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
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