NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P1
SEPTEMBER 2014
MEMORANDUM
MARKS: 150
This memorandum consists of 16 pages.
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NOTE:



If a candidate answered a question TWICE, mark the FIRST attempt ONLY.
If a candidate crossed out an attempt of a question and did not redo the question, mark
the crossed out question.
Consistent accuracy (CA) applies in ALL aspects of the memorandum.
QUESTION 1
1.1
1.1.1
1.1.2
1.1.3
2𝑥 2 − 7𝑥 + 3 = 0
(2𝑥 − 1)(𝑥 − 3) = 0
2𝑥 = 1
or 𝑥 = 3
1
𝑥=
2
 Factors
1
 𝑥=
2
 𝑥=3
(3)
2
17𝑥 − 8 = 3𝑥
0 = 3𝑥 2 − 17𝑥 + 8
−𝑏±√𝑏2 −4𝑎𝑐
𝑥=
2𝑎
−(−17)±√(−17)2 −4(3)(8)
=
2(3)
17±√193
=
6
𝑥 = 5,15 or 𝑥 = 0,52
(2𝑥 − 3)(4 − 𝑥) ≥ 0
3
Critical values: 𝑥 = 2 or 𝑥 = 4
 Standard form
 Substitution into
formula
 Simplify
 Each value of x

(5)
 Critical values
− − −0 + + + + + + + + + 0 − − −
3
2
4
3
∴
≤𝑥 ≤4
2
3
 𝑥≥2
 𝑥≤4
(3)
OR
(2𝑥 − 3)(4 − 𝑥) ≥ 0
3
Critical values: 𝑥 = 2 or 𝑥 = 4
 Critical values
3
2
∴
4
3
2
≤𝑥 ≤4
3
 𝑥≥2
 𝑥≤4
OR
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(2𝑥 − 3)(𝑥 − 4) ≤ 0
3
Critical values: 𝑥 = 2 or 𝑥 = 4
NW/September 2014
 Critical values
o
3
2
∴
4
3
2
3
 𝑥≥2
 𝑥≤4
≤𝑥 ≤4
(3)
1.2
1.2.1
1.2.2
4𝑎+𝑏 = 2𝑏+4
22𝑎+2𝑏 = 2𝑏+4
∴ 2𝑎 + 2𝑏 = 𝑏 + 4
𝑏 = 4 − 2𝑎
2
2𝑎 − 3𝑎𝑏 = −4
Substitute 𝑏 = 4 − 2𝑎
2𝑎2 − 3𝑎(4 − 2𝑎) = −4
2𝑎2 − 12𝑎 + 6𝑎2 = −4
8𝑎2 − 12𝑎 + 4 = 0
2𝑎2 − 3𝑎 + 1 = 0
(2𝑎 − 1)(𝑎 − 1) = 0
1
𝑎 = 2 or 𝑎 = 1
1
𝑏 = 4 − 2 (2) or
𝑏=3
or
1.3
 Exponential law
 Equate exponents
(2)
 Substitute 𝑏 =
4 − 2𝑎
 Standard form
 Factors
 Both values of a
𝑏 = 4 − 2(1)
𝑏=2
 Both values of b
(5)
(0,04)0,5
4
=(
100
1
2
)
1
=√
=
=
4
100
OR
2 2 2
[(10) ]
 Simplify
2
10
1
 Answer
(2)
5
1
(If ± 5, penalise 1 mark)
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1.4
 Squaring
𝑦 2 = (√2 − 1)(√2 − 1)
𝑦 2 = 2 − 2√2 + 1
𝑦 2 = 3 − 2√2
∴ 𝑦 = √3 − 2√2
∴𝑦=𝑥
 Expand
 Simplify
OR
𝑥 = √3 − 2√2
= √2 − 2√2 + 1
= √(√2 − 1)
 Expand
 Bracket squared
2
= √2 − 1
∴ 𝑥 = 𝑦 = √2 − 1
 Simplify
(3)
[23]
QUESTION 2
 Expansion
20
2.1
∑(15 − 4𝑛) = 3 + (−1) + (−5)+ . . .
