2017 DSE Math Paper 2 solution
2017 DSE 數學卷二答案題解
01. A D A D D A B A C C
11 B C B B C D D A D D
21 C D A A C A B C B B
31 D D C D B C C A A B
41 D B C B A
DSE 數學卷二答案及分析: 有15條MC， 是初中時學的，反映出若學生在初中時有打好數學基礎， 則DSE Maths 可穩當取得及格 甚至成績
( 我們可以在 公共圖書館找到 DSE數學卷二試題 )
現先附上解答, 括號內數字為DSE數學卷二考生答對百分率。
Q1. A (89%)
3m2 - 5mn + 2n2 + m – n
= (m-n) (3m-2n) + (m-n)
= (m-n) (3m-2n+1)
Q2. D (72%)
Q3. A (73%)
Times each term by 2a (每項乘以2a),
a + 4b = 2(2a) + (2a)b/a
a + 4b = 4a + 2b
a = 2b/3
Q4. D (90%)
1/π4 = 0.010265982
= 0.010266 (6 d.p.)
Q5. D (87%)
6–x < 2x–3 or 或 7–3x > 1
9 < 3x or 或 6 > 3x
x > 3 or 或 x < 2
x < 2 or 或 x > 3
Q6. A (73%)
f(x) = 2x2 - 5x + k
f(2) - f(-2)
= [(2)(4)-5(2)+k] - [(2)(4)+5(2)+k]
= -20
Q7. B (76%)
p(x) = 2x2 - 11x + c
p(7) = 0;
2(49) – 11(7) + c = 0;
c = -21
p(-1/2)
= 2(-1/2)2 - 11(-1/2) - 21
= -15
n = -8
Q9. C (44%)
y = (px + 5)2 + q
The vertex is at left side of y-axis and below x-axis.
頂點位於y軸的左側，在x軸下方.
The vertex 頂點 can be (-5, -1)
（MC方法): (代入p=1 & q=-1) 或 (p=-1 & q=-1)
case 1: substitute代入 p=1 & q=-1)
y = (px + 5)2 + q
y = (x + 5)2 - 1
so vertex頂點 = ( -5, -1 )
It is true as the shape of the graph.
(這是正確的因為附合圖表的形狀。)
(So we need not try case 2 不需要代入p=-1)
so所以 p=1 and q=-1
so所以 p > 0 and q < 0
Q10. C (76%)
2000 (1 + 2.5%)8 - 2000
= 437
to TOP
Q11. B (56%)
actual area 實際面積
= 4 x 20000 cm x 20000 cm
= 4 x 200m x 200m
= 16 x 10000 m2
= 1.6 x 105 m2
Q12. C (84%)
y = ax2 + c
so a + c = 7
4a + c = 13
so a = 2; c = 5
when 當 x = 3,
y = 2(3)2 + 5 = 23
Q13. B (86%)
The nth pattern has 2n more dots than the (n-1)th pattern.
第n個圖案比第（n-1）個圖案多2n個點
The 7th pattern 第7個圖案
= 1 + 2(2+3+4+5+6+7)
= 55
Q14. B (71%)
Let A = area of ΔBCD (設A =ΔBCD的面積)
so A + 24 = area of ΔABD
A + A + 24 = 84, so A = 30
CDx12/2 = 30, so CD = 5
AD = 14 - 5 = 9
AB = √(92 + 122 ) = 15
BC = √(52 + 122 ) = 13
perimeter 周界 = 15 + 13 + 14
= 42cm
Q15. C (53%)
circular cone 圓錐體: 2r, h; cylinder 圓柱體: r, 3h
volume of cone 圓錐體積
= (1/3)π(2r)2(h)
= (1/3)π(4r2h) = 36π
so r2h = 27
volume of cylinder 圓柱體積
= πr2(3h)
= 27x3π = 81π
Q16. D (32%)
（MC方法）To simplify, assume DE ┴ CA and FB ┴ CA.
