MathematicsT STPM Baharu PENANG 2012

2 0 1 2 T R IA L S T P M

B A H A R U

M A T H E M A T IC S T

S M J K

J IT S IN

S e c t t i o n A [ 4 5 m a r k s ] A n s w e r a l l q u e s t i o n s i n t h i s s e c t i o n . 2 1 . F i n d t h e s e t v a l u e s o f x f o r w h i c h 1 − 3 x > . x 2 . ( a ) G i v e n t h a t y = l o g 2 x a n d l o g

y

2

2

x − l o g

x

8 + lo g

2

2

k

+ k l o g

+ k y + 2 k − 3 = 0 .

( b ) S o l v e t h e e q u a t i o n 2

2 x + 1

= 3 ⋅2

x

− 1 .

[ 5 m x

a r k s ]

4 = 0 , s h o w t h a t [ 3 m

a r k s ]

[ 4 m

a r k s ]

− 4 ⎞ ⎟ 3 . G i v e n t h a t M 2 1 5 4 ⎟ . F i n d t h e m a t r i x N – 6 M 1 4 1 6 ⎠⎟ s h o w t h a t M ( N – 6 M ) = k I w h e r e k i i s a n i n t e g e r a n d I i s a 3 × 3 m a t r i x . ⎛ 2 ⎜ = 0 ⎜ ⎜ − 1 ⎝

0

− 1 ⎞ ⎛ 1 5 ⎟ ⎜ 1 ⎟ a n d N = ⎜ − 1 ⎜ − 4 2 ⎠⎟ ⎝

,P E N A N G

− 1

S t a t e t h e v a l u e o f k a a n d h e n c e f i n d t h e i n v e r s e o f m a t r i x M . 4 . S o l v e t h e f o l l o w i n g s y s t e m

o f lin e a r e q x – 2 2 y y + + z z = = 2 x x + y y – – 3 z z = = 4 x x – 7 y y + + z z = =

[ 7 m

u a t i o n s u s i n g G a u s s i a n e l i m i n a t i o n : 0 5 [ 8 m – 1

a n d

a r k s ]

a r k s ]

5 . T h e f u n c t i o n s f f a n d g a a r e d e f i n e d a s :

f : x → g : x →

x      x ≥ 0 3 l n x    x > 0

( a ) S k e t c h t h e g r a p h o f f a n d s t a t e w h e t h e r f − 1 e x i s t . G i v e a r e a s o n f o r y o u r a n s w e r . ( b ) F i n d g − 1 a n d s t a t e i t s d o m a i n . ( c ) F i n d t h e c o m p o s i t e f u n c t i o n f g − 1 a n d s t a t e i t s r a n g e .

[ 9 m

a r k s ]

− 7 x + 8 [ 4 m a r k s ] i n p a r t i a l f r a c t i o n s . 2 ( x + 2 ) ( 1 − 3 x ) ( b ) T h e r e m a i n d e r o b t a i n e d w h e n 3 x 3 + m x 2 − 4 x − 2 i s d i v i d e d b y x + 1 i s t w i c e t h e r e m a i n d e r o b t a i n e d w h e n t h e s a m e e x p r e s s i o n i s d i v i d e d b y x − 2 . F i n d t h e v a l u e

6 . ( a ) E x p r e s s

o f m m .

6 x

2

[ 5 m

a r k s ]

9 5 4 / 1 * T h i s q u e s t i o n p a p e r i s C O N F I D E N T I A L u n t i l t h e e x a m i n a t i o n i s o v e r . C O N F I D E N T I A L * 2

