Form 5 Mathematics II(Core)Answers Final Exam.2015 1. C

266)6( xxxxx == •

2. C When ,0=x 8)40)(10( +−+=y 84 +−= 4= 3. A

321

3 aaaa ×=

a

a

a

a

=

=

=

=×

21

31

23

323

4. A

25−=−=+

αββα

292

)]2(2)5[(2

]2)[(2

2222

2

2

22

−=−

−−−=

−+=

+=+

αβαββα

αββα

αβ

βα

5. C For II, )5)(1(432 −−−=∆

0

11<−=

∴ 532 −+−= xxy does not have any x-intercept.

For III, when y = 0, 01

12=

+x

,012 =

which is impossible. 6. B

072162

21

13

=+−+=−

=+−

yxxy

xy

Slope)2(

1−

−=

21

=

7. D

Slope of 723 =+ yx is23

− .

If the line is perpendicular to ,723 =+ yx its slope must be

32

231

=

−÷−=

For A, slope23

−= For B, slope32

−=

For C, slope23

=

For D, slope32

=

8. B Let ∠OBC = x and ∠OCB = y.

xOBCOBA =∠=∠ (tangent properties) yOCBOCA =∠=∠ (tangent properties)

In ∆ABC, °=++++° 18068 yyxx (∠ sum of ∆)

°=+°=+

5611222

yxyx

In ∆OBC, °=++∠ 180yxBOC (∠ sum of ∆)

°=∠°=°+∠

12418056

BOCBOC

9. D PA = PB (tangent properties) = 5 cm

ACPA ⊥ (tangent ⊥ radius)

cm 12cm 144

cm 5)85( 22

==

−+=AC

10 B )4()42)(3( 2 −−−+ xxx = 412642 23 +−−+− xxxx = 8562 23 −−+ xxx 11. B

x =34

−

3x = −4 3x + 4= 0 ∴ 3x + 4 is a factor of f (x).

12. D +− ++ 3 1 24 xx 3 = 0

+− •• xx 2844 3= 0 Let xu 2= , the equation becomes −24u 8u + 3 = 0 (2u − 1)(2u − 3)= 0

u =21 or

23

Since xu 2= , we have

212 =x or

232 =x

12−= 23log2log =x

∴ 1−=x 2log23log

=x

= 0.585 (cor. to 3 sig. fig.) 13. A

=+=+−

)2.(..........18)1()1.(..........04

yxyx

From (1), we have y = x + 4……….(3) Substituting (3) into (2), we have (x + 1)(x + 4) = 18 +2x 5x + 4 = 18 +2x 5x − 14 = 0 (x + 7)(x − 2) = 0 x = −7 or 2 When x = −7, y = −7 + 4 = −3. When x = 2, y = 2 + 4 = 6. ∴ The solutions are (−7, −3) and (2, 6).

14. A

+−=

+=

)2(..........15

)1........(..........32 xxy

kxy

Substituting (1) into (2), we have 153 2 +−=+ xxkx 0)1(82 =−+− kxx Since the straight line touches the parabola at one point only, the discriminant of (3) equals to zero,

−− 2)8( 4(1)(1 − k) = 0 64 − 4 + 4k = 0 4k = −60 k = −15 15. A Let kxxxP ++= 52)( 2 .

12

0)4(5)4(2

0)4(2

−==+−+−

=−

kk

P

)32)(4(

1252)( 2

−+=−+=xx

xxxP

∴ It is also divisible by 2x − 3. 16 D

0)3)(2(06

6

)6log(log

)6log(log22log2

)6log(2log

log4log

)6log(2log

log)6(loglog

2

2

2

42

=−+=−−

+=

+=

+=

+=

+=

+=

xxxx

xx

xx

xx

xx

xxxx

x = 3 or −2 (rejected)

17. C Let the three numbers be x − 4, x − 2 and x respectively.

).........(069386

69386

693)86(

693)4)(2(

23

23

2

∗=−+−

=+−

=+−

=−−

xxx

xxx

xxx

xxx

Let −+−= xxxxP 86)( 23 693.

0693)11(8)11(611)11( 23

=−+−=P

∴ x − 11 is a factor of P(x). By long division,

63 5

69363 69363

555

6938 5

11

6938 6 11

2

2

2

23

23

++

−−

−

−+

−

−+−−xx

xxxx

xx

xx

xxxx

)635)(11()( 2 ++−= xxxxP )(∗ becomes =++− )635)(11( 2 xxx 0.

Since 06352 =++ xx is no real solutions, x = 11 only. ∴ The largest number is 11. 18.B Let x km/h be the original speed.

0)80)(90(0720010

7200)10(7201

)10(10

12061

)10(10

6010

10120120

2

=−+=−+

=+

=+

×=

+−+

=+

−

xxxx

xxxx

xxxx

xx

x = 80 or −90 (rejected) The new speed= (80 + 10) km/h = 90 km/h

Percentage change=80

8090 − × 100%

= 12.5% ∴ The speed is increased by 12.5%.

19. B

xkcy += , where c and k are non-zero constants.

y will tend to c as x tends to infinity. 20. D

Since 2yxz ∝ , we have

2ykxz = , where k ≠ 0.

New value of x = (1 + 0.12)x = 1.12x New value of y = (1 − 0.08)y = 0.92y

New value of z = 2)92.0()12.1(

yxk

= 2529700

ykx

≈ 1.323z Percentage change in z

≈z

zz −323.1 × 100%

= 32.3% (cor. to 3 sig. fig.) ∴ z will increased by 32.3%. 21. A Since 2xy ∝ , we have 2kxy = , where k ≠ 0.

y

kx

xky1

=

=

∴ k

1 is a constant.

