Marking Scheme Add Math Zone C Kuching 2008

1

Marking Scheme of Additional Mathematic Zone C Common Paper Paper 1 1.

(a)

{(1, a), (5, a), (5, u), (9, i)}

(b)

Many to many 2

2.

m 8 1  8 2 1 m  8   (8) 2 m4

3.

x a  f −1 ( x)    3 3 2 Either b = − or a = −2 is correct. 3 2 Both b = − and a = −2 are correct 3

3

3

4.

3 + k = −1 k = −4 3k = −p p = 12 4

5.

x − 4x + 3 > 0 2

( x − 1)( x − 3) > 0 x < 1, x > 3

6.

3

(a)

x=4

(b)

Either minimum value = −4 or (x – 4)2 is correct f(x) = (x − 4) 2 − 4 3

7.

log 6 6

log 6 (2 t 1)

log 6 2t  1  log 6 11

 log 6 11

2t  1  11

t 5

4

2 8.

4(3n )(32 ) − 7(3n ) +

3n 3

1 (36 − 7 + )(3n ) 3 88(3n1 ) . Since 88 is a multiple of 8 and (3n – 1) is an integer for all positive integer of n(n = 1, 2, 3, ….) , then 4(3n+2) – 7(3n) + 3n – 1 is divisible by 8.

9.

1(2

3

n−1

s10 =

) = 512 and n = 10

1(210 − 1) = 1023 2 −1

10.

h = 2 + 0.42 + 0.0042 + 0.000042 + ... or a = 0.42, r = 0.01 k h 0.42 h 2.4 = 2+ or = k 1 − 0.01 k 1 − 0.01 h 80 = , h = 80, k = 33 k 33

11.

(a)

2

3 T1 = 2(1) + 3 = 5 or T2 = 2(2) + 3 = 7 d = T2 − T1 = 7 − 5 = 2 (b)

10  2  S10 − S 2 =  (2(5) + (10 − 1)(2))  −  (2(5) + (2 − 1)(2))  2  2  128 *Alternative: use T3 = 9 as first term and find S8 4

12.

13.

1 2 1 1 = −2 y 2x 1 1− 4x = y 2x 2x y= 1− 4x

m=−

4

2PR =3RQ 2 ( x − 2) 2 + [ y − ( −1)]2 = 3 ( x − 3) 2 + ( y − 5) 2 5x2 – 38x + 5y2 – 98y + 286 = 0 3

14.

6 (a)    2 (b) 3i  5 j % %

2

3 15.

r  35a  12b % % % h  35 k  23

16.

17.

3

uuu r uur uuur (a) SQ  SP  PQ uuu r SQ  5 x  4 y % % uur uuu r uuu r 1 uuu r uuu r (b) TR  TQ  QR  SQ  PS 4 uur 1 5 TR  (5 x  4 y )  4 y  x  y 4 % 4% % 3(2sin x cos x)  5cos x  0

4

cos x(6sin x  5)  0 5 6 x  90, 270, 236.44,303.56

cos x  0 or sin x 

18.

Length of arc AB = 12 cm

6  6  6( )  24

or

4

  2 radian



19.

2 x 180 = 11435' 

f '( x )  20( x  3)

3

3

f '(2)  20(2  3)3  20 2

20.

2

1 x  (1  x)  , y  1 3 , 2 2   dy  (1  x)   3   dx 2   p

When p =6, x  1  2(6)  11 ,

dy  (1  (11))   3    18 dx 2  

dy  18(4)  72 units s -1 dt 4

21.

3

 4x  − 3x  = 4 2 k k 2 2 [2(3) – 3(3)] – [2k – 3k] = 4 3

∫ (4 x − 3)dx = 4 = 

2

(2k – 5)(x + 1) = 0, k =

5 ( k > 0) 2 3

4 22.

Either (2 + x) < (12 + 8 + 7 + 3) or (2 + x + 12) > (8 + 7 + 3) Both (2 + x) < (12 + 8 + 7 + 3) and (2 + x + 12) > (8 + 7 + 3) 4 < x < 28 3

23.

(a)

C3 × C4

8

15

= 76440 (b)

3

P3 × 4 P4

=144 4

24.

4 3 3 2 5 4 × or × or × (any one correct) 12 11 12 11 12 11 4 3 3 2 5 4 × + × + × (all correct with the summation) 12 11 12 11 12 11 19 66

25.

