2022年澳门四高校联合入学考试数学正卷 附加卷（pdf版 含答案）

澳門四高校聯合入學考試（語言科及數學科）
Joint Admission Examination for Macao Four Higher Education Institutions
(Languages and Mathematics)
2022 年試題及參考答案
2022 Examination Paper and Suggested Answer
數學附加卷 Mathematics Supplementary Paper
注意事項：
1. 考生獲發文件如下:
1.1 本考卷包括封面共 22 版
1.2 草稿紙一張
2. 請於本考卷封面填寫聯考編號、考場、樓宇、考室及座號。
3. 本考卷共有五條解答題，每題二十分，任擇三題作答。全卷滿分為六十分。
4. 必須在考卷內提供的橫間頁內作答，寫在其他地方的答案將不會獲評分。
5. 必須將解題步驟清楚寫出。只當答案和所有步驟正確而清楚地表示出來，考生方可獲
得滿分。
6. 本考卷的圖形並非按比例繪畫。
7. 考試中不可使用任何形式的計算機。
8. 請用藍色或黑色原子筆作答。
9. 考試完畢，考生須交回本考卷及草稿紙。
Instructions:
1. Each candidate is provided with the following documents:
1.1 Question paper including cover page – 22 pages
1.2 One sheet of draft paper
2. Fill in your JAE No., campus, building, room and seat no. on the front page of the examination
paper.
3. There are 5 questions in this paper, each carries 20 marks. Answer any 3 questions. Full mark
of this paper is 60.
4. Put your answers in the lined pages provided. Answers put elsewhere will not be marked.
5. Show all your steps in getting to the answer. Full credits will be given only if the answer and
all the steps are correct and clearly shown.
6. The diagrams in this examination paper are not drawn to scale.
7. Calculators of any kind are not allowed in the examination.
8. Answer the questions with a blue or black ball pen.
9. Candidates must return the question paper and draft paper at the end of the examination.
2
任擇三題作答，每題二十分。請把答案寫在緊接每條題目之後的 3 頁橫間頁內。
Answer any 3 questions, each carries 20 marks. Write down the answers on the 3 lined pages
following each question.
1.
P
C
E
B
A
D

如上圖所示，在三棱錐 P-ABC 中，PC 垂直 ABC，|| = 3，∠ = ，

D 和 E 分別是 AB 和 BC 上的點，且 | | = | | = √2，|| = 2，|| = 1。
(a) 求 ||。[提示: 設 M 為 CE 的中點。] (6 分)

(b) (i) 證明 ∠ = 。 (2 分)

(ii) 求 ∠，答案以 cos 表示。 (5 分)
(c) 求二面角 P-AB-C，答案以 tan 表示。 (7 分)
[提示: 設 X 為 AB 上一點，使得 CX 與 AB 垂直。證明 PX 與 AB 垂直。]
As shown in the above figure, P-ABC is a triangular pyramid, PC is perpendicular to

ABC, || = 3, ∠ = . The points D and E are on AB and BC, respectively, and

| | = | | = √2, || = 2,|| = 1.
(a) Find ||. [Hint. Let M be the midpoint of CE.] (6 marks)

(b) (i) Show that ∠ = . (2 marks)

(ii) Find ∠ . Express your answer in terms of cos. (5 marks)
(c) Find the dihedral angle P-AB-C. Express you answer in terms of tan.
[Hint: Let X be the point on AB such that CX and AB are perpendicular.
Show that PX and AB are perpendicular.] (7 marks)
3
2. (a) 已知函數 ()= 3 + 2，且 (1)= 0。
(i) 求方程 ()= 0 的所有解。 (3 分)
(ii) 求 ′() 及 ′(′)。 (2 分)
(iii) 求 f (x) 的局部極大值和局部極小值。 (3 分)
(iv) 求曲線 = () 的拐點。 (2 分)
(v) 運用 (i) – (iv) 的結果，繪出曲線 = ()。 (3 分)
(b) 求由曲線 = 8 + 24 及 = 8 所包圍的區域的面積。 (7 分)
(a) Given function ()= 3 + 2, and that (1)= 0.
(i) Find all the solution(s) of the equation ()= 0. (3 marks)
(ii) Find ′() and ′(′). (2 marks)
(iii) Find the local maximum and local minimum values of (). (3 marks)
(iv) Find the inflection point(s) of the curve = (). (2 marks)
(v) Using the results in (i) – (iv), sketch the curve = (). (3 marks)
(b) Find the area of the region bounded by the curves = 8 + 24 and
= 8 . (7 marks)
4
3. 已知定點 ( 1,0) 和 (1,0)。曲線 C 上任一點 (,) 都有 = 。
(a) 證明 C 是拋物線 = 4。 (4 分)
(b) 若直線 = + 與 C 相切，證明 ab= 1。 (4 分)
(c) 設 > 0。
(i) 除原點以外，求直線 : = 與 C 的交點 P。答案以 m 表示。 (2 分)
(ii) 求曲線 C 在 P 的切線 的斜率。答案以 m 表示。 (4 分)
(iii) 求 m 的值使得 L1 與 L2 的夾角為 tan 。 (6 分)
Given fixed points ( 1,0) and (1,0). Any point (,) on the curve C satisfies
= .
(a) Show that C is the parabola = 4. (4 marks)
(b) Suppose that the straight line = + is tangent with C. Show that ab= 1. (4 marks)
(c) Suppose > 0.
(i) Find, besides the origin, the intersection point P of the straight line
: = and C. Express your answer in terms of m. (2 marks)
(ii) Find the slope of the tangent line of C at P. Express your answer
in terms of m. (4 marks)

