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

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

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

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

7

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

10

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

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

Skema Add Math P2 Trial SPM ZON a 2012

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