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1. (a) (i) 8P5 or 8C5×5!

(ii) 4C3× 4C2×5!

(b) (i) 64×52×6!

(ii) ‘9’ from first set only: 5C3× 4C2×6!

‘9’ from second set only: 5C4× 4C1×6!

no ‘9’ chosen: 5C4× 4C2×6!

total = 5C3× 4C2+ 5C4× 4C1+ 5C4× 4C2×6! [8]

2. (a) (i) in+2=in×i2=-i×-1=i

(ii) i2n+1=in2×i=-i2×i=-i

(b) 1-ii+2i×ii=i-i2i2+2=1+i-1+2=1+i

(c) 1+i4-1-i4

=1+i2+1-i2×1+i2-1-i2

=1+2i-1+1-2i-1×1+2i-1-1+2i+1

=0×4i=0 [7]

3. (a) 53=5+1×10×6+1×4×15=5+60+60=125

(b) 5C2 , 4C1

(c) n3=n+ n-1C0× n+1C1× nC2+ nC0× n-1C1× n+1C2 [5]

4. (a) (i) 0-1111-1110×110-1-110-11=100010001

A and B are inverses of each other.

(ii) 0-1111-1110×xyz=-683

xyz=0-1111-1110-1×-683=110-1-110-11×-683=21-5

∴x=2, y=1, z=-5

4. (b) (i) T1 performs a rotation of 180°

∴y=-x-12

(ii) Reflection of the line y=x

T2=0110

(iii) New Area =T3×1023=6×323=64 units2 [12]

5. (a) (i) y=-3cos4x-π12=-3cos4x-π3

∴ A=-3 ω=4 θ=-π3

(ii) y=-3sin4x+π6

(b)

Scale factor (y=2 at x=5π24) Period of π2 Vertical Asymptote at x=π12 Scale factor (y=2 at x=5π24) Period of π2 Vertical Asymptote at x=π12

[8]

6. (a) (i) 2×8=16=4

(ii) “If ab is irrational, then both a and b must be irrational”

(b) A ⇒ B: If the triangle has two equal sides, then it is isosceles,

and therefore it has two congruent sides.

∴ A ⇒ B is valid and True.

B ⇒ A: If the triangle has two congruent sides, then it is isosceles,

and therefore it has two equal sides.

∴ B ⇒ A is valid and True.

∴ A ⇔ B

(c) ∀x∈P, ∃ y∈ P :xy∈Q [6]

7. Assume n is even and that n3 is odd.

Let m∈N:n=2m is even.

n3=2m3=8m3=24m3

Since n3 cannot be both even and odd simultaneously, then this

is a contradiction. And therefore n must be odd. [4]