1415&1516數學A

74

2015/2016 學年 入學考試試題

第一部分: 七題全部作答。 1. (a) 求在 _(1x)4_(1x2₎4 _{的展開式中 x}5 _{的係數。 (2 分)} (b) 設 ≥ 2 為整數。在 _(1x)n _{的展開式中，x 的係數是}4 _{x 的係數的 6}2 倍， 求 n 的值。 (4 分) 2. 設函數 ) 6 8 ln( 1 ) ( 2_{ }   x x x f 。 (a) 求 f x( ) 的定義域。 (4 分) (b) 求 f x( ) 的值域。 (4 分) 3. (a) 用數學歸納法，證明對任意正整數 n，有 2 2 3 3 3 _{( 1)} 4 1 2 1  _n  n n 。 (4 分) (b) 用 (a) 的結果，對任意正整數 n，求以下和的公式 : 3 3 1 3 3 3 3 _{2 3 4 ( 1) (2 )} 1 _{     } r r _{ } n   。 (4 分) 4. 設 A 和 B 為實數，且 3 2   B

A 及 cosAcosB2 2cosAcosB。

(a) 證明 2cos( ) 2 2 2 cosAB   AB 。 (3 分) (b) 求 2 cosAB的值。 (4 分) 5. 解不等式 4 3 2 2 2 5 4      x x x 。 (8 分)

75 6. 設 {a_n}_n1 為一等差數列，其公差為負數。設 a₁9 _及 a₁、a₂1、4a₃ 為一等比數列的連續三項。 (a) 求此等差數列的通項 an。 (4 分) (b) 對任意正整數 n，求 |a₁||a₂|_|a_n |。 (4 分) 7. 盒子中有 9 張卡片，紅色、藍色、白色各 3 張，每種顏色的卡片分別標號 為 1、2、3。 (a) 從盒中隨機抽出卡片 3 張，求這 3 張卡片的標號之和小於 5 的概率。(3 分) (b) 從盒中隨機抽出卡片 2 張，求這 2 張卡片顏色不同且標號之和 小於 4 的概率。 (4 分)

76

第二部分:任擇三題作答，每題十六分。 8.

如圖I，在四棱錐 P-ABCD 中，底 ABCD 是菱形，且 ∠ = 。PDC 是邊長 為2 的等邊三角形，且與 ABCD 垂直。設 M 為 PB 的中點。 (a) 求 PA 與 ABCD 所成角的大小。 (6 分) (b) (i) 設 N 為 PA 的中點及 E 為 CD 的中點。證明 MNEC 為一長方形， 且與 PA 垂直。 (6 分) (ii) 求三棱錐 P-DMC 的體積。 (4 分) 9. (a) 一正圓柱的底半徑和高之和為 36 cm。設此正圓柱的底半徑為 x cm 及 體積為 V( x) cm3_{，其中 V( x) 為 x 的函數。} (i) 求 V( x)。 (2 分) (ii) 求 V' x( ) _及 V '' x( )。 (2 分) (iii) 繪出曲線yV( x)。圖中給出 V( x) 的局部極大點、局部極小點和拐點。 (7 分) (b) 求由曲線yx2x 和直線 y 3 x 所包圍的區域的面積。 (5 分) D A B P 圖I C M

77 10. 已知圓 C: + − 2 + 4 − 4 = 0 。設直線 ∶ = + 與 C 交於不 同的兩點 ( , ) 及 ( , )。 (a) 求 C 的圓心及半徑。 (2 分) (b) (i) 證明 和 滿足方程 2 + (2 + 2) + ( + 4 − 4) = 0。 (1 分) (ii) 以 b 表 + 、x x 、y + y 和 。 (4 分) (c) 求 A 與 B 的距離，答案以 b 表示。 (5 分) (d) 若以線段 AB 為直徑的圓通過原點 O，求 b 的值。 (4 分) 11. 設 i 1。 (a) 設複數 z 滿足 4 + 2 ̅ = 3√3 + 。

(i) 求 z，並按極式 r(cosisin) 表示，其中    。 (5 分) (ii) 設 = sin + cos ，  。求 | − | 的取值範圍。(4 分) (b) 設複數 w 滿足 = 1， ≠ 1 及 w 的輻角為 。

(i) 證明 1 + + + + + + = 0。 (1 分)

(ii) 證明對任意正整數 n， + = 2 cos( )。 (3 分) (iii) 求 cos + cos(2 ) + cos(3 ) 的值。 (3 分)

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) 給出以 x、y 和 z 為未知量的方程組 (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) 對使得 (E) 有多於一個解的 p 及 q 的值，求 (E) 的通解。 (6 分) 全卷完

79

2015/2016 ADMISSION EXAMINATION PAPER

Section I. Answer all 7 questions.

