NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P2 FEBRUARY/MARCH 2011 MEMORANDUM

MARKS: 150

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2 NSC – Memorandum

DBE/Feb. – Mar. 2011

QUESTION 1 1.1

55 + 55 + 50 + 47 + 42 + 3x = 48,375 8 249 + 3 x = 48,375 8 3 x = 138 x = 46

249 + 3x = 48,375 8 9 3 x = 138 (2) 9

9 max and min 9 median 9 Q1 and Q3 9 whiskers

1.2

42

46,5

QUESTION 2 2.1

Mass (kg) 60 ≤ x < 70 70 ≤ x < 80 80 ≤ x < 90 90 ≤ x < 100 100 ≤ x < 110

2.2

52,5 54 55 56

Frequency

5 7 7 4 2

99 Frequencies 99 Cumulative Frequencies

Cumulative Frequency 5 12 19 23 25

(4)

Cumulative Frequency Curve

99 plotting points 1 mark: 3 – 5 points correctly 0 marks : 2 or less points correctly plotted

30 25 Cumulative Frequency

(4) [6]

20

9 graph

15

(3)

10 5 0 40

50

60

70

80

90

100

110

120

Mass (kg)

2.3

Mean = 79,28

99 answer

(2)

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2.4



Standard Deviation = 11,02 79,28 – 11,02 = 68,26 79,28+11,02 = 90,3 17 players lie in this interval. 17 = 68% 25

999 sd = 11,02 9 17 players 9 68%

(5) [14]

QUESTION 3

Question 3.1

3.1 & 3.2

Scatter Plot showing Arm Span vs Height

4 marks: All points plotted correctly. 3 marks: 9 – 11 points correct 2 marks: 6 – 8 points correct 1 marks: 3 – 5 points correct 0 marks if less than 3 points plotted correctly. (4)

200 195 190

Height (cms)

185 180 175 170 165 160 155 150 150

160

170

180

190

200

Arm Span (cm)

3.3

Yes. The relationship between arm span and height is a positive, linear one so we can expect a person with below average arm span to have below average in height.

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Question 3.2 99 linear best fit Line (2)

9 Yes 9 Reason

(2) [8]

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QUESTION 4

4

P

y

2

R

-5

Q

5

x

-2

-4

4.1

Let β be the angle of inclination of PQ. tan β = m PQ

9 tan β = mPQ 9 tan β = 4

2 − (−2) − 1 − (−2) tan β = 4 β = 75,96°

tan β =

9 answer

(3) 4.2

4.3

⎛ −1+ 3 2 + 0 ⎞ ; M⎜ ⎟ 2 ⎠ ⎝ 2 M (1 ; 1)

9 x-value 9 y-value

(2) 9 substitution into correct formula 9 answer

PQ = (−1 + 2) 2 + (2 + 2) 2 = 17 PR = (−1 − 3) 2 + (2 − 0) 2 =

20

QR = (0 − (−2)) + (3 − (−2)) 2

=

4.4

2

29

Perimeter = 29 + 20 + 17 = 13,98 units = 14 to the nearest whole number y − 1 = 4( x − 1) y = 4x − 3

9 answer 9 sum 9 answer

(5)

9m=4 9 substitution of (1 ; 1) 9 answer

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QUESTION 5

5.1.1

x 2 + y 2 − 8x + 6 y

= (2) 2 + (−9) 2 − 8(2) + 6(−9) = 4 + 81 − 16 − 54 = 15 Hence, the point lies on the circumference of the circle. OR x 2 + y 2 − 8 x + 6 y = 15

9 substitution 9 answer

(2)


( x − 4) 2 + ( y + 3) 2 = 15 + 16 + 9

(2)

( x − 4) + ( y + 3) = 40 2

2

( x − 4) 2 + ( y + 3) 2 = (2 − 4) 2 + (−9 + 3) 2

5.1.2

= 22 + 62 = 40 ∴The point lies on the circumference of the circle. x 2 + y 2 − 8 x + 6 y = 15 ( x − 4) 2 + ( y + 3) 2 = 15 + 16 + 9 ( x − 4) 2 + ( y + 3) 2 = 40 Circle centre (4 ; − 3) − 3 − (−9) mrad = 4−2 mrad = 3