為簡化起見，假設DE┴CA和FB┴CA。
For rotational symmetry 因旋轉對稱性,
area of HGBE 面積 = area of DFGH 面積.
Now CE : EB = 3 : 2
Let 設 CH=3h and HG=2h, so GA=CH=3h
Let 設 HE=3k and GB=5k
Area of ΔABG 面積 = (5k)(3h)/2 = 135
So, kh = 18
Area of HGBE 面積 = area of DFGH 面積
= (3k+5k)(2h)/2
= 8kh = 8x18
= 144
Q17. D (36%)
Let 設 a =∠ACD , b=∠EDB
∠ACD + ∠CAD = ∠CDB (ext. ∠ of Δ) (Δ的外角)
a + 60° = 60° + b
a = b
ΔDBE ~ ΔCAD (AAA)
BE = 12
4 16
BE = 33
CE = 16 – 3 = 13
Q18. A (45%)
∠CDA = 180° - 124° = 56°
∠CAD = 180° - 2(56°) = 68°
∠BCA = 68° (alt.∠s, 內錯角, AE//BC）
∠ABC = 180° - 2(68°) = 44°
Q19. D (68%)
AH2 = (9-5+2)2 + (11-6+4-1)2
= 62 + 82
AH = 10
Q20. D (15%)
Without a diagram, this question becomes very difficult. 因為沒有圖表，這個問題顯得非常困難。
Given 已知: AE = ED,
∠ABE = ∠CBD = ∠DBE = x .
If 如果 ∠ABC < 90°,
then 則 AE < ED (impossible 不可能).
Thus 因此 ∠ABC > 90°
Since 因為 BC // AD,
∠BDE = ∠DBC = x (alt.∠s, 內錯角 BC//AD)
so ∠BDE = ∠DBE = x
so BE = ED (sides opp. eq.∠s)
so AE = ED = BE
Treating 以 AE = ED = BE = radius 半徑,
then E is the centre of circle ABD,
所以E是圓ABD的中心，
and ∠ABD = 90° (∠ in semi-circle 半圓上的圓周角).
so ∠ABE = ∠DBE = ∠CBD = 45° .
so ∠ABC = 135°, so II is true, II是對的.
so ∠BDE = 45° and ∠BEA = 90° .
so ΔABE ≡ ΔDBE III is true, III是對的.
and AB = BD I is true, I是對的.
so I, II and III are true.
to TOP
Let 設 ∠BED = x.
Then 因此 ∠BCD = 180° – x .
As AD is diameter 直徑, ∠ACD = 90°
∠CDB = ∠CBD
= 110° – 90° = 20°
In 在 ΔDCD,
200° + 20° + 180° – x = 180°
x = 40°
Q22. D (58%)
Let 設 x = EC
x/2 = tan40°
x = 2tan40°
Let 設 y = ∠AED
tan y = 2/(3-2tan40°)
= DA/DE
y = 56.5° = 57°
Q23. A (30%)
Suppose假設 L1 passes thru通過 (1,0) & (0,-1)
and L2 passes thru通過 (-1,0) & (0,2)
L1: y = x - 1
1 = x - y
m = -1, n = 1.
L2: y = 2x + 2
y/2 = x + 1
-1 = x - y/2
q = x + py
q = -1, p = -1/2.
I. m < p, -1 < -1/2 , I is correct, I是對的
II. n > q, 1 > -1 , II is correct, II是對的
III. n+m < p+q,
1 + (-1) < -1/2 + (-1) , III is wrong, III是錯的
Q24. A (59%)
9x−5y+45=0
9x+45=5y
9x/5 + 9 = y
Let m be slope of required line.
設m為所需線的斜率
m(9/5) = -1
m = -5/9
(y-0)/(x+3) = -5/9
5x+15 = -9y
5x+9y+15=0
Q25. C (57%)
∠POR= 340° – 160° = 180°
POR is a straight line 直線.
