2006-2007 F.6 Pure Mathematics Final Examination

Marking Scheme

1.(a)Putting x= 1 in (1+x)2n+1 = .

22n+1 =

Since

1M+1A

(b)=

On the other hand,

=

For mnkm, consider the coefficient of xk, we have =.

1M+1A

2.(a)(i)=

= 1M+1A

(ii)Differentiate the above expression with respect to x, we have

(b)

= 1M

= 2A

HKALE 2004 I Q3

3.(a)(i)R = 1A

R6 = 1A

(ii)∵det A =  0

A1 exists and A1= 1A

A1RA=

= in which all elements are integers.1A

(b)By (a)(ii) we have

= (A1RA)6

= A1R6A = A1A = I(by (a)(i))1A

ChooseM = n which all the elements are integers and

M3 = = I

Here M = =

1M+1A

HKALE 1997 I Q7

4.(a)(det A)[ det(A1xI) ]= det [ A(A1xI) ]

= det (I xA)

= det [ x (A x1I ) ]

= (x)3 det (A x1I)

det (A1xI) = 1M+1A

(b)(i)det (A xI)=

= x3 + 8x2 17x+ 4

= (4 x)(x2 4x + 1)1A

4 is a root of det(A xI) = 0 and the other roots are 2 .

(ii)det A = 4.

By (a),det (A1xI)=

= (4x 1)(1  4x+ x2)1

roots of det (A1xI) = 0 are and 1

5.f : R\{1} R\{1} is defined by .

(a)yR\{1}, let x be such that .

y (x 1) = x + 1

R\{1}( x 1 as y + 1 y 1)

f is a surjective function.1M+1A

(b)(i)f1: R\{1} R\{1} is defined by .1A

(ii)f1(f1(x))= f1()

= 1A

1M+1A

HKALE 1990 I Q6

6.Case 1x2

| x 1 |  | x+ 2 |= (x 1) + (x+ 2)= 3  2

x2 is the solution set.

Case 22 x 1

| x 1 |  | x+ 2 |= (x 1)  (x+ 2)= 2x 1

The inequality becomes 2x 1  2

i.e.x1M

2 x is the solution set.1A

Case 3x 1

| x 1 |  | x+ 2 |= (x 1)  (x+ 2)= 3  2

no solution in this range.

The solution set is {xR: x}1A

97f6e31 Q10

7.(a)(I) :

(1)  (2) : y + 2z = 4y = 42z

From (1) ,x = 3 + yz = 73z

Solutions of (I) are of the form (73t , 42t , t), tR.

1M + 1A

(b)(II) is solvable iff there exists t R such that (73t, 42t , t) satisfies
x + ypz = q 2p

1M + 1A

(7  3t) + (4  2t) pt = q

(5 + p)t = 11q------(*)

Case 1

If p =  5 and q = 11, then (*) holds for any tR. Hence (II) has infinite number of solutions.

Case 2

If p5 and qR. Then (*) holds iff t = . Hence (II) has a unique solution.

(c)Put q = 11 in (II). By (b), (II) is always solvable.

Case 1p = 5, (x, y, z) = (73t , 42t , t)

(III) is solvable(73t)2 + (42t)2 + t2 = 21

for some tR

14t2 58t + 44 = 0

t = 1 or 1

(x , y, z) = ( 4 ,2 , 1) or ()

(c)Case 2p5.

In this case, (II) has the solution (7 , 4 , 0) only and it doesn't satisfy the equation x2 + y2 + z2 = 21.

(III) is not solvable.

Intell Q&A F.6 QA1

8.(a)(i)

1M+1A

(ii)

= 1M+1A

(iii)From (ii), 1

Furthermore,

(As  2  r  n)1M

= < 31M+1A



(b)For n = 6,1M

the statement is true for n = 1.

Assuming that the statement is true for n = k, k  1.

i.e.1M

For n = k + 1,

(k + 1)!= (k + 1) k!

= 1A

[ From (a), ]

= and1A

(k + 1)!= (k + 1) k!

[ From (a), ]

1M+1A

The statement is true for n = k + 1 and hence by the principle of induction, the statement holds for all n  6.

HKALE 2007 Q10

9.(a)an+4an+2=

= 1M

= 1A

(b)(i)(1)

= [By (a)]

= = …

= 1M+1A

1A

(2)

= [By (a)]

= = …

1M+1A

(3)

1M+1A

(ii)By (b)(i)(1), {a2n1} is monotonic increasing and by (b)(i)(2), {a2n} is monotonic decreasing.

