QUEEN’S COLLEGE

Yearly Examination, 2006-2007

Pure Mathematics

Secondary 6E, SDate : 14/6/2007 Time : 8:30 - 11:30 am

Instructions :

(1) Answer ALLquestions in Section A and Section B.

(2) All workings must be clearly shown.

(3) Unless otherwise specified, numerical answers must be exact.

(4) This paper consists of 4 pages with 200 marks.

FORMULA FOR REFERENCE

Section A :Answer ALL questions in this section. (80 marks)

1.Let  1 be a complex root of x3– 1 = 0.

(a)Show that 2 +  + 1 = 0.

(b)Let f(x) = .

Show that x =  is a root of the equation f(x) = 0.

Hence factorize f(x) into two non-linear factors with integral coefficients.

(12 marks)

2.Let , ,  be the roots of the equation x3– 17x2 + kx – 125 = 0, where .

If , ,  are in geometric sequence, solve the given equation and also find the value of k.

(10 marks)

3.(a)Expand in ascending powers of x as far as the term in x3.

(b)By considering the derivative of a geometric series and using (a),

find n if the coefficients of x and x2 in the expression of

1 + 2(2x + 1) + 3(2x + 1)2 + … + n(2x + 1)n-1

are in the ratio of 1 : 6.

(12 marks)

4.(a)By putting x = i in the identity:

x4+ax3– x + 2 (x2 + 1) Q(x) + 2x + b, where Q(x) is a polynomial in x,

determine the numerical values of the real constants a and b.

(b)Find the remainder when x2007 + 1 is divided by (x2 + 1)(x + 1).

(12 marks)

5.A complex number, represented by z = x + y i with x, y R , may be visualized as a

2  2 matrix Z = .

(a)Verify that such matrix operations are valid for both addition and multiplication in correspondence to that of two complex numbers, z1 = x1 + y1i , z2 = x2 + y2i .

(b)Write down without proof, the matrix representation corresponding to (x + y i)-1 ?

(c)What is the complex number that can be multiplied to x + y i in order to rotate it counter-clockwisely by 90o with respect to the origin ? Write down also its corresponding matrix of rotation.

(12 marks)

6.Prove by mathematical induction that

is true  n N .

(10 marks)

7.(a)Prove that , k N .

(b)Use (a) to prove that , k N .

(c)Use (b) to deduce that :

, n N / {1} .

(12 marks)

Section B :Answer ALL questions in this section. (120 marks)

8.(a)By considering the function f(x) =(ai x– bi)2, prove the Cauchy-Schwarz Inequality:

“Let a1, a2, …., an ; b1, b2, …., bn be two sets of non-zero real numbers.

Equality sign holds iff . ”(14 marks)

(b)By (a), prove that :

(i) a, b, c > 0 such that a + b + c = 1,

(ii) a, b, c, x, y, z > 0(16 marks)

9.(a)By using de Moivre’s Theorem write in the form

, where a, b, c, d, e are constants.

(10 marks)

(b)Use (a) to find an equation with roots .

(10 marks)

(c)Use (b) to find the values of :

(i)

(ii)(10 marks)

10.(a)Factorize into 3 linear factors in terms of a, b.(6 marks)

(b)Use (a) to solve the system of equation:

, where a  1, b  1, a + b –1 .(12 marks)

(c)Use (a) to solve the equation: for x .

Give your general solutions in radian forms.(12 marks)

11.(a)For any two square matrices A and B of the same order, is it true that

(A + B)2 = A2 + 2AB + B2 ?

If yes, prove it. If no, disprove it by using a counterexample with matrices of order 2.

Tell also the condition on A and B when the equality holds.(7 marks)

(b)Show that, ifand

then (A + B)2 = A2 + 2AB + B2.(7 marks)

(c)Let

(i)Find P2, P3, P4 and guess a formula for Pn, where n N.

(ii)Prove your guess in (c)(i) by mathematical induction.

(iii)Use (b) to find the inverse of Pn .(16 marks)

< End of Paper >

Solution

1.(a)Since  is the complex root of 3– 1 = 0.3 = 1….(1)

2 +  + 1 = ,  1.(4M)

(b)f() = 7 + 6 + 4 + 22 + 1

=(3)2  + (3)2 + 3 + 22 + 1

= + 1 +  + 22 + 1

=2 (2 +  + 1)

=0, by (1).(4M)

By Factor Theorem, x =  is a root of the equation f(x) = 0.

By (a), f(x) has a factor x2 + x + 1 .

By division, f(x) = (x2 + x + 1) (x5– x3 + 2x2– x + 1) .(4A)

2.By Vieta’s theorem,(2M)

Since , ,  are in geometric sequence,  = 2….(4)(2M)

Subst. (4) in (3), 3 = 125,  = 5….(5)(2A)

Subst. (5) in (1), + 5 +  = 17,  = 12 –….(6)

Subst. (5), (6) in (4),(12 –) = 52, 2– 12 + 25 = 0

Solving, we get  = ( is rejected since )(1A)

From (6), = .(1A)

Subst. (5) in x3– 17x2 + kx – 125 = 0,53– 17(52) + k(5) – 125 = 0,

 k = 85(2A)

(Ans)(, , ) = (, 5, ) , k = 85.(2A)

3.(a)(2M)

(4A)

(b)1 + 2(2x + 1) + 3(2x + 1)2 + … + n(2x + 1)n-1

=(2M)