1. (a) In the expansion of _(1x)4_(1x2₎4_{, find the coefficient of x .} 5 _{(2 marks)}

(b) Let n ≥ 2 be an integer. In the expansion of _(1x)n_{, the coefficient of} 4

x is

6 times of the coefficient of _x2_{. 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 f(x). (4 marks)

(b) Find the range of f x( ). (4 marks)

3. (a) Use mathematical induction to show that for any positive integer n, 3 3 3 2_{( 1)}2

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 3 4 ( 1) (2 )} 1 _{     } r r _{ } n   . (4 marks)

4. Let A and B be real numbers such that

3 2   B

A and cosAcosB2 2cosAcosB_.

(a) Show that 2cos( )

2 2 2

cosAB   AB . (3 marks)

(b) Find the value of . 2

cosAB (4 marks)

5. Solve the inequality 4. 3 2 2 2 5 4 _     x x x (8 marks)

80

6. Let {a_n}_n1 be an arithmetic progression with negative common difference. Suppose 9

1

a and a1,a21,4a3 are in geometric progression.

(a) Find the general term an of this arithmetic progression. (4 marks)

(b) For any positive integer n, find |a₁||a₂|_|a_n |. (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

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 V( x) cm3_{, where V( x)}

is a function in x.

(i) Find V( x). (2 marks)

(ii) Find V' x( ) _and V '' x( ). (2 marks) (iii) Sketch the curve yV( x). In the graph, give the local maximum

points, local minimum points and inflection points of V( x). (7 marks) (b) Find the area of the region bounded by the curve yx2x

and the line

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 i 1.

(a) Suppose z is a complex number and satisfies 4z + 2z = 3√3 + i .

(i) Find z, and express z in polar form r(cosisin),   . (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)}

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)

84

2015/2016 學年 參考答案 MODEL ANSWER

1. (a) 40 (b) = 11 2. (a) { : 1 < < 7} (b) { : > _√ } 3. (a) For LHS = 1 = RHS.

Suppose the result is true for nk. Then

( 1) ( 2) . 4 1 ) 1 ( ) 1 ( 4 1 ) 1 ( 1 3_{ k}3_{ k} 3 _{ k}2 _k 2_{ k} 3 _{ k} 2 _k 2 

Hence the result is true for n k 1. By the principle of mathematical induction, the result is true for all positive integers.

(b) _n2(4_n3) 4. (a) 2cos( ) 2 2 2 cosAB   AB (b) 2 2 5. 1 xlog₂3 or xlog₂7 6. (a) a_n 134n

(b) Write S_n | a₁|_|a_n |. Then S₁9, S₂ 14, S₃ 15, and _{S }2_n2_11_n30 n for n4. 7. (a) 42 5 (b) 4 1

8. (a) Given PDCis 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 |_| |_|= .

85 (b) (i) In DEA, _|DE|2 _|AE|2_12 _{( 3)}2 _22 _|AD|2 _{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) _V(_x)_(36_x2_x3)_{, 0 x36.}

(ii) _V(_x)_3_(24_xx2)_{, V(x)6(12x)}

(iii) The curveyV(x)attains a local maximum point at x24, and an inflection point at x12. 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_{ }  



 x x x dx

10. (a) (i) Center is (1,-2), radius is 3

(b) (i) Putting yxb in the equation _x2 _{ y}2 _2_x4_y4_0_{, the result}

follows. (ii) x₁x₂ b1, 2 4 4 2 2 1   b b x x , y₁ y₂ (x₁b)(x₂b)b1, ( )( ) ₂2 4. 2 2 1 2 1       x b x b b b y y

