Post on 16-May-2018
QUEEN’S COLLEGEYearly Examination, 2010-2011Form 5 Mathematics Paper II
Solutions
1. Answer: A
, y intercept = C.
2. Answer: A
3. Answer: C
x 1 2 4 6
x2 1 4 16 36
y 3 12 48 108
3 3 3 3
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: DLet .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. x 2 + y 2 – 14 x – 12 y + 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 = 0The 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: DLet .
By sine formula,
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 = 25In △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 that andIn
In
19. Answer: D
I: Median of A = $50
Median of B = $44Median 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 = $46Inter-quartile range = $(46 – 42)
= $4Inter-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, = 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: CII.
new
III.
24. Answer: D
C : x2 + y2 8x 6y + 12 = 0 (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 = 0Since 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: CPG PH
The equation of the locus of P is x 2 + y 2 – 5 x – 3 y = 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
= 83III 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
–1OR is not perpendicular to PQ.
III is not correct.
Only I and II must be correct.
36. Answer: A
37. Answer: A
38. Answer: CPF = 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
By the cosine formula,
Since MN > 0, MN is minimum when is minimum.
Minimum of