(iii) Find the value(s) of m such that the angle between L1 and L2 is tan
. (6 marks)

5
4. 設 = √ 1。
(a) (i) 設 = 3 + 5 及 = 5 + 。若 = + 滿足 | | = | |，
其中 x 和 y 為實數，求 x 和 y 的關係式。 (4 分)
(ii) 在阿根圖中繪出 ， 和 z 的軌跡。 (2 分)
(iii) 求 | | 的最小值。 (2 分)

(b) 設 ω = cos + sin 。

(i) 證明 = 1，並推導出 1 + + + + = 0。 (4 分)

(ii) 證明 + = 2cos ， = 1,2,3,…. 。 (2 分)


(ii) 用 (i) 和 (ii) 的結果，求 cos + cos + cos 的值。 (6 分)

Let = √ 1.
(a) (i) Let = 3 + 5 and = 5 + . Suppose = + satisfies | | = | |,
where x and y are real. Find a relation between x and y. (4 marks)
(ii) In the Argand diagram, plot the points , and the locus of z. (2 marks)
(iii) Find the minimum value of | |. (2 marks)

(b) Let ω = cos + sin .

(i) Show that = 1. Deduce that 1 + + + + = 0. (4 marks)

(ii) Show that + = 2cos , = 1,2,3,….. (2 marks)

(ii) Using the results in (i) and (ii), find the value of

cos + cos + cos . (6 marks)

6
+ +
5. (a) 因式分解行列式 + + 。 (8 分)
+ +
(b) 已知以 x、y 和 z 為未知量的方程組:
+ + = 1
(): + + = ，
+ + =
其中 p 和 q 為常數。
(i) 求 p 的取值範圍，使得 (E) 有唯一解。 (4 分)
(ii) 對使得 (E) 有多於一個解的 p 及 q 的值，求 (E) 的通解。 (8 分)
+ +
(a) Factorize the determinant + + . (8 marks)
+ +
(b) Given the system of equations with unknowns x, y and z:
+ + = 1
(): + + = ,
+ + =
where p and q are constants.
(i) Find the range of p such that (E) has a unique solution. (4 marks)
(ii) Find the general solution of (E) for those values of p and q such that (E) has more
than one solution. (8 marks)
7
參考答案：
1. (a) 設 M 為 CE 的中點。因 | | = | |，Δ 是一等腰三角形，故 ⊥ 。
| | | |
已知 ∠ = ，故 Δ 和 Δ 相似，從而得 = 。
| | | |
| |
因 | | = | | | | = 1，故 || = || = 。
| |

(b) (i) 因 || = 4 = | | + | |，故 ∠ = 。

(ii) 由 (a) 的解中，有 | | = 1。故 | | = | | + | | = √5。因此，
| |
| || | ∠ = cos = cos = cos 。
| || | √√ √
(c) 設 X 為 AB 上一點，使得 ⊥ 。連同 PC⊥ (因 ⊥ )，得
||
⊥ 。因而得知二面角 P-AB-C 與 ∠ 相等及 ∠ = tan 。
||

