Transcript
Page 1: Skema Add Math P2 Trial SPM ZON a 2012

3472/2MatematikTambahanKertas 22 ½  jamSept 2012

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2012

MATEMATIK TAMBAHAN

Kertas   2

Dua   jam   tiga   puluh   minit

JANGAN  BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

MARKING SCHEME

Skema Pemarkahan ini  mengandungi   13   halaman  bercetak

Page 2: Skema Add Math P2 Trial SPM ZON a 2012

2

ADDITIONAL MATHEMATICS MARKING SCHEME

TRIAL SPM exam Zon A Kuching 2012 –  PAPER 2

QUESTIONNO.

SOLUTION MARKS

1 x = 1+ 2 y                                          P1

2            2                                                                           Eliminate  x or  y(1+2y)  − 5 y  + (1+ 2 y) y − 6 = 0        K1

Solve the quadratic2                                  K1        equation by quadratic− (5)   ±       (5)     −   4(1)( − 5)

y =                                                                                                formula @ completing the

2(1)                                                                square must be shown

y = 0.854, −5.854      N1                                                         Note      :     

or                                                                        OW− 1  if the working of solving

x = 2.708, −10.708    N1                    quadratic equation is not shown.

5

5

2

(a)

(b)

2                             2      K1f(x) = b − ax  − 2apx − ap

a =2      N1

2             K1−2ap = −6  or   b − ap  = 9

N1                                                                                                                                                                  3  b = 27/2                                                N1    p =  2

         3     27     −   ,           y   2   2 

·                                 N1      (−3/2, 27/2)max and(0, 9)

9N1      The shape of the

function.

x

5

2

7

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4

QUESTIONNO.

SOLUTION MARKS

3

(a)

(b)

a = 5, d = 8               K1

77 = 5 + (n −1)(8)                 K1

n = 10            N1

3

a = 5π , d = 8π            K1

10S10 =    [2(5π ) + (10 −1)(8π )]   K1

2

= 410π      N1

∴ the total cost  = RM90.20          N1

4

7

4

(a)

(b)(i)

sin   2 x

cos 2x          K11     −1

cos 2x

2sin x cos x=                            2                   K1

1− (1− 2sin  x)

= cot x               N1

3

y

y = 2 cos 3x2

x0                             π                         π

2

−2

5

QUESTIONNO.

SOLUTION MARKS

(ii)

Shape of cosine curve      P1

Amplitude                        P1

Period                              P1

2 cos3x = −2k             K1

k = −1, 1     N18

5(a)

∑ fx = 30.5(3) + 50.5(7) + 70.5(6) + 90.5(2) +110.5(2) = 1270

K11270                  K1

x =   20= 63.5                                            N1

3

(b)2                          2                           2                           2                             2

2            (30.5)      (3)      +      (50.5)      (7)      +      (70.5)      (6)      +      (90.5)      (2)      +      (110.5)      (2)   σ   =                                    3 + 7 + 6 + 2 + 2      K1

2

− (63.50)91265                  K1

=           − 4032.2520

= 531                                N1

3

6

6

(a)OP = OK + KP      K1

=   a    + h(−a +   b   )         K1

OP = (1− h)a + hb          N1

3

QUESTIONNO.

SOLUTION MARKS

(b) 3          h        K1=

4h + 8    1− h

(4h −1)(h + 3) = 0                   K1

1h =             N1

4

PL : KL  = 3 : 4     N1

4

7

Page 4: Skema Add Math P2 Trial SPM ZON a 2012

0.2 0.3 0.4 0.5

= 1.5 ⇒ n = 3 N1

5

Correct both axes (Uniform scale) K1All points are plotted correctly N1Line of best fit N1

log10 y Q7

(a) Each set of values correct  (log10 y must be at least2 decimal places)  N1, N1

2.2log 10 y = n

2log 10 x  + log 10 (p + 1)        K1

where Y = log   10 y, X = log 10 x,

2.0 m = n  and  c = log   10 (p + 1)(c)  (i) X = log 10 6.9 = 0.8388 = 0.84

Y = 1.56 = log 10 y ⇒ y = 36.31          N1

1.8

1.6

1.4

1.2

n (ii)    = gradient

2n

2         0.8 − 0

(iii)  log 10 (p + 1) = Y-interceptlog 10 (p + 1) = 0.30  K1p = 0.9953               N1

×

×

×

×

.(0.80, 1.50)

×

1.0

0.8×

0.6

0.4

(0, 0.30)

0.2

log10 x

0 0.1                                                                                     0.6            0.7           0.8 0.9           1.0

log10 x 0.30 0.60 0.70 0.78 0.90 1.00 N1

log10 y 0.75 1.19 1.36 1.46 1.65 61.80 N1

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8

9

QUESTIONNO.

