2020 DSE Math Paper 1 Mock (Q.12-19)
2020 DSE 數學卷一模擬考試 (題解及Marking Scheme)
2020 DSE數學考試臨近，我根據hkdse math paper 1 考試新趨勢，為考生提供免費的 2020 DSE Math Paper 1 Mock Exam， 附上 詳細題解 及 Marking Scheme (在網頁底部)，希望對考生有幫助!
大多數考生認為 Q16, Q17, Q18, Q19 是很困難的。一些考生認為 Q13, Q14 也困難。
在 Q13 的 Marking Scheme，我用顏色突顯了關鍵字，以便考生更容易理解統計學 的概念。另外，我更正了一些打字。
如果您未完成Q.1-11，
請先完成Q.1-11 (請點擊下面圖片鏈接)。
2020 DSE MATH PAPER 1 MOCK - SECTION A(2) (con't)
2020 DSE Math Paper 1 Mock Q12-14
12. Let f(x) be a cubic polynomial. When f(x) is divided by x – 2 , the remainder is 16 . When f(x) is divided by x + 1 , the remainder is – 65 . It is given that f(x) is divisible by 2x2 – 5x + 6 .
設 f(x) 為三次多項式。當 f(x) 除以 x – 2 時，餘數為16。當 f(x) 除以 x + 1 時，餘數為 – 65。 已知 f(x) 能被 2x2 – 5x + 6 整除。
(a) Find the quotient when f(x) is divided by 2x2 – 5x + 6 .
求f(x) 除以2x2 – 5x + 6的商。 (3 marks)
(b) How many rational roots does the equation f(x) = 0 have?
Explain your answer.
方程 f(x) = 0 有多少個有理根？試解釋你的答案。 (3 marks)
13. The stem-and-leaf diagram below shows the distribution of the hourly wages (in dollars) of the workers in a group.
下面的圖顯示某組的工人的時菥(以元為單位)的分佈。
Stem (tens) Leaf (units)
幹(十位) 葉(個位)
4 r 2
5 1 s s 3 4 4 6 7 9
6 7 7 8
7 4 t
It is given that the inter-quartile range of the distribution is $14.
已知該分佈的四分位數間距為 $14。
(a) Find s. 求 s。 (2 marks)
(b) It is given that the mean of the distribution is $58 and the range of the distribution exceeds $36 .
已知該分佈的平均值為 $58，且該分佈的分佈域超過 $36。
(i) Find r and t. 求 r 及 t 。
(ii) Two more workers now join the group. It is found that both of the mean and the range of the distribution of the hourly wages are increased by $1. Find the hourly wages of these two workers.
現再有兩名工人加入該組。得知上述時菥的分佈的平均值及
分佈域均增加 $1。求這兩名工人各人的時菥。 (6 marks)
14. The coordinates of the points P and Q are (−1, 7) and (3, 1) respectively.
點 P 及點 Q 的坐標分別為 (−1, 7) 及 (3, 1) 。
(a) Let L be the perpendicular bisector of PQ. 設 L 為 PQ 的垂直平分線。
(i) Find the equation of L. 求 L 的方程。
(ii) Suppose that G is a point lying on L. Denote the x-coordinate of G by h. Let C be the circle which is centred at G and passes through P and Q.
假定 G 為 L 上的一點。將 G 的 x 坐標記為 h。
設 C 為一圓，其圓心為 G 且通過 P 及 Q。
Prove that the equation of C is 證明 C 的方程為 :
3x2 + 3y2 – 6hx – (4h+20)y + 22h – 10 = 0 . (6 marks)
b) The coordinates of the point R are (−3, −3). 點 R 的坐標為 (−3, −3).
Using (a)(ii), or otherwise, find the area (in terms of π) of the circle which passes through P, Q and R.
利用(a)(ii)，或其他方法，求通過 P、Q 及 R 的圓的面積 (答案以π表示)。 (3 marks)
SECTION B (35 marks)
2020 DSE Math Paper 1 Mock Q15-19
15. An nine-digit password is formed by a permutation of 1, 2, 3, 4, 5, 6, 7, 8 and 9 .
一個九位密碼由 1、2、3、4、5、6、7、8 和 9 的排列所組成。
(a) How many different nine-digit passwords can be formed?
可組成多少個不同的九位密碼？ (1 mark)
(b) If the first digit and the second digit of a nine-digit password are even numbers, how many different nine-digit passwords can be formed?
