PDF《2015/2016 学年 入学考试试题》

P a s t E x a m i n a t i o n P a p e r s

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74 2015/2016 学年 入学考试试题 第一部分: 七题全部作答.

1. (a) 求在

4 2

4 )

1 ( )

1 ( x x ? ? 的展开式中 x5 的系数. (2 分) (b) 设≥2为整数.在nx)

1 ( ? 的展开式中,

4 x 的系数是

2 x 的系数的

6 倍, 求n的值. (4 分) 2. 设函数 )

6 8 ln(

1 ) (

2 ? ? ? ? x x x f . (a) 求fx()的定义域. (4 分) (b) 求fx()的值域. (4 分) 3. (a) 用数学归纳法,证明对任意正整数 n,有22333)1(4121?????nnn?.(4 分) (b) 用(a) 的结果,对任意正整数 n,求以下和的公式 :

3 3

1 3

3 3

3 )

2 ( )

1 (

4 3

2 1 n r r ? ? ? ? ? ? ? ? ? ? ? . (4 分) 4. 设A和B为实数,且32? ? ? B A 及BABAcos cos

2 2 cos cos ? ? ? . (a) 证明 ) cos(

2 2

2 2 cos B A B A ? ? ? ? . (3 分) (b) 求2cos B A? 的值. (4 分) 5. 解不等式

4 3

2 2

2 5

4 ? ? ? ? ? x x x . (8 分) S y l l a b u s

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75 6. 设??1 } { n n a 为一等差数列,其公差为负数.设91?a及1a、12?a、34a 为一等比数列的连续三项. (a) 求此等差数列的通项 n a . (4 分) (b) 对任意正整数 n,求||||||21naaa????.(4 分) 7. 盒子中有

9 张卡片,红色、蓝色、白色各

3 张,每种颜色的卡片分别标号 为

1、

2、3. (a) 从盒中随机抽出卡片

3 张,求这

3 张卡片的标号之和小於

5 的概率.(3 分) (b) 从盒中随机抽出卡片

2 张,求这

2 张卡片颜色不同且标号之和 小於

4 的概率. (4 分) P a s t E x a m i n a t i o n P a p e r s

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76 第二部分:任择三题作答,每题十六分. 8. 如图 I,在四棱锥 P-ABCD 中,底ABCD 是菱形,且∠=.PDC 是边长 为2的等边三角形,且与 ABCD 垂直.设M为PB 的中点. (a) 求PA 与ABCD 所成角的大小. (6 分) (b) (i) 设N为PA 的中点及 E 为CD 的中点.证明 MNEC 为一长方形, 且与 PA 垂直. (6 分) (ii) 求三棱锥 P-DMC 的体积. (4 分) 9. (a) 一正圆柱的底半径和高之和为

36 cm.设此正圆柱的底半径为 x cm 及 体积为 ) (x V cm3 ,其中 ) (x V 为x的函数. (i) 求)(x V . (2 分) (ii) 求)('

xV及)('

'

xV.(2 分) (iii) 绘出曲线 ) (x V y ? .图中给出 ) (x V 的局部极大点、局部极小点和拐点. (7 分) (b) 求由曲线 x x y ? ?

2 和直线 x y ? ?

3 所包围的区域的面积. (5 分) D A B P 图ICMSyllabus2016/201777 10. 已知圆 C: + ?

2 +

4 ?

4 =

0 .设直线 ∶ = + 与C交於不 同的两点 ( , ) 及(,). (a) 求C的圆心及半径. (2 分) (b) (i) 证明 和 满足方程

2 + (2 + 2) + ( +

4 ? 4) = 0. (1 分) (ii) 以b表+、x x 、y + y 和.(4 分) (c) 求A与B的距离,答案以 b 表示. (5 分) (d) 若以线段 AB 为直径的圆通过原点 O,求b的值. (4 分) 11. 设1??i.(a) 设复数 z 满足

4 +

2 ? = 3√3 + . (i) 求z,并按极式 ) sin (cos ? ? i r ? 表示,其中 ? ? ? ? ? ? . (5 分) (ii) 设=sin + cos , ? ? ? ? ? ? .求|?|的取值围.(4 分) (b) 设复数 w 满足 = 1, ≠

1 及w的辐角为 . (i) 证明

1 0. (1 分) (ii) 证明对任意正整数 n, + =

2 cos( ). (3 分) (iii) 求cos + cos(2 ) + cos(3 ) 的值. (3 分) P a s t E x a m i n a t i o n P a p e r s

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78 12. (a) 因式分解行列式

2 2

2 2

2 2

2 2

2 )

2 ( )

2 ( )

2 ( )

1 ( )

1 ( )

1 ( c b a c b a c b a ? ? ? ? ? ? . (6 分) (b) 给出以 x、y 和z为未知量的方程组 (E): ? ? ? ? ? ? ? ? ? ? ? ? ? ?