𝑛=3
 Substitute a and
d into correct
formula
 𝑛 = 18
𝑎 = 3 , 𝑑 = −4 , 𝑛 = 18
S18 =
2.2
2.2.1
18
2
[2(3) + 17(−4)]
= 9[−62]
= −558
5
216 𝑙 ; 6 (216)𝑙 ;
On 2nd day: T2 =
5
6
 Answer
5
5
( ) 216 𝑙
6 6
( 216) = 180 𝑙
5
 6 of 216
 Answer
(2)
OR
Day 2: T2 = 216 −
2.2.2
(4)
1
6
(216) = 180 𝑙
𝑎 = 216;
On the 7th day: Tn = 𝑎𝑟 𝑛−1
 216 −
1
6
 Answer
5
𝑟=
6
(216)
(2)
 a
 r
5 6
T7 = (216) ( 6 )
= 72,337 . . .
≈ 72,34 𝑙
 Answer (2 dec.)
(3)
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2.3.1
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2(3𝑥 − 1) + 2(3𝑥 − 1)2 + 2(3𝑥 − 1)3 + . . .
Converge if −1 < 𝑟 < 1
∴ −1 < 3𝑥 − 1 < 1
0 < 3𝑥 < 2
2
0<𝑥 < 3
NW/September 2014
 −1 < 𝑟 < 1
 𝑟 = 3𝑥 − 1
 Answer
(3)
2.3.2
 Value of a
 Value of r
1
𝑎 = 2 (3 (2) − 1) = 1
1
1
𝑟 = 3 (2) − 1 = 2
𝑆∞ =
=
𝑎
1−𝑟
 Substitution into
correct formula
1
1
1−(2)
 Answer
(4)
= 2
2.4
2
;
𝑥−2
x
;
12
12 − 𝑥
12 − 𝑥 − (𝑥 − 2)
;
y
𝑦 − 12
𝑦 − 12 − (12 − 𝑥)
2nd difference = 6
12 − 𝑥 − (𝑥 − 2) = 6
−2𝑥 + 14 = 6
−2𝑥 = −8
𝑥=4
Substitute 𝑥 = 4 into
𝑦 − 12 − (12 − 𝑥) = 6
𝑦 − 12 − (12 − 4) = 6
𝑦 − 20 = 6
𝑦 = 26
 2nd difference
 Equation in terms
of x
 𝑥=4
 Substitute into
2nd equation
 𝑦 = 26
(5)
[21]
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QUESTION 3
12
ℎ(𝑥) =
+6 , 𝑥 >0
𝑥−4
Vertical asymptote: 𝑥 = 4
Horizontal asymptote: 𝑦 = 6
y-intercept: 𝑥 = 0
𝑦=3
12
x-intercept: 𝑦 = 0
+6=0
𝑥−4
12 = −6(𝑥 − 4)
6𝑥 = 12
𝑥=2
y
h
 Both asymptotes
 Shape
𝑦=6
 𝑦≠3
3o
 𝑥=2
2
0
x
(4)
𝑥=4
3.2
𝑘(𝑥) =
−12
𝑥−4
 −12, −4
 −6
 𝑥>0
– 6 , 𝑥>0
(3)
[7]
4.1
4.1.1
QUESTION 4
𝑓(𝑥) = 2𝑥 2 − 6𝑥 − 20
−𝑏
−(−6)
3
Turning point: 𝑥 = 2𝑎 = 2(2) = 2 OR f’(x) = 4x – 6
 Value of x
0 = 4x – 6
3
x = 2
3
3
3
−49
1