SOLUTION MARKS

8

(a)

(b)

(c)

3      or          3 1      3  or   mnormal  =            K1

1            1 +c         K1

+       N1

3

B(4, 0)      K1

4    1   1      K1

=2   N1

=1    unit

3

volume of revolution                        K1                K1

=                           1     4    1=                              K1

= 1      unit 3N1

4

10

QUESTIONNO.

SOLUTION MARKS

9(a) r=5      K1

PR = 8.660     K1

23.66 cm   N1

3

(b)                   π                    K1s = 5 π −           3 

=10.47          N1

2

(c) Area of shaded region of semi circle1         12                         2

=    (10) (π ) −   (5) (π )                                  K12                 2

2

=117.825 / 117.8096 cm

Area of shaded region PQR

1          2    π             1  =   (10)       −   (5)(10) sin 60°      K1

2          3    2                                                         K12

=30.716 / 30.7093 cm

Total area of shaded region

=117.825+30.716                  K12

=148.541 / 148.5192 cm                       N1

5

10

QUESTIONNO.

SOLUTION MARKS

10

(a)(i)

(ii)

2( − 4)   +   3(0)           2(0)   +   3(8)                 K1or

3 + 2                    3 + 2   −   8     24    

C =      ,         N1  5    5 

8                         24 −   + x                       + y

5                         5                                   K1= −2         or                           or     (−2, 0)2                          2

         12         24           N1A −     , −         5       5 

4

(b) 1       24         12                   24         8                 12         24                     8       24 =     0(−      ) + (−     )(0) + (−4)(     ) + (−   )(0) − 0(−     ) − (−      )(−4) − (0)(−   ) − (     )(0)

2          5             5                        5            5                    5             5                          5         5K1

N1   = 19.2

2

(c) mAB = −3    K1

y = −3x − 12    N1

(d) 2                     2( x − (−4))  + ( y − 0)        K1

2         2

x  + y  + 8x = 0    N1

10

QUESTIONNO.

SOLUTION MARKS

11(a)

(i)

(ii)

(b)

(i)

(ii)

(iii)

8                           2     3         3     5           K1C3(   ) (   )

5        5= 0.2787        N1

8                           2     6         3     2       8                           2     7         3                                            2     8      K1                  K11 −  C6(   ) (   )  −   C7(   ) (   )  − (   )

5        5                        5        5              5

0.9052        N1

5

0.75    P1

h   −   163  = −0.6      K116

h = 153.4    N1

P(−1 ≤ z ≤ 0.75)   or  1 − P(Z > 1) − P(Z > 0.75)  or  R(−1) − R(0.75)

K1

0.6147   N1

5

10

QUESTIONNO.

SOLUTION MARKS

12(a) −6t + 16 = 0       K1

2          N1t =2 3

2             K1s = −3t  + 16t

2   2                              2     K1s = − 3(   )  + 16(   )

3             3

1         N1= 21   m

3

5

(b) t(−3t + 16) = 0        K1

1       N1t =5 3

2

(c) | s83 − s0 | + | s4 − s83 |         K1

2                                                                       2                    K1= | 21 3 − 0 | + |16 − 21 3 |

= 26 23   N1

3

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11

12

13

Answer for question 15(a)       I. x  +  y  ≤ 18 N1

II. 4x  +  3 y  ≥  24   N1y

QUESTIONNO.

SOLUTION MARKS

13

(a)

(b)

x = 135             N1

y = 7.20    N1

2

K1

414.75n + 829.5 = 420n + 819                       K1

n =2    N1

3

(c) RM 8.80       K1.RM 6.37N1

2

(d) I   11    = 155    K108

155K1    I11 =            ×100

10    138.25

112.12    N1

3

10

QUESTIONNO.

SOLUTION MARKS

14

(a)

(b)

(c)

9.7              9=                   K1

sin ∠ACB    sin 60°

∠ACB = 68.97° / 68°58'           N1

2

∠ECD =180° − 68.97° =111.03° /111°2′      K1

2               2         2ED  = 4.5  + 9  − 2(4.5)(9)cos 111.03°   K1

ED = 11.42 cm     N1

3

∠BAC = 180° − 60° − 68.97° = 51.03° / 51°2′            P1

1 Area of  triangle ABC =     ×9.7×9×sin 51.03°        K1

2= 33.94

1                                                    K1Area of  triangle CDE =    ×9×4.5×sin 111.03°

2= 18.90

Area of  quadrilateral ABCD =33.94 + 18.90   K1

52.84      N1

5

10

Page 7: Skema Add Math P2 Trial SPM ZON a 2012

2 4 6 8 10 12 14

III. y  ≤  2xN1

(b) Refer to the graph,

20

18

1 or 2  graph(s) correct3 graphs correct

Correct area

N1(c) (i)  x = 8

N1

N1

K1

16 (ii)   (6, 12) P1

Maximum Profit = RM [1.50(6) + 2(12)] K1

14

12 (6, 12) ·

=  RM 33.00 N1

10

10

8

6

R4

2

x0 16 18


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