現考慮 Δ 的面積以求||。從 |||| = |||| ，得


| || | ( )
|| = = = 。因此， 二面角 P-AB-C 是 tan = tan
√5。
| |



√
√

8
2. (a) (i) 因 (1)= 0，故 1是 () 的因式。計算得 ()= ( 1)( 2 2)。
故 ()= 0 = 1 或 2 2 = 0 = 1 或 = 1± √3。
(ii) ()= 3 6 ， ()= 6 6。
(iii) ()= 0 = 0 或 = 2。
當 < 0， ()> 0 ，故 () 是遞增的。
當 0 < < 2， ()< 0 ，故 () 是遞減的。
當 2 < ，()> 0 ，故 () 是遞增的。
因此， (0)= 2 是一局部極大值， (2)= 2 是一局部極小值。
(iv) ()= 0 = 1。當 < 1， ()< 0; 當 > 1， ()> 0。
因此，(1,0)是曲線 = () 的拐點。
(v)
y
4
3
y=f(x)
2
1
0 x
-2 -1 0 1 2 3 4
-1
-2
-3
= 8 + 24
(b) 解 ，得 = 2 或 = 6。
= 8
當 2 < < 6 ，曲線 = 8 在曲線 = 8 + 24 之上。
因此，所求面積為

∫ (8 ) ( 8 + 24) = ∫ 2 + 16 24


= + 8 24


= 。

9
3. (a) 因 = ( + 1,) ， = (2,0) 及 = ( 1,) ，故
= 2( + 1)= 2( 1) + = 4。
= 4
(b) 由 ， 得 + (2 4) + = 0。 (1)
= +
因直線 = + 與 C 相切，(1)有重根，其判別式為 0，
即 (2 4) 4 = 0 。因此， = 1。
= 4
(c) (i) 解 ，得 = 0 或 = 。故除原點外的交點 P 是 , 。
=





(ii) 用 (b)，設 為 = + 。 因 P 是 上的一點，故 = + ，

並由此得 4 4 + = 0。 因此， = 。


(iii) 因 > 0，設 tan = 及 tan = 分別為 及 的斜率，

其中 0 < < < 。由

1 tan tan
= tan(
2
)= = = 4 1 + tan tan 1 + ( ) 2 + 2
得 4 + 2 = 0。因此， = 2± √2。
10
4. (a) (i) 設 = + ，則
| | = | |
( 3) + ( 5) = ( 5) + ( 1)
( 3) + ( 5) = ( 5) + ( 1)
2 + 2 = 0。
(ii)
6
z1 2 + 2 = 0
5
4
3
2
1 z2
0
0 1 2 3 4 5 6 7 8
(iii) 當 z 為 和 的中點時，即 = 4 + 3，| | 達到其最小值。
此時，| | = √5。

(b) (i) = (cos + sin ) = cos + sin = 1。

1 = 0 ( 1)( + + + 1)= 0
+ + + 1 = 0 (因 ≠ 1)。
2 2 2( ) 2( )
(ii) + = (cos + sin )+ (cos + sin )
7 7 7 7

= (cos + sin )+ (cos sin )


= 2cos 。


(iii) cos + cos + cos



= + +


= [( + 2 + )+ ( + 2 + )+ ( + 2 + )]


= [( + 2 + )+ ( + 2 + )+ ( + 2 + )]


= [5 + (1 + + + + )]


= 。

11
+ + + +
5. (a) + + =
+ +
+ +
= ( )( ) 1 1 +
1 1 +
+ +
= ( )( ) 1 1
0 0
+ + + +
= ( )( ) 0 1
0 0
= ( )( )( )( + + )。
1 1
(b) (i) (E) 有唯一解當且僅當 1 1 ≠ 0，即 ≠ 1 及 ≠ 2。
1 1
+ + = 1
當 = 1，(E) 變成 + + = 1 。故 = 1 及其解為
+ + =
= 1 ， = ， = ， ， ∈ 。
+ 2 = 1
(ii) 當 = 2，(E) 變成 2 + + = 2。由這三方程之和得 0 = 1，故 = 1。
2 + =
+ 2 = 1
然後，解 得 = 1 + ， = ， = ， ∈ 。
2 + + = 2
12
Suggested Answer:
1. (a) Let M be the mid-point of CE. As Δ is an isosceles triangle with | | = | |,

we have ⊥ . Given ∠ = and so Δ and Δ are similar. Thus,

| | | | | |
= . As | | = | | | | = 1, we get || = || = .
| | | | | |

(b) (i) As || = 4 = | | + | |, we get ∠ = .