62 − 65 (a) z = 15 z = −0.2

3

70 − 65 ) 15 = 0.3696

(b) P ( z >

4

5

Paper 2 y = 8 − 2 x or equivalent (Express x or y as the subject)

1.

2 x 2 + (8 − 2 x) 2 − x(8 − 2 x ) − 18 = 0 (use substitution method to eliminate x or y) 4 x 2 − 20 x + 23 = 0 −(−20) ± (−20) − 4(4)(23) 2(4) (use formula correctly to solve quadratic equation) x=

x = 1.793 or 3.207 y = −4.414 or 2.

(a)

52 =

− 1.586

416 x (use the formula, x  ) n n n=8

2  y2 x 2 2 2 49 =  (52) (use the formula,   x ) 8 n

y (b)

2

 8 [ 49  (52) 2 ]  22024

New mean = 3(52) + 5 = 161 New standard deviation = 3 ( 49 )  21

3.

(a)

a  S1  3(1) 2  5(1)  8 T2  S2  S1  [3(2) 2  5(2)]  8  14 (find T2 ) d  14  8  6

(b)

T10  8  9(6) (use Tn  a  (n  1)d ) T10  62

(c)

3n 2  5n  1062 n  18

6 4.

(a)

: The shape of cos. : The amplitude is 3 : One period for the range 0  x  2 : Modulus of the graph. (b)

x . 2 x : the straight line of y  1  in the graph. 2 (give K1 if the equation is wrong, straight line is drawn correctly according to the wrong equation, but no N1 for answer, even the answer is 4 solutions) : find the equation y  1 

: 4 solutions (cannot give N1 if equation is wrong or the straight is drawn wrongly) 5.

(a)

uuu r uur uuur uuur uuur uur SQ  SP  PQ or QR  QT  TR or (use triangle law)

uuur uuur uuur PU  PQ  QU

uuu r SQ  4a  6b % % uuur QR  4a  6b % % uuur uuur 1 uuur uuur 1 PU  PQ  QT or PU  4a  (6b) 2 % 2 %

(b)

uuur PU  4a  3b % % uuu r uuu r PR  8a  6b or UR  4a  3b r uu u r % % % uuu % (Find vector PR or UR ) uuu r uuur Show PR  2 PU or

uuur uuu r PU  UR

7 6.

dy  (6 x  4) dx ; dx 

dy  3 x 2  4 x  c1 dx

11  3(2) 2  4(2)  c1 dy  3 x 2  4 x  15 dx

c1  15; y   (3 x 2  4 x 15) dx ;

y  x 3  2 x 2 15 x  c2

y  x 3  2 x 2 15 x 5 dy x   or x = 3 (use 3x2 – 4x – 15 = 0 ; = 0) 3 dx 5 5 5 400 y = (  )3 – 2(  )2 – 15(  ) = 3 3 3 27 2 400 d y 5 Maximum value of y =  6( )  4  14 ; 2 27 dx 3 y = (3)3 – 2(3)2 – 15(3) = – 36 d2y  6(3)  4  14 ; dx 2 7.

(a)

xy x

2

4.3 1

8.8 4

Minimum value of y = 36

16.5 9

27.2 16

41.0 25

57.6 36

correct axes and scale All point plotted correctly

(Refer graph on page 8) (b)

xy  ax 2 

a=

Line of best fit

b (express equation in linear form) a

57.6  16.5 (Equate a to the gradient) 36  9 a 1.522

b b  2.78 (Equate to the xy-intercept) a a b  4.186 y  4.9

8

9 8.

(a)

(i) 

3  1  2 p p  2   1  3 5  ( 1)  , ,   (Midpoint of AC = Midpoint of BM) 2 2   2 2     

p (ii) M AD

5 2

5  (1) 7  2  1  3 8

7 y  2   ( x  2) 8

8 y  7 x  30  0 (b)



5   2(1)  1( x) 2( 2 )  1( y )  ,  1,5    1 2 1 2     5 2( )  1( y ) 2(1)  1( x) 2 solve  1 or 5 1 2 1 2

P   5, 10  (c)

1 2



5 3 5 3     (1)( )  ( 1)( 1)  (3)( )  (5)(5)   ( 1)(5)  (3)( )  ( 1)(5)  (1)( )  2 2 2 2     = 17 unit 2

10 9.