2 0 1 2 T R IA L S T P M

B A H A R U


S M J K

J IT S IN

,P E N A N G

S e c t t i o n B [ 1 5 m a r k s ] A n s w e r a n y o n e q u e s t i o n i n t h i s s e c t i o n . 7 . R r F ( ( S l F

e l a t i v e t o a f i x e d o r i g i n O , t h e p o i n t s A A , B a n d C h h a v e p o s i t i o n v e c t o r s g i v e n b = 5 i – 2 jj + 3 k , c = 4 i + jj – 2 k e s p e c t i v e l y b y a = 2 i + 3 jj – k , in d (i) t h e l e n g t h o f A B , c o r r e c t t o 3 s i g n i f i c a n t f i g u r e s , i i ) a n g l e B A C , c o r r e c t t o t h e n e a r e s t d e g r e e , i i i ) t h e a r e a o f t r i a n g l e A B C , c o r r e c t t o 3 s i g n i f i c a n t f i g u r e s . h o w t h a t , f o r a l l t h e r e a l v a l u e s o f t h e p a r a m e t e r t , t h e p o i n t P w i t h p o s i t i o n v e c t o r i e s o n t h e l i n e t h r o u g h A a n d B . [ 1 5 m a r k s ] i n d p s u c h t h a t O P i s p e r p e n d i c u l a r t o A B .

8 . T h e p o i n t s A A a n d B h a v e p o s i t i o n v e c t o r s 3 i + r e l a t i v e t o t h e o r i g i n O . T h e p o i n t C i i s o n t h e 2 O O A . T h e p o i n t D i i s o n O B p r o d u c e d a n d i s s t h a t OO C X D i s a p a r a l l e l o g r a m . S h o w t h a t t h e F i n d ( i ) i n t h e f o r m r = u + t v , t h e e q u a t i o n s o f t h (ii) th e p o s itio n v e c to r o f th e p o in t o f in te rs e ( i i i ) t h e a n g l e B A X . ( i v ) t h e a r e a o f t h e p a r a l l e l o g r a m O C X D .

2 jj + k a n l i n e O A p p u c h t h a t B l i n e A X i s

d i + ro d u D = p a ra

2 jj + 3 k , r c e d a n d is O B . T h e p lle l to th e

e s p s u c o in v e c

e c t i v e l y , h t h a t A C = = t X i s s u c h t o r i + jj + k .

e l i n e A X a n d C D . c t i o n b e t w e e n t h e l i n e s A A X a n d C D . [ 1 5 m

a r k s ]


2 0 1 2 T R IA L S T P M

B A H A R U


S M J K

J IT S IN

,P E N A N G

A N S W E R S C H E M E S M J K J I T S I N − S T P M T r i a l E x a m m i n a t i o n 2 0 1 2 M a r k i n g s c h e m e f o r M a t h e m a t i c s T P a p e r 1

N o 1

S e c t t i o n A [ 4 5 m a r k s ] o r k i n g / A n s w e r

W

1 − 3 x x >

2 x

P a r t i a l m a r k s

, x ≠ 0 y

y = 1 − 3 x

G ra p h : V

s h a p e 1 ( , 0 ) : D 3 R e c ip r o c D

y = 3 x − 1

• A ( 1 , 2 ) y =

1

2

T o t a l m a r k s 5

& 1

a l : 1

x

P o in t A

: B 1

x

0 1 1 3

T h e s e t o f v a l u e s o f x i s { x x | x ∈ R , x < 0 o r x > 1 } .

A n s : M 1 A 1

O R 1 − 3 x >

2 x

, x ≠ 0

⇔ 1 − 3 x > 1 − 3 x −

2 x

× ( − 1 ) ,

x

o r 1 − 3 x < −

2 x

M 1

> 0

x − 3 x 2 − 2 x

2

> 0

3 x 2 − x + 2 x

< 0


2 0 1 2 T R IA L S T P M

B A H A R U

3 x 2 − x + 2 = 3 ( x 2 −

2 ( a )

x


S M J K

lo g

2

x −

2

x − l o g lo g lo g

2 2

8 x

x

8 + lo g

+ k lo g

2

2

k

+ k l o g ⎛ lo g

x

4 = 0

4 ⎞ ⎟ = 0 l o g x 2 ⎝ ⎠

2 2 + k ⎜

2

⎛ 2 ⎞ + k + k ⎜ ⎟ = 0 , y ⎝ y ⎠ l o g 2 8 = l o g 2 2 3 = 3 , l o g 2 4 = l o g 2 2 2 = 2 y 2 − 3 + k y + 2 k = 0 y 2 + k y + 2 k − 3 = 0 y −

