 

= βˆ’6π‘₯Β Β 

Answer ONLY: 0 marks

βœ“ 1 βˆ’ 3π‘₯2 βˆ’ 6π‘₯β„Ž βˆ’ 3β„Ž2

βœ“ substitution

βœ“ common factor

βœ“ answer

(4)

7.2 Β 12

βœ“ y = x + 2 + x -1

βœ“1Β βœ“ βˆ’π‘₯βˆ’2

(3)

7.3

y = 3x2 - 2x +1
y' = 6x - 2

6x - 2 = 4
6x = 6
x = 1
y = 2Β (1; 2)

βœ“ y ' = 6x - 2

βœ“ 6x - 2 = 4

βœ“ x-coordinate

βœ“ answer

(4)

Β  Β  Β [11]

Β 

QUESTION 8

8.1.1

f (x) = a(x + 2)(x -Β 2/3 )(x - 3)
-16 = a(2 + 2)(2 -2/3 )(2 - 3)
-16 = - 16 a
Β  Β  Β  Β  Β  Β  3
a = 3

f (x) = 3(x + 2) (Β x - 2/3 )(x - 3)
= 3(Β x - 2/3 )(x2 - x - 6)
= 3(x3 - x2 - 6x - 2/3 x2 + 2/3 x + 4)
= 3x3 - 5x2 -16x +12

βœ“ substitution of x-coordinates

βœ“ substitution of point

βœ“ value of a

βœ“ substitution

βœ“ removing brackets

(5)

8.1.2

𝑓(π‘₯) = 3π‘₯3 βˆ’ 5π‘₯2 βˆ’ 16π‘₯ + 12

𝑓′(π‘₯)Β = 9π‘₯2 βˆ’ 10π‘₯ βˆ’ 16 = 0 (9π‘₯ + 8)(π‘₯ βˆ’ 2) = 0
9π‘₯ + 8 = 0 π‘œπ‘Ÿ/π‘œπ‘“ π‘₯ βˆ’ 2 = 0
π‘₯ = βˆ’ 8Β  π‘œπ‘ŸΒ  π‘₯ = 2
Β  Β  Β  Β  Β 9

𝑦 = 4900 (20,16)Β Β 
Β  Β  Β  Β 243

B(βˆ’Β 8Β ; Β 20,16)
Β  Β  Β  9

βœ“ 𝑓′(π‘₯)Β = 0

βœ“ factors

βœ“ x-values

βœ“ coordinates of P

(4)

8.1.3

13
βœ“x-intercepts

βœ“y-intercept

βœ“turning pts

βœ“ shape

(4)

8.2.1

f (x) = ax3 + bx2 + 3x + 3 Β and g(x) = Β f n(x) = 12x + 4

f '(x) = 3ax2 + 2bx + 3
f nΒ (x) = 6ax + 2b

6a = 12Β  and 2b = 4
a = 2Β  Β andΒ  b = 2

βœ“ a = 2

βœ“ b = 2

(2)

8.2.2

12x + 4 = 0

x =- 1
Β  Β  Β  Β 3
∴ x > - 1
Β  Β  Β  Β  Β  3

βœ“ x = - 1
Β  Β  Β  Β  Β  Β 3

βœ“ answer

(2)

8.2.3

f nΒ (x) <Β 0Β  for x <Β -Β 1Β Β Β Β  (concavedown )
Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  3
f n(x) >Β 0Β  Β for Β x >Β -Β 1Β Β Β Β  (concaveups )
Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β 3
x =- 1 is the x-coordinateof the point of inflection 3 where the concavity changes.
Β  Β  Β  Β 3

βœ“ answer

βœ“ answer

(2)

Β  Β  [19]

Β 

QUESTION 9

9.1

EF = a - 2x
DE = tan 600
BE

DE = x tan 600
= √3x

Area = l x b
= √3x x (a - 2x)
= √3ax - 2√3x2

No mark for the answer

βœ“ EF = (a – 2x)

βœ“ tan ratio

βœ“ answer

βœ“ substitution

(4)

9.2

A(x) = √3ax - 2√3x2

A ' (x) = √3a - 4√3x = 0
x = -√3a
      -4√3

x = aΒ 
Β  Β  Β  4

14

βœ“ derivative

βœ“ f ' (x) =Β 0

βœ“ answer

βœ“ substitution

βœ“ answer

(5)

Β 

Β 

[9]

Β 

QUESTION 10

10.1.1

P(F and S) =Β  67Β  / 0, 28
Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β 236

βœ“ answer

(1)

10.1.2

P(M ) Β΄ P(not S )
=Β 120Β xΒ 100
Β  Β  236Β  Β 236
= 0, 22

P(M ∩ not S ) =  51 
Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  236
= 0, 22
∴ P(M ) x P(not S ) = P(M ∩ not S )
β†’ events are independent

βœ“ P(M) x P(not S)

βœ“ answer

βœ“ answer

βœ“ conclusion

(4)

10.2.1

Β 15

βœ“ 1st branchΒ 

βœ“ 2nd branchΒ 

βœ“ outcomes

(3)

10.2.2 Β 16

βœ“ setting up equation

βœ“ standard form

βœ“ answer

(3)

Β  Β  [11]

TOTAL : 150

Last modified on Thursday, 12 August 2021 11:18