Queen’s College
Mock Examination 2005-2006
PURE MATHEMATICS, PAPER I
1.(a) (1)Since Cr = Cn – r , (2)
Consider the derivative:
(3) / 1A
1M
1M
1A / for (1)
for using differentiation
for using product rule or chain rule
for correct result
(b) Comparing the coefficient of xn-2 on both sides of (3) in (a),
/ 1M
1A / for comparing coefficient
for writing both sides correctly
2. a , b are the roots of x2 – x – 1 = 0, a > b
By Vieta’s theorem , a + b = 1, a b = – 1 ……(1)
Let P(n) the proposition be :
For P(0), a0 = 0 = . \ P(0) is true.
For P(1), L.H.S. = a1 = 1
R.H.S. = \ P(1) is true.
Assume P(k-1) and P(k) are true for some integer k, that is,
………(2)
For P(k + 1),
, by (2)
, by (1)
\ P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true for all natural numbers, n. / 1A
1A
1A
1M
1A
1A / for Vieta’s theorem or
for P(0), P(1)
for inductive hypothesis
for second principle
for using Vieta’s Theorem
for deducing correctly
3. (a) cos 5q + i sin 5q = (cos q + i sin q)5
= cos5 q + 5 cos4 q (i sin q) + 10 cos3 q (i sin q)2 + 10 cos2 q (i sin q)3 +
5 cos q (i sin q)4 + (i sin q)5
= cos5 q - 10 cos3 q sin2 q + 5 cos q sin4 q
+ i (5 cos4 q sin q - 10 cos2 q sin3 q + sin5 q)
By comparing real parts,
cos 5q = cos5 q - 10 cos3 q sin2 q + 5 cos q sin4 q
= cos5 q - 10 cos3 q (1 – cos2 q) + 5 cos q (1 – cos2 q)2
= / 1M
1M
1A
1M / for de Moivre’s Theorem
for Binomial Theorem
for correct expansion
for sin2 q = 1 – cos2 q
(b) By setting cos 5q = cos q (16 cos4 q - 20 cos2q + 5) = 0
The principal roots are
The principal roots of 16 cos4 q - 20 cos2q + 5 = 0 are .
Let x = cos q , by comparion, the roots of 16x4 – 20 x2 + 5 = 0
are / 1A
1M
1A / for roots of cos 5q = 0
for deleting the extra root
for answers
4. (a) x3 – 2x2 + 5x – 3 = 0
a + b + g = 2
ab + bg +ga = 5 ….(1)
abg = 3 / 1A / for all values correct
(b) a2 + b2 + g2 = (a + b + g)2 – 2(ab + bg +ga)
= 22 – 2 (5) , by (1)
= –6 /
1M
1A / for completing squarefor answer
(c) a , b and g be the roots of x3 – 2x2 + 5x – 3 = 0
…..(2)
…..(3)
…..(4)
(2) + (3) + (4),
/
1M
1A / for property of rootsfor answer
5.
= (b – c)(b – x)(c – x)(bcx – a3)
, where bc ¹ 0 / 1M
1M
1M
1A
1A / for subtraction
for taking out factors
for Sarrus Rule of expansion
for correct factorization
for roots
6. (a)
/ 1M
1A / for joining fractions
for answer
(b)
/ 1M
1M
1A / for using v
for difference method
for answer
7. 9x4 – 27x3 + 8x2 – 27x + 9 = 0
Divide the whole equation by x2, we have
….(1)
Let
The equation (1) becomes:
9(y2 – 2) – 27 y + 8 = 0
9y2 – 27 y – 10 = 0
(3y – 10) (y + 1) = 0
\ / 1M
1A
1A / for dividing by x2
for calculating
for y
(i) When ,
\ 3x2 – 10x + 3 = 0
\ (3x – 1)(x – 3) = 0
\
(ii) When
\ 3x2 + x + 3 = 0
\
The four roots of the equations are . / 1A
1A / for answers of x.
for answers of x.
8. (a) (i)
\ P is a matrix of rotation through an angle anticlockwisely. / 1A
1A / for answer
for describing the angle of rotation
8. (a) (ii)
/ 1A
1A
1A
1A / may be omitted if last answer is correct
(b) (i) Let . Then
= (ax2 + 2bxy +dy2 + l) = (0)
\ a = 5, b = c = -13, d = 5 (since M is symetric) and l = 72
\ M = = Q and l = 72 /
1M
1A
1A
1A
/ for letting Mfor the simplified matrix
for l
for M
(b) (ii) Let , the image of under the rotation through an angle of about the origin. \
The image of (*) under the rotation is:
since PT = P-1
, by (a)(ii)
\ -8X2 + 18Y2 + 72 = 0
\ The curve (*) is a hyperbola. / 1A
1A
1A
1A
9. (a) Let
Since f(x) has alternating signs in the coefficients, the roots of f(x) = 0 are positive.