99 ( x − 4) 2 + ( y + 3) 2 = 40 9 centre 9 gradient of radius

1 3 1 y + 9 = − (x − 2) 3 1 25 y =− x− 3 3

m tan = −

9 gradient of tangent 9substitution 9 answer

(7)

5.2

A(6 ; 4) B 10 (3 ; – 1)

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Radius AB = 10 Distance from A to centre of circle is =

(6 − 3)2 + (4 + 1)2

= 9 + 25 = 34 AB 2 = 34 − 10 AB 2 = 24 AB = 24


9 subs into distance formula

9 34 9 AB 2 = 34 − 10 9 answer

(5)

AB = 2 6 AB = 4,90

OR r 2 = 10 r = 10 Radius ⊥ tangent By Pythagoras AB 2 = (6 − 3) 2 + (4 + 1) 2 − 10

= 24 AB = 4,90

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9 r = 10 99 AB2 = (6 − 3) 2 + (4 + 1) 2 − 10 9 AB = 4,90 (5) [14]

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QUESTION 6 y

C

B

x

A

6.1

9x=0

9 + ( y + 2) 2 = 25 ( y + 2) 2 = 16 y + 2 = ±4 y = 2 or y = – 6 B(0 ; 2) OR x=0 (0) 2 − 6(0) + y 2 + 4 y = 12

9 factors 9 answers 9 answer for B

(4) 9x=0

y 2 + 4 y − 12 = 0 ( y + 6)( y − 2) = 0 y = – 6 or y = 2 B(0 ; 2)

9 factors 9 answers 9 answer for B

(4) 6.2

99 answer

C(6 ; 2)

(2) 6.3

2

2

3⎞ ⎛ 3⎞ ⎛ ⎛ 3⎞ ⎜ x − 3× ⎟ + ⎜ y + 2 × ⎟ = ⎜5× ⎟ 2⎠ ⎝ 2⎠ ⎝ ⎝ 2⎠ 2

9⎞ ⎛ ⎛ 15 ⎞ 2 ⎜ x − ⎟ + ( y + 3) = ⎜ ⎟ 2⎠ ⎝ ⎝2⎠

2

9 each part ×

2

3 2

2

9⎞ ⎛ 2 ⎜ x − ⎟ + ( y + 3) = 56,25 2⎠ ⎝

6.4.1

AB = (12 − 3) + (10 − (−2)) 2

9 answer

(2) 2

= 9 2 + 12 2 = 15 6.4.2

The radii are 5 and 10. rA + rB = 5 + 10


(2) 9 addition of radii

= 15 = AB The circles will only intersect at one point.

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9 answer

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QUESTION 7

− 3 = x cos150° − 2 sin 150° − 3 = − x.

9 expansion 9 substitution

3 1 − 2. 2 2

3 x=2 2 4 x= 3 y = x. sin 150° + 2. cos150°

9 answer 9 expansion

⎛ 4 1 3⎞ ⎟ . + 2.⎜⎜ − ⎟ 2 3 2 ⎝ ⎠ 2 = . 3− 3 3 3 =− 3

y=

9 answer [5]

QUESTION 8

8.1 y 7

N

6

/

999 coordinates of new points (3)

5

M

N

4

/

3

M

2 1

x -9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

-1 -2 -3 -4

P

-5 -6 -7

8.2.1 8.2.2 8.2.3

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/

99

MN 2 = / / 3 M N area ΔMNP 4 = / / / 9 area ΔM N P area ΔMNP ⎛4⎞ =⎜ ⎟ // // // area ΔM N P ⎝9⎠

P

(2) 99

(2) n +1

99

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QUESTION 9

9.1

A/ (– 12; – 6)

9.2

x = x cos α − y sin α − 12 cos α − 6 sin α = −12 − 2 cos α − sin α = −2.........(1)