∠QOR= 340° – 280° = 60°
Let 設 d = distance required 所需距離
d / 4 = sin60° = √3 / 2
d = 2√3
Q26. A (53%)
C1: x2+y2+8x−4y−5=0
centre 圓心 G1=(−4,2),
(radius半徑)2=(-4)2+(2)2-(-5) =25
radius半徑= 5
C2: 2x2+2y2+8x−4y−5=0
We need to divide each term by 2 so that coefficients of x2 & y2 become 1.
我們需要將每項除以2，以使x2和y2的係數變為1。
x2+y2+4x−2y−2.5=0
centre圓心 G2=(−2,1),
(radius半徑)2= (-2)2+(1)2-( −2.5) =7.5
radius半徑= 2.7386
I. slope斜率 of OG1 = (2-0)/(-4-0) = -1/2
slope斜率 of OG2 = (1-0)/(-2-0) = -1/2
slope斜率 of OG1 = slope of OG2
G1, G2 & O are collinear 共線.
I is correct, I是對的
II. radius半徑 of C1 = 5 > 2.7386 (radius半徑 of C2)
II is correct, II是對的
III. As由於 O=(0,0), G1=(−4,2), G2=(−2,1),
we note that G1 is more distant from O.
我們注意到G1離O更遠。
Thus因此 OG1 > OG2.
III is wrong, III是錯的
Q27. B (39%)
AB is a chord 弦. As由於 AP = BP,
locus of P is perpendicular bisector of AB.
P的軌跡是AB的垂直平分線。
centre 圓心 = (3,2)
putting (3,2) into equation of straight line,
代（3,2）入直線方程，
3 + 2(2) + k = 0
7 + k = 0
k = −7
Q28. C (84%)
Note注意 5+20 = 25
5+20+15+10+10 = 60
Probability 概率
= 25/60 = 5/12
Q29. B (76%)
From the figure, lower quartile = 15
從圖中可以看出，下四分位數= 15
Q30. B (51%)
Note mean 注意平均值 = 5
2+3+4+6+7+9+10+m+n = 9x5 = 45
We have 因此, m + n = 45 - 41 = 4
As m & n are positive integers,
由於m＆n是正整數，
there are 3 cases of values of m & n.
m＆n有3個值。
(m,n) = (1,3) or (2,2) or (3,1)
I) For m=1, n=3, mode 衆數 =3.
Thus I may not be true. I可能不是真的.
II) For above 3 cases, median 中位數=4.
Thus II must be true. II必是真的.
III) For m=1, n=3, range 分佈域=10−1=9
In this case, range 分佈域=8 is wrong 錯.
Thus III may not be true.
III可能不是真的.
to TOP
Transformation of graph of function:
函數圖像的變換：
The graph is enlarged to 2 times along x-axis.
圖像沿x軸放大到2倍。
The y-intercept will not be changed.
y軸截距不會改變。
Q32. D (50%)
方法 1: Calculator: press 按 MODE 3 (BASE),
press 按 DEC, enter 輸入 8x8x8 EXE.
We get 得到 51210.
press 按 HEX, We get 得到 20016.
only只有 (D) has the ending 結尾 with 20016.
(Remember to press MODE 1 (COMP) to return normal computation.)
（請記住按MODE 1（COMP）返回正常計算。）
方法 2: 8x8x8 = 2x2x2x8x8
= 2(2x8)(2x8)
= 2x16x16 = 20016.
Q33. C (53%)
√ y - 0 = (8-0)/(0-4)(x-4)
√ y = -2(x-4) = -2x+8
y = (-2x+8)2
y = 4x2 - 32x + 64
Q34. D (41%)
log9 y = x - 3 ...(1)
2(log9 y)2 = 4 - x ...(2)
(1)+(2),
2(log9 y)2 + (log9 y) = 1
2(log9 y)2 + (log9 y) - 1 = 0
log9 y = 0.5 or log9 y = -1
y = 90.5 = 3 or y = 1/9
x = log9 y + 3 ...(3)
when當 y=3, x= 0.5 + 3 = 3.5
when當 y=1/9, x= -1 + 3 = 2
Q35. B (45%)
Let 設 x = 5/(2-i) + ki.
