−

1−

2014-1

-PE

N-J

IT S

IN

Mark

ing sch

eme :

No

Work

ing/A

nsw

er P

artia

l

ma

rks

Tota

l

ma

rks

1

(a) 2

2

2

)x

1(

C

x1

B

x1

A

)x

1)(

x1(

xx

92

−+

−+

+≡

−+

−+

Try

to so

lve fo

r A(−

2), B

(−1) an

d C

(5) : 2

+ 9

−1 =

C(1

+1)

or 2

− 9

− 1

= A

(1 +

1)2

or −

1 =

A −

B o

r equiv

alent

22

2

)x

1(

5

x1

1

x1

2

)x

1)(

x1(

xx

92

−+

−−

+−

≡−

+

−+

B1

M1

A1

8

(b) f is o

ne to

one fu

nctio

n

Try

to fin

d in

verse

4

3x

)x(

f2

1−

=−

),

0[D

1f

∞=

−,

),

4 3[

R1

f∞

−=

−

B1

M1

A1

B1B

1

2

θθ

sin3

cos

3+

=

θα

θα

sinsin

rco

sco

sr

+

3co

sr

=α

, 3

sinr

=α

r2 =

32 +

2

)3

(

tan α

=

3 3

θθ

sin3

cos

3+

=

)6

cos(

32

πθ

−

B1

M1

M1

7

(a) −1 ≤

)

6co

s(π

θ−

≤ 1

32

)6

cos(

32

32

≤−

≤−

πθ

32

sin3

cos

33

2≤

+≤

−θ

θ

Max

imum

valu

e is 3

2, M

inim

um

valu

e is 3

2−

(b)

6 πθ

−=

6 π

±,

62

ππ

−

ππ

θ2

,3

,0

=

M1

A1(b

oth

)

M1

A1

3

(a) 4

4lo

g

log

log

2 22

=−

yx

(chan

gin

g b

ase)

82

2=

y x (elim

inate lo

g)

yy

256

)32

2(2

=+

(elimin

ate one o

f the v

ariables)

y =

16, x

= 6

4

M1

M1

M1

A1A

1

5

(b) 4

x −

1 >

3 −

x o

r 4x −

1 <

−(3

− x

) (defin

ition)

M1

4

−

2−

5

1 4 3

17

23

83

2

32

1

−−

→

→−

1 2 3

17

23

21

0

32

12

12

2R

RR

− −

− −→

→−

8 2 3

84

0

21

0

32

13

13

3R

RR

−−

→

→−

0 2 3

00

0

21

0

32

13

23

4R

RR

∴ x

= −

1 −

7t, y

= 2

t + 2

, z = t w

here t ∈

R

B1

M1

M1

M1

A1

5

6

(a) f(y) =

y3 −

y3 =

0

x −

y is a facto

r of x

3 − y

3 [factor th

eorem

]

x3 −

y3 =

(x −

y)(x

2 + x

y +

y2)

Puttin

g y

= −

y

M1

A1

B1

5

x >

5 4

or x

<

3 2−

∴ set o

f valu

es of x

= {

x | x

∈R

, x <

3 2

− o

r x >

5 4

}

A1A

1

A1

4

1 −

2sin

2θ >

3sin

θ +

2

(2sin

θ +

1)(sin

θ +

1) <

0

2 1sin

1−

<<

−θ

OR

1

sin−

>θ

, 2 1

sin−

<θ

[allow

separate in

equalities w

ithout “an

d”.]