𝑦 = 𝑓 ( ) = 2( )2 − 6 ( ) − 20 =
/−242
2
2
2
2
OR 𝑦 =
3
4𝑎𝑐−𝑏2
D(2 ;
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4𝑎
−49
2
=
4(2)(−20)−(−6)2
4(2)
=
−49
2
 Value of y
(2)
1
/−242
3
) OR D ( 2 ; −24,5)
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4.1.3
4.1.4
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2𝑥 2 − 6𝑥 − 20 = 0
𝑥 2 − 3𝑥 − 10 = 0
(𝑥 − 5)(𝑥 + 2) = 0
𝑥 = 5 or 𝑥 = −2
A(−2 ; 0 ) B(5 ; 0 )
𝑔(𝑥) = −2𝑥 + 𝑘
Substitute (−2 ; 0):
0 = −2(−2) + 𝑘
𝑘 = −4
2𝑥 2 − 6𝑥 + 𝑝 = 0
2𝑥 2 − 6𝑥 − 20 = −𝑝 − 20
∴ −𝑝 − 20 < −24,5
−𝑝 < −4,5
𝑝 > 4,5
NW/September 2014
 𝑓(𝑥) = 0
 Factors
 Both coordinates
(3)
 Substitute 𝑥 =
−2
 Answer
(2)
 Change to 𝑓(𝑥)
 Answer
(2)
OR
2𝑥 2 − 6𝑥 + 𝑝 = 0
For non-real roots ∆< 0
𝑏 2 − 4𝑎𝑐 < 0
(−6)2 − 4(2)(𝑝) < 0
36 − 8𝑝 < 0
−8𝑝 < −36
𝑝 > 4,5
 36 − 8𝑝 < 0
 Answer
(2)
OR
4.1.5
For no point of intersection f must be shifted more
than 24,5 units upwards. ∴ y-intercept will move 20 + 4,5
units upwards .
∴ 𝑝 > 4,5
𝑓(𝑥). 𝑔(𝑥) ≤ 0
𝑥 = −2 or 𝑥 ≥ 5
 Answer only

(2)
 𝑥 = −2
 𝑥≥5
(2)
4.1.6
𝑦 = −2𝑥 + 𝑡
−2𝑥 + 𝑡 = 2𝑥 2 − 6𝑥 − 20
0 = 2𝑥 2 − 4𝑥 − 20 − 𝑡
Tangent if roots are equal: ∆= 0
(−4)2 − 4(2)(−20 − 𝑡) = 0
16 + 160 + 8𝑡 = 0
8𝑡 = −176
𝑡 = −22
OR
𝑓 ′ (𝑥) = 4𝑥 − 6
𝑚𝑡 = −2 ∴ 4𝑥 − 6 = −2
4𝑥 = 4
𝑥=1
𝑦 = 𝑓(1) = −24
∴ −24 = −2(1) + 𝑡
𝑡 = −22
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 Equate
 Standard form
 ∆= 0
 Answer
(4)
 𝑓 ′ (𝑥) = −2
 𝑥=1
 𝑦 = −24
 𝑡 = −22
(4)
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4.2.1
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𝑝(𝑥) = 𝑥 2 + 1 and 𝑟(𝑥) = 𝑥 2 + 2𝑥
𝑝(𝑥 + 1) − 2 = (𝑥 + 1)2 + 1 − 2
= 𝑥 2 + 2𝑥 + 1 + 1 − 2
= 𝑥 2 + 2𝑥
= 𝑟(𝑥)
Shift p 1 unit to the left and 2 units downwards.