Calculator: press 按 MODE 2 (CMPLX)
Enter 輸入 5 / (2 - i) EXE.
We get 得到 2 + i.
( The real part 實部 is 2. Press 按 shift EXE, the imaginary part 虛部 is 1i. )
so x = 2 + i + ki = 2 + (1+k) i.
As由於 x is a real number 實數, k + 1 = 0
k = -1
(Remember to press MODE 1 (COMP) to return normal computation.)
（請記住按MODE 1（COMP）返回正常計算。）
Q36. C (66%)
I. obviously wrong, I 明顯錯了 (not A.S.)
II. 60π - 45π = 15π
= 45π - 30π
II is correct, II是對的
II. (π-60) - (π-45) = -15
= (π-45) - (π-30)
III is correct, III是對的
Q37. C (45%)
y ≤ 9 (solution below & including line y = 9)
x - y - 9 ≤ 0
x - 9 ≤ y, y ≥ x - 9
(solution above & including line y = x - 9)
x + y - 9 ≥ 0, y ≥ -x + 9
(solution above & including line y = -x + 9)
Test測試 all 3 corners.
Greatest value occurs at (9, 0).
最大值出現在（9,0）。
x - 2y + 43 = 9 -2(0) + 43 = 52
Q38. A (49%)
note 留意 ∠CAB = ∠DCA
(alt.∠s 內錯角, AB//DC)
Let 設 x = ∠CAB = ∠DCA
tan x = 21/28 = 3/4
cos x = 4/5
DE2 = 52 + 282 - 2(5)(28)(4/5)
= 585 = 9(65)
DE = 3√65
Q39. A (34%)
BD2 = 252 - 152
BD = 20
BD2 + CD2 = 202+ 212
= 841 = 292 = BC2
so ∠BDC = 90°
tan ∠BAD = BD/AD = 20 / 15
∠BAD = 53°
Q40. B (62%)
∠OAB = 90° - 68° = 22°
∠ABO = 22°
∠CBO = 26°
∠ABC = ∠ABO + ∠CBO
= 22° + 26° = 48°
to TOP
3x + 4y = 3p
Let 設 p = 4, 3x + 4y = 12
When y=0, x=4. so p=4, OP = 4.
Let 設 M(0,c) be y-intercept y-截距 of 3x + 4y = 12
Substitute 代入 (0,c) in 3x + 4y = 12, we get 得到 4c = 12.
so c =3 , OM = 3
Let 設 a be ∠MPO
tan a = OM / OP = 3 / 4
a = 36.86989765
tan 2a = 3.428571429 = 24 / 7
q / p = 24 / 7
p : q = 7 : 24
(This is a very good method as we need not find the coordinates of the in-centre.)
（這是一個非常好的方法，因為我們不需要找內心的坐標。）
Q42. B (71%)
13C5 x 6C4 =19305
Q43. C (59%)
P(at most 3 times 最多3次)
= 1 - P(4 times 4次)
= 1 - 0.74
= 0.7599
Q44. B (78%)
Let 設 SD = standard deviation 標準差.
standard score 標準分
= (33 - 45) / SD = -2
so, SD = (33 - 45)/(-2)
= 6
Q45. A (48%)
New data 新數據 = 8 x old data 舊數據.
so, m2 = 8m1 ; r2 = 8r1
so, I and II are correct, I和II是正確
since 因為 variance 方差= SD2
v2 = 82 v1 = 64 v1
so, III is wrong, III是錯誤
..........................................................