,P E N A N G

) + 2

3 1 1 = 3 ( x − ) 2 − 3 ( − ) 2 + 2 6 6 1 1 1 = 3 ( x − ) 2 + 1 > 0 6 1 2 S i n c e 3 x 2 − x + 2 > 0 ⇒ x < 0 2 1 − 3 x < − , x ≠ 0 x 2 1 − 3 x + < 0 x x − 3 x 2 + 2 < 0 x 3 x 2 − x − 2 × ( − 1 ) , > 0 x ( 3 x + 2 ) ( x − 1 ) > 0 x L e t 3 x + 2 > 0 , x − 1 > 0 , x > 0 2 x > − , x > 1 , x > 0 3 u s e n u m b e r l i n e , … 2 − < x < 0 o r x > 1 3 ∴ t h e s e t o f v a l u e s o f x i s { x x | x ∈ R , x < 0 o r x > 1 } G i v e n t h a t y = l o g 2 x lo g

J IT S IN

3

M 1 ( e i t h e r )

A 1

A 1 A 1 3

M

1 ( c h a n g i n g b a s e )

M

1 ( s u b s t . c o r r e c t y y , l o g 2 8 = 3 , l o g 2 4 = 2 )

A 1


2 0 1 2 T R IA L S T P M 2 ( b )

B A H A R U


J IT S IN

,P E N A N G

2 x + 1

= 3 ⋅ 2 x − 1 . 2 ( 2 x ) 2 − 3 ⋅ 2 x + 1 = 0 ( 2 ⋅ 2 x − 1 ) ( 2 x − 1 ) = 0 T h e n 2 ⋅ 2 x − 1 = 0 a n d 2 2

S M J K

2 x =

1

4 M

x

2

2 2 x = 2 − 1 x = − 1

M 1 ( f a c t o r i z e ) A 1 ( b o t h )

− 1 = 0

x

1 ( q u a d r a t i c f o r m )

= 1

2 x = 2 0 x = 0

A 1 ( b o t h )

∴ x = − 1 , 0 3

G iv e n th a t M

⎛ 1 5 ⎜ N – 6 M = − 1 ⎜ ⎜ − 4 ⎝ ⎛ 3 ⎜ = ⎜ − 1 ⎜ 2 ⎝

⎛ 2 ⎜ = 0 ⎜ ⎜ − 1 ⎝ − 1 1 5 4

− 1 3

− 2

0 2 1

− 1 ⎞ ⎛ 1 5 ⎟ ⎜ 1 a n d N = − 1 ⎟ ⎜ ⎟ ⎜ − 4 2 ⎠ ⎝

− 4 ⎞ ⎟ 4 ⎟ – 6 1 6 ⎠⎟ 2 ⎞ ⎟ − 2 ⎟ 4 ⎠⎟

⎛ 2 ⎜ 0 ⎜ ⎜ − 1 ⎝

0 2 1

− 1 ⎞ ⎟ 1 ⎟ 2 ⎠⎟

0 2 ⎞ − 1 ⎞ ⎛ 3 − 1 ⎛ 2 ⎜ ⎟ ⎜ ⎟ M ( N – 6 M ) = 0 2 1 − 1 3 − 2 ⎜ ⎟ ⎜ ⎟ ⎜ − 1 1 ⎟ ⎜ 2 ⎠ ⎝ 2 4 ⎠⎟ − 2 ⎝ ⎛ 1 0 0 ⎞ ⎛ 4 0 0 ⎞ ⎜ 0 4 0 ⎟ = 4 ⎜ 0 1 0 ⎟ = ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ 0 0 4 ⎟ ⎝ ⎠ ⎝ 0 0 1 ⎠ ∴ M ( N – 6 M ) = 4 I s h o w n = 4 ∴ k =

− 1 1 5 4

− 4 ⎞ ⎟ 4 . ⎟ 1 6 ⎠⎟

M

7

1

A 1

M

1

A A

1 1

M

1

M (N

– 6 M ) = 4 I N − 6 M M ( ) = I 4


2 0 1 2 T R IA L S T P M − 1

∴ M

=

1 4

B A H A R U


S M J K

J IT S IN

,P E N A N G

( N − 6 M )

