This memorandum consists of 15 pages. NATIONAL. SENIOR CERTIFICATE.
GRADE 12. MATHEMATICS P1. FEBRUARY/MARCH 2009. MEMORANDUM. M
...
NATIONAL SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P1 FEBRUARY/MARCH 2009 MEMORANDUM
MARKS: 150
This memorandum consists of 15 pages.
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Mathematics/P1
2 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 1 1.1.1
1.1.2
1 =4 x 3x 2 − 4 x + 1 = 0 (3x − 1)( x − 1) = 0 1 x= or x = 1 3 3x +
3 standard form 3 factors 33 answers (4)
5 x ( x − 3) = 2
3 standard form 3 substitution into formula
5 x 2 − 15 x − 2 = 0
x=
− (−15) ± (−15) 2 − 4(5)(−2) 2(5)
3 265
15 ± 265 x= 10 x = 3,13 or x = – 0,13
33 answers
(5) 1.1.3
x − 2x > 3 x2 − 2x − 3 > 0
3 standard form
( x − 3)( x + 1) > 0
3 critical values
2
+
–
0
–1
0
+
OR
3
3 33 answers
x < – 1 or x > 3 1.2
–1
(4)
x − 3y = 1 x = 1 + 3y
(1 + 3 y ) − 2(1 + 3 y ) y + 9 y = 17
3 x = 1 + 3y 3 substitution
1 + 6 y + 9 y 2 − 2 y − 6 y 2 + 9 y 2 − 17 = 0
3 simplification
2
2
12 y 2 + 4 y − 16 = 0 3y2 + y − 4 = 0 (3 y + 4)( y − 1) = 0 4 y = − or y = 1 3 x = −3 or x = 4
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3 factors 3 answers 3 3answers
(7
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3 NSC – Memorandum
DoE/Febr. – March 2009
OR x − 3y = 1 x −1 y= 3
⎛ x −1⎞ ⎛ x −1⎞ x − 2 x⎜ ⎟ = 17 ⎟ + 9⎜ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎛ x 2 − 2x + 1⎞ 2x 2 − 2x ⎟⎟ = 17 x2 − + 9⎜⎜ 3 9 ⎝ ⎠ 2
2
2x 2 − 2x + x 2 − 2 x + 1 = 17 3 2 2 3x − 2 x + 2 x + 3 x 2 − 6 x + 3 = 51 x2 −
4 x 2 − 4 x − 48 = 0 x 2 − x − 12 = 0 ( x + 3)( x − 4) = 0 x = −3 or x = 4 4 y = − or y = 1 3 1.3
let 1234567893 = n Then 1234567893 × 1234567894 – 1234567895 × 1234567892 = n(n + 1) − (n + 2)(n − 1)
= n2 + n − n2 − n + 2 =2
3 method 3 simplification 3 answer
(3)
OR
Full marks for (3 × 4) – (5 × 2) = 2
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[23]
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4 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 2
2.1.1
2.1.2
2.1.3
2.2
2.3
1 1 + 1× 2 2 × 3 1 1 = + 2 6 2 = 3 S2 =
1 1 1 + + 1× 2 2 × 3 3 × 4 2 1 = + 3 12 9 = 12 3 = 4
3 answer
(1)
S3 =
3 answer
(1)
1 1 1 1 S4 = + + + 1× 2 2 × 3 3 × 4 4 × 5 3 1 = + 4 20 16 = 20 3 answer 4 = 5 If n represents the number of terms then the sum will be n divided by 3 3 answer n + 1. OR n Sn = n +1 3 answer 2008 Sn = 2009
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(1) (2)
(1) [6]
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5 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 3
3.1
(2 p − 3) − (1 − p ) = ( p + 5) − (2 p − 3) 2 p − 3 −1+ p = p + 5 − 2 p + 3 3p − 4 = − p + 8 4 p = 12 p=3