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01. A D A D D A B A C C
11 B C B B C D D A D D
21 C D A A C A B C B B
31 D D C D B C C A A B
41 D B C B A
DSE 數學卷二答案及分析: 有15條MC， 是初中時學的，反映出若學生在初中時有打好數學基礎， 則DSE Maths 可穩當取得及格 甚至成績
( 我們可以在 公共圖書館找到 DSE數學卷二試題 )
現先附上解答, 括號內數字為DSE數學卷二考生答對百分率。
Q1. A (89%)
3m2 - 5mn + 2n2 + m – n
= (m-n) (3m-2n) + (m-n)
= (m-n) (3m-2n+1)
Q2. D (72%)
Q3. A (73%)
Times each term by 2a (每項乘以2a),
a + 4b = 2(2a) + (2a)b/a
a + 4b = 4a + 2b
a = 2b/3
Q4. D (90%)
1/π4 = 0.010265982
= 0.010266 (6 d.p.)
Q5. D (87%)
6–x < 2x–3 or 或 7–3x > 1
9 < 3x or 或 6 > 3x
x > 3 or 或 x < 2
x < 2 or 或 x > 3
Q6. A (73%)
f(x) = 2x2 - 5x + k
f(2) - f(-2)
= [(2)(4)-5(2)+k] - [(2)(4)+5(2)+k]
= -20
Q7. B (76%)
p(x) = 2x2 - 11x + c
p(7) = 0;
2(49) – 11(7) + c = 0;
c = -21
p(-1/2)
= 2(-1/2)2 - 11(-1/2) - 21
= -15
Q8. A (62%)
4x2 + m(x+1) + 28
= 4x2 + mx + (m + 28)
= mx(x+3) + n(x-4)
m = 4;
4 + 28 = - 4n = 32
4x2 + m(x+1) + 28
= 4x2 + mx + (m + 28)
= mx(x+3) + n(x-4)
m = 4;
4 + 28 = - 4n = 32
n = -8
Q9. C (44%)
y = (px + 5)2 + q
The vertex is at left side of y-axis and below x-axis.
頂點位於y軸的左側，在x軸下方.
The vertex 頂點 can be (-5, -1)
（MC方法): (代入p=1 & q=-1) 或 (p=-1 & q=-1)
case 1: substitute代入 p=1 & q=-1)
y = (px + 5)2 + q
y = (x + 5)2 - 1
so vertex頂點 = ( -5, -1 )
It is true as the shape of the graph.
(這是正確的因為附合圖表的形狀。)
(So we need not try case 2 不需要代入p=-1)
so所以 p=1 and q=-1
so所以 p > 0 and q < 0
Q10. C (76%)
2000 (1 + 2.5%)8 - 2000
= 437
to TOP
Q11. B (56%)
actual area 實際面積
= 4 x 20000 cm x 20000 cm
= 4 x 200m x 200m
= 16 x 10000 m2
= 1.6 x 105 m2
Q12. C (84%)
y = ax2 + c
so a + c = 7
4a + c = 13
so a = 2; c = 5
when 當 x = 3,
y = 2(3)2 + 5 = 23
Q13. B (86%)
The nth pattern has 2n more dots than the (n-1)th pattern.
第n個圖案比第（n-1）個圖案多2n個點
The 7th pattern 第7個圖案
= 1 + 2(2+3+4+5+6+7)
= 55
Q14. B (71%)
Let A = area of ΔBCD (設A =ΔBCD的面積)
so A + 24 = area of ΔABD
A + A + 24 = 84, so A = 30
CDx12/2 = 30, so CD = 5
AD = 14 - 5 = 9
AB = √(92 + 122 ) = 15
BC = √(52 + 122 ) = 13
perimeter 周界 = 15 + 13 + 14
= 42cm
Q15. C (53%)
circular cone 圓錐體: 2r, h; cylinder 圓柱體: r, 3h
volume of cone 圓錐體積
= (1/3)π(2r)2(h)
= (1/3)π(4r2h) = 36π
so r2h = 27
volume of cylinder 圓柱體積
= πr2(3h)
= 27x3π = 81π
Q16. D (32%)
（MC方法）To simplify, assume DE ┴ CA and FB ┴ CA.