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 y x x y x y x y y x x y x y x y y x x AB                             (d) We have OAOB and so 1 2 2 1 1 _{ } x y x y , i.e., x₁x₂ yy_{1 2} 0 Thus, 0. 2 4 2 2 4 4 2 2       b b b b Solving, we get b = 1 or b4. 11. (a) (i) 6 sin 6 cos 2 1 2 3   i i z   

(ii) We have (cos sin ) (sin cos )2 2 2sin( ₆) 6 2 6    _{        }   w z and thus 0 zw 2. (b) (i) As w1, . 0 1 0 ) 1 )( 1 ( 0 1 6 5 6 5 7 _{            }   w w w w w w (ii) As _w7 _1_{, we get} 7 ₁  w and hence w 1.

Let wcos isin. Then wn _cos(n)_isin(n) _and

also wn _cos(n)_isin(n)_{. The result follows.}

(iii) Note that as _w7 _1, _{we have w}6 _{ w}1_{, w}5 _{ w}2 _{and w}4 _{ w}3_{. Thus,}

2 1 3 3 2 2 1 5 6 ) 3 cos( ) 2 cos( cos 1 ) 3 cos( 2 ) 2 cos( 2 cos 2 1 ) ( ) ( ) ( 0 1                                w w w w w w w w _ 12. (a) 4(ab)(ac)(bc) (b) (i) We need 0. 1 1 1  p p p p p p Thus, p1 and . 2 1   p

87 (ii) When p1, we must have q1 and the system of equations becomes x yz1. The general solution is x1st, ys, zt where s

and t are any real numbers. When _p21_{, we must have q}_{2 (by}

considering the sum of all the three equations). The system of equations becomes            2 1 2 1 2 1 2 1 2 1 z y x z y x

88

2014/2015 學年 入學考試試題

第一部分: 七題全部作答。 1. 解 logx274log9 x 1。 (6 分) 2. 設 1 1 ) ( ₂ 2    _xx e e x f 。 (a) 證明 f(x) 在定義域上是增函數。 (2 分) (b) 求 f x( ) 的值域。 (3 分) (c) 求 _      2 ) ( arccos f x 的值域。 (3 分) 3. 設 {a_n}_n1 為一正項數列，使得對任意正整數 n，有 1 4 ) 1 ( 2 1      n n a a a _ 。 用數學歸納法，證明對任意正整數n，有 an  n2 1。 (8 分) 4. (a) 設 k 為正整數，求 ) ] 4 1 3 tan[( ] ) 6 1 cos[(k  k  的值。 (3 分) (b) 求 3sin2xcos2x 2 的通解。 (5 分) 5. 解不等式 | − 4 + 2| < 1。 (6 分) 6. 設 {an}n1 為一等比數列，其公比 r1，且 a1 a2 a3 21 及 64 3 2 1a a  a _。 (a) 求此等比數列的通項 a 。 _n (5 分)

89 (b) 求最小的 m 使得 a1a2 am654321。 (3 分) 7. 一袋内有红球 6 個及白球 4 個。 (a) 從袋中隨機逐一把球抽出，直至 4 個白球都被抽出。問第 8 次抽球剛好 把 4 個白球都抽出的概率是多少? (4 分) (b) 從袋中隨機逐一把球抽出，當中若抽出的是红球，就馬上把它放回袋內， 然後再繼續抽球，直至 4 個白球都被抽出。問第 5 次抽球剛好把 4 個 白球都抽出的概率是多少? (4 分) 第二部分:任擇三題作答，每題十六分。 8. (a)

如圖I，在三棱錐 AEXG 中，AEEXG，P 為 AX 上一點，F 為 P 在 XE

的垂足。

(i) 證明 PF EXG 。 (2 分)