3 equating differences 3 simplification 3 answer
(3) 3.2.1
First term = 1 – p = 1 – 3 = – 2
3–2
3.2.2
−2 ;3;8 Common difference = 5
3 answer
3.3
(1) (1)
OR 3 p − 4 = 3(3) − 4 = 5 After the first term -2, all the other terms end in either a 3 or an 8. Perfect squares never end in a 3 or an 8. 33 explanation
(2) [7] QUESTION 4
4.1
4.2
6 ; 6 ; 2 ; – 6 ; – 18 ;… First difference: 0, – 4, – 8, – 12, … Therefore the next term is: – 18 – 16 = – 34 6 6 2 –6 0 –4 2a = −4
–4
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(1)
–8 –4
a = −2 Tn = −2n 2 + bn + c 6 = −2 + b + c 8=b+c ...(i ) 6 = –8 + 2b + c 14 = 2b + c …(ii) (ii) – (i): 6=b ∴c=2 Tn = −2n 2 + 6n + 2 OR
3 – 34
9 a=–2
9 solving simultaneously 9b=6 9c=2 9 general term
(5)
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2
6
6
2
DoE/Febr. – March 2009
–6 9 a=–2
4
0
–4 2a = −4
–4 –4
–8 –4
a = −2 T0 = c = 2 T1 = −2 + b + 2 = 6 b=6 Tn = −2n 2 + 6n + 2 OR
3c=2 3 substitution 3b=6 3 general term
(5)
3obtaining differences
T2 − T1 = 0 T3 − T2 = −4 T4 − T3 = −8
3 adding both sides
M Tn − Tn −1 = −4(n − 2) = 8 − 4n Adding both sides Tn − 6 = −4[1 + 2 + 3 + ....(n − 2)] ⎡ (n − 2)(n − 1) ⎤ Tn = −4 ⎢ ⎥+6 2 ⎣ ⎦ 2 Tn = −2n + 6n + 2
4.3
3 substituting T1 = 6 3 expression for sum of terms 3simplification
(5)
− 6838 = −2n 2 + 6n + 2 − 2n 2 + 6n + 6840 = 0 n 2 − 3n − 3420 = 0
n=
3±
(− 3)2 − 4(1)(−3420) 2
3 ± 117 2 n = 60 or n = - 57 not possible
3 standard form 3 substitution
n=
∴ T60 = −6838
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3 values of n 3 answer
(4) [10]
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QUESTION 5
5.1
Area of unshaded square = 1=
1 16
15 16
31 3
1 16
3 answer
(2) 5.2
Sum of the unshaded areas of the first seven squares 1 ⎞ 1 ⎞ ⎛ ⎛ 1⎞ ⎛ = (1 − 1) + ⎜1 − ⎟ + ⎜1 − 2 ⎟ + L + ⎜1 − 6 ⎟ ⎝ 4 ⎠ ⎝ 4⎠ ⎝ 4 ⎠ 1 ⎞ ⎛ 1 1 = 7 − ⎜1 + + 2 + L + 6 ⎟ 4 ⎠ ⎝ 4 4 7 ⎛ ⎛ ⎛1⎞ ⎞⎞ ⎜ 1⎜1 − ⎜ ⎟ ⎟ ⎟ ⎜ ⎜ ⎝4⎠ ⎟⎟ ⎠⎟ = 7−⎜ ⎝ 1 ⎜ 1− ⎟ ⎜⎜ ⎟⎟ 4 ⎝ ⎠ = 7 − 1,333251953... = 5,666748047... = 5,67
3expression for sum of unshaded areas 3 simplification 3 answer
(5)
OR Sum of the unshaded areas of the first seven squares 1 ⎞ 1 ⎞ ⎛ 1⎞ ⎛ ⎛ = (1 − 1) + ⎜1 − ⎟ + ⎜1 − 2 ⎟ + L + ⎜1 − 6 ⎟ ⎝ 4⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ 1 ⎞ ⎛1 1 = 6 − ⎜ + 2 +L+ 6 ⎟ 4 ⎠ ⎝4 4 6 1 ⎛⎜ ⎛ 1 ⎞ ⎞⎟ 1− ⎜ ⎟ 4 ⎜ ⎝ 4 ⎠ ⎟⎠ = 6− ⎝ 1 1− 4 = 6 − 0,332519531... = 5,666748047... = 5,67
[7]
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8 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 6
6.1
3b=1 3c=2 3 substitution of (0 ; 0) 3 a=2
a +2 x −1 a 0= +2 0 −1 0 = −a + 2 y=
a=2 y=
2 +2 x −1
(4)