為簡化起見，假設DE┴CA和FB┴CA。
For rotational symmetry 因旋轉對稱性,
area of HGBE 面積 = area of DFGH 面積.
Now CE : EB = 3 : 2
Let 設 CH=3h and HG=2h, so GA=CH=3h
Let 設 HE=3k and GB=5k
Area of ΔABG 面積 = (5k)(3h)/2 = 135
So, kh = 18
Area of HGBE 面積 = area of DFGH 面積
= (3k+5k)(2h)/2
= 8kh = 8x18
= 144
Q17. D (36%)
Let 設 a =∠ACD , b=∠EDB
∠ACD + ∠CAD = ∠CDB (ext. ∠ of Δ) (Δ的外角)
a + 60° = 60° + b
a = b
ΔDBE ~ ΔCAD (AAA)
BE = 12
4 16
BE = 33
CE = 16 – 3 = 13
Q18. A (45%)
∠CDA = 180° - 124° = 56°
∠CAD = 180° - 2(56°) = 68°
∠BCA = 68° (alt.∠s, 內錯角, AE//BC）
∠ABC = 180° - 2(68°) = 44°
Q19. D (68%)
AH2 = (9-5+2)2 + (11-6+4-1)2
= 62 + 82
AH = 10
Q20. D (15%)
Without a diagram, this question becomes very difficult. 因為沒有圖表，這個問題顯得非常困難。
Given 已知: AE = ED,
∠ABE = ∠CBD = ∠DBE = x .
If 如果 ∠ABC < 90°,
then 則 AE < ED (impossible 不可能).
Thus 因此 ∠ABC > 90°
Since 因為 BC // AD,
∠BDE = ∠DBC = x (alt.∠s, 內錯角 BC//AD)
so ∠BDE = ∠DBE = x
so BE = ED (sides opp. eq.∠s)
so AE = ED = BE
Treating 以 AE = ED = BE = radius 半徑,
then E is the centre of circle ABD,
所以E是圓ABD的中心，
and ∠ABD = 90° (∠ in semi-circle 半圓上的圓周角).
so ∠ABE = ∠DBE = ∠CBD = 45° .
so ∠ABC = 135°, so II is true, II是對的.
so ∠BDE = 45° and ∠BEA = 90° .
so ΔABE ≡ ΔDBE III is true, III是對的.
and AB = BD I is true, I是對的.
so I, II and III are true.
to TOP
Q21. C (59%)
Let 設 ∠BED = x.
Then 因此 ∠BCD = 180° – x .
As AD is diameter 直徑, ∠ACD = 90°
∠CDB = ∠CBD
= 110° – 90° = 20°
In 在 ΔDCD,
200° + 20° + 180° – x = 180°
x = 40°
Q22. D (58%)
Let 設 x = EC
x/2 = tan40°
x = 2tan40°
Let 設 y = ∠AED
tan y = 2/(3-2tan40°)
= DA/DE
y = 56.5° = 57°
Q23. A (30%)
Suppose假設 L1 passes thru通過 (1,0) & (0,-1)
and L2 passes thru通過 (-1,0) & (0,2)
L1: y = x - 1
1 = x - y
m = -1, n = 1.
L2: y = 2x + 2
y/2 = x + 1
-1 = x - y/2
q = x + py
q = -1, p = -1/2.
I. m < p, -1 < -1/2 , I is correct, I是對的
II. n > q, 1 > -1 , II is correct, II是對的
III. n+m < p+q,
1 + (-1) < -1/2 + (-1) , III is wrong, III是錯的
Q24. A (59%)
9x−5y+45=0
9x+45=5y
9x/5 + 9 = y
Let m be slope of required line.
設m為所需線的斜率
m(9/5) = -1
m = -5/9
(y-0)/(x+3) = -5/9
5x+15 = -9y
5x+9y+15=0
Q25. C (57%)
∠POR= 340° – 160° = 180°
POR is a straight line 直線.
