13

14

-1 S

MK

TIN

GG

I PO

RT

DIC

KS

ON

MA

RK

ING

SC

HE

ME

1.

a) f -1(x) =

x –

1

B

1

g h

as no

inv

erse becau

se it is no

t a on

e-to-o

ne fu

nctio

n. B

1

gof

b) g

o f = │

x +

1 │

D1

(correct v

ertex (-1

,0) an

d

y-in

tercept (0

, 1))

x D

1(V

shap

e and

abo

ve x

-axis)

c) │x

+ 1

│=

│x│

,

M1

solu

tion

set is { x

: x ≥

0}

A1

2.

a) ����� =

��2� � � �� ����� �� �� ������� ��

M1

= ��2� � ���� =

con

stant →

Un is th

e nth

term o

f a GP

. A

1 A

1

b) d

= T

2 – T

1

= 1

0 –

�� � ��� M

1 �� � �� ��2� ���5

B1

r = ��2� �� � ��� � A

1

� ∑�� �

����� � �!���

� � � � M

1 A

1

3.

a) (i) Sh

ow

L.H

.S. =

R.H

.S. =

26

B1

(ii) z1 z

2 = –

13

+ 1

3 i B

1

Arg

(z1 z

2 ) = tan

-1(13

/-13

) M1

= �" π

A1

b) √ 16�30( =

a + b

i

a2 –

b2 =

16

2ab

= –

30

M1

(a2 +

9)(a

2 –

25

) = 0

M1

a2 =

-9 an

d a =

±5

, b=)3

A1

(reject)

√ 16�30( = 5

– 3

i , –5

+3

i A1

(bo

th an

swers g

iven

)

4.

a) * 12

�32

6�11

1�2

7 , -./ 0 →→

do

ER

O →* 1

2�3

02

�50

00 ,

-.�2-/12.�5- 0

M1

A1

� for eq

uatio

n to

hav

e solu

tion

s r + 2

q –

5p

= 0

. A1

Eq

uatio

ns in

system

s are ind

epen

den

t and

there are in

finitely

man

y so

lutio

ns. B

1B

1

b) x

+ 2

y –

3z =

1

2y –

5z =

0

B1

(for b

oth

)

Let z =

t wh

ere t 2 3,� y

= �5�

and

x =

1 –

5t +

3t=

1 –

2t. M

1

i.e. solu

tion

s of sy

stem is (1

– 2

t, �5� , t) w

here t 2 3

. A1

5.

a) Fo

r grap

h , D

1 (co

rrect vertices) ,D

1(co

rrect shap

e)

b) M

idp

oin

t of P

N is ( 6���

, �� √1�7 � ) B

1 B

1

Let X

= 6��� an

d Y

= �� √1�7 �

x =

2X

-2 th

en Y

= �� 8 1��27�2� �

M1

4(X

– 1

)2 +

4Y

2 = 1

i.e. equ

ation

of lo

cus o

f mid

po

int o

f PN

wh

en P

mo

ves o

n th

e ellipse is

4(x

– 1

)2 +

4y

2 = 1

A1

(x –

1)2 +

y2 =

( �� )2 i.e. lo

cus is eq

uatio

n o

f a circle with

centre (1

,0) an

d

radiu

s ½. M

1 A

1A

1

6.

(1+

x2)

p= 1

+ p

x2+

…

(1 –

x )

-q = 1

+ q

x +

½ (q

2+q

)x2 +

9: �: ;�:���6 <=

+ …

B1

Su

bstitu

te abo

ve in

to ���6 � >���6� ?

= 1

+ q

x +

[q(q

+1

)+p

]x2 +

[ : �:�� ��:�� ��=@:=

]x3 +

…

M

1

A1

b) C

oefficien

t of x

3 is : �:���=

B1

0 ≤

: �:���=

≤ A�����=

M1

Larg

est po

ssible co

efficient o

f x3 is 1

62

. A1

Sectio

n B

7.

3 –

2p

+7

= 0

M1

(-1)3 +

p(-1

)2 +

7(-1

)+ q

= - 1

6 M

1

p =

5 A

1

q =

–1

3 A

1

a) f(1

) = 0

→ (x

– 1

) is a factor o

f f(x) B

1

f(x) =

(x –

1 )(x

2+6x

+1

3) =

0 M

1 A

1

Fo

r x2+

6x

+1

3 =

0, b

2 - 4

ac = –

16

< 0

→ it h

as no

real roo

ts. M1

� f(x

) = 0

has o

nly

on

e real roo

d, i.e., x

= 1

. A1

Wh

en f(x

)>0

, (x –

1 )(x

2+6

x+

13

) >0

, M1

(g

et answ

er from

sketch

ing g

raph o

r nu

mb

er line)

set of x

is {x

:x>

1, x23

}

A

1

b)

x +

9 =

A(x

2 + 6

x+

13

) + (B

x+

C)(x

– 1

) M1

So

lve…

. A=

½ , B

= ½

and

C =

– 5

/2 M

1 A

1

� 6�BC�6� � ���6��� �

6����6 �=6���� A

1

8.

a) a +

b+

c = 3

i + 6

k B

1

un

it vecto

r is D��E√� M

1 A

1

b)

cos θ

= * ��� 0. * � = 0√ �√ "�

M1

=

B8 ��"�� M1

θ =

39

.2o A

1

c) a

x b

= (3

-2)i –

(3 –

1)j +

(2 –

1)k

M1

= i –

2j +

k A

1

d)

BD

= B

A +

BC

M1

OD

= O

A +

OC

– O

B

= i –

4j A

1

e ) BA

= -j –

2k

and

BC

= -5

j – k

Area o

f parallelo

gram

AB

CD

= │

BA

X B

C│

=

│- 9

i │

=

9 u

nit 2