(ii) 設平面 GAX 與平面 GEX 所夾的二面角為 α。證明

∆ = ∆ cos ， 其中 ∆ 及 ∆ 分別 代表 ∆ 及 ∆ 的面積。 (7 分) X A F P 圖I E G

90 (b) 如圖II，ABCD-EFGH 為一正立方體，P 為 BF 的中點。用 (a)(ii) 的結果， 或用其他方法，求平面GAP 與平面 GEF 所夾的二面角。 (7 分) 9. (a) 已知函數 _f(_x)_{ x}3 _6_x2 _9_x2_。 (i) 求 f (x) 及 f (x)。 ( 2 分 ) (ii) 求 f(x) 的局部極大點、局部極小點和拐點。 ( 4 分 ) (iii) 繪出曲線y f( x)。 ( 2 分 ) (iv) 繪出曲線 y f(x1)。 ( 2 分 ) (b) 求在第一、二像限內由 x-軸，直線 yx 及曲線y6 x 2所包圍的區域 之面積。 ( 6 分 ) 10. 已知拋物線 _P: _y2 _{ xa}( _1) , > 0, 與直線 L:x yb 交於不同的兩點 ) , (x1 y1 H 和 K(x2,y2)。 (a) (i) 求以 x1 和 x2 為根的二次方程。 (1 分) (ii) 證明 a b4 40。 (2 分) (iii) 證明 x x b2 a 2 1 及 y1y2 aab。 (4 分) A B D C 圖II P F E H G

91 (b) 設直線 OH 與直線 OK 互相垂直，其中 O 為原點。 (i) 證明 2 2   b b a 。 (3 分) (ii) 若直線 L 與原點 O 的距離為 a，求 b。 (6 分) 11. 設 i 1及 − ≤ ≤ 。

(a) (i) 按極式 r(cosisin) 表示 1 + cos + sin 。 (4 分) (ii) 由此，簡化     sin cos ) sin cos 1 ( 8 i i    為極式。 (3 分) (b) (i) 用棣美弗定理，或其他方法，證明

cos7 _64cos7_112cos5_56cos3 _7cos_{。 (3 分)} (ii) 用 (i) 的結果，證明 64 − 112 + 56 − 7 = 0 的根為 14 5 cos , 14 3 cos , 14 cos      。 (3 分) (iii) 由 (ii) 的結果，求 14 5 cos 14 3 cos 14 cos   的值。 (3 分)

92 12. (a) 因式分解行列式 ab b a c ca a c b bc c b a 2 2 2 2 2 2 ) ( ) ( ) (    。 (7 分) (b) (i) 求以 X, Y, Z 為變量的方程組           c X Y b X Z a Y Z 的解。 (3 分) (ii) 設 a, b, c 為正數。證明以 x, y, z 為變量的方程組           c yz xz b yz xy a xz xy 有解當且僅當 ( + − ) > 0, ( + − ) > 0, ( + − ) > 0。 (6 分) 全卷完

93

2014/2015 ADMISSION EXAMINATION PAPER

Section I. Answer all 7 questions.

1. Solve log_x274log₉ x1. (6 marks)

2. Let 1 1 ) ( ₂ 2    _xx e e x f .

(a) Show that f(x) is an increasing function on its domain. (2 marks) (b) Find the range of f x( ). (3 marks)

(c) Find the range of _      2 ) ( arccos f x . (3 marks)

3. Let {an}n1 be a sequence of positive numbers such that for any positive integer n,

1 4 ) 1 ( 2 1      n n a a

a _ . Use mathematical induction to show that, for any positive integer n, a_n  n2 1. (8 marks)

4. (a) Let k be a positive integer. Find the value of ] ) 4 1 3 tan[( ] ) 6 1 cos[(k  k  . (3 marks)

(b) Find the general solution of 3sin2xcos2x 2. (5 marks)

94

6. Let {a_n}_n1 be a geometric sequence with common ratio r1. Suppose 21

3 2

1a a 

a and a1a2a3 64_.

(a) Find the general term a of _n the sequence. (5 marks) (b) Find the smallest m such that a₁a₂ _a_m 654321. (3 marks)

7. A bag contains 6 red balls and 4 white balls.

(a) Balls are drawn randomly one by one from the bag until all the 4 white balls are drawn. What is the probability that all the four white balls are drawn at the 8th _draw?

(4 marks)

(b) Balls are drawn randomly one by one from the bag, until all the 4 white balls are drawn. If a red ball is drawn, it is returned to the bag before the next draw. What is the probability that all the four white balls are drawn at the 5th _draw?

95 Section II Answer any three question. Each question carries 16 marks.

8. (a)

In Figure I, AEXG is a triangular pyramid with AEEXG. P is a point on AX,

F is the foot of perpendicular from P to XE.