NW/September 2014
 (𝑥 + 1)
 −2
 Explanation
(3)
4.2.2
(If you replace x with 𝑥 + 1 you are shifting the graph 1
unit to the left and when subtracting 2 you are shifting the
graph 2 units down)
𝑦 ∈ [1 ; ∞)
 𝑦≥1
(1)
OR
𝑦 ≥ 1 ,𝑦 ∈ 𝑅
[19]
5.1
QUESTION 5
(1; 4) or (2; 16)
 Any one point
(1)
5.2
𝑓(𝑥) = 𝑎𝑥 2
Substitute inverse point:
(1; 4)
OR
2
4 = 𝑎(1)
𝑎=4
(2; 16)
16 = 𝑎(2)2
4𝑎 = 16
𝑎=4
∴ 𝑓(𝑥) = 4𝑥 2 , 𝑥 ≥ 0
 Substitute correct
coordinate
 Equation 𝑓(𝑥)
 𝑥≥0
(3)
OR
Substitute (4; 1) or (16; 2) into 𝑥 = 𝑎𝑦 2 :
4 = 𝑎(1)2
OR
16 = 𝑎(2)2
4=𝑎
16 = 𝑎. 4
4=𝑎
∴ 𝑓(𝑥) = 4𝑥 2 , 𝑥 ≥ 0
 Substitute correct
coordinates
 Equation 𝑓(𝑥)
 𝑥≥0
(3)
5.3
5.4
𝑔(𝑥 + 2) = 16
4𝑥+2 = 16
4𝑥+2 = 42
∴ 𝑥+2=2
𝑥=0
2<𝑥≤3
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 Substitution into
𝑔(𝑥)
 Same base
 𝑥=0
(3)
 𝑥>2
 𝑥≤3
(2)
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QUESTION 6
6.1
(1 +
𝑖(4) 4
)
4
1+
𝑖(4)
4
𝑖(4)
4
= (1 +
0,14 12
)
12
4
0,14 12
)
12
4
12
= √(1 +
0,14
= √(1 +
)
12
𝑖(4)
 Substitution into
formula
−1
 Simplification
= 0,03540 . . .
4
𝑖4
= 0,141639 . . .
Quarterly rate = 14,16 %
 Answer
(3)
6.2
R3 500
+ R5 700
T0
T1
T2
7% p.a. quarterly
T3
T4
8% p.a. monthly
A
T5
 (1 +
𝐴 = 𝑃(1 + 𝑖)𝑛
= 3 500 (1 +
0,07 8
4
) (1 +
0,08 36
12
)
+ 5 700 (1 +
= 5 107,73273. . . + 6 685,4612 . . .
= 𝑅 11 793,1939 . . .
≈ 𝑅 11 793,19
0,08 24
12
)
 (1 +
0,07 8
)
4
0,08 36
12
)
 5 700 (1 +
0,08 24
)
12
 Answer
(4)
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QUESTION 7
Calculate monthly payment for Option 1:
P= 950 000
x
T0
T1
T240
𝑖=
Option 1:
𝑃 =
x
0,12
12
= 0,01
𝑥[1−(1+𝑖)−𝑛 ]
𝑖
950 000 =
𝑥=
=
𝑥[1−(1,01)−240 ]
0,01
950 000 . 0,01
[1−(1,01)−240 ]
 Simplify
9 500
0,908 184 . . .
= 10 460,3182. . .
≈ 𝑅 10 460,32
Total cost for Option 1:
Option 1 = 10 460,32 × 240 + 6 000
= 𝑅 2 516 476,80
7.2
 Substitution into
correct formula
0,12
 𝑖 = 12 = 0,01
 𝑛 = 240
Total cost for Option 2:
Option 2 = 10 328,16 × 240 + 200 × 240
= 𝑅 2 526 758,40
Option 1 is the best. By choosing option 1 he will save
R10 281,60.
 Answer
 Calculation
 Answer
(7)
 200 × 240
 Answer
 Conclusion
(3)
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QUESTION 8
Note: Penalise only once for notation.