∠QOR= 340° – 280° = 60°
Let 設 d = distance required 所需距離
d / 4 = sin60° = √3 / 2
d = 2√3
Q26. A (53%)
C1: x2+y2+8x−4y−5=0
centre 圓心 G1=(−4,2),
(radius半徑)2=(-4)2+(2)2-(-5) =25
radius半徑= 5
C2: 2x2+2y2+8x−4y−5=0
We need to divide each term by 2 so that coefficients of x2 & y2 become 1.
我們需要將每項除以2，以使x2和y2的係數變為1。
x2+y2+4x−2y−2.5=0
centre圓心 G2=(−2,1),
(radius半徑)2= (-2)2+(1)2-( −2.5) =7.5
radius半徑= 2.7386
I. slope斜率 of OG1 = (2-0)/(-4-0) = -1/2
slope斜率 of OG2 = (1-0)/(-2-0) = -1/2
slope斜率 of OG1 = slope of OG2
G1, G2 & O are collinear 共線.
I is correct, I是對的
II. radius半徑 of C1 = 5 > 2.7386 (radius半徑 of C2)
II is correct, II是對的
III. As由於 O=(0,0), G1=(−4,2), G2=(−2,1),
we note that G1 is more distant from O.
我們注意到G1離O更遠。
Thus因此 OG1 > OG2.
III is wrong, III是錯的
Q27. B (39%)
AB is a chord 弦. As由於 AP = BP,
locus of P is perpendicular bisector of AB.
P的軌跡是AB的垂直平分線。
centre 圓心 = (3,2)
putting (3,2) into equation of straight line,
代（3,2）入直線方程，
3 + 2(2) + k = 0
7 + k = 0
k = −7
Q28. C (84%)
Note注意 5+20 = 25
5+20+15+10+10 = 60
Probability 概率
= 25/60 = 5/12
Q29. B (76%)
From the figure, lower quartile = 15
從圖中可以看出，下四分位數= 15
Q30. B (51%)
Note mean 注意平均值 = 5
2+3+4+6+7+9+10+m+n = 9x5 = 45
We have 因此, m + n = 45 - 41 = 4
As m & n are positive integers,
由於m＆n是正整數，
there are 3 cases of values of m & n.
m＆n有3個值。
(m,n) = (1,3) or (2,2) or (3,1)
I) For m=1, n=3, mode 衆數 =3.
Thus I may not be true. I可能不是真的.
II) For above 3 cases, median 中位數=4.
Thus II must be true. II必是真的.
III) For m=1, n=3, range 分佈域=10−1=9
In this case, range 分佈域=8 is wrong 錯.
Thus III may not be true.
III可能不是真的.
to TOP
Q31. B (42%)
Transformation of graph of function:
函數圖像的變換：
The graph is enlarged to 2 times along x-axis.
圖像沿x軸放大到2倍。
The y-intercept will not be changed.
y軸截距不會改變。
Q32. D (50%)
方法 1: Calculator: press 按 MODE 3 (BASE),
press 按 DEC, enter 輸入 8x8x8 EXE.
We get 得到 51210.
press 按 HEX, We get 得到 20016.
only只有 (D) has the ending 結尾 with 20016.
(Remember to press MODE 1 (COMP) to return normal computation.)
（請記住按MODE 1（COMP）返回正常計算。）
方法 2: 8x8x8 = 2x2x2x8x8
= 2(2x8)(2x8)
= 2x16x16 = 20016.