如果九位密碼的第一個位和第二個位是偶數，則可以組成多少個不同的九位密碼？ (2 marks)
16. Let a and b be real numbers such that log3(3a+2b) = 3 and log3(6a+7b) = 4 .
設 a 和 b 為實數，以使 log3(3a+2b) = 3 和 log3(6a+7b) = 4。
(a) Find a and b . 求 a 和 b 。 (2 marks)
(b) The 1st term and the 2nd term of a geometric sequence are log a and log b respectively. Find the least value of n such that the sum of the (n+1) th term and the (2n+1) th term of the sequence is greater than 88888(log3) .
幾何序列的 第1項 和 第1項 分別為 log a 和 log b 。求 n 的最小值，以使序列的 第(n+1)項 和 第(2n+1)項 之和 大於 88888(log3) 。 (4 marks)
17.(a) Express 1/(3+4i) in the form of a + bi , where a and b are real numbers.
將 1/(3+4i) 表成 a + bi 的形式，其中 a 及 b 均為實數。 (2 marks)
(b) The roots of the quadratic equation x2 + px + q = 0 are 50/(3+4i) and 50/(3−4i) . 二次方程 x2 + px + q = 0 的根為 50/(3+4i) 及 50/(3−4i) 。
(i) Find p and q , 求 p 及 q ，
(ii) Find the range of values of r such that the quadratic equation x2 + px + q = r2 has no real roots.
求 r 值的範圍使得二次方程 x2 + px + q = r2 沒有實根。 (5 marks)
18. The Figure (a) shows a piece of paper card of a quadrilateral ABCD . It is given that AD = 30 cm, BC = 34 cm, CD = 54 cm, ∠ABD = 58o and ∠ADB = 65o .
圖(a) 顯示了一張四邊形ABCD的紙卡。 已知 AD = 30 cm, BC = 34 cm, CD = 54 cm, ∠ABD = 58o 及 ∠ADB = 65o .
(a) Find AB . 求 AB . (2 marks)
(b) The paper card in Figure (a) is folded along BD such that the ΔABC lies on the horizontal ground and ∠ABC = 116o as shown in Figure (b).
然後將四邊形紙卡沿 BD 折疊，使 ΔABC 處於水平地面上， ∠ABC = 116o ，如圖（b）所示。
(i) Find AC . 求 AC .
(ii) Let M be a point lying on AD such that BM is perpendicular to AD. David claims that ∠BMC is the angle between the face ABD and the face ACD. Do you agree? Explain your answer.
設M為AD上的一點，以使BM垂直於AD。 大衛聲稱∠BMC是平面ABD和平面ACD的交角。 你是否同意？ 試解釋你的答案。 (5 marks)
19. Let f(x) = 2x2 – (8–4k)x + 2k2 – 5k + 10 , where k is a positive constant. R is the vertex of the graph of y = f(x) . 設 f(x) = 2x2 – (8–4k)x + 2k2 – 5k + 10，其中k是一個正常數。 R是 y = f(x) 的圖的頂點。
(a) Using the method of completing the square, express the coordinates of R in terms of k . 使用 方法，以k表示R的坐標。 (3 mark)
(b) The graph of y = g(x) is obtained by reflecting the graph of y = f(x) with respect to the x-axis and then translating the resulting graph upwards by 16 units. Let S be the vertex of the graph of y = g(x) . Denote the origin by O . y = g(x) 的圖是通過相對於x軸反射 y = f(x) 的圖，然後將所得圖向上平移16個單位而獲得的。設S為 y = g(x) 的圖的頂點。用O表示原點。
(i) Express the coordinates of S in terms of k . 用k表示S的坐標。
(ii) Denote the point (12, 16) by H . 用H表示點 (12, 16)。
Find k such that the area of the circle passing through H, O and R is the least. 求 k 使通過H，O和R的圓的面積最小。
Are H, O, R and S concyclic? Explain your answer.
H，O，R和S是共圓嗎？試解釋你的答案。
Find the coordinates of the orthocentre of ΔHOR and the coordinates of the orthocentre of ΔHOS .
求ΔHOR的垂心的坐標 和 ΔHOS的垂心的坐標。 (9 mark)
2020 DSE Math Paper 1 Mock: Solutions and Marking Scheme are given below:
12(a) Let ax + b be the required quotient. 1M
Then, f(x) = (ax + b)(2x2 – 5x + 6)
Note that f(2) = 16 and f(-1) = -65 . 1M (for either one)
(a(2) + b) (2(2)2 – 5(2) + 6) = 16 and
(a(-1) + b) (2(-1)2 – 5(-1) + 6) = -65
So 2a + b = 4 and -a + b = -5
Solving, a = 3 and b = -2 1A (for both correct)
Thus, required quotient is 3x – 2 .