8.1
𝑓 ′ (𝑥) = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ→0
= lim
 Formula
ℎ
3−(𝑥+ℎ)2 − (3− 𝑥 2 )
 Substitution
ℎ
ℎ→0
2
3−𝑥2 −2𝑥ℎ−ℎ −3 + 𝑥2
ℎ
ℎ→0
= lim
2
=
−2𝑥ℎ− ℎ
lim
ℎ
ℎ→0
=
ℎ(−2𝑥− ℎ)
lim
ℎ
ℎ→0
 Simplify
 Factorise
= lim (−2𝑥 − ℎ)
ℎ→0
 Answer
(5)
= −2𝑥
8.2
8.2.1
𝑦=
2
3
√𝑥
= 2𝑥
𝑑𝑦
𝑑𝑥
− 𝜋𝑥
−1
3
= 2 .−
= −
8.2.2
8.3
 2𝑥
− 𝜋𝑥
2
3
1
3
. 𝑥
−4
3
−1
3
− 𝜋
4
𝑥− 3 – 𝜋
 −
2
3
 −𝜋
4
𝑥− 3
(3)
2
𝑥𝑦 − 𝑦 = 𝑥 − 1
𝑦(𝑥 − 1) = 𝑥 2 − 1
(𝑥−1)(𝑥+1)
𝑦=
(𝑥−1)
𝑦=𝑥+1
𝑑𝑦
=1
𝑑𝑥
𝑓(𝑥) = 𝑥 2 − 10𝑥 + 25
𝑚𝑡 = 𝑓 ′ (𝑥)
−8 = 2𝑥 − 10
2𝑥 = 2
𝑥=1
 Factorise LHS
 Factorise RHS
 Simplify
 Answer
(4)
 Multiply brackets
 𝑓 ′ (𝑥)
 𝑚𝑡 = −8
 Answer
(4)
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9.1.1
12
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QUESTION 9
𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 + 𝑞𝑥 + 30
At turning point 𝑓 ′ (𝑥) = 0:
3𝑥 2 + 2𝑝𝑥 + 𝑞 = 0
Substitute 𝑥 = −1:
3(−1)2 + 2𝑝(−1) + 𝑞 = 0
3 − 2𝑝 + 𝑞 = 0
−2𝑝 + 𝑞 = −3 - - - (i)
 𝑓′(𝑥)
 Substitute 𝑥 =
−1
 Equation (i)
Substitute (−1; 36) in 𝑓(𝑥):
(−1)3 + 𝑝(−1)2 + 𝑞(−1) + 30 = 36
−1 + 𝑝 − 𝑞 + 30 = 36
𝑝 − 𝑞 = 7 - - - (ii)
−𝑝 = 4
𝑝 = −4
Substitute 𝑝 = −4 into (ii):
−4 − 𝑞 = 7
𝑞 = −11
(i) + (ii):
9.1.2
9.1.3
NW/September 2014
 Substitute
(−1; 36)
 Equation (ii)
 𝑝 = −4
 𝑞 = −11
(7)
At C: 𝑓(𝑥) = 𝑔(𝑥) = 36
𝑥 3 − 4𝑥 2 − 11𝑥 + 30 = 36
𝑥 3 − 4𝑥 2 − 11𝑥 − 6 = 0
But 𝑥 = −1 at A
∴ (𝑥 + 1)(𝑥 2 + 𝑘𝑥 − 6) = 0
𝑘 + 1 = −4 OR touching point at A
𝑘 = −5
∴ (𝑥 + 1)(𝑥 + 1)(𝑥 − 6) = 0
∴ (𝑥 + 1)(𝑥 2 − 5𝑥 − 6) = 0
∴ (𝑥 + 1)(𝑥 + 1)(𝑥 − 6) = 0
𝑥=6
C (6; 36)
 𝑓(𝑥) = 36
 𝑥 + 1 factor
 𝑥=6
(3)
 𝑥 = −1
 𝑦 = 26
Turning point of k: (−1; 26)
(2)
9.2
(5; 4)
x
ℎ
(2; -3)
2
5
x
𝑦 = ℎ′ (𝑥)
 Shape
 𝑥=2
 𝑥=5
(3)
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QUESTION 10
10.1
y
h
0
P
S
Q
6 R
P(𝑥; 𝑥 2 ) and Q (𝑥; 0)
QR = 6 − 𝑥
PQ = 𝑥 2
Area of PQRS = PQ × QR
= 𝑥 2 × (6 − 𝑥)
= 6𝑥 2 − 𝑥 3
10.2
For maximum area:
𝑑𝐴
𝑑𝑥
x
 QR = 6 − 𝑥
 PQ = 𝑥 2
 Method
(3)
=0
12𝑥 − 3𝑥 2 = 0
3𝑥(4 − 𝑥) = 0
𝑥 = 0 or 𝑥 = 4
N/A
Maximum area = 6(4)2 − (4)3
= 32 units2
 12𝑥 − 3𝑥 2
𝑑𝐴

=0
𝑑𝑥
 𝑥=4
 Answer
(4)
[7]
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11.1
14
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NW/September 2014
QUESTION 11
C- correct W- wrong
C
0,7
C
0,6
0,3
 Tree diagram
W
C
0,4
0,4
W
0,6
W
11.2
P(2nd answer correct)
= P(CC) + P(WC)
= (0,6 × 0,7) + (0,4 × 0,4)
