HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 4

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第31～45題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第1～10題答案，請參閱題解Part 1。

如欲查閱第11～20題答案，請參閱題解Part 2。

如欲查閱第21～30題答案，請參閱題解Part 3。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/07/hkdse-solutions-to-hkdse-maths.html）

31. The answer is B.

Let x=1.2.

The given expression= 1.5625

Option A= -4.6875

Option B= 1.5625

Option C= 0.9375

Option D= -0.9375

32. The answer is D.

From the graph,

when x=0, log₅y=2.

y=25

Therefore, 25=ab⁰

a=25

33. The answer is A.

For every digit '1',

The rightmost digit= 2⁰

The 2nd digit= 2¹

The 3rd digit= 2²

…

The nth digit= 2^n-1

Therefore,

the given expression

=2¹²+2¹⁰+2⁷+2³+2⁰

=2¹²+2¹⁰+137

34. The answer is A.

The given expression

=[4ki-(6+ki)]÷i

=(3ki-6)÷i

=3k-6÷i

=3k+6(i²)÷i

=3k+6i

35. The answer is D.

Try (1,3) on different constraints.

It satisfies all the three inequalities.

As (1,3) only lies in triangle ODF but not other triangles,

triangle ODF is the only solution to the inequality.

36. The answer is C.

Arithmetic sequence: same difference between every two consecutive terms

Therefore, the common difference is=(-6-18)÷3=-8

The second term is 18-(-8)=26

37. The answer is C.

The graph of y=g(x) can be obtained by translating y=f(x) 2 units to the left and 3 units downwards.

Translating 2 units to the left: x+2

Translating 3 units downwards: -3

Therefore g(x)=f(x+2)-3

38. The answer is A.

By sine law,

y÷sin(180°-47°-56°)=x÷sin56°

y÷sin77°=x÷sin56°

y=xsin77°÷sin56°

39. The answer is A.

Interest rate p.a.=6%

Interest rate per month=0.5%

Let $x the amount deposited every month.
x(1+0.5%)¹²+x(1.05%)¹¹+...+x(1+0.5%)²+x(1+0.5%)=10000
x=10000÷(1.005+1.005²+1.005³+...+1.005¹²)=806.63

40. The answer is C.

Let X be a point on AC such that EX is perpendicular to AC and DX is perpendicular to AC.

AC²=AB²+BC² (Pyth. Theorem)

AC²=3²+4²

AC=5cm

Area of triangle ADC= AD×DC÷2= DX×AC÷2

AD×DC=DX×AC

4×3=DX×5

DX=2.4cm

tantheta=DE÷DX=6÷2.4=2.5

41. The answer is B.

Let O be the centre of the circle.

OA is perpendicular to TA. (tangent+radius)

As ∠CTA+∠TAO=90°+90°=180°

CT//OA (int. ∠s supp.)

∠TCA=∠OAC=20° (alt. ∠s, CT//OA)

Since OA=OC (radii)

∠ACO=∠OAC=20° (base ∠s, isos. triangle)

∠BCO=20°+20°=40°

OB=OC (radii)

∠CBO=∠BCO (base ∠s, isos. triangle)

∠BOC=180°-40°-40°=100° (∠ sum of triangle)

BC÷sin100°=OB÷sin40° (sine law)

Radius of the circle

=OB

=6÷sin100°×sin40°

=3.9cm

42. The answer is B.

As x=140, y=3.

For option A:

LHS=3, RHS=0.52

For option B:

LHS=3, RHS=3

For option C:

LHS=3, RHS=-0.52

For option D:

LHS=3, RHS=-3

43. The answer is D.

The probability

=P(red, non-red)+P(green, non-green)+P(white, non-white)

=2/9×7/9+3/9×6/9+4/9×1

=68/81

44. The answer is B.

The probability

=(2×₆P₆)÷₇P₇

=2/7

45. The answer is D.

The new mode= (32+3)×2= 70

The new interquartile range

=(Old Q₃+3)×2-(Old Q₁+3)×2

=2×(Old Q₃-Old Q₁)

=2×27

=54

Assume there are only two numbers in the set.

Let M be the old mean.

The new mean= 2M+6

Let X and Y be the old values of the two data.

The new value of X= 2X+6

The new value of Y= 2Y+6

The old variance= [(M-X)²+(M-Y)²]÷2= 25

The new variance

={[(2M+6)-(2X+6)]²+[(2M+6)-(2Y+6)]²}÷2

=[(2M-2X)²+(2M-2Y)²]÷2

=[4(M-X)²+4(M-Y)²]÷2

=4[(M-X)²+(M-Y)²]÷2

=4×25

=100

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