NATIONAL SENIOR CERTIFICATE GRADE 11

[Pages:12]NATIONAL SENIOR CERTIFICATE

GRADE 11

MATHEMATICS P1 NOVEMBER 2007 MEMORANDUM

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Mathematics/P1

QUESTION 1 1.1.1(a) (x + 3)(x - 1) = -x + 1

2 NSC MEMORANDUM

x2 + 2x - 3 = -x +1

(x + 3)(x -1) + (x -1) = 0

x2 + 3x - 4 = 0

OR (x -1)(x + 4) = 0

(x + 4)(x - 1) = 0

x = 1 or x = -4

x = -4 or x = 1

1.1.1(b) x2 + 3x - 4 < 0

(x + 4)(x -1) < 0

1.1.2

-4< x 0

x > -5 2

x 0

2.3

Yes

2 2007.52000 = 2 7.2 .5 2000 2000

= 128.(2.5) 2000

= 128.(10) 2000

the sum of the digits is 1 + 2 + 8 = 11

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DoE/November 2007

9substitution 9multiplication

completing the square (9+ 4; 9 -1)

(4) [25]

9 5x 2 9 3x2 9 6x 2 98x2

9 substitution

9 simplification

9M= 5 6

92x + 5 > 0

9 x>-5 2

9x 0

9 yes 9 27.22000 (exponential law) 9 128 9 grouping bases

with same exponents 9sum of digits

(4)

(3) (3)

(5) [15]

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5 NSC MEMORANDUM

QUESTION 3

3.1

29

3.2

Tn = an2 + bn + c

From (1 ; 1) 1 = a + b + c

c = 1 ? a ? b

From (2 ; 5) 5 = 4a + 2b + c

5 = 4a + 2b + 1 ? a ? b

4 = 3a + b

(i)

From (3; 11) 11 = 9a + 3b + c

11 = 9a + 3b + 1 ? a ? b

10 = 8a + 2b

(ii)

Solving (i) and (ii) simultaneously: 8 = 6a + 2b 10 = 8a + 2b 2 = 2a a = 1 b = 1 c = ? 1

Tn = n2 + n ? 1

OR

Tn = an2 + bn + c

From (1 ; 1) 1 = a + b + c

(i)

From (2 ; 5) 5 = 4a + 2b + c

(ii)

From (3; 11) 11 = 9a + 3b + c (iii)

(ii) ? (i)

3a + b = 4

(iv)

(iii) ? (ii) 5a + b = 6

(v)

(v) ? (iv)

2a = 2

a = 1

b = 1

c = ? 1

Tn = n2 + n ? 1

3.3

Pn = n2 + n ? 1

P100 = 1002 + 100 ? 1 = 10 099

OR

Pn = 100(101) -1 Pn = 10 099

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DoE/November 2007

9 answer (1)

9 c=

9 (i)

9 (ii) 9 9 a 9 b 9 c

99 eqns (i),(ii),(iii) 9 eqn (iv) 9 eqn (v)

9 a=1 9 b=1 9 c = ? 1

(7)

9 substitution 9 answer

9 substitution 9 answer

(2) [10]

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6 NSC MEMORANDUM

QUESTION 4

4.1

2; 1; 1 ; 1 ; 1 ; 1

2 4 8 16

Height reached during the 6th bounce = 1 16

4.2

1st bounce : 4 1 2

= 22.2-1

= 22-1 = 21

2nd bounce : 4 1 1 = 22.2-1.2-1 2 2

= 22-2 = 20

3rd bounce: 4 1 1 1 = 22.2-1.2-1.2-1 = 22-3 = 2? 1 2 2 2

nth bounce : 4 1 n = 2(2-n) 2

OR

1ste bounce: 2de bounce: 3de bounce:

4de bounce:

. . nde bounce :

2 = 21 1 = 20 1 = 2-1 2 1 = 2-2 4

2 2-n

4.3

22-n = 1

512

22-n = 2-9

2 - n = -9

n = 11

during the 11th bounce

DoE/November 2007

99 Answer

(2) 99 (for 1st bounce 2nd bounce and 3rd bounce) 99nth bounce

(4) 9 substitution 9 2-9 9 equating exponents 9 answer

(4) [10]

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QUESTION 5

5.1

A = P(1 - i) n

21500 = 86000(1 - i)4

0.25 = (1 - i)4

4 0.25 = 1 - i

i = 1 - 0.707

i = 0.2928932188 percentage rate = 29,29%

7 NSC MEMORANDUM

5.2.1

1

+

i

=

1 +

i(m) m

m

1 + i = 1 + 0,12 12 12

1 + i = 1,126825

i = 12,68%

5.2.2

T0

T1

T2

T3 T4

T5

i(12i)12==001.1,2122 = 0.01 12

i(2) i2

= =

0.14 0,124

=

0.07

2

Original investment 75000 = P1 + 0,12 2?12 1 + 0,14 3?2

12 2

75 000 = 1,905529326 P P = R 39 359,14

DoE/November 2007

9formula 921500 9substitution

9answer 9answer as

percentage

(5)

9 Formula

9 Substitution

9Simplification

9 answer

12,68%

(4)

9 i 9 next i. 9formula 99substitution 9answer

(6)

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8 NSC MEMORANDUM

DoE/November 2007

5.3.1

E (in billions of rands)

y

5.5 5

4.5 4

3.5 3

2.5 2

1.5 1

0.5

O

1

2

3

4

5

-0.5 t (in years)

5.3.2

The expenditure is increasing by 0,5 billion rand each year. OR The graph is a straight line OR Linear appreciation

5.3.3 E = 0,5t + 2

5.3.4 E = 0,5(8) + 2 E = R 6 billion

99 plotting of points

(one mark for plotting one or (2) two correct points; two marks for plotting three or four correct points)

x

9 0,5 per year

(1)

OR

9 straight line

92

(2)

90,5t

(1)

9 answer

[21]

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