OR ( x − 1)( y − 2) = k But g passes through the origin ∴ (−1)(−2) = k = 2
∴ ( x − 1)( y − 2) = 2
6.2
g ( x) = ( x − 1) + q 2
0 = (2,5 − 1) 2 + q 9 q=− 4 9⎞ ⎛ Turning point ⎜1 ; − ⎟ 4⎠ ⎝ 6.3
6.4
y=2 x=2 h( x) = −( x − 1) 2 +
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3 equation 3 through origin 3 substitution 3 equation 3p=1 3 substitution of (2,5 ; 0) 9 3 q=− 4
(4)
3 answer
(4) 3 answer 3 answer
(2) 9 4
3 answer
(1) [11]
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9 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 7
7.1
7.2
7.3
Any base raised to the power 0 is 1 which means the y-intercept of the graph h( x) = a x will be (0 ; 1) therefore Q(0 ; 1) OR h(0) = a 0 = 1 ∴ Q(0 ; 1) 1 a −1 = 2 1 1 = a 2 a=2 2y = x
3 substitution 3 answer
(2) 3 interchanging x and y 3 answer (2) 3 point (0,5 ; – 1) or any other valid point 3 point (1 ; 0) 3 shape
y = log 2 x 7.4
3 y-intercept 3 any base raised to power 0 is 1 (2)
y 4
3
(3)
2
h -1 1
(2 ; 1)
x -1
O
1
2
3
4
5
6
7
8
-1
-2
-3
7.5
7.6
x > 0,5
3 reading off from graph 3 answer
∴ 2 + x log 3 = x log 2 2 ∴x = = − 11.36 log 2 3
(2) 3 equating 3 logs both sides 3 answer (3)
X
⎛2⎞ OR ⎜ ⎟ = 100 ⎝3⎠ ∴ x log (2 3 ) = 2 ∴x =
[14]
2 = − 11.36 log 2 3
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10 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 8 3 shape 3 amplitude
y
8.1
(2)
2
1
x -180
0
-90
90
180
270
360
-1
-2
8.2
y ∈ [−4 ; 4]
33 answer
720°
33 answer
Shift f 90° to the left to get 2cos x. ∴θ = 90° + n.360°
33 answer Accept 90°
(2) 8.3
(2) 8.4
(2) [8]
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QUESTION 9
9.1
18
i ⎞ ⎛ 2000⎜1 + ⎟ = 2860 ⎝ 12 ⎠ 18
i ⎞ ⎛ ⎜1 + ⎟ ⎝ 12 ⎠ i 1+ 12 i 12 i i
= 1,43
3 substitution
18
i ⎞ ⎛ 3 ⎜1 + ⎟ = 1,43 ⎝ 12 ⎠
= 1,020069541... = 0,020069541... = 0,24083... = 24,08%
3 i = 0,24083… 3 answer
(4) 9.2 ⎡⎛ 0,08 ⎞12 ⎤ 100 ⎢⎜1 + ⎟ − 1⎥ 12 ⎠ ⎢⎣⎝ ⎥⎦ Fv = 0,08 12 = R1 244,99
The accumulated amount is less than R1 300 required to buy the bike. Farouk will not be able to buy the bike on 1 January 2009. QUESTION 10 10.1 15 Loan = 125000 − × 125000 100 Loan = R 106 250 OR Loan = 0,85 × 125 000 Loan = R 106 250 10.2 ⎡ ⎛ 0,125 ⎞ −6×12 ⎤ ⎟ ⎢1 − ⎜1 + ⎥ 12 ⎠ ⎝ ⎢ ⎥ 106250 = x ⎢ ⎥ 0,125 ⎢ ⎥ 12 ⎢⎣ ⎥⎦
⎛ ⎛ 0,125 ⎞ −6×12 ⎞ ⎟ 1106,770833 = x⎜1 − ⎜1 + ⎟ ⎜ ⎝ ⎟ 12 ⎠ ⎝ ⎠ x = R 2104,94
3 formula 3 substitution
33 answer 3 conclusion
(5) [9]
3 answer
(1)
3 substitution 3 – 72 ⎛ 0,125 ⎞ 3⎜ ⎟ ⎝ 12 ⎠ 3 simplification
3 answer
(5) [6]
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12 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 11
11.1