1sin

−>

θ ⇒

π

θπ

<<

−, θ

≠

2 π−

2 1sin

−<

θ ⇒

6

6 5π

θπ

−<

<−

[all correct]

[can b

e implied

]

∴ th

e solu

tion set =

}

62

,2

6 5:

{π

θπ

πθ

πθ

−<

<−

−<

<−

M1

M1

M1

M1

A1

6

−π

π

6 5π

−

6 π−

2 π−

0

−

3−

x3 +

y3 =

(x +

y)(x

2 − x

y +

y2)

x6 −

y6 =

(x3 −

y3)(x

3 + y

3)

x6 −

y6 =

(x −

y)(x

+ y

) (x2 +

xy +

y2)(x

2 − x

y +

y2)

B1

B1

(b) f(x

) = (x

− 2

)(x2 +

kx +

4)

f(x) =

0

(x −

2)(x

2 + k

x +

4) =

0

x =

2 o

r x2 +

kx

+ 4

= 0

, let q(x

) = x

2 + k

x +

4

f(x) h

as at least 2 real d

istinct ro

ots,

x2 +

kx

+ 4

= 0

has tw

o real ro

ots an

d q

(2) ≠

0

b2 −

4ac ≥

0 an

d q

(2) ≠

0

k2 −

4(1

)(4) ≥

0 an

d 4

+ 2

k +

4 ≠

0

k2 ≥

16 an

d 2

k +

8 ≠

0

4≥

k an

d 2

k ≠

−8

k ≥

4, k

≤ −

4 an

d k

≠ −

4

∴ th

e set of v

alues o

f k =

{k | k

∈R

, k ≥

4 o

r k <

−4}

B1

M1

A1A

1

A1

5

−

4−

7

AB

=

−

−

54

1

21

3

42

1

−

−−

−

76

11

10

917

86

13

=

−

−

−

30

0

03

0

00

3

AB

=

−

10

0

01

0

00

1

3 =

−3

I

A−

1AB

= A

−1(−

3I)

IB =

−3

A−

1

−3A

−1 =

B

A−

1 =

B3 1

−

−=

−

−

13 1 8

54

1

21

3

42

1

z y x

AX

= C

A−

1AX

= A

−1C

IX =

A−

1C

X =

A−

1C

−

−

−−

−

−=

13 1 8

76

11

10

917

86

13

3 1

z y x

−=

3 1 2

z y x

∴ x

= 2

, y =

−1, z =

3

(b) A

B =

−3

I

AB

B−

1 = −

3IB

−1

AI =

−3

B−

1

−3B

−1 =

A

B−

1 =

A3 1

−

− −

=

−

−−

−

14

20 13

76

11

10

917

86

13

r q p

BY

= D

B−

1BY

= B

−1D

IY =

B−

1D

Y =

B−

1D

− −

−

−

−=

14

20 13

54

1

21

3

42

1

3 1

r q p

B1

B1

B1

M1

M1

A1

A2, 1

, 0

B1

B1

M1

M1

8

−

5−

−=

1 3 1

r q p

∴ p

= 1

, q =

−3, r =

1

A1

A2,1

,0

8

(a) )

5()

5()

4)(

4)(

3)(

3(1

22

−=

xx

x

M1

4

−

6−

24

0)

12 25

(=

x 2

40

lg)

12 25

lg(

=x

x =

7.4

67 (4

sf)

M1

M1

A1

(b) f(1

) = 0

6 −

7 +

a + b

− 1

2 =

0

a + b

= 1

3 …

(1)

f(−1) =

−50

6 −

7(−

1) +

a + b

(−1) −

12 =

−50

a − b

= −

51 …

(2)

a = −

19, b

= 3

2

(a) f(−2) =

0 ⇒

x +

2 is a facto

r of f(x

)

(b) f(x

) = (x

− 1

)(x +

2)(6

x2 +

Ax

+ 6

)

= (x

2 +

x −

2)(6

x2 +

Ax

+ 6

)

By lo

ng d

ivisio

n o

r com

parin

g co

efficients o

r any v

alid

meth

ods,

A =

−13

f(x) =

(x −

1)(x

+ 2

)(2x

− 3

)(3x

− 2

)

(c) (x −

1)(x

+ 2

)(2x

− 3

)(3x

− 2

) > 0

By g

raphical o

r num

ber lin

e meth

od o

r any v

alid m

ethod,

Set o

f valu

es of x

= {

x | x

∈R

, x <

−2 o

r 1

3 2<

<x

or

2 3>

x}

M1

M1

A1(b

oth

)

M1A

1

M1

M1A

1

A1

M1A

1

11