10  POQ = 12 (use s  r )

(a)

 POQ = 1.2 radians (b)

Identify the right angle triangle, know OPQ = 0.6 radian and radius OP is the hypotenuse

cos 0.6rad  (alternative: use sin 

5 a (use cos  ) radius h

o and follow by Pythagoras theorem) h

5 , radius = 6.058 cm cos 0.6rad 1 1 2 2  10   1.2  or  6.058  2.4  2 2

radius  (c)

1 2

(use A  r 2 to find area of sector PQTS or OQRS) 1 1 2 2  10   sin1.2rad  or  6.058  sin 2.4rad  2 2 1 absin C (use 2 to find the area of tringle PQS or ORS) 60 – 46.6 or 44.04 – 12.39 (find the area of segment QTS or QRS) = 31.65 – 13.4 (use area of segment QRS – area of segment QTS) = 18.25 cm2

11 10.

(a)

dy  3x 2  2 x  1 dx Gradient of tangent  3(2) 2  2(2)  1  9 Gradient of normal   1 y  5   ( x  2) ; 9

(b)

(i)

x  9 y  47

1 or y  3 3 (solve simultaneous equations, not necessary getting the correct answers)

8 y  3( y 2  1)  0 = (3 y  1)( y  3)  y  

A = ( 8, 3) (ii)

1 9



3 64 3   know volume is equal to    y 2 dy  minus   y 4  2 y 2  1 dy 1  0 9 

 64 3  π  27 y 

3

0

 y5 2 y3  −π  − + y 3 5 

3

(do integration correctly)

1

1 2  64   243  π  (27) − 0 − π ( − 18 + 3) − ( − + 1) 5 3  27   5  (do substitution correctly) = 30

14 π unit 3 15



12 11.

(a)

(i)

7

C1 × .251 × .756

or

7

C0 × .250 × .757

n r nr (use Cr p q correctly)

1  7C1  .251  .756  7C0  .250  .757 (knowing p ( x  2)  1  P( x  1)  P ( x  0) ) = 0.5551 (ii)

n(0.25)(0.75) = 375 n = 2000

(b)

(i)

53  61 or 11

72  61 x  (use z  correctly) 11 

1 – 0.23352*  0.15866* or equivalent 0.60782 (ii)

n(0.60782) = 200 n = 329

13 12.

AB2 = 7.92 + 62 − 2×7.9×6 cos ∠ 130° (Use cosine rule in ΔABC)

(a)

AB = 12.62 cm sin ∠ADC sin 50 = (Use sine rule in ΔADC) 7.9 6.4 sin ADC  0.9456

ADC  108.99( ADC is obtuse) DAC  180  50  108.99  21.01 DC 6.4  (Use sine rule in ΔADC) sin 21.01 sin 50 DC = 3 cm 1 (6.4)(3  6) sin108.99 or equivalent 2 (use area of triangle = 12 ab sin C ) = 27.23 cm2 13.

(a)

120 

RM 5.40 130 x 110 = I x 100 or equivalent (Use I 07 / 06  I 06 / 04  I 07 / 04  100 )

(b)

(c)

Q P1 x 100 (Use I  1  100 ) Q0 4.5

143  I iWi = 120  3  135(2 y )  130  6  150( y ) (Find  IiWi correctly and use x = 2y)

(i)

I 

120  3  135(2 y )  130  6  150( y ) (divide  IiWi by  Wi correctly) 3  2y  6  y 120  3  135(2 y )  130  6  150( y )  132 (Equate I 3  2y  6  y

x = 4, y = 2 (ii)

33 Q X 100  132 (Use I  1  100 ) Q05 Q0 RM25

to 132)

14 14.

(a)

a

dy  4t  12 (find a using differentiation dx

4t  12  0 (use

dy  0 when v maximum) dx

vmax  2(3) 2  12(3)  18 ms-1 (b)





s   2t 2  12t dt   s = 0 when t = 0, thus c = 0.  s  

sA  

2(3)3  6(3) 2 3

or

sB  

(c)

Solve 

2t 3  6t 2 (find s using integration) 3

2(2)3  6(2)2 (substitute t = 2 or t = 3 is s) 3

 sB

sA

2t 3  6t 2  c 3

 17

1 3

2t 3  6t 2  0 (use s = 0) 3 t  9s

(d)

Solve 2t 2  12t  0 (use v = 0) t  6s

15.

x + y ≤ 80 or equivalent

(a)

y ≤ 3x or equivalent y ≥ 30 or equivalent (b) (Refer graph on page 13) At least one straight line is drawn correctly (from inequalities involving x and y) All the three straight lines are drawn Correctly Region is correctly shaded (c)

(i)

45

(ii)

Optimum point (20,60) Substitute (20,60) in 45x + 60y RM4500

15