(2A)

3n – 6 = 24

n = 10.(2A)

4.(a)x4+ ax3– x + 2  (x2 + 1) Q(x) + 2x + b….(1)

Put x = i in (1),

1 – ai – i + 2 = 2i + b3 –(a + 1) i = 2i + b(3A)

a = –3, b = 3.(3A)

(b)x2007 + 1 = (x2 + 1)(x + 1) q(x) + Ax2 + Bx + C….(2)(2M)

Put x = –1 in (2), A – B + C = 0(1A)

Put x = i in (2), i2007 + 1 = A + Bi + C  i4501+3 + 1 = (C – A) + Bi(1M)

–i + 1 = (C – A) + Bi

C – A = 1….(3)

B = –1….(4)

Solving (2), (3), (4) (A, B, C) = (–1, –1, 0)

The remainder is – x2–x.(2A)

5.(a)z1 = x1 + y1i , z2 = x2 + y2i .

z1 + z2 = (x1 + x2 ) + (y1 + y2 ) i , z1 z2 = (x1 x2 – y1 y2) + (x1y1 + x2 y2 ) i(2A)

Z1 = ,Z2 =

Z1 + Z2 = + = (2A)

Z1 Z2 = =(2A)

(b)Z-1 =(2A)

(c)–y + x i = (0 + i) (x + y i)(2M)

(2A)

6.Let , n N .

Let P(n) be the proposition : un = vn .(2M)

For P(1),

 u1 = v1 and P(1) is true.(2A)

Assume P(k) is true for some k N , i.e. uk = vk….(*)(2A)

For P(k + 1),

,by (*)

= vk+1 .(4M)

P(k + 1) is true.

By the Principle of Mathematical Induction, P(n) is true  n N .

7.(a)(2M + 2M)

(b)By (a),(2M)

(c)By (b), put k = 2, 3, …, n ,(2M)

::

Adding the (n – 1) inequalities and add 1 to both sides, (2M)

.for correct cancellation(2A)

8.(a)(ai x - bi)2 0i = 1, 2, …, n

f(x) =xR(4M)

For the eq. f(x) = 0, we have  0and (for which the equality holds)(4M)

(2A)

Equality sign holds (ai x - bi)2 = 0,i = 1, 2, …, n

ai x - bi = 0,i = 1, 2, …, n

(4M)

(b)(i)Apply Cauchy-Schwarz Inequality to the numbers :

(3M)

(3A)

 [6(a + b + c) + 3]  3(3M)

= 27 since a + b + c = 1.

(ii)Apply Cauchy-Schwarz Inequality to the numbers :

(4M)

(3A)

9.(a) (3M)

sin 9= 9 cos8 sin – 84 cos6 sin3 + 126 cos4  sin5 – 36 cos2sin7+ sin9(2A)

= 9 cos8 - 84 cos6(1 – cos2) + 126 cos4  (1 – cos2 )2 - 36 cos2 (1 – cos2)3+ (1 – cos2)4

= 256 cos8– 448 cos6 + 240 cos4 – 40 cos2 + 1(2M) + (3A)

(b)For , 1  k  4, k N .

.(4M)

(2A)

 The equation with roots is

256 x4 – 448 x3 + 240 x2 – 40 x + 1 = 0(4A)

(c)(i)Sum of roots of the equation in (a) ,

(2M) +(2A)

(ii)Put y = 1/x in 256 x4 – 448 x3 + 240 x2 – 40 x + 1 = 0(2M)

are roots of y4– 40y3 + 240 y2– 448 y + 256 = 0

Sum of roots = =40(2A)

= 40 – 4 = 36(2A)

10.(a)(3M)

= (3A)

(b)

(2M)

(2M)

(2M)

x = (2A)

y = (2A)

z = (2A)

(c)(sin x – sin 2x)(sin x – sin 3x)(sin x + sin 2x + sin 3x) = 0(2M)

(i)sin x – sin 2x = 0  sin (1 – 2 cos x) = 0

 x = n or x = , where n Z .(2A)

(ii)sin x – sin 3x = 0 - 2 sin x cos 2x = 0

 x = n or x = , where n Z .(2A)

(iii)sin x + sin 2x + sin 3x = 0(2M)

sin 2x = 0 or

 x = or x = (2A)

Hence the roots of the equations are: , , ,

where n Z .(2A)

11.(a)Not true. (2A)

Counterexample: We need to find any two matrices not commune, AB  BA .

(3A)

Condition for equality:

(A + B)2 = (A + B) (A + B) = A2 + AB + BA + B2 = A2 + 2AB + B2

or AB = BA(2A)

(b)By (a), it is sufficient to show that AB = BA.(3M)

AB = (2A)

BA = (2A)

So AB = BA .

(c) (i)P2 = (1A)

P3 = (1A)

P4 = (1A)

We guess that n N.(2A)

(ii)Let P(n) be the proposition in (i) above,

P(1) is obviously true.(1A)

Assume P(k) is true for some kN, (2M)

For P(k + 1),

Pk+1 = PkP = (2M)

(1A)

P(k + 1) is true .

By the Principle of mathematical induction, P(n) is true n N.

(iii)In (b), take A = Pn, then B is the inverse of A . i.e. B = (Pn)-1.

 a = and a – n2 + b = 0  b = (2A)

 (Pn)-1 = (3A)

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