= 0,42 + 0,16
= 0,58 OR
= 58%
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵)
0,1 = (𝑥 + 0,1) × (0,1 + 0,3)
0,1 = (𝑥 + 0,1) × (0,4)
0,1 = 0,4𝑥 + 0,04
0,06 = 0,4𝑥
0,06
𝑥=
0,4
 Sum of
probabilities
 Substitution
 Answer
 𝑃(𝐴 ∩ 𝐵) =
𝑃(𝐴) × 𝑃(𝐵)
 Substitution
𝑥 = 0,15
 Value of x
𝑥 + 0,1 + 0,3 + 𝑦 = 1
∴ 0,15 + 0,1 + 0,3 + 𝑦 = 1
𝑦 = 0,45
11.3.1
11.3.2
11.3.3
12! = 12 × 11 × 10 ×. . . × 3 × 2 × 1
= 479 001 600
8 and 11 can be arranged in 2! different ways. Consider
8 and 11 as an entity. 11 objects can be arranged in 11!
different ways.
Total arrangements
= 2! × 11!
= 2 × (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
= 2 × 39 916 800
= 79 833 600
4! 4×3×2×1
=
= 4 × 3 = 12
2!
2×1
OR
4 × 3 = 12
TOTAL:
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(4)
 Sum of
probabilities is 1
 Value of y
(5)
 12!
 Answer
(2)
 2!
 11!
 Answer
(3)
 4! ÷ 2!
 Answer
(2)
Answer only, full
marks.
[16]
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NW/September 2014
ANALYSIS OF QUESTION PAPER: MATHEMATICS PAPER: 1
Question
Nr
Concept
1.1
1.1.1
1.1.2
1.1.3
Equations
Factorisation
Formula
Inequality
1.2
1.2.1
1.2.2
Simultaneous Equations
Exponential Equation
Simultaneous Equations
1.3
1.4
Exponents
Surds
2
2.1
2.2.1
2.2.2
2.3.1
2.3.2
2.4
Sequences and Series
Sigma notation
Geometric sequence
3
3.1
3.2
Hyperbola
Sketch
Reflection
4
4.1.1
4.1.2
4.1.3
4.1.4
4.1.5
4.1.6
4.2
4.2.1
4.2.2
Quadratic function
Turning point
X-intercept
Equation line
Infinite series
Level 1
Knowledge
Level 2
Routine
Procedures
DATE: Sept 2014
Level 3
Complex
Procedures
Level 4
Problem
Solving
3
5
3
2
5
2
3
4
2
3
3
4
Quadratic number pattern
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5
4
3
2
3
2
2
2
4
3
1
Please turn over
Mathematics/P1
Question
Nr
Concept
5
5.1
5.2
5.3
5.4
Inverse
6
6.1
6.2
Financial Mathematics
7.1
7.2
Loans
8
8.1
8.2.1
8.2.2
8.3
Differential Calculus
9
9.1.1
9.1.2
9.1.3
9.2
Cubic Function
Equation
Coordinate
Transformation
Application
10.1
10.2
Application
11
11.1
11.2
11.3.1
11.3.2
11.3.3
Probability
16
NSC – Memorandum
Level 1
Knowledge
NW/September 2014
Level 2
Routine
Procedures
Level 3
Complex
Procedures
Level 4
Problem
Solving
1
3
3
2
3
4
7
3
5
3
4
4
7
3
2
3
3
4
4
5
2
3
2
TOTAL:
%
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26
17,3
52
34,7
51
34
21
14