1 1 ) ( z py px q pz y px pz py x E . (i) 求p的值使得 (E) 有唯一解. (4 分) (ii) 对使得 (E) 有多於一个解的 p 及q的值,求(E) 的通解. (6 分) 全卷完 S y l l a b u s

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79 2015/2016 ADMISSION EXAMINATION PAPER Section I. Answer all

7 questions. 1. (a) In the expansion of

4 2

4 )

1 ( )

1 ( x x ? ? , find the coefficient of

5 x . (2 marks) (b) Let n ≥

2 be an integer. In the expansion of n x)

1 ( ? , the coefficient of

4 x is

6 times of the coefficient of

2 x . Find the value of n. (4 marks) 2. Let )

6 8 ln(

1 ) (

2 ? ? ? ? x x x f be a function. (a) Find the domain of ) (x f . (4 marks) (b) Find the range of f x ( ) . (4 marks) 3. (a) Use mathematical induction to show that for any positive integer n,

2 2

3 3

3 )

1 (

4 1

2 1 ? ? ? ? ? n n n ? . (4 marks) (b) Using the result of (a), for any positive integer n, find a formula for the following sum:

3 3

1 3

3 3

3 )

2 ( )

1 (

4 3

2 1 n r r ? ? ? ? ? ? ? ? ? ? ? . (4 marks) 4. Let A and B be real numbers such that

3 2? ? ? B A and B A B A cos cos

2 2 cos cos ? ? ? . (a) Show that ) cos(

2 2

2 2 cos B A B A ? ? ? ? . (3 marks) (b) Find the value of .

2 cos B A? (4 marks) 5. Solve the inequality .

4 3

2 2

2 5

4 ? ? ? ? ? x x x (8 marks) P a s t E x a m i n a t i o n P a p e r s

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80 6. Let ? ?1 } { n n a be an arithmetic progression with negative common difference. Suppose

9 1 ? a and

1 a ,

1 2 ? a ,

3 4a are in geometric progression. (a) Find the general term n a of this arithmetic progression. (4 marks) (b) For any positive integer n, find | | | | | |

2 1 n a a a ? ? ? ? . (4 marks) 7. A box contains

9 cards. Among them,

3 cards are red,

3 cards are blue and

3 cards are white. The cards of each colour are labeled 1, 2, 3. (a) Three cards are drawn randomly from the box. Find the probability that the sum of the numbers on these three cards is less than 5. (3 marks) (b) Two cards are drawn randomly from the box. Find the probability that the two cardshave different colours and the sum of the numbers on these two cards is less than 4. (4 marks) S y l l a b u s

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81 Section II Answer any three questions. Each question carries

16 marks. 8. In Figure I, P-ABCD is a pyramid. Base ABCD is a rhombus with ∠ = . PDC is an equilateral triangle with side length

2 and is perpendicular to ABCD. Let M be the midpoint of PB. (a) Find the angle between PA and ABCD. (6 marks) (b) (i) Let N be the midpoint of PA and E be the midpoint of CD. Show that MNEC is a rectangle and is perpendicular to PA. (6 marks) (ii) Find the volume of the triangular pyramid P-DMC. (4 marks) 9. (a) The sum of the base radius and the height of a circular cylinder is

36 cm. Suppose the base radius is x cm and the volume is ) (x V cm3 , where ) (x V is a function in x. (i) Find ) (x V . (2 marks) (ii) Find ) ( '

x V and ) ( '

'

x V . (2 marks) (iii) Sketch the curve ) (x V y ? . In the graph, give the local maximum points, local minimum points and inflection points of ) (x V . (7 marks) (b) Find the area of the region bounded by the curve x x y ? ?

2 and the line x y ? ?

3 . (5 marks) P a s t E x a m i n a t i o n P a p e r s

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82 10. Given the circle C : + ?

2 +

4 ?

4 = 0. Suppose that the straight line ∶ = + intersects with C at two distinct points ( , ) and ( , ). (a) Find the center and radius of C. (2 marks) (b) (i) Show that x and x satisfy the equation

2 + (2 + 2) + ( +

4 ? 4) = 0. (1 marks) (ii) Express + , , y + y and in terms of b. (4 marks) (c) Find the distance between A and B. Give your answer in terms of b. (5 marks) (d) Suppose that the circle having the segment AB as a diameter passes through the origin O. Find the value(s) of b. (4 marks) 11. Let

1 ? ? i . (a) Suppose z is a complex number and satisfies 4z + 2z = 3√3 + i . (i) Find z, and express z in polar form ), sin (cos ? ? i r ? . ? ? ? ? ? ? (5 marks) (ii) Let u = sin θ + i cos θ, ? ? ? ? ? ? . Find the range of | ? |. (4 marks) (b) Suppose w is a complex number and satisfies w = 1, ≠ 1. Let β be the argument of w. (i) Show that