(ii) From the solution of (a), we have | | = 1. So, | | = | | + | | = √5.
Hence,
| || || |
∠ = cos = cos = cos .
| || | √√ √
(c) Let X be the point on AB such that ⊥ . Then, as PC⊥ (since ⊥ ), we get
||
⊥ . So, the dihedral angle P-AB-C is equal to ∠, and ∠ = tan .
||

To find ||, consider the area of ∠. We have |||| = |||| and hence


| || | ( ) || = = = . Hence, the dihedral angle is tan = tan
√5.
| | √

√

2. (a) (i) As (1)= 0, we know that 1 is a factor of (). By direct calculation,
()= ( 1)( 2 2).
Hence, ()= 0 = 1 or 2 2 = 0 = 1 or = 1± √3.
(ii) ()= 3 6, ()= 6 6.
(iii) ()= 0 = 0 or = 2.
When < 0, ()> 0 and so () is increasing.
When 0 < < 2, ()< 0 and so () is decreasing.
When 2 < , ()> 0 and so () is increasing.
Hence, (0)= 2 is a local maximum value, (2)= 2 is a local minimum value.
(iv) ()= 0 = 1. When < 1, ()< 0; when > 1, ()> 0.
Hence, is (1,0) is the inflection point of the curve = ().
(v)
y
4
3
y=f(x)
2
1
0 x
-2 -1 0 1 2 3 4
-1
-2
-3
13
= 8 + 24
(b) Solving , we obtain = 2 or = 6. = 8
For 2 < < 6, the curve = 8 is above the curve = 8 + 24.
Hence, the required area is

∫ (8 ) ( 8 + 24) = ∫ 2 + 16 24


= + 8 24


= .

3. (a) We have = ( + 1,), = (2,0) and = ( 1,). Hence,
= 2( + 1)= 2( 1) + = 4.
= 4
(b) From , we get + (2 4) + = 0. (1)
= +
As the line = + is tangent to C, (1) has a double root, its discriminant is 0,
i.e., (2 4) 4 = 0. Hence, = 1.
= 4
(c) (i) Solving , we get = 0 or = . As the intersection point P is not the origin,
=

it is , .


(ii) By (b), let be = + . As P is a point on , we get = + which gives

4 4 + = 0. Hence, = .


(iii) As > 0, let tan = and tan = be the slopes of and , respectively,

where 0 < < < . Then,

1 tan tan
= tan( )= = 2 = ,
4 1 + tan tan 1 + ( ) 2 + 2
from which we get 4 + 2 = 0. Hence, = 2± √2.
14
4. (a) (i) Let = + . Then,
| | = | |
( 3) + ( 5) = ( 5) + ( 1)
( 3) + ( 5) = ( 5) + ( 1)
2 + 2 = 0.
(i)i
6
z1 2 + 2 = 0
5
4
3
2
1 z2
0
0 1 2 3 4 5 6 7 8
(iii) | | attains it minimum when z is the mid-point of and , i.e., = 4 + 3.
In this case, | | = √5.
(b) (i) = (cos + sin ) = cos + sin = 1.

1 = 0 ( 1)( + + + 1)= 0
+ + + 1 = 0 (since ≠ 1).
2 2 2( ) 2( )
(ii) + = (cos + sin )+ (cos + sin )
7 7 7 7

= (cos + sin )+ (cos sin )


= 2cos .


(iii) cos + cos + cos





= + +


= [( + 2 + )+ ( + 2 + )+ ( + 2 + )]


= [( + 2 + )+ ( + 2 + )+ ( + 2 + )]


= [5 + (1 + + + + )]


= .