9 substitution

y / = y cos α + x sin α

9 substitution

9 answer

(1) /

6 cos α − 12 sin α = −6 cos α = 2 sin α − 1

9 simplification

… (2) 9 substitution

Substitute (2) into (1) − 2(2 sin α − 1) − sin α = −2 − 4 sin α + 2 − sin α = −2 − 5 sin α = −4 4 sin α = 5 α = 53,13°

9 simplification 9 answer

(6)

OR y

(– 12 ; 6)

θ

x

α

(– 12 ; –6)

1 2 θ = 26,565° α = 2(26,565°) α = 53,13°

tan θ =

1 2 9 θ = 26,565° 99 α = 2(26,565°) 9 answer 99 tan θ =

(6) [7]

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QUESTION 10

10.1.1

cos 28° = 1 − sin 2 28°

9 1 − sin 2 28° 9answer

= 1− a2 10.1.2

(2)

cos 64° = cos 2(32°) = 2 cos 2 32° − 1

9 cos 2(32°) 9 2 cos 2 32° − 1 9answer

= 2b 2 − 1 10.1.3

sin 4° = sin(32° − 28°) = sin 32° cos 28° − cos 32° sin 28° = 1 − b 2 . 1 − a 2 − ab

(3) 9 sin(32° − 28°) 9 expansion 99answer

(4)

OR sin 4° = sin(60° − 2 × 28°) = sin 60° cos(2 × 28°) − cos 60° sin( 2 × 28°)

(

)

3 1 1 − 2a 2 − ( 2a ) 1 − a 2 2 2 3 = − 3a 2 − a 1 − a 2 2 =

OR sin 4° = sin( 2 × 32° − 60°) = sin( 2 × 32°) cos 60° − cos(2 × 32°). sin 60°

(

)

1 3 = 2.b 1 − b 2 . − 2b 2 − 1 2 2 3 = b 1 − b 2 − 3b 2 + 2

OR Using sin(A+B) + sin(A – B) = 2.sinA.cosB With A = 28° and B = 32° sin 60° + sin( −4°) = 2ab sin 4° =

3 − 2ab 2

OR

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Using sin(A+B) + sin(A – B) = 2.sinA.cosB With A = 32° and B = 28° sin 60° + sin( 4°) = 2 1 − b 2 . 1 − a 2 sin 4° = 2 1 − b 2 . 1 − a 2 −

OR Using sin 4° = 2 sin 2°. cos 2°

3 2

( ( ( (

) )

1 1 − a 2 − 3a 2 1 3 1− b2 − b and sin 2° = sin(32° − 30°) = 2 1 3 1− a2 + a and cos 2° = cos(30° − 28°) = 2 1 and cos 2° = cos(32° − 30°) = 3b + 1 − b 2 2 then 1 sin 4° = 3b 1 − a 2 − 3ab + 1 − a 2 . 1 − b 2 − 3a 1 − b 2 2 OR 1 sin 4° = 3 1 − b 2 1 − a 2 + 3a 1 − b 2 − 3b 1 − a 2 − ab 2 and sin 2° = sin(30° − 28°) =

10.2

10.3.1

) )

{

}

{

}

b 1− a2 − a 1− b2 = cos 32°. 1 − sin 2 28° − sin 28° 1 − cos 2 32° = cos 32°. cos 28° − sin 28°. sin 32° = cos(32° + 28°) = cos 60° 1 = 2 sin 130°. tan 60° cos 540°. tan 230°.sin 400° sin 50° × tan 60° = cos180° × tan 50° × sin 40° sin 50° × 3 sin 50° − 1× × cos 50° cos 50° 3 cos 50° =− cos 50° = − 3

=

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9 substitution 9 cos 28° 9 sin 32° 9 compound angle formula

(4) 9 sin 50° 9 tan 50° 9sin40o 9cos50o sin 50° 9 cos 50° 9 −1 9answer

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10.3.2


(1 − 2 sin 75°)(1 + 2 sin 75°) = 1 − 2 sin 2 75° = cos150° =

− 3 2


9simplification 9 1 − 2 sin 2 75° 9 cos150° 9answer

(4)