By rational zero theorem, he possible rational roots are 1, 3, 5, 7, 15, 21, 35, 105.
f(3) = 33 – 15(32) + 71(3) – 105 = 0
By factor theroem, x – 3 is a factor of f(x).
By division, f(x) = (x – 3)(x2 – 12x + 35)
= (x – 3)(x – 5)(x – 7)
The roots of f(x) = 0 are 3, 5, 7. / 1M
1A
1A / for factor theorem
for factorization
for roots
(b) (i)
= -p (1)
(2)
(3)
= (-S1)2 – 2q = S12 – 2q
\ (4) / 1M
1M
1A+1A / for Vieta’s Theorem
for expansion
for answer
(b) (ii) Denoting S the symmetric sum,
(5)
(6)
(5) – 3(6), , by (3)
(7)
. / 1A
1A
1A+1A / correct expansion of S1
correct expansion of S2
Note: may use the method as Q.4 (c)
for values of m, l
(c)
Put
From (4), (7)
The cubic equation form by the roots is .
By (a), the roots are 3, 5, 7.
\ (x, y, z) = (3,5,7) or (3,7,5) or (5,7,3) or (5,7,3) or (7,3,5) or (7,5,3) / 1A
1M
2A / for correct evaluation
for forming equation
deduct 1A for
(x, y, z) = (3,5,7)
10. (a) A = , |A| = 3a2 + a3 = a2 (a + 3).
Min A = , Cof A =
Adj A =
Inverse of A = , where a0, -3.
/ 1A
1A
1A
1A / for determinant
for minor and cofactor
(may be omitted if adj A is correct)
for adjoint
for inverse
(b) (i) is equivalent to
When a0, -3, inverse of A exists,
/ 1M
1M
1A / for writing in inverse form
for multiplication of matrices
for answer
(b) (ii) When a = 0, the system of equation becomes:
When b = 1 , the system becomes x + y + z = 1
and has infinite number of solution: (x, y, z) = (1 – t1 – t2, t1, t2).
When b ¹ 1 , the system is inconsistent and has no solution. / 1A
1A
1A
1A / for reducing the system
for reducing more
for answer
for answer
(b) (iii) When a = -3, the system of equation becomes:
Adding the 3 equations, b2 + b +1 = 0.
Since D = 12 – 4(1)(1) < 0. The quadratic equation has no real solution.
\ The system of equations is inconsistent and has no solution. / 1A
1M
1M
1A / for reducing the system
for adding equations
for knowing no real solution
for answer
11. (a) f(x) = ln x – x + 1 , x > 0
When 0 < x < 1, f’(x) > 0.
When x > 1, f’(x) < 0
\ f(x) attains its maximum value at x = 1.
\ f(x) £ f(1) = ln 1 – 1 + 1 = 0 " x > 0.
\ f(x) £ 0 " x > 0. / 1A+1M
1M
1M
1A / for correct derivative and setting it to 0 to find x = 1
for first derivative test
for max
for comparing f(x) and f(1)
(b) Let .
Then for i = 1, 2, …, n,
/ 1A / for using (a)
The equality holds if
i.e. a1 = a2 = …. = an = A. / 1A
1M
1A
1M / for taking sum
for property of ln
for correct result
for proof
(c)
\ ….(1)
Replace in (1), we have
….(2)
(1) ´ (2), we get ….(3)
Put ai = i in (3),
/ 1A
1A
1M
1M
1M / for
for
for multiplication
for ai = i
for Si
12. (a) (1 + z)2n + (1 – z)2n = 0
, where k = 0, 1, 2, …, (2n – 1).
\
, where k = 0, 1, 2, …, (2n – 1).
Since
where k = 0, 1, …., n – 1.
Hence the roots are , where k = 0, 1, …., n – 1. / 1M
1M
1A
1M
1A
1M / for transposing term
for using de Moirvre’s Th.
for change of subject
for using half-angle formula
for getting answers
for rewriting answers
12. (b) By (a), are roots of (1 + z)2n + (1 – z)2n = 0
Equating the coefficient z2n on both sides, we get A = 2.
/ 1M
1A
1M
1A / for writing factors corresponding to roots
for writing the leading coeff.
for finding the leading coeff.
for final answer
12. (c) ….(1)
Coefficient of z2n – 2 on L.H.S. =
= 4n2 – 2n
Coefficient of z2n – 2 on R.H.S. =
Equating Coefficient of z2n – 2 on both sides,
/ 1M
1A
1A
1M
1A / for using combination
for correct coeff.
for correct coeff.
for changing to sec
for correct answer
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