GRADE 11 NATIONAL SENIOR CERTIFICATE

[Pages:14]GRADE 11

NATIONAL SENIOR CERTIFICATE

MATHEMATICS P2 NOVEMBER 2007 MEMORANDUM

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QUESTION 1

1.1

m AB

=

17 - 5 12 - 3

= 12 9

= 4 3

1.2

m BC

=

-

3 4

2 NSC MEMORANDUM

1.3

20 -17 = - 3

x -12 4

3 =-3 x -12 4

3x - 36 = -12

3x = 24

x=8

C(8; 20)

20 - 17 x - 12

.

4 3

=

-1

OR

x

3 - 12

.

4 3

=

-1

x - 12 = -4

x=8

1.4

AB = (xa - xb )2 + (ya - yb )2

= (17 - 5)2 + (12 - 3)2

= 225 = 15

BC 2 + AB2 = AC2

AC2 = (8 - 3)2 + (20 - 5)2

25 + 225 = 250

OR

= 250

AC = 5 10

= 5 10

Perimeter = 15 + 5 + 5 10 = 20 + 5 10 units

DoE/November 2007

9 substitution

9 value (2)

9 value (1)

9 substitution

9 simplification 9 x = 8

(3)

9 formula 9 substitution 9 value

9 correct use of Pythagoras' Theorem 9 value AC OR 9 formula and substitution 9 value AC

9 value perimeter (6) [11]

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3 NSC MEMORANDUM

QUESTION 2

2.1

Midpoint AC x1 + x2 ; y1 + y2

2

2

Midpoint AC 6 + 6 ; 4 + (-2)

2

2

Midpoint AC (6 ; 1)

2.2

m BD

=

yb xb

- -

yd xd

= 2-1 -2-6

= -1 8

equation of BD:

y=-1x+c 8

2 = - 1 (- 2) + c

OR

8

c=7 4

y - y1 = m(x - x1 )

y - 2 = - 1 (x + 2)

8 y=-1x-1 +2

84 y=-1x+ 7

84

equation of BD :

y = - 1 x + 7 or x + 8y -14 = 0 84

2.3

y = ?2 x + 8

equation :

y = ?2x + c

4 = ?2(6) + c

c = 16

y = ?2x + 16

2.4

m BC

=

2 - (-2) -2-6

mBC

=

4 -8

mBC

=

-1 2

tan = - 1 2

reference angle : 26, 6? = 180? ? 26,6? (obtuse angle) = 153,4?

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DoE/November 2007

9 substitution into correct formula 9 y?coordinate 9 x?coordinate

(3)

9 formula and substitution 9 value of gradient

9 formula 9 substitution

9equation of BD (any form accepted)

(5)

9 gradient 9 substitution 9 value

(3)

9 substitution

9

m BC

=

-1 2

9 tan 9 ref angle

9 value (5)

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4 NSC MEMORANDUM

2.5

AC x-axis

C^ = - 90? (exterior angle of triangle)

= 153,4? - 90?

= 63,4?

DoE/November 2007

9 AC x-axis 9 substitution 9 value

(3) [19]

QUESTION 3 3.1.1 Reflection in the y-axis

3.1.2

P

P/

4

9reflection 9y-axis

(2)

-5

R

P //

2

Q Q/

O

Q //

-2

-4

R //

5

3

9 image

99 correct coordinates

(3)

-6

3.1.3 R (1; ? 4)

3.1.4 (x; y) (? y; x)

9 coordinates (1)

9 x-coordinate 9 y-coordinate

(2)

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3.2.1

18

&

16

3.2.2

14

5 NSC MEMORANDUM

12

10

A'''(9; 9) 8

6

C/ (6; 6)

4

2 A(1; 1)

O

A'' (3; 3) 5

10

15

3.3

40 = 2 and 80 = 2

20

40

k=2

DoE/November 2007

9 first enlargement 9 second enlargement 99 enlarged figures centred at the origin

(4) 9 C / coordinates 9 B// coordinates

(2)

B// (18; 9)

99 value (2) [16]

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QUESTION 4

4.1

3cos150?.sin 270?

tan(- 45?) + cos600?

= 3 -

3 2

(-

1)

(-1) + - 1

2

33

=2 -3 2

=- 3

6 NSC MEMORANDUM

4.2

tan(180? - x).sin(90? + x) - sin y.cos(90? - y)

sin(- x)

=

(-

tan

(-

x).(cos sin x)

x)

-

sin

y .sin

y

= sin x . cos x - sin2 y cos x sin x

= 1 - sin2 y

= cos2 y

DoE/November 2007

9 cos150? = - 3 2

9 sin 270? = -1

9 tan(- 45?) = -1

9 cos 600? = - 1 2

9 simplification (5)

9 sin y 9?tan x 9 cos x 9 ? sin x 9 sin x identity

cos x 9 1 9 identity

(7)

[12]

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QUESTION 5

7 NSC MEMORANDUM

5.1.1

k.cos = ? 2 and k.sin = 3 cos < 0 and sin > 0 quadrant II (90?; 180?)

5.1.2

tan = sin

cos 3 =k -2 k =3 -2

OR tan = - 3

2

k 3

-2

5.1.3

cos2 + sin 2 = 1 - 2 2 + 3 2 = 1 k k 4+9 = k2 k 2 = 13

k = 13

OR

k 2 = 32 + (-2)2 =9+4 = 13

k = 13

k 3

-2

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DoE/November 2007

9 cos < 0 9 sin > 0 9conclusion of quadrant

(3)

9 use correct ratio

9 value (2)

9 sketch 9 value

(2)

9 identity 9 substitution 9 multiplication by LCD 9 k2 = 13

99 sketch 9 use of Pythagoras statement 9 k2 = 13

(4)

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5.2

5tan x = 125

8 NSC MEMORANDUM

5 tan x = 53

tan x = 3

reference angle = 71,6?

x = 71,6?

or x = 180? + 71,6? = 251,6?

DoE/November 2007

9 53 9 tan x = 3 9 ref angle = 71,6? 9 x = 71,6?

9 x = 251,6? (5)

5.3

sin x(2cos x -1) = 0

sin x = 0 or cos x = 1 2

9 sin x = 0 en cos x = 1 2

x = 0? + k.360? or x = 180? + k.360? or x = 60? + k360? 99 x = k.180

or x = 300? + k.360?

9 x = 60? + k360? 9 x = 300? + k.360?

x = k.180?

9 kZ

kZ

9 general solution notation

(7)

[21]

QUESTION 6

6.1.1

CS = 64? sin 64? = 15

SG SG = 15

sin 64?

SG = 16,69 m

9definition 9 substitution

9SG

(3)

6.1.2

SH2 = (16,69)2 + (7,32)2 ? 2(16,69)(7,32).cos 116? = 439,2508074...

SH = 20,96 m

OR

CG = 16,692 -152 = 7,32 SH = 152 + 14,642 = 20,96

9 cos rule or

pythagoras 9 substitution 9 value

(3)

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