Maths P1 SG Nov 2007 Final MEMO 231007

[Pages:12]SENIOR CERTIFICATE EXAMINATION - 2007

MATHEMATICS P1 STANDARD GRADE OCTOBER/NOVEMBER 2007

MARKING MEMORANDUM

This memorandum consists of 12 pages. Final version: 23 October 2007

External Moderator: .......................................... External Moderator: .......................................... Internal Moderator: ..........................................

Copyright reserved

Please turn over

Mathematics SG/P1

2 Marking Memorandum

1.1 1.1.1

(x - 3)(x + 2) = 6

x2 + 2x - 3x - 6 = 6

x2 - x -12 = 0

(x - 4)(x + 3) = 0

x = 4 or x = -3

1.1.2 5x2 + 7x - 2 = 0

x = - b ? b2 - 4ac 2a

= - (7) ?

(7)2 - 4(5)(- 2) 2(5)

= - 7 ? 89 10

= ?1,64 or 0,24

1.1.3

2x - 6 = 20 - x2

4x2 - 24x + 36 = 20 - x2

5x2 - 24x + 16 = 0

(5x - 4)(x - 4) = 0

x= 4

5

check: 2x - 6 0

or x = 4

x 3 hence x = 4

OR

- 22 484 = 22 x 4

5

25 5

5

2 = 20 -16 x = 4

DoE/Oct-Nov/2007

finding product correctly standard form factors (subst. in formula) both x-values (CA from factors) If write (x - 3) = 6 or (x + 2) = 6 (4) Break down BD 0/4

formula

substitution into formula

simplification

each value of x [NOTE: Minus 1 mark for incorrect rounding off on the whole paper] If write wrong formula then get x = -6,06 or - 7,94 (5) Max. 3/5

for squaring for 4x2 - 24x + 36 standard form factors both solutions

rejecting x = 4

5

(various methods of checking) CA applies

(6)

Copyright reserved

Please turn over

Mathematics SG/P1

3

Marking Memorandum

1.2

2x - y = 7 ......... .. (1)

x2 + xy + y 2 = 7 ........... (2)

From (1): y = 2x - 7 ....... (3)

Substitute (3) in (2):

x2 + x(2x - 7) + (2x - 7)2 = 7

x2 + 2x2 - 7x + 4x2 - 28x + 49 = 7

7x2 - 35x + 42 = 0 ?7: x2 -5x+6 = 0 OR (7x-14)(x-3) = 0

(x - 2)(x - 3) = 0

x = 2 or x = 3

y = 2(2) - 7 y = 2(3) - 7

From (3)

or

y = -3

y = -1

OR x = y + 7 y + 7 2 + y y + 7 + y 2 = 7

2 2 2

y 2 + 14 y + 49 + y 2 + 7 y + y 2 = 7

4

2

? 4 : y 2 + 14 y + 49 + 2 y 2 + 14 y + 4 y 2 = 28

7 y 2 + 28y + 21 = 0

?7 : y2 + 4y +3 = 0

(y + 3)(y + 1) = 0

y = -3 or y = -1

x = 2 or x = 3

DoE/Oct-Nov/2007

y-subject of formula substitution expansion simplification factors both values of x

each value of y OR

x-subject of formula substitution expansion

simplification

factors both values of y

each value of x (8) (CA applies) [23]

Copyright reserved

Please turn over

Mathematics SG/P1

2.1 x = 4 + 3k k -6 1 12

2.2 k 2 x 2 - kx - 7 = 0 = b2 - 4ac

= (- k )2 - 4(k 2 )(-7)

= k 2 + 28k 2 = 29k 2 0 for all values of k Answer 2.2.1

4 Marking Memorandum

x is ... Unreal Irrational rational

2.3 f (x) = x3 - x + 3

( ) ( ) ( ) f

-1

2

=

-1

2

3

-

-1

2

+3

=

-

1 8

+

1 2

+

3

=

3

3 8

or

27 8

or 3,375 or 3,38 or 3,4

The

remainder

is

3

3 8

2.4 g(x) = x3 + m2x2 -11x -15m g(3) = 0

(3)3 + m2(3)2 -11(3) -15m = 0

27 + 9m2 - 33 -15m = 0 9m2 -15m - 6 = 0

? 3 : 3m2 - 5m - 2 = 0

(3m +1)(m - 2) = 0

m = - 1 or m = 2

3

DoE/Oct-Nov/2007

x unreal x irrational x rational (3) Only one word per line

formula for substitution

= 29k 2

answer 2.2.1 Full marks if substitute values for k and correct conclusion. (4) Correct answer only 1/4

f (- 1) 2

substitution

answer Accept other methods (3) eg. long division

g(3) = 0 or implied

substitution

standard form

factors

each value of m (7) (CA applies) [17]

Copyright reserved

Please turn over

Mathematics SG/P1

5 Marking Memorandum

DoE/Oct-Nov/2007

3.1 y = x 2 - 4x - 5 3.1.1 A & B: y = 0 = x2 - 4x - 5

0 = (x +1)(x - 5)

x = -1 or x = 5

A (-1;0) B (5;0)

3.1.2 (a): OC = 5 units

(b):

ED:

x = 5 + (- 1)

2

x=2

OR

x

=

-

b 2a

=

-

(- 4) 2(1)