15
+ + + +
5. (a) + + =
+ +
+ +
= ( )( ) 1 1 +
1 1 +
+ +
= ( )( ) 1 1
0 0
+ + + +
= ( )( ) 0 1
0 0
= ( )( )( )( + + ).
1 1
(b) (i) (E) has a unique solution if and only if 1 1 ≠ 0, i.e., ≠ 1 and ≠ 2.
1 1
+ + = 1
When = 1, (E) becomes + + = 1. So, = 1 and the solution is
+ + =
= 1 , = , = , , ∈ .
+ 2 = 1
(ii) When = 2, (E) becomes 2 + + = 2. The sum of all the three equations gives
2 + =
0 = 1.
+ 2 = 1
Hence, = 1. Then, solving , we get = 1 + , = , = , ∈ .
2 + + = 2
16澳門四高校聯合入學考試 (語言科及數學科)
Joint Admission Examination for Macao Four Higher Education Institutions
(Languages and Mathematics)
2022 試題及參考答案
2022 Examination Paper and Suggested Answer
數學正卷 Mathematics Standard Paper
1
第一部份 選擇題。請選出每題之最佳答案。
1. 在下列集合中，表示空集的是
A. {0} B. {x : sin x cos x 3} C. {x : x2 1 0, x }
D. { } E. {(x, y) : x2 y2 0, x , y }
2. 若 a > b > 0 及 m > 0，下列哪一個不等式是正確的？
A. b b m B. a a m C. b b m
a a m b b m a a m
D. a a m
b b E. 以上皆非 m
3. 若多項式 x 3 3x 2 ax b 分別被 x 2 及 x 1 除，餘數相等。求 a 之值。
A. 4 B. 3 C. 4 D. 9 E. 6
4. 7 4 3
A. 3 2 B. 2 3 C. 3 2
D. 2 6 E. 2 2 3
5. 若方程 px2 2(p 3)x p 1 0 有實根，求 p 的取值範圍。
A. 0 p 5 B. p 5 C. 5 p 1
7 7 7
D. p 9 E. 以上皆非
7
6. 已知 2a 5b 10 ，則 1 1 之值是多少？
a b
A. 2 B. 1 C. 2 D. 2 E. 1
2 2
7. 右圖中三個圓及線段 AB 互切。兩個大圓半徑分別為 3
單位及 4 單位。求小圓的半徑。
A. 84 48 3 B. 2 1 C. 2 2 2
D. 6 4 2 E. 42 24 3
8. 約翰在一次遊戲中勝出的概率是 1 。他在連續三次遊戲中勝出至少一次的概率是多少？
4
A. 1 B. 27 C. 35 D. 37 E. 43
64 64 64 64 64
2
y 2x 4
9. 下圖中，哪一區域是不等式組 x y 5 的解集？

y 2 0
A. B. C. III D. IV E. 以上皆非
10. 已知直線 L 穿過點 (1, 2)，並與直線 2x 3y 4 垂直。直線 L 和 y 軸相交於
A. 0, 7 B. 0, 5 C. (0, 3) D. 0, 1 E. 0, 3
2 2 2 2
11. 已知某室內表演場所觀眾席有 6400 張座椅，場內每行都有座椅 32 張。為了配合現行防疫社交
距離措施，同一行最多可有連續 4 張座椅被佔用。根據以上要求，每場表演的入座人數最多為
A. 5000 B. 5200 C. 5400 D. 5600 E. 5800
12. 已知方程式 3x 2 8x m 0 1 1的兩個實根為 x1 和 x2 。若 和 的算術平均數為 2，則 m 之值x1 x2
是多少？
A. 2 B. 1 C. 4 D. 1 E. 2
13. 從 1, 2, 3, 4, 5 中先後選取兩個不同數，分別作為一個兩位數的十位和個位，此兩位數小於 40 的
概率為
A. 1 B. 2 C. 3 D. 4 E. 1
5 5 5 5
14. 方程 6x 2x 2x 1 6x 1 2x 2 的解為
A. x 1 B. x 1 C. x 0 D. x 1 E. x 1
2 2
15. 彼得先面向正東方走了 n 公里，然後他向右轉了 150 度並走了 3 公里，結果他離出發點恰好為
3 公里。求 n 之值。
A. 3 3 B. 3 C. 1 D. 3 E. 2 3 或 3
3 3
3
第二部份 解答題。
1. 已知二次函數 f (x) ax2 bx c 的圖像 C 通過點 (5, 0)，對稱軸為 x 2，且 f (x) 有最小值 9。
(a) 求 a、b 和 c 之值。 (3 分)
(b) 將圖像 C 向左平移 3 個單位，再向上平移 3 個單位，求所得圖像的函數表達式。 (2 分)
(c) 設函數 g(x) f (3sin x)，求函數 g(x) 的最大值和最小值。 (3 分)
na
2. 設 {a } 是首項為 1的等比數列。數列 {b } 滿足 b n n n 1 n n 1 n 。已知 a1 , 4a2 , 16a4 3
成等差數列。
(a) 求 {an}n 1 和 {bn}n 1 的通項公式。 (3 分)
(b) 求 {an}n 1 的前 n 項和 Sn 及 {bn}n 1 的前 n 項和 Tn。 (5 分)
2 y 2 2
3. 已知橢圓 C : x 1 (a b 0) 的離心率為 ，且點
2 2 (2 2, 4) 在 C 上。 b a 2
(a) 求 C 的方程。 (3 分)
(b) 設直線 L : y k1x b 不通過原點 O 且不平行於坐標軸。直線 L 與 C 有兩個交點 A 和 B，線
段AB 的中點為 M，直線 OM 的斜率為 k2 。證明 k1k2 2。 (5 分)
4. (a) 把 3cos sin 以 r cos( ) 形式表示，並求 3cos sin 的取值範圍。 (2 分)
(b) 求 3cos sin 1 在 0 2 的解。答案要以弧度表示。 (3 分)
(c) 如果 3cos sin 1 ，且 0 ，求 sin 。 (3 分)
2 2 12
5. 用數學歸納法證明對任意正整數 n，34n 2 52n 1 能被 14 整除。 (8 分)
4
參考答案
第一部份 選擇題。
題目編號 最佳答案
1 B
2 C
3 E
4 B
5 D
6 A
7 A
8 D
9 D
10 A
11 B
12 E
13 C
14 C
15 E
(第二部份答案由下頁開始)
5
第二部份 解答題。
1. (a) 由題意知
25a 5b c 0