(i) Show that PF EXG . (2 marks) (ii) Suppose the angle between the planes GAX and GEX is α. Show that ∆ = ∆ cos , where ∆ and ∆ denote the area of ∆

and the area of ∆ , respectively. (7 marks) (b)

In Figure II, ABCD-EFGH is a cube. P is the mid-point of BF. Using the result in (a)(ii),

or otherwise, find the angle between the planes GAP and GEF. (7 marks)

X A F P Figure I E G A B D C Figure II P F E H G

96

9. (a) Given function _f(_x)_{ x}3 _6_x2 _9_x2_.

(i) Find f (x) and f (x). (2 marks) (ii) Find the local maximum, local minimum and inflection points of f(x)

(4 marks)

(iii) Sketch the curve y f( x). (2 marks) (iv) Sketch the curve y f(x1). (2 marks) (b) Find the area of the region in the first and second quadrants bounded by the

x-axis, the line yx and the curve _{y 6 x} 2_{. (6 marks)}

10. Given that the parabola _P:_y2 _{ xa}( _1)_{, > 0, and the straight line L:x yb} intersect at two distinct pointsH(x1,y1) and K(x2,y2).

(a) (i) Find a quadratic equation whose roots are x1 and x2. (1 marks)

(ii) Show that a b4 40. (2 marks) (iii) Show that x x b2 a

2

1 and y1y2 aab. (4 marks)

(b) Suppose the lines OH and OK are perpendicular, where O is the origin. (i) Show that

2 2   b b a . (3 marks)

(ii) If the distance between the line L and the origin O is a, find b.

(6 marks)

11. Let i 1 and − ≤ ≤ .

(a) (i) Express 1 + cos + sin in polar form r(cos isin) (4 marks) (ii) Hence, simplify

    sin cos ) sin cos 1 ( 8 i i   

to polar form. (3 marks)

(b) (i) Using De Movire’s theorem, or otherwise, show that

97 (ii) Using the result in (i), show that the roots of

64 − 112 + 56 − 7 = 0 are 14 5 cos , 14 3 cos , 14 cos      . (3 marks)

(iii) Using the result in (ii), find the value of

14 5 cos 14 3 cos 14 cos    . (3 marks)

12. (a) Factorize the determinant

ab b a c ca a c b bc c b a 2 2 2 2 2 2 ) ( ) ( ) (    . (7 marks)

(b) (i) Find the solution of the system of equations           c X Y b X Z a Y Z in

variables X,Y, Z. (3 marks)

(ii) Suppose a, b, c are positive numbers. Show that system of equations

          c yz xz b yz xy a xz xy

in variables x, y, z has a solution if and only if

( + − ) > 0, ( + − ) > 0, ( + − ) > 0. (6 marks)

98

2014/2015 學年 參考答案 MODEL ANSWER

1. x3 or x332

2. (a) Suppose uv. Then 0

) 1 )( 1 ( ) ( 2 ) ( ) ( _{2 2} 2 2       _u u _vv e e e e v f u f . (b) }{y:1 y1 (c)       _{ } 2 3 : z  z 3. For ,n1 we have 1 4 ) 1 ( 2 1 1    a

a . Solving, since a₁ is positive,

1 ) 1 ( 2 3 1   a .

Thus the result is true for n1.

Suppose the result is true for n1,...,k. Then

0 ) 3 8 4 ( 2 1 4 ) 1 ( 2 ) 1 ( 2 1 4 ) 1 ( )] 1 2 ( 3 [ 1 4 ) 1 ( 2 1 2 1 2 1 1 2 1 1 2 1 1 1                                    k k a a a a k k k a a k a a a a k k k k k k k k k  

Solving, as a is positive, we get _k1 a_k12(k1)1. The result follows. 4. (a) 2 3 ) 1 (_ k1 (b) 12 8 ) 1 ( 2    _{ } k _ k , where k is an integer 5. 2 3x1 or 3 x2 3 6. (a) _4n1 n a (b) 11 7. (a) 6 1 (b) 88200 1207

8. (a) (i) In AXE , AE //PF because

2     PFX AEX . With EXG

99 (ii) Let M be the point on GX such thatAME . ThenAM GX ,

GX

EM  and ( cos) cos

2 1 2 1 AGX EGX EM GX AM GX S S_    _ .