Furthermore, a2 a4… a2n a2n1[By (b)(i)(3)]

 a2n3 … a3 a11A

{a2n+1} is bounded from above by a2 and {a2n} is bounded from below by a1.

 and both exist.1M

(c)If a2 < a1, then following the same argument in (b), we can prove that , and .

a1 > a3… > a2n1 > a2n > a2n2… > a2

{a2n1} is monotonic decreasing and bounded from below by a2 while {a2n} is monotonic increasing and bounded from above by a1.

 and both exist.

HKAL 1982 I Q7

10.(a)A v = vhas non-zero solutions

(AI) v = 0has non-zero solutions

det (A I) = 01M

1A

23 + 2 = 0 = 1 or 2.

1 = 1 , 2 = 22A

(b)SinceAv1 = v1andA v2 = 2 v2

we haveAV =



2v11 = v21andv12 = v22------(*)

1M+1A

Assume the contrary that V is singular. Then det V = 0

v11 v22v12 v21 = 0

v11 v22v22 (vx11) = 0[ From (*) ]

v11 v22 = 0

v11 = 0 or v22 = 0

v1 = 0 or v2 = 0contradiction !

V is non-singular.

1M+1A

(c)By (b),AV = 1A

As V is non-singular,

A = 1M

An = 1A

Finding v1 and v2 :

Put  = 1 in Av = v , we have

2v11 = v21

 is one of the solutions

Put  = 2 in Av = v , we have

v12 = v22

 is one of the solutions

LetV = . Then V-1 = 1M+1A

= An =

= 1A

Kiangsu-ChekiangCollege (Shatin)

Form 6 Pure Mathematics Paper I

Final Examination 2006-2007

Students’ Performance

Section A

Question Number / Performance in General
1 / (a)Fair. Some students were not aware that and some even didn’t know they have to consider the binomial expansion of (1 + x)2n+1.
(b)Poor. The students either fail to set up the identity or they didn’t have the intention to consider the coefficient of xk.
2 / Good. Only a few students who failed to get the correct answer.
3 / (a)Fair. The students were not aware that R6 is a rotation with angle of rotation 6  60.
(b)Satisfactory. Some of the students could not find A1 correctly.
(c)Poor. Most of the students were not aware that M could be deduced from part (b). Actually M could be (A1R2A) or (A1RA).
4 / (a)Good.
(b)(i)Good.
(ii)Fair. Some students could not set up the connection between (b)(i) and (b)(ii).
5 / (a)Poor. Most of the students mixed up the definitions of injective function and surjective functions.
(b)(i)Satisfactory.
(ii)Poor. The students were not aware that they have to consider ths cases n is even and n is odd separately.
6 / Very Good. Only one or two students made some careless computational mistakes.

Section B

Question Number / Performance in General
7 / (a)Good. Only a few students who could not get the general solution.
(b)Poor. Some students did not know how to tackle this part and someobtained the equation (5 + p)t = 11q, but they did not know how to consider the different cases precisely.
(c)Satisfactory. The case when p 5 was treated worse.
8 / (a)(i)Good. Some students proved the statement by MI. Only a few students failed to complete this part.
(ii)Very good. (Book work)
(iii)Good but a few students made a serious mistake. They argued that for all k = 2, 3, …, n and hence the sum was < 1 as well.
(b)Very good.
9 / Poor. Only two students have attempted this question and their performance are not satisfactory. The students were weak at simplification of complicated expressions.
10 / (a)Fair. Most of the students who attempted this part found the value of  by solving the simultaneous equations Av = v.
(b)Fair. The students were not aware that V is singular if det V is 0.
(c)Poor. The students have difficulty in finding V1 because they did not aware that they could substitute suitable values for v1 and v2.

Kiangsu-ChekiangCollege (Shatin)

Form 6 Pure Mathematics Paper 1

Final Examination 2006-2007

Consolidation Exercises

Paper 1

1.(a)Prove that

(b)Prove that .

2.Prove that for any positive integers m and n,

(a) 1  r  min { m , n }

(b)Hence deduce that

 0  s  n.

3.Let T be the rotation in the Cartesian plane anticlockwise about the origin by an angle , where 0 < 2. It is given that P1, P2, P3,... are the points in the Cartesian plane, where P1= (5,12), P2= (12, 5) and T transforms Pkto Pk+lfor each positive integer k.

(a)Find  .

(b)Let A be the matrix representing T. Find A2007 .

(c)Write down the coordinates of Pnfor all positive integers n.

4.Let P be a non-singular 2  2 real matrix and Q =, where and are two distinct real numbers. Define M = P1QP and denote the 2  2 identity matrix by I.

(a)Find real numbers and  ,in terms of and  , such that M2 = M + I .

(b)Prove that det(M2 +  I) =  ( + )2 .

-The End-