1 + w + w + w + w + w + w = 0. (1 marks) (ii) Show that for any positive integer n, w + w- = 2cos(nβ). (3 marks) (iii) Find the value of cos β + cos(2β) + cos(3β). (3 marks) S y l l a b u s

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83 12. (a) Factorize the determinant

2 2

2 2

2 2

2 2

2 )

2 ( )

2 ( )

2 ( )

1 ( )

1 ( )

1 ( c b a c b a c b a ? ? ? ? ? ? . (6 marks) (b) Given a system of equation (E) with unknowns x, y and z: ? ? ? ? ? ? ? ? ? ? ? ? ? ?

1 1 ) ( z py px q pz y px pz py x E . (i) Find the values 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. (6 marks) End of Paper P a s t E x a m i n a t i o n P a p e r s

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84 2015/2016 学年 参考答案 MODEL ANSWER 1. (a)

40 (b) =

11 2. (a) { :

1 <

<

7} (b) { : >

√ } 3. (a) For LHS =

1 = RHS. Suppose the result is true for . k n ? Then . )

2 ( )

1 (

4 1 )

1 ( )

1 (

4 1 )

1 (

1 2

2 3

2 2

3 3

3 ? ? ? ? ? ? ? ? ? ? ? k k k k k k k ? Hence the result is true for .

1 ? ? k n By the principle of mathematical induction, the result is true for all positive integers. (b) )

3 4 (

2 ? ? n n 4. (a) ) cos(

2 2

2 2 cos B A B A ? ? ? ? (b)

2 2 5.

3 log

1 2 ? ? x or

7 log2 ? x 6. (a) n an

4 13 ? ? (b) Write . | | | |

1 n n a a S ? ? ? ? Then ,

9 1 ? S ,

14 2 ? S ,

15 3 ? S and

30 11

2 2 ? ? ? n n Sn for

4 ? n . 7. (a)

42 5 (b)

4 1 8. (a) Given PDC ? is equilateral and so ⊥ . With ⊥ , we get ⊥ . Hence, the angle between PA and ABCD is ∠ . From PDC ? , | | =

2 sin = √3. From 2| || | cos = 3. Hence,3 . Thus, ∠ = tan | | | | = . S y l l a b u s

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85 (b) (i) In , DEA ?

2 2

2 2

2 2 | |

2 )

3 (

1 | | | | AD AE DE ? ? ? ? ? and so ∠ = . Thus, ⊥ . Together with ⊥ , we have ⊥ ? and hence ∠ = . In ? , as M and N are midpoints of PB and PA, respectively, we get ∥ . In rhombus , ∥ . Hence ∥ . In ? , we have 1. Together with | | = 1, we have that MNEC is a rectangle. From (a)i.e., ? is isosceles. As N is the midpoint of PA, we have ⊥ . From above, ∥ and ⊥ ? , we have ⊥ . Thus, ⊥ . (ii) We have sin ∠ = √3 sin = √ . Hence the area of ? is | || | = √ . We also have cos ∠ = √3 cos = √ . Hence the volume is * (area of 9. (a) (i) )

36 ( ) (

3 2 x x x V ? ? ? , .

36 0 ? ? x (ii) )

24 (

3 ) (

2 x x x V ? ? ? ? , )

12 (

6 ) ( x x V ? ? ? ? ? (iii) The curve ) (x V y ? attains a local maximum point at

24 ? x , and an inflection point at

12 ? x . It is increasing on [0,24], decreasing on [24,36], convex on [0,12] and concave on [12,36]. (b)

3 32 ) ( )

3 (

1 3

2 ? ? ? ? ?? dx x x x 10. (a) (i) Center is (1,-2), radius is

3 (b) (i) Putting b x y ? ? in the equation

0 4

4 2

2 2 ? ? ? ? ? y x y x , the result follows. (ii) ,

1 2

1 ? ? ? ? b x x

2 4

4 2

2 1 ? ? ? b b x x , ,

1 ) ( ) (

2 1

2 1 ? ? ? ? ? ? ? b b x b x y y .

2 4

2 ) )( (

2 2

1 2

1 ? ? ? ? ? ? b b b x b x y y P a s t E x a m i n a t i o n P a p e r s

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86 (c)

2 2

1 2

1 2

1 2

1 2

1 2

1 2

2 1

1 2

1 2

1 2

2 2

2 2

1 2

1 2

2 1

2 2

1 2

12 18

2 2

8 ) (

4 ) (

2 2

2 )

4 4

2 ( )

4 4

2 (

2 2 ) ( ) ( ) ( ) ( b b y y x x y y x x y ........