b 4a ，

4a 2b c 9
解得
a 1, b 4, c 5。
因此所求函數解析式為
f (x) x2 4x 5。
(b) y (x 1)2 6 x2 2x 5 。
(c) g(x) (3sin x)2 4(3sin x) 5 (3sin x 2)2 9。
由於 3 3sin x 3， 2 [ 3,3]，函數 g(x) 在3sin x 3處取得最大值，即
Max g(x) f ( 3) 16。
函數 g(x) 在3sin x 2處取得最小值, 即
Min g(x) f (2) 9。
2. (a) 設 an a1q
n 1。由 a1, 4a2,16a3 成等差數列知16a3 a1 8a2 ，
即16a1q
2 a1 8a
1
1q。由於 a1 1，可得 q 。因此 4
n 1 na 1n ，b 。 4 n 4n
(b) 由等比數列前 n 項和公式得，
S 4 1n 1 3 4n 。
由
T 1 2 3 4 n 1 nn 4 42 43 44 4n
， (1)
1 4n
我們有
4Tn 1
2 32
4
3
n 。 (2)
4 4 4 4n 1
(2) 式減 (1) 式得
3T 1 1 1 1 1 n 4 1 nn 4 42 43 4n 1 4n 1 3 4n n 。 4
因此
T 4 1 nn 1 9 4n 3 4n 。
6
3. (a) 由題意知 a 和 b 滿足
a2 b2 2

a 2 ，
8 16
b2 a2
1
解得 a2 32及b2 16。因此所求橢圓方程為
x2 y2 1。
16 32
x2 y2(b) 聯立直線 y k1x b及橢圓 1得 16 32
2 (k 2x 1x b) 1。
16 32
整理得
(2 k 21 )x
2 2k1bx b
2 32 0。
線段 AB 的中點M (x0, y0)的坐標為
k bx 10 ， y
2b 。
2 k 2 01 2 k
2
1
於是我們得到直線 OM 的斜率為
k 22 。 k1
因此， k1k2 2。
3 1
4. (a) 3 cos sin 2( cos sin ) 2(cos cos sin sin ) 2cos( )。
2 2 6 6 6
由此知 2 3cos sin 2。

(b) 由 (a) 可知 3 cos sin 2cos( )。
6
1
由 3 cos sin 1，得 cos( ) 。
6 2
2因為 0 2 ，我們有 4或 ，即 7 或 。
6 3 3 2 6
1 1 1
(c) 因為 3 cos sin ，所以 2cos( ) ，即cos( ) 。
2 6 2 6 4
1 5
由二倍角公式得 1 2sin2( ) ，因此 sin2 ( ) 。由於 0 ，從而有
2 12 4 2 12 8
5 10
0 ，因此 sin( ) 。
2 12 2 12 8 4
7
5. 證明：設 S(n) 表示命題“34n 2 52n 1 能被 14 整除”。
(1) 當 n 1時，原式 34 2 52 1 36 53 729 125 854 ，854 14 16， S(1)成立。
(2) 假設 n k (k ) 時， S(n) 成立，即34k 2 52k 1能被14整除。
當 n k 1(k ) 時，
34(k 1) 2 52(k 1) 1
34k 4 2 52k 2 1
34 34k 2 52 52k 1
81 34k 2 25 52k 1
56 34k 2 25 34k 2 25 52k 1
=56 34k 2 25（ 34k 2 52k 1）,
上式中 56 34k 2 可以被14整除，且 25 (34k 2 52k 1) 根據假設也可以被 14 整除， S(k 1)也
成立。
由 (1)，(2) 及數學歸納法，對任意正整數 n，34n 2 52n 1 能被 14 整除。
8
Part I Multiple choice questions. Choose the best answer for each question.
1. Which of the following is an empty set
A. {0} B. {x : sin x cos x 3} C. {x : x2 1 0, x }
D. { } E. {(x, y) : x2 y2 0, x , y }
2. If a > b > 0 and m > 0, which of the following inequalities is true
A. b b m B. a a m C. b b m
a a m b b m a a m
D. a a m E. none of the above b b m
3. If the polynomial x 3 3x 2 ax b is divided by x 2 and x 1 respectively, the remainders are equal.
Find the value of a.
A. 4 B. 3 C. 4 D. 9 E. 6
4. 7 4 3
A. 3 2 B. 2 3 C. 3 2
D. 2 6 E. 2 2 3
5. If the equation px2 2(p 3)x p 1 0 has real roots, find the range of p.
A. 0 p 5 B. p 5 C. 5 p 1
7 7 7
D. p 9 E. none of the above
7
6. Given 2a 5b 10 , what is the value of 1 1
a b
A. 2 B. 1 C. 2 D. 2 E. 1
2 2
7. In the right figure, the three circles and the line segment AB
touch each other. The two larger circles have radii 3 units
and 4 units, respectively. Find the radius of the smallest
circle.
A. 84 48 3 B. 2 1 C. 2 2 2
D. 6 4 2 E. 42 24 3
8. John has probability 1 of winning a game. What is the probability that he wins at least one game in
4
three consecutive games
A. 1 B. 27 C. 35 D. 37 E. 43
64 64 64 64 64
9
y 2x 4
9. In the below figures, which of the regions is the solution set to the system of inequalities x y 5