∴ yx ∝

22. A θθθ 23 sincoscos + = )sin(coscos 22 θθθ + = )1(cosθ = θcos 23. B

°=

=

=

=

303

1tan

31

cossin

cossin3

θ

θ

θθ

θθ

24. B

0sintan =θθ 0tan =θ or 0sin =θ θ = 0°, 180°, 360° or θ = 0°, 180°, 360° θ = 0°, 180° or 360° ∴ The equation has 3 roots. 25. D

053sin

03sin5

<−=

=+

θ

θ

∴ θ lies in quadrant III or IV. 26 A For A, when x = 180°, y = 1 + °180cos = 1 − 1 = 0 ∴ The graph passes through (180°, 0).

27. A ∠BOC = 25° × 2 (∠ at centre twice ∠ at⊙ ce ) = 50° Area of the sector OBC

= 22 cm 360508π ××

= 2cm 9π80

Area of ∆OBC

= 2cm 50sin8821

°×××

= 2cm 50sin32 ° Area of the shaded region

= 2cm 50sin329π80

°−

= 2cm 41.3 (cor. to 3 sig. fig.) 28. C Let PQ = 2x and AQ = y. In ∆PAQ,

)1.(....................

226tan

226tan

°=

=°

yx

yx

In ∆MAQ, Since the angle of depression from M to A is equal to ∠MAQ, we have

)2.(..........tan

tan

MAQyxyxMAQ

∠=

=∠

Substituting (1) into (2), we have

2439.0tan

226tantan

=∠

°=∠

MAQ

yMAQy

∠MAQ = 13.7° (cor. to 3 sig. fig.) ∴ The angle of depression from M to A is 13.7°.

29. A Let the length of each side of the cube be 1 unit. Then EG = AF = AH = FH = 22 11 + = 2 . For A,

EGAEAGE =∠tan

=2

1

∠AGE = 35.3° (cor. to 3 sig. fig.) For B,

EHAEAHE =∠tan

= 1 ∠AHE = 45° For C, ∠AFG = 90° For D, AF = AH = FH = 2 ∴ ∆AFH is an equilateral triangle. ∴ ∠AFH = 60°

30. C °+° 80sin10sin 22 = °+° 80sin80cos 22 = 1 °+° 70sin20sin 22 = °+° 70sin70cos 22 = 1 ∴ +°+° 20sin10sin 22 °+°+ 90sin80sin 22 = )70sin20(sin)80sin10(sin 2222 °+°+°+° °++ 90sin 2 = 1 × 4 + 1 = 5 31. D −1≤ θsin ≤ 1 −2≤ θsin2 ≤ 2 −3≤ 1sin2 −θ ≤ 1 0 ≤ 2)1sin2( −θ ≤ 9 1 ≤ 1)1sin2( 2 +−θ ≤ 10 The maximum value is 10.

32 A Let ∠APC = x, ∠APB = 180° − x, ∠BAP = ∠CAP = y. In ∆ABP, by the sine formula,

)180sin(sin xAB

yBP

−°=

xy

ABBP

sinsin

= )sin)180sin(( xx =−°

In ∆ACP, by the sine formula,

xy

ACPC

xAC

yPC

sinsinsinsin

=

=

∴ ACPC

ABBP

=

43

=

=

ACAB

ACAB

PCBP

∴ AB : AC = 3 : 4 33. A AF = 22 68 + cm (Pyth. theorem) = 10 cm AH = 22 126 + cm (Pyth. theorem) = 180 cm FH = 22 812 + cm (Pyth. theorem) = 208 cm By the cosine formula,

512072

)10)(180(2208100180

))((2cos

222

=

−+=

−+=∠

AFAHFHAFAHFAH

∠FAH = 74.4° (cor. to 3 sig. fig.)

34. B T(5)= 90 − 2)5(3 = 15 35. A

242

644

)6()2(262

26log2log

)2log()6log(log)2log(

22

2

==

+=++

+=+++

=+

++

=+

+−+=−+

xx

xxxx

xxxxx

xx

xx

xx

xxxx

36. C The number of ways = 3

253 CC = 30

37. B The number of ways = 7

4C = 35 38. A For A, P \ Q = {1, 7, 9} For B, Q \ P = {2, 5} For C, P ∩ Q = {4, 8} For D, P ∪ Q = {1, 2, 4, 5, 7, 8, 9} 39. D P(late for meeting exactly once) = 0.2 × 0.8 × 0.8 + 0.8 × 0.2 × 0.8 + 0.8 × 0.8 × 0.2 = 0.384

40. A Median = 12

Mean =3114

Standard deviation = 6.82 (cor. to 3 sig. fig.) Inter-quartile range = 10 After the greatest data is removed,

new median =2

1212 + = 12

new mean = 12.375 new standard deviation = 4.21 (cor. to 3 sig. fig.) new inter-quartile range = 6.5 41. C For I,

2.022.0

11011

22.0222.0

1.12.0

22.0

≠=

=

∴ It is not a geometric sequence.

For II,

92

992

11111

992

9992

111

92

992

÷≠=÷

=÷

∴ It is not a geometric sequence.

For III,

9999100

999910

99991000

9999100

÷=÷

=999910

99991

÷ = 0.1

∴ It is a geometric sequence.

42. A There is only form to arrange them: B G G G G G B, where B is a boy and G is a girl.

The number of ways = 2! × 5! 43. A P(Amy sits opposite to Charlie) = P(Amy takes a seat) × P(Charlies takes the seat opposite to Amy | Amy takes a seat)

=31

44×

=31

44.

C

Let µ be the mean marks. Andy’s marks = µ + 2 × 6 = µ + 12 Cherry’s marks = µ + 2.5 × 6 = µ + 15 The difference = (µ + 15) − (µ + 12) = 3 45. D