=

2

OR

y = 0 or 0 = x2 - 4x - 5 can be awarded at A & B

factors x-values

coordinate form (accept A = -1; B = 5) Answer only 4/4 (4) (2 for A ;2 for B)

OC = 5 (must be positive if negative no mark)

method x = 2

dy = 2x - 4 = 0 dx x=2

Subst. x = 2 into y = x 2 - 4x - 5

y = (2)2 - 4(2) - 5

y = -9 ED = 9 units OR: y = 4ac - b2

4a = 4(1)(-5) - (-4)2

4(1) = ?9 ED = 9

Substitute x = 2

ED = 9 (must be positive if negative no mark) OR:

(c): BC = OC 2 + OB 2

BC using Pythagoras

= 52 + 52 OR (5 - 0)2 + (0 + 5)2

BC = 50 CA applies

= 50 or 5 2 or 7,07

(7)

3.1.3 g : y = 1x - 5

gradient y-intercept

(3) Answer only full marks

Copyright reserved

Please turn over

Mathematics SG/P1

6 Marking Memorandum

3.2 3.2.1 a = 8 a

a2 = 8

r = OP = a2 + a2 = 2a2

OP = 8 + 8 = 16 = 4 OR:

OP = 2k

= 2(8) =4

3.2.2 y

f g x

4

g

4.1 4.1.1

5 x+1.4 x 2 x+ 2.10 x

=

5 x +1.22 x 2 x + 2.2 x.5 x

= 5 x +1- x.22x - x -2- x

= 51.2-2

= 5 = 1,25 = 1 1

4

4

OR

5x+1 . 4 x 2 x+2 .10x

=

5.5 x .4 x 22.2x.10 x

= 5(20) x 4(20) x

=5 4

OR

5 x +1.4 x 2 x + 2.10 x

=

5x.5.2x.2x 2x.22.2x.5x

=

5 4

DoE/Oct-Nov/2007

Substitute a

Cross multiply

r = OP

OP = 4 (accept 2a 2 ) OR:

formula Substitute a answer (4)

f: semi circle above x - axis radius

g: hyperbola shape (should not cut axes) quadrants

for touching

(If on separate set of axes deduct mark for not touching).

(5) [23]

2 2x 2 x.5x

exponential law

5 4

x

5 .5

x

( 20 )

5 4

2 x

2 .2

5x.5 & 2 x.22 2 x.2 x

(4)

2 x.5 x 5 4

Copyright reserved

Please turn over

Mathematics SG/P1

7 Marking Memorandum

( ) 4.1.2

5 - 3 2 + 2 15

= 5 - 2 5 3 + 3 + 2 15 = 8

4.1.3

log 125 + 3 log 2

= log125 + log 23

= log125(8)

= log1000 = 3 OR: 3log 5 + 3log 2 = 3(log 5 + log 2) = 3log10

=3

4.2 4.2.1 6x - 6x.6-1 = 180

( ) 6x 1 - 6-1 = 180 OR:

6x 5 = 180 6

6 x = 180 ? 6 5

6 x = 216 = 63

x =3 OR:

6x-1(6 -1) = 180

6x-1 = 36 6x-1 = 62 x -1= 2 x =3

4.2.2 log x - log(x - 5) = 2

log x = log100 OR: log x = 2

x-5

x-5

x = 100 x-5

x = 102 x-5

x = 100x - 500

99x = 500

x = 500 = 5,05 99

DoE/Oct-Nov/2007

( ) for knowing a 2 = a

middle term answer (3) Answer only 0/3

log law log law

log1000 = 3

log law common factor 3

log10 = 1 (3) Answer only 0/3

6x.6-1 (common factor/ split)

( ) 6x 1 - 6-1

5 6 216 answer OR: common factor divide by 5 36 = 62 equating exponents answer (Answer with checking 3/5; (5) answer only 0/5)

log law 2 = log100 or 102 = 100

removing logs

adding similar terms (99x)

answer (5) (CA Applies)

Copyright reserved

Please turn over

Mathematics SG/P1

4.3 5x = 3 log 5x = log 3 x log 5 = log 3

x = log 3 log 5

= 0,68 OR x = log5 3 x = log 3

log 5 = 0,68

8 Marking Memorandum

DoE/Oct-Nov/2007

apply logs both sides log 5x = x log 5 x-subject of formula

answer

OR log-form

change of base

answer (4) [24]

5.1 5.1.1

Tn = a + (n -1)d

62 = 2 + (n -1)(3)

62 = 2 + 3n - 3 63 = 3n

n = 21

OR: The terms are: -1 + 3?1;-1 + 3? 2;-1 + 3? 3;....;-1 + 3? 21

21 terms

formula substitution Tn = 62 substitution for a and d

answer [NOTE: Wrong formula max ? for substitution of Tn or a & d] OR

Pattern /terms answer (4) [Answer only 4/4]

5.1.2

Sn

=

n 2

(a

+ l)

S 21

=

21 (2 +

2

62)

= 21(32)

= 672

OR

Sn

=

n 2

[2a

+

(n

-1)d ]

S 21

=

21[2(2) +

2

(21 - 1)(3)]

= 21(32)

= 672

formula for S n substitution (CA applies)

answer

OR formula for S n substitution

(CA applies)

answer [Answer only 3/3 with evidence in 5.1.1] [Wrong formula max : 1/3 (3) for substituting n = 21]

Copyright reserved

Please turn over

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download