OR sin 75° = sin( 45° + 30°) = sin 45°. cos 30° + cos 45°. sin 30° 2 3 2 1 . . + 2 2 2 2 3 1 2 sin 75° = + =a 2 2 (1 − 2 sin 75°)(1 + 2 sin 75°) =

= (1 − a )(1 + a ) = 1− a2 ⎛3 1 3 1⎞ . ⎟ = 1 − ⎜⎜ + + 2. 2 2 ⎟⎠ ⎝4 4 =− 10.4

3 2

sin x + cos 2 x − cos x = 0

9answer

(4)

2

sin 2 x + (cos 2 x − sin 2 x) − cos x = 0 cos 2 x − cos x = 0 cos x(cos x − 1) = 0 cos x = 0 or cos x = 1 x = ±90° + k .360° or x = 0° + k .360° k ∈ Z = k .360°

10.5.1

9simplification 9 1 − 2 sin 2 75° 9 cos150°

(i.e. x = 90° + k .180° or x = k .360° ± 90°, k ∈ Z ) x = 0°; 90°; 180°

9 (cos 2 x − sin 2 x) 9 cos 2 x − cos x = 0 9factors 9 cos x = 0 or cos x = 1 9 90° + k .360° 9 k .360° 9 x = −90° + k .360° (7) 999 each value

(3)

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10.5.2


sin x (cos 2 x − sin 2 x). cos 2 x. tan x cos x = 2 2 sin x sin x 2 2 cos x − sin x = cos x. sin x cos x sin x − = sin x cos x cos x − tan x = sin x


9 (cos 2 x − sin 2 x) sin x 9 cos x 9 answer 9

cos x sin x − sin x cos x

9answer

(5) [39]

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QUESTION 11

11.1

ˆ EC 2 = DE 2 + DC 2 − 2DE.DC cos C = (7,5) 2 + (9,4) 2 − 2.(7,5)(9,4) cos 32° = 25,03521844... EC = 5,0 metres

11.2

11.3

ˆ E sin 32° sin DC = 7,5 5,0 ˆ E = 7,5. sin 32° sin DC 5,0 = 0,7948788963 ˆ E = 52,6° DC Area of ΔDEC 1 ˆ = DE.DC sin D 2 1 = (7,5)(9,4) sin 32° 2 = 18,7m 2

9substitution into cosine rule 9 25,03521844... 9answer (3) 9 sin rule 9 0,7948788963 9answer

(3) 9 substitution 9answer

(2)

OR Area of ΔDEC 1 = CE.DC sin 52,6° 2 1 = (5,0)(9,4) sin 52,6° 2 = 18,7m 2 11.4

EG 7,5 EG = 7,5. sin 32°

sin 32° =

9ratio 9substitution

= 4,0 EF = (4 + 3,5) = 7,5 metres

9answer

(3) [11]

OR EG = EC.sin 52,6° = (5,0).sin 52,6° = 4,0 EF = 4,0 + 3,5 = 7,5 OR Copyright reserved

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1 .DC.EG = area Δ DEC 2 1 (9,4)EG = 18,7 2 18,7 × 2 ∴ EG = 9,4 = 4,0 EF = 4,0 + 3,5 = 7,5

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QUESTION 12

12.1

Period = 360°

9answer

(1) 12.2

Amplitude =

99answer

1 2

(2)

12.3

9shape 9x intercepts 9 amplitude

y

2

(3)

f

1

g

x -180

-150

-120

-90

-60

-30

30

60

90

120

150

180

-1

-2

12.4

2 solutions

9answer

(1) 12.5

− 60° ≤ x ≤ 120° or x ∈ [−60° ; 120°]

9 − 60°; 120° 9 notation

12.6

− 90° < x < 30° or x ∈ (−90° ; 30°)

99 − 90°; 30° 9 notation

(2)

(3) [12] TOTAL:

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150