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