Similarly, let N be the point on GX such thatPNF . Then GX PN  , FN GX and  ) cos cos ( 2 1 2 1 PGX FGX FN GX PN GX S S_    _ . Hence,

S_GEF S_EGX S_FGX S_AGX cos S_PGX cos S_GAPcos.

(b) Suppose the length of one side of the cube is a. Then, we can get 2

2 1 a S_GEF  and 2 8 3 a

S_GAP  . Hence, the required angle is

3 2 cos1 _.

9. (a) (i) _f(_x)_3_x2 _12_x9_{, f (x)6x12}

(ii) & (iii) The curvey f(x)attains a local maximum point at x1, a local minimum point at x3, and an inflection point at x2. (iv) Replace the curve y f(x), x1, by the mirror image of y f(x),

1 

x about the line x1. (b) 3 22 6 4 6 6 2 0 2 0 6 2 _{    } 





 x dx x xdx 10. (a) (i) _x2 _(_a2_b)_x(_b2_a)_0

(ii) The equation in (i) has real and distinct roots. Its discriminant is positive, i.e., [_(_a2_b)]2 _4(1)(_b2_a)_0_{. The result follows.}

(ii) Product of roots: x x b2 a b2 a

2 1 ₁ ; ab a a b b a b b x x x x b b x b x b y y                ) ( ) 2 ( ) ( ) )( ( 2 2 2 1 2 1 2 2 1 2 1

(b) (i) Suppose x₁x₂ 0. As OH OK, we have 1

2 1 2 1 _ x x y y . The result follows from

(a) (ii). If x₁x₂ 0, then by (a) (ii), _{a . Also, OH and OK are the b}2

two axes and so H or K is the point (1,0). Hence b1 and the result follows.

100

(b) (ii) Distance from O to L is

2 | | 1 1 | 0 0 | 2 2 b b     . Hence we have |b| 2a. Using (i), we get b22 2 if b0; and b22 2 if b0. 11. (a) (i) In the question,  should satisfy   .

Let 1cos isin r(cos isin) where r0. Then, 2 cos 2 ) cos 1 ( 2 sin ) cos 1 ( _{ } 2 _ 2_{    }   r (as 2 2 2    _{ }  )and 2 tan cos 2 cos sin 2 cos 1 sin tan 2 2 2 2       _     . Hence, 2    and thus       _    2 sin 2 cos 2 cos 2 sin cos 1  i    i  . (b) ) 5 sin 5 (cos 2 cos 2 ) sin( ) cos( ) 4 sin 4 (cos 2 cos 2 ) sin( ) cos( 2 sin 2 cos 2 cos 2 sin cos ) sin cos 1 ( 8 8 8 8 8 8                  i i i i i i i                     _    

(b) (i) By comparing the real parts of both sides of (cos _isin)7 _cos7 _isin7 _, the result follows.

(ii) Putting xcos , we have 0 cos and 0 7 cos 0 7 cos 56 cos 112 cos 64 0 7 56 112 64 2 4 6 2 4 6                  x x x For cos7 0, 14 7  

 k  , where k is an integer. Excluding those  such that cos 0, the result follows.

(iii) Product of roots is 64 7  and hence 64 7 14 5 cos 14 3 cos 14 cos2 2 2 _      .

101 12. (a) )_(_a2 _b2 _c2)(_abc)(_ab)(_bc)(_ca (b) (i) 2 a c b X    , 2 b c a Y    , 2 c b a Z    (ii) From (i), we have

2 a c b yz   , 2 b c a xz   , 2 c b a xy   . Suppose abc0, acb0 and bca0. Then, ) ( 2 ) )( ( ) ( ) )( ( 2 a c b b c a c b a yz xz xy x         .(*) Hence ) ( 2 ) )( ( a c b b c a c b a x        , (and similarly) ) ( 2 ) )( ( b c a a c b c b a y        and ) ( 2 ) )( ( c b a a c b b c a z        form a solution.

Conversely, if there is a solution then x, y, z are nonzero. From (*), either 0 or 2 of the factors on the right-hand side are negative. If there are 2 negative factors, say, abc0 and acb0, then 0 2 2         xy xz a b c a c b

a and this gives a contradiction. Hence all the 3 factors are positive.