f ( x + h) − f ( x) h →0 h 2 ( x + h) − 2( x + h) − x 2 + 2 x = lim h →0 h 2 2 x + 2 xh + h − 2 x − 2h − x 2 + 2 x = lim h →0 h 2 2 xh + h − 2h = lim h →0 h h(2 x + h − 2) = lim h →0 h = lim(2 x − 2 + h)
9 method
= 2x − 2
9 answer
f ′( x) = lim
9 substitution 9 simplification
9 factorising
h →0
(5) 11.2.1
D x [( x − 3) ] 3
2
[
= Dx x 6 − 6 x 3 + 9 = 6 x 5 − 18 x 2
]
3 simplification 33 answer
(3) 11.2.2
y=
4 x −
− 1 2
3
x 9
1 y = 4x − x3 9 3 − dy x2 = −2 x 2 − dx 3
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3 power form 33 answer
(3) [11]
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13 NSC – Memorandum
DoE/Febr. – March 2009
QUESTION 12
12.1
3 h′( x ) 3 substitution of x = – 1 3 h′( x) = 0
h′( x) = −3 x 2 + 2ax + b h′(−1) = −3(−1) 2 + 2a(−1) + b 0 = −3 − 2 a + b 2a − b = −3 K (i) h′(2) = −3(2) 2 + 2a(2) + b 0 = −12 + 4a + b 4a + b = 12 K (ii) (ii) + (i): 6a = 9 3 a= 2 ⎛3⎞ 2⎜ ⎟ − b = −3 ∴ ⎝2⎠ b=6 OR
h(−1) = −(−1) 3 + a (−1) 2 + b(−1) = −9 2 2a − 2b = −9
∴a − b =
L (i)
3 simplification 3 substitution 3 solving simultaneously
(6)
−7 2
3 substitution of x = -1 −7 3 h(−1) = 2 3 simplification
h(2) = −(2) + a(2) 2 + b(2) = 10 4a + 2b = 18 L (ii)
3 substitution of x = 2 and h( 2) = 10 3 simplification
(i) + (ii):
3 solving simultaneously
3
6a = 9 3 a= 2
−9 ⎛3⎞ ⎜ ⎟−b = 2 ⎝ 2⎠ b=6
12.2
Average Gradient 10 − (−3,5) = 2 − (−1) 13,5 = 3 9 = 2
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(6)
3 substitution
3 answer
(2)
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12.3
h′( x) = −3x 2 + 3 x + 6 h′(−2) = −3(−2) 2 + 3(−2) + 6 h′(−2) = −12 Point of contact (−2 ; 2) y − 2 = −12( x + 2) y = −12 x − 22
14 NSC – Memorandum
DoE/Febr. – March 2009
3 h′( x ) 3 substitution 3 gradient 3 point 3 answer
(5) 12.4
12.5
h′( x) = −3 x 2 + 3 x + 6 h ′′( x) = −6 x + 3 − 6x + 3 = 0 1 x= 2 OR −1+ 2 x= 2 1 x= 2 p > 3,5 or p < −10
9 second derivative 9=0 9 answer
(3)
33 answer
(2) [19]
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QUESTION 13
13.1
3substitution 3 simplification
AB 2 = (a 2 − 0) 2 + (a − 3) 2 AB 2 = a 4 + a 2 − 6a + 9
13.2
(2) d AB2 da d 3 AB 2 = 0 da 3 simplification 3 factorisation 3 answer
d AB 2 = 4a 3 + 2a − 6 da 0 = 4a 3 + 2a − 6
3
0 = 2a 3 + a − 3 0 = (a − 1)(2a 2 + 2a + 3) a =1 There is no real solution for 2a 2 + 2a + 3 = 0
(5) [7]
QUESTION 14
14.1
x ≥ 200 x + y ≤ 600 50 x + 100 y ≤ 45000
3 answer 3 answer 3 answer
(3) 14.2 and 14.3
3 600 3450 3 200 3 600 3 900
y 700
600
(5)
Number of trousers
500
3 feasible region
(1) 400
300
200
100
x 100
200
300
400
500
600
700
800
900
Number of white shirts
14.4
P = 30x + 40y
33 answer
14.5
Maximum at (300 ; 300)
3 search line 3 answer
(2) (2) [13]
TOTAL: 150
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