Q33. C (53%)
√ y - 0 = (8-0)/(0-4)(x-4)
√ y = -2(x-4) = -2x+8
y = (-2x+8)2
y = 4x2 - 32x + 64
Q34. D (41%)
log9 y = x - 3 ...(1)
2(log9 y)2 = 4 - x ...(2)
(1)+(2),
2(log9 y)2 + (log9 y) = 1
2(log9 y)2 + (log9 y) - 1 = 0
log9 y = 0.5 or log9 y = -1
y = 90.5 = 3 or y = 1/9
x = log9 y + 3 ...(3)
when當 y=3, x= 0.5 + 3 = 3.5
when當 y=1/9, x= -1 + 3 = 2
Q35. B (45%)
Let 設 x = 5/(2-i) + ki.
Calculator: press 按 MODE 2 (CMPLX)
Enter 輸入 5 / (2 - i) EXE.
We get 得到 2 + i.
( The real part 實部 is 2. Press 按 shift EXE, the imaginary part 虛部 is 1i. )
so x = 2 + i + ki = 2 + (1+k) i.
As由於 x is a real number 實數, k + 1 = 0
k = -1
(Remember to press MODE 1 (COMP) to return normal computation.)
（請記住按MODE 1（COMP）返回正常計算。）
Q36. C (66%)
I. obviously wrong, I 明顯錯了 (not A.S.)
II. 60π - 45π = 15π
= 45π - 30π
II is correct, II是對的
II. (π-60) - (π-45) = -15
= (π-45) - (π-30)
III is correct, III是對的
Q37. C (45%)
y ≤ 9 (solution below & including line y = 9)
x - y - 9 ≤ 0
x - 9 ≤ y, y ≥ x - 9
(solution above & including line y = x - 9)
x + y - 9 ≥ 0, y ≥ -x + 9
(solution above & including line y = -x + 9)
Test測試 all 3 corners.
Greatest value occurs at (9, 0).
最大值出現在（9,0）。
x - 2y + 43 = 9 -2(0) + 43 = 52
Q38. A (49%)
note 留意 ∠CAB = ∠DCA
(alt.∠s 內錯角, AB//DC)
Let 設 x = ∠CAB = ∠DCA
tan x = 21/28 = 3/4
cos x = 4/5
DE2 = 52 + 282 - 2(5)(28)(4/5)
= 585 = 9(65)
DE = 3√65
Q39. A (34%)
BD2 = 252 - 152
BD = 20
BD2 + CD2 = 202+ 212
= 841 = 292 = BC2
so ∠BDC = 90°
tan ∠BAD = BD/AD = 20 / 15
∠BAD = 53°
Q40. B (62%)
∠OAB = 90° - 68° = 22°
∠ABO = 22°
∠CBO = 26°
∠ABC = ∠ABO + ∠CBO
= 22° + 26° = 48°
to TOP
Q41. D (21%)
3x + 4y = 3p
Let 設 p = 4, 3x + 4y = 12
When y=0, x=4. so p=4, OP = 4.
Let 設 M(0,c) be y-intercept y-截距 of 3x + 4y = 12
Substitute 代入 (0,c) in 3x + 4y = 12, we get 得到 4c = 12.
so c =3 , OM = 3
Let 設 a be ∠MPO
tan a = OM / OP = 3 / 4
a = 36.86989765
tan 2a = 3.428571429 = 24 / 7
q / p = 24 / 7
p : q = 7 : 24
(This is a very good method as we need not find the coordinates of the in-centre.)
（這是一個非常好的方法，因為我們不需要找內心的坐標。）
Q42. B (71%)
13C5 x 6C4 =19305
Q43. C (59%)
P(at most 3 times 最多3次)
= 1 - P(4 times 4次)
= 1 - 0.74
= 0.7599
Q44. B (78%)
Let 設 SD = standard deviation 標準差.
standard score 標準分
= (33 - 45) / SD = -2
so, SD = (33 - 45)/(-2)
= 6
Q45. A (48%)
New data 新數據 = 8 x old data 舊數據.
so, m2 = 8m1 ; r2 = 8r1
so, I and II are correct, I和II是正確
since 因為 variance 方差= SD2
v2 = 82 v1 = 64 v1
so, III is wrong, III是錯誤
..........................................................
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