(b) f(x) = 0
(3x – 2)(2x2 – 5x + 6) = 0 ( by (a) )
3x – 2 = 0 or 2x2 – 5x + 6 = 0
(-5)2 - 4(2)(6) 1M
= -23 < 0
So the equation 2x2 – 5x + 6 = 0 does not have real roots. 1M
Note that 2/3 is a rational root of f(x) = 0 .
Thus the equation f(x) = 0 has 1 rational root. 1A
13(a) Inter-quartile range = $14
67 – (50+s) =14 1M
s = 17 – 14 = 3 1A
(b) (i) Range > $36
(70 + t) – (40 + r) > 36 1M (for either one)
So, t – r > 6 … (1)
58(16) = (40+r)+42+51+2(53)+53+2(54)
+56+57+59+2(67)+68+74+(70+t)
So, r + t = 10 … (2)
(1)+(2), 2t > 16, so t > 8
So, t = 9 and r = 1 1A (for both correct)
(ii) Let $a and $b be hourly wages of the two workers, where a ≤ b .
Range = 79 - 41 = $38
New range = 38 + 1 = $39
a + b = (58+1)(16+2) – 58(16) 1M
a + b = 134
case 1: a = 40 1M
Since a + b = 134 ,
so b = 134 – 40 = 94
so new range = 94 – 40 = 54
But it is impossible.
case 2: 41 ≤ a ≤ 80
In this case, b = 80 1A
Since a + b = 134 ,
so a = 134 – 80 = 54 1A
Thus, the hourly wages of the 2 workers are $54 and $80.
14(a)(i) Mid-point of PQ = (1, 4)
Slope of PQ = (7–1)/(–1–3) 1M
= –3/2
Equation of L is
y – 4 = (2/3)(x – 1) 1M
2x – 2 = 3y – 12
2x – 3y + 10 = 0 1A
(ii) Let k be the y-coordinate of G .
By (a)(i), 2h – 3k + 10 = 0
k = (2h + 10)/3 1M
The equation of C is
(x – h)2 + (y – k)2 = (3 – h)2 + (1 – k)2 1M
x2 - 2hx + h2 + y2 - 2ky + k2 = 9 - 6h + h2 + 1 - 2k + k2
x2 + y2 - 2hx - 2y(2h + 10)/3 = 9 - 6h + 1 - 2(2h + 10)/3
3x2 + 3y2 – 6hx – (4h+20)y + 22h – 10 = 0 . 1 (must give reasons)
(b) Denote the circle which passes through P, Q and R by C.
Centre of C lies on the perpendicular bisector of PQ.
Let h be the x-coordinate of the centre of C.
By (a)(ii),
3(−3)2 + 3(−3)2 – 6h(−3) – (4h+20)(−3) + 22h – 10 = 0 . 1M (for using a(ii))
54 + 18h + 12h + 60 + 22h - 10 = 0
So, h = −2
Equation of C is x2 + y2 + 4x – 4y – 18 = 0 .
Required area = π [ (-2)2 + 22 –(–18) ] 1M
= 26π 1A
15(a) The required number
= 9P9 = 362 880 1A
(b) The required number
= (4P2)(7P7) 1M
= 60 480 1A
16(a) 3a + 2b = 33 = 27 and 6a + 7b = 34 = 81 1M (for either one)
a = 3 and b = 9 1A (for both correct)
(b) Let T(n) be nth term of the geometric sequence.
T(1) = log3 and T(2) = log9 = 2log3 1M (for either one)
The common ratio is 2 .
(log3)(2)n + (log3)(2)2n > 88888(log3)
(2n)2 + (2n) – 88888 > 0 1M
2n < -298.6413255 or 2n > 297.6413255
log2n > log(297.6413255) 1M
nlog2 > log(297.6413255)
n > 8.217431039
Note that n is an integer.