y 2 0
A. B. C. III D. IV E. none of the above
10. Suppose that the line L passes through the point (1, 2), and is perpendicular to the line 2x 3y 4 .
L and the y-axis will intersect at
A. 0, 7 B. 0, 5 C. (0, 3) D. 0, 1 E. 0, 3
2 2 2 2
11. There are 6400 chairs in an indoor stadium. Each row of the venue has 32 chairs. Current social
distancing measure requires that no more than 4 consecutive chairs could be occupied in the same row.
To comply with this requirement, the maximum number of chairs that could be occupied in each
performance is
A. 5000 B. 5200 C. 5400 D. 5600 E. 5800
12. Suppose the two real roots of the equation 3x 2 8x m 0 are x1 and x2 . If the arithmetic mean of
1 and 1 is 2, what is the value of m
x1 x2
A. 2 B. 1 C. 4 D. 1 E. 2
13. Two different numbers are picked up from 1, 2, 3, 4, and 5 sequentially to form the tens and unit digits
of a two-digit number. The probability that the two-digit number is less than 40 is
A. 1 B. 2 C. 3 D. 4 E. 1
5 5 5 5
14. The solution of the equation 6x 2x 2x 1 6x 1 2x 2 is
A. x 1 B. x 1 C. x 0 D. x 1 E. x 1
2 2
15. Peter first faced east and walked for n kilometers, and then he turned 150 degrees to the right and walked
3 kilometers. Now he was 3 kilometers away from the starting point. Find the value of n.
A. 3 3 B. 3 C. 1 D. 3 E. 2 3 or 3
3 3
10
Part II Problem-solving questions.
1. Given that the graph C of the quadratic function f (x) ax2 bx c passes through the point (5, 0). Its
axis of symmetry is x 2 , and the minimum value of f (x) is 9.
(a) Find the values of a, b and c. (3 marks)
(b) Find the expression of the function after shifting the graph C to the left by 3 units, and then shifting
it up by 3 units. (2 marks)
(c) Let g(x) f (3sin x) . Find the maximum and minimum values of g(x) . (3 marks)
na
2. Let {an}n 1 be a geometric sequence with 1 as the first term, and sequence {bn}n 1 is given as b
n
n . 4
Suppose that a1 , 4a2 , 16a3 form an arithmetic sequence.
(a) Find the general terms for {an}n 1 and {bn}n 1 . (3 marks)
(b) Find Sn , sum of the first n terms for {an}n 1 , and Tn , sum of the first n terms for {bn}n 1 .
(5 marks)
2 y 2 2
3. Suppose the eccentricity of the ellipse C : x 1 (a b 0) is , and (2 2, 4) is a point lying
b 2 a 2 2
on C.
(a) Find the equation of C. (3 marks)
(b) Let L : y k1x b be one line which does not pass through the origin O and is not parallel to the
coordinate axes. There are two intersections A and B for the line L and the ellipse C. The midpoint
of the line segment AB is M, and the slope of the line OM is k2 . Show that k1k2 2 . (5 marks)
4. (a) Express 3cos sin in the form of r cos( ) , and find the range of 3cos sin .
(2 marks)
(b) Solve 3cos sin 1 for 0 2 . Answer is to be presented in radians. (3 marks)
(c) If 3cos sin 1 , and 0 , find sin . (3 marks)
2 2 12
5. Prove by mathematical induction that 34n 2 52n 1 is divisible by 14 for any positive integer n.
(8 marks)
11
Suggested Answer
Part I Multiple choice questions.
Question Number Best Answer
1 B
2 C
3 E
4 B
5 D
6 A
7 A
8 D
9 D
10 A
11 B
12 E
13 C
14 C
15 E
(Answers for Part II start from next page)
12
Part II Problem-solving questions.
1. (a) The coefficients of the quadratic function satisfy the following system
25a 5b c 0

