QUEEN’S COLLEGE
Yearly Examination, 2010-2011
Form 5 Mathematics Paper II
Solutions
1. Answer: A
, y intercept = C.
2. Answer: A
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3. Answer: C
x / 1 / 2 / 4 / 6x2 / 1 / 4 / 16 / 36
y / 3 / 12 / 48 / 108
/ 3 / 3 / 3 / 3
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4. Answer: C
By substituting (2) into (1), we have
∵ The simultaneous equations have real solutions.
∴ (3) has real roots.
∴
i.e.
∴ The minimum value of k is –5.
5. Solution: B
C : x2 + y2 + 4x + 16y + 28 = 0
P lies on the circle C.
Q lies outside the circle C.
R lies inside the circle C.
S lies on the circle C.
6. Answer: D
Let .
By substituting into , we have
∴ The graph of passes through (-1, 3).
∴ The graph of is obtained by translating the graph of in the positive direction of the x-axis by 1 unit.
∴
∴ The required symbolic representation is .
7. Answer: A
x = or x = -3 (rejected)
8. Answer: B
radius ^ y-axis
radius = 7
The equation of the circle is
(x – 7)2 + (y – 6)2 = 72
x2 – 14x + 49 + y2 – 12y + 36 = 49
i.e. x2 + y2 – 14x – 12y + 36 = 0
9. Answer: C
10. Answer: A
P(less than 3 trials) = P(‘1st trial’ or ‘2nd trial’)
= P(1st trial) + P(2nd trial)
= P(1st trial) + P(1st trial fail) ´ P(2nd trial | 1st trial fail)
11. Answer: D
The locus of P is the angle bisectors of the angles between L1 and L2.
Condition 1: L3
Distance between point P and the line L1 = distance between point P and the line L2
y + 1 = x – 1
x – y = 2
Condition 2: L4
Distance between point P and the line L1 = distance between point P and the line L2
y + 1 = 1 – x
x + y = 0
The equation of the locus of P is x – y = 2, x + y = 0.
12. Answer: C
Consider the data 13, 17, 17, 19, 21, 23.
Largest datum = 23
Smallest datum = 13
Median
13. Answer: D
Let .
By sine formula,
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\
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14. Answer: C
In △PQT,
(Pyth. theorem)
In △QRT,
RT2 = QR2 – TQ2 (Pyth. theorem)
In △RSP, by the sine formula,
, cor. to 3 sig. fig.
15. Answer: C
In △CFE,
∠CFE + 15° = 65°
∠CFE = 50°
∠AFB =∠CFE = 50°
In △AED,
∠EAD + 65° = 90°
∠EAD = 25°
In △AFB, by the sine formula,
, cor. to 3 sig. fig.
The length of the shadow BF is 5.52 m.
16. Answer: C
∠BAC = 200° – 140° = 60°
∠ABC = 180° – (245° – 180°) – (180° – 140°)
= 180° – 65° – 40°
= 75°
In △ABC, by the sine formula,
, cor. to 3 sig. fig.
The distance between A and C is 11.2 km.
17. Answer: B
The required angle is ∠VCM.
In △MBC,
In △VMC,
tan∠VCM =
=
∠VCM = 46°, cor. to the nearest degree
The angle between VC and the plane ABCD is 46°.
18. Answer: C
Let M be a point on DF such thatand
In
In
19. Answer: D
I: Median of A = $50
Median of B = $44
Median of A > median of B
II: Range of A = $(56 – 36)
= $20
Range of B = $(56 – 40)
= $16
Range of A > range of B
III: For A:
Q1 = $48
Q3 = $52
Inter-quartile range = $(52 – 48)
= $4
For B:
Q1 = $42
Q3 = $46
Inter-quartile range = $(46 – 42)
= $4
Inter-quartile range of A = inter-quartile range of B
I, II and III must be correct.
20. Answer: B
From the question, = 2 400 cm, s = 17.2 cm.