2 ⎞ − 1 ⎛ 3 ⎟ 1 3 2 = − − ⎟ 4 ⎜⎜ − 2 4 ⎠⎟ ⎝ 2 1 1 ⎞ ⎛ 3 − ⎜ 4 4 2 ⎟ ⎜ ⎟ 1 3 1 = ⎜ − − ⎟ ⎜ 4 4 2 ⎟ ⎜ 1 ⎟ 1 ⎜⎜ 1 ⎟ ⎟ − 2 ⎝ 2 ⎠ 1 ⎜

4

A 1

G i v e n t h a t x x – 2 2 y y + + z = = 0 2 x x + y y – – 3 z z = = 5 4 x x – 7 y y + + z = = – 1

⎛ 1 ⎜ ⎜ 2 ⎜ 4 ⎝

− 2 1

− 7

R R

2 3

3

0 ⎞

1

⎟ − 3   5 ⎟ 1 − 1 ⎠⎟

B 1

− 2

+ ( − 2 R 1 ) → R 2 + ( − 4 R 1 ) → R 3

⎛ 1 ⎜ ⎜ 0 ⎜ 0 ⎝

− 2

↔ R 3

⎛ 1 ⎜ ⎜ 0 ⎜ 0 ⎝

− 2

+ ( − 5 R 2 ) → R 3

⎛ 1 ⎜ ⎜ 0 ⎜ 0 ⎝

R

R

8

2

5 1

1 5

1 0

0 ⎞

1

⎟ − 5   5 ⎟ − 3 − 1 ⎠⎟ 1

1 ( o n e o p e r a t i o n ) 1 ( o n e o p e r a t i o n )

0 ⎞

⎟ − 3   − 1 ⎟ 5 ⎠⎟ − 5 1

0 ⎞

⎟ − 3   − 1 ⎟ 1 0 1 0 ⎠⎟

[ e c h e l o n f o r m T h u s ,

M M

M 1 A 1 ( o n e o p e r a t i o n )

]

1 0 z = = 1 0 … … . . ( 1 ) y – – 3 z z = = – 1 … … . . ( 2 ) x – 2 y y + + z = = 0 … … … ( 3 )

A 1


2 0 1 2 T R IA L S T P M fro m

B A H A R U

( 1 ) ,


S M J K

J IT S IN

,P E N A N G

z = = 1

s u b s t . z = = 1 i n t o ( 2 ) , y y – – 3 ( 1 ) = – 1 y = = 2 s u b s t . z = = 1 a n d y = = 2 i n t o ( 3 ) , x x – 2 ( 2 ) + ( 1 ) = 0 x = 3

M

1

T h e r e f o r e , x = 3 , y = = 2 a n d z = = 1 A 1 5 ( a )

G i v e n t h a t f : x →

2

x      x ≥ 0

y f ( x ) = x

x

0 S in c e a n y h o riz y = f(x ) a t o n ly o n e f u n c tio n a s f − 1  e x i s t  b e c a u o n e , a n d d e fin e 5 ( b )

G r a p h : D 1

t a l l i n e y = k f o r k ≥ 0 c u t s t h e g r a p h B 1 e p o i n t , t h e r e f o r e y = f ( x ) i s o n e t o c h f − 1 e x i s t s .  f o r  t h e  g i v e n  d o m a i n  f ( x ) i s o n e t o d f o r a l l v a l u e s o f x . o n o n s u s e

g ( x ) = 3 l n x , D g = ( 0 , ∞ ) , R g = ( − ∞ , ∞ ) L e t y = g − 1 ( x )

∴

∴ g

x = g ( y ) ) = 3 l n y x l n y = 3 − 1

x

f g

− 1

M

1

A 1

( x ) = e 3

D o m a i n o f g − 1 ( x ) = R g = { x x | x ∈ ℜ } 5 ( c )

3

A 1

4

x

( x ) = f ( e ) 3

M

1


2 0 1 2 T R IA L S T P M

B A H A R U


S M J K

J IT S IN

,P E N A N G

x

= e 3 A 1 A 1

x

= e 6 , T h e r a n g e o f f g 6 ( a )