b 4a .

4a 2b c 9
Solving the equation system yields
a 1, b 4, c 5 .
The function is
f (x) x2 4x 5 .
(b) y (x 1)2 6 x2 2x 5 .
(c) We have g(x) (3sin x)2 4(3sin x) 5 (3sin x 2)2 9 .
Since 3 3sin x 3, 2 [ 3,3], the function obtains its maximum value at 3sin x 3 , that is,
Max g(x) f ( 3) 16 .
The function attains its minimum value at 3sin x 2, that is,
Min g(x) f (2) 9 .
2. (a) Let an a1q
n 1 . As a1, 4a2,16a3 are in arithmetic progression, so 16a3 a1 8a2 , i.e.,
16a q21 a1 8a1q .
Since a1 1 , we have q
1 . Thus
4
n 1 n
a 1n , bn . 4 4n
(b) From the geometric sum formula, we get Sn
4 1 1n . 3 4
Since
T 1 2 3 4n 2 3 4
n 1 n , (1) 4 4 4 4 4n 1 4n
we have
4Tn 1
2 3 42 3
n
. (2) 4 4 4 4n 1
Subtracting (1) from (2) gives
3T 1 1 1 1 1 n 4 1 1n 2 n . 4 4 43 4n 1 4n 3 4n 4n
Hence
T 4n 1 1 nn . 9 4 3 4n
13
3. (a) a and b satisfy the system
a2 b2 2

a 2 .
8
2
16
b a2
1
This implies that a2 32 , b2 16 . Thus the equation is
x2 y2 1 .
16 32
2 y2
(b) Combining the equations of the line y k1x b and the ellipse
x 1 , we get
16 32
x2 (k1x b)
2
1.
16 32
That is, (2 k 2 21 )x 2k
2
1bx b 32 0 . Then the coordinates of M are
x k1b0 2 , y0
2b .
2 k1 2 k
2
1
Therefore, the slope of the line segment OM is
k 22 . k1
Consequently, k1k2 2 .
3 1
4. (a) 3 cos sin 2( cos sin ) 2(cos cos sin sin ) 2cos( ) .
2 2 6 6 6
It follows from the above that 2 3cos sin 2 .

(b) From the result of (a), we get 3 cos sin 2cos( ) .
6
1
Since 3 cos sin 1, so cos( ) .
6 2
2 4 7
Since 0 2 , we have or , that is or .
6 3 3 2 6
1 1 1(c) Since 3 cos sin , therefore 2cos( ) , that is cos( ) .
2 6 2 6 4
2 1 5By the double-angle formula, we have 1 2sin ( ) , so sin2 ( ) . And since
2 12 4 2 12 8
0 , we get 0 . 2 12
Therefore
5 10
sin( ) .
2 12 8 4
14
5. Proof: Let S(n) be the statement “34n 2 52n 1 is divisible by 14”.
(1). When n 1 ,
34+2+52+1=36+53=729+125=854 , 854 14 16 .
Therefore, S(1) is true.
(2). Assume that S(n) is true for n k (k ) , i.e., 34k 2 52k 1 is divisible by 14, where k .
So when n k 1(k ) ,
34(k 1) 2 52(k 1) 1
34k 4 2 52k 2 1
34 34k 2 52 52k 1
81 34k 2 25 52k 1
56 34k 2 25 34k 2 25 52k 1
=56 34k 2 25（ 34k 2 52k 1）,
Where 56 34k 2 is divisible by 14, and according to the assumption, 25 (34k 2 52k 1) is also
divisible by 14.
In other words, S(k 1) is also true.
By (1), (2) and the Principle of Mathematical Induction, the statement is true for all positive integers
n.
15