2 365.6 cm = (2 400 – 2 × 17.2) cm
=
The percentage of the rolls of toilet paper with lengths less than 2 365.6 cm
= the percentage of the rolls of toilet paper with lengths less than
=
21. Answer: D
For group A,
mean
standard deviation
For group B,
mean
Standard deviation
∴ m1 > m2 and s1 = s2
22. Answer: A
new
Percentage change
23. Answer: C
II.
new
III.
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24. Answer: D
C : x2 + y2 - 8x - 6y + 12 = 0 ¼¼ (1)
By substituting y = 0 into (1), we have
x = 2 or x = 6
Coordinates of A = (2, 0)
L divides circle C into two equal parts.
L passes through the centre of circle C.
Slope of L
The equation of L is
25. Answer: B
Consider the simultaneous equations:
Substitute (i) into (ii):
x2 + (mx + 6)2 = 12
x2 + m2x2 + 12mx + 36 = 12
(1 + m2)x2 + 12mx + 24 = 0
Since the straight line touches the circle, we have = 0,
i.e.
26. Answer: B
The area of OACB is the sum of the areas of △OAB and △ABC. Only when the locus of C is parallel to AB, the height of △ABC is fixed and hence the area of △ABC is fixed. Since the area of △OAB is a constant, the locus of C must be parallel to AB.
27. Answer: C
PG ^ PH
The equation of the locus of P is x2 + y2 – 5x – 3y = 0.
28. Answer: B
In △ABD,
∠BDA = 180° – ∠DAB – ∠ABD (∠ sum of △)
= 180° – 110° – 32°
= 38°
By the sine formula,
BD = 21.279, cor. to 5 sig. fig.
In △BCD, by the cosine formula,
, cor. to 5 sig. fig.
, cor. to 3 sig. fig.
29. Answer: A
Area of quadrilateral ABCE = area of △ABD – area of △CDE
, cor. to 3 sig. fig.
30. Answer: B
Let M and N be the mid-points of BC and EF respectively.
Let the length of each side of the cube be .
Join AC.
In △XMC,
31. Answer: D
From the diagram:
The length between two ends of both box-and-whisker diagrams are the same.
∴ I is correct.
The length of the box of Chinese examination is smaller than that of English examination.
∴ II is correct.
The median mark of Chinese examination is higher than that in English examination.
∴ III is correct.
32. Answer: B
Wai Ming’s standard score
i.e.
, cor. to the nearest integer
33. Answer: D
I: 101 is greater than the mean 98.
After deleting the datum 101, new mean < 98
I must be correct.
II: 101 is closer to the original mean.
After deleting the datum 101, the standard deviation will increase.
II is not correct.
III: Since 101 is not the largest or the smallest datum in the set of data,
after deleting the datum 101, the range will remain unchanged.
Range = 133 – 50
= 83
III must be correct.
Only I and III must be correct.
34. Answer: C
35. Answer: B
I: ∠POQ = 90°
PQ is the diameter of the circle. (converse of ∠ in semi-circle)
Coordinates of the centre = the mid-point of PQ
=
= (–2 , 1.5)
I must be correct.
II: The equation of PQ is
PQ is the diameter of the circle.
R lies on the straight line.
II must be correct.
III: Slope of OR
Slope of PQ
Slope of OR ´ slope of PQ = –0.75 ´ 0.75
= –0.562 5
¹ –1
OR is not perpendicular to PQ.
III is not correct.
Only I and II must be correct.
36. Answer: A
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37. Answer: A
38. Answer: C
PF = distance between point P and the y-axis
As it is given that the equation of the locus of P is y2 = 4x – 4,
comparing the coefficients of the two equations, we have
i.e. 2k = 4 and k2 = 4
k = 2 and k = ±2
39. Answer: B
In △AFD,
Let G lies on BC such that BC ^ GD.
∠FDG = 40°
In △FDG,
Area of △BDC
, cor. to 3 sig. fig.
40. Answer: A
Area of
Area of
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By the cosine formula,
Since MN > 0, MN is minimum when is minimum.
\ Minimum of