PERCUBAAN SPM 2011 NEGERI TERENGGANU

MATHEMATICS PAPER 2SECTION B

1449/2

PANITIA MATEMATIK SMSTSP 2011

12(a) Xx 1.5 2 3 4 6 8 10 12 14

y 16 12 8 6 4 3 2.4 2 1.7

√ K1 √ K1 (b). graph

(c). (i) 6.5 ≤ x ≤ 6.9

√ P1 axes drawn in corect direction, uniform scale √ K2 * 9 points correctly plotted * curve passes trough all 9 points √ N1 *smooth continous curve

without any straight line using the given scalar.

√ p1 (c). (i) 8.55 ≤ y ≤ 8.59 √ p1

(d). y = -x +13

(b). graph

y = -x +13 √ K1 straight line corectly drawn

√ K1 corect equation of straight line

2.0 ≤ X1 ≤ 2.4 10.8 ≤ X1 ≤ 11.0 √ N1 √ N1

Q13. (a) (i) ( 0, 5 )

(ii) (3, 6)

√ P1

√ P2

(b) (i) T = Rotation, 900 anti clockwise, centre ( 0, 0 )

√ P3 S = Enlargement, scale factor of 2, centre ( 0, -1 ) √ P3

(c) Area JHKL = (2)2 X 7 √ K2

= 28 unit2 √ N1

Score Frequency Midpoint

10 – 19 3 14.5

20 – 29 6 24.5

30 – 39 10 34.5

40 – 49 15 44.5

50 – 59 12 54.5

60 – 69 9 64.5

70 – 79 5 74.5

Q14. (a)

√ P1 √ P1

(b) (i) 40 - 49 √ P1(b) (ii) √ K2

= 46.83 √ N1

freq

uenc

y

14.5 44.534.524.5 74.564.554.5

√ P1 axes drawn in corect direction, uniform scale √ K2 * 7 bars correctly drawn

* curve passes trough all 9 points √ N1 *smooth continous clines using the given scalar.

Q15. (a) plan

6 cm

3 cm

√ K1 Correct shape with rectangle ABCD, all solid lines.√ K1 AB > CB√ N1 measurement correct to ± 0.2 cm (one way) and all angles <A , <B, <C, <D. = 90° ± 1°

(b) (i)

6 cm

8 cm

√ K1 Correct shape with triangle BCF and trapezium HIML. All solid lines.

√ K1 BC = HI < HL< IM

√ N2 measurement correct to ± 0.2 cm (one way) and all angles <B , <H, <I, <C. = 90° ± 1°

(b) (ii) 5 cm

7 cm

8 cm2 cm

4 cm

√ K1 Correct shape with rectangle JINM and CDEF , all solid lines.

√ K1 CH < CF = FK < AB < FE < MI√ N2 measurement correct to ± 0.2 cm (one way) and all angles <A, <B , < C, <H, <I, <J. = 90° ± 1°

√ K1 dotted line KG

Q16 (a)

PQ√ P1

R

R = ( 60°S, 160 )

√ P1

√ P2

(b) θ X 60 = 60 X 60

3600 n.m√ K1

√ N1(c) (i) speed = distance / time

√ K1√ N1

(c) (ii) Distance of parallel of latitiude = 3600 n.m

θ = 120°

Longitude T = 120º - 20º = 100º W

√ K2

√ K1√ N 1