Thus, the least value of n is 9 . 1A
17(a) 1 / (3+4i)
= 1 / (3+4i) x (3−4i) / (3−4i) 1M
= 3/25 − 4i/25 1A
(b)(i) 留意 50/(3+4i) = 6−8i 及 50/(3−4i) = 6+8i 。
兩根之和
= 50/(3+4i) + 50/(3−4i) 1M (任何一項)
= (6-8i) + (6+8i)
= 12 1A (任何一項)
兩根之積
= 50/(3+4i) x 50/(3−4i) = 100
因此，可得 p = −12 及 q = 100 1A 給兩項均正確
(ii) 當方程 x2 + px + q = r2 ,
i.e. x2 − 12x + 100 = r2 沒有實根時，可得 Δ < 0
故此，可得 (-12)2 – 4(1)(100 – r2) < 0 1M
36 < 100 – r2
r2 < 64
因此，可得 –8 < r < 8 1A
18(a) AB / sin∠ADB = AD / sin∠ABD 1M
AB / sin65o = 30 / sin58o
AB ≈ 32.06095708
≈ 32.1 cm 1A
(b)(i) AC2 = AB2 + BC2 – 2(AB)(BC)(cos∠ABC) 1M
≈ (32.06095708)2 + 342 – 2(32.06095708)(34)(cos∠116o)
AC ≈ 56.0322913
≈ 56.0 cm 1A
(ii) In ΔABD, ∠DAB = 180o – 58o – 65o = 57o
BM = AB sin∠DAB
≈ 32.06095708sin57o ≈ 26.88858108 cm
AM = ABcos∠DAB
≈ 32.06095708cos57o ≈ 17.46164872 cm
(We need to find ∠CDA first, in order to find CM.)
In ΔACD,
cos∠CAD = (AD2 + AC2 – CD2) / ( 2(AD)(AC) )
≈ (302 + 56.03229132 – 542) / (2 x 30 x 56.0322913) 1M
∠CAD ≈ 70.47505057o
CM2 = AM2 + AC2 – 2(AM)(AC)cos∠CAD
≈ (17.46164872)2 + (56.0322913)2 – 2(17.46164872)(56.0322913)(cos70.47505057o)
CM ≈ 52.825369 cm
CM2 + AM 2 ≈ (52.825369) 2 + (17.46164872) 2
≈ 3095.428786
AC2 ≈ (56.0322913)2 ≈ 3139.617668
So, CM2 + AM2 ≠ AC 2
So, ∠AMC is not a right angle. 1M
So, ∠BMC is not the angle between the face ABD and the face ACD.
Thus, the claim is not agreed. 1A (must give all reasons)
19(a) f(x) = 2x2 – (8–4k)x + 2k2 – 5k + 10
f(x) = 2[x2 – (4–2k)x + (2–k) 2 – (2–k)2] + 2k2 – 5k + 10
f(x) = 2[x2 – (2–k) 2]2 – 2(2–k)2 + 2k2 – 5k + 10
1M (for completing the square)
= 2( x – (2–k) ) 2 + 3k + 2 1M
Coordinates of R are (2–k , 3k + 2) 1A
(b)(i) Note that -3k-2+16 = 14-3k
Coordinates of S are (2–k , 14–3k) 1A
(ii) Note that the area of circle passing through H and O is the least when HO is a diameter of the circle. 1M
If R lies on the circle, then ∠HRO = 90o .
[(3k + 2 – 0) / (2 – k – 0)] x [(3k + 2 – 16) / (2 – k – 12)] = –1 1M+1A
[(3k + 2) (3k – 14)] / [(2 – k) (– k – 10)] = –1
(3k + 2) (3k – 14) = (2 – k) (k + 10)
10k2 – 28k – 48 = 0
k = 4 or k = –6/5 (rejected) 1A
Thus, the area of the circle passing through H, O and R is the least when k = 4 .
When k = 4, coordinates of S are (–2 , 2) .
The product of slope of OS and slope of HS
= [(2 – 0) / (– 2 – 0)] x [(2 – 16) / (– 2 – 12)] 1M
= –1
So, ∠HSO = 90o
Therfore, when k = 4, ∠HRO = 90o, and HO is a diameter of the circle passing through H, O and R .
So, when k = 4, S lies on the circle passing through H, O and R .
So, H, O, R and S are concyclic. 1A (must give all reasons)
Note that coordinates of R are (–2 , 14) when k = 4 .
In ΔHOR, ∠HRO = 90o, so R is the orthocentre of ΔHOR . 1M (for either one)
So, coordinates of the orthocentre of ΔHOR are (–2 , 14) .
In ΔHOS, ∠HSO = 90o, so S is the orthocentre of ΔHOS .
So, coordinates of the orthocentre of ΔHOS are (–2 , 2) .
1A (for both correct)
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