6 x

L e t

( x

2

D f g − 1 = { x | x ∈ R } − 1

A 1

( x ) i s { y y : : y > > 0 }

A B x + c − 7 x + 8 ≡ + 2 x + 2 + 2 ) (1 − 3 x ) 1 − 3 x

4

2

+ 2 ) + ( 1 − 3 x ) ( B x + C ) 2 ⎛ ⎛ 1 ⎞ 2 ⎞ 1 ⎛ 1 ⎞ ⎛ 1 ⎞ S u b s t . x = , 6 ⎜ ⎟ − 7 ⎜ ⎟ + 8 = A ⎜ ⎜ ⎟ + 2 ⎟ ⎜ ⎝ 3 ⎠ ⎟ 3 ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ ⎠ 2 7 ⎛ 1 ⎞ − + 8 = A ⎜ + 2 ⎟ 3 3 ⎝ 9 ⎠ 2

6 x

− 7 x + 8 ≡ A (x

M

2

1 9

1 9 A 3 9 A = 3

1

A 1

=

M

1

C o m p a r i n g c o e f f i c i e n t s o f x 2 , 6 = A − 3 B − 3 B = 3 B = − 1 C o m p a r in g th e c o n s ta n t te r m

∴

6 ( b )

3 x

6 x ( x

3

2

: 8 = 2 A + C C = 2

− 7 x + 8 3 ( − x + 2 ) ≡ + 2 x + 2 + 2 ) (1 − 3 x ) 1 − 3 x 2

A 1

+ m x 2 − 4 x − 2

h e n x = – 1 , f ( − 1 ) = 3 x 3 + m x

5

W

2

− 4 x − 2 = 3 ( − 1 ) 3 + m ( − 1 ) 2 − 4 ( − 1 ) − 2


2 0 1 2 T R IA L S T P M

B A H A R U


= − 3 + m + 4 − 2 = − 1 + m h e n x = 2 , f ( 2 ) = 3 x 3 + m x

S M J K

J IT S IN

,P E N A N G

B 1

W

f(– 1 ) − 1 + m − 1 + m − 2 9

2

− 4 x − 2 = 3 ( 2 ) 3 + m ( 2 ) 2 − 4 ( 2 ) − 2 = 2 4 + 4 m − 8 − 2 B 1 = 4 m + 1 4

m = −

1 1

A 1

7

N o

7

M M

= 2 f ( 2 ) = 2 ( 4 m + 1 4 ) = 8 m + 2 8 = 7 m m 2 9

S e c t t i o n B [ 1 5 m a r k s ] o r k i n g / A n s w e r

W

P a r t t i a l m a r k s

   

( i ) A B = b − a = ( 5 i – 2 jj + 3 k ) – ( 2 i + 3 jj – k ) = 3 i – 5 jj + 4 k L e n g t h o f A B = 3 2 + ( − 5 ) 2 + 4 2

M

1

M

1

T o t a l m a r k s 1 5

= 5 0 = 7 . 0 7 u n i t s ( 3 s i g . f i g . )

A 1

   

( i i ) A C

= c − a

= ( 4 i + jj – 2 k ) – ( 2 i + 3 jj – k ) = 2 i – 2 jj – k    

L e n g t h o f A C =

2

2

+ 2

2

B 1

+ 1 2

= 3

c o s ∠ B A C =

=

  

   

  

  

A B

A C

A B ⋅ A C

( 3 i − 5 j + 4 k ) ⋅ ( 2 i − 2 j − k )

1

M

1

(

=

5 0 ) ( 3 ) 6 + 1 0 − 4

M

1 5

2

9 5 4 / 1 * T h i s q u e s t i o n p a p e r i s C O N F I D E N T I A L u n t i l t h e e x a m i n a t i o n i s o v e r . C O N F I D E N T I A L * 1 0

2 0 1 2 T R IA L S T P M

B A H A R U 4

=

2

=

5

M A T H E M A T IC S T 2

J IT S IN

,P E N A N G

A 1

5

2

S M J K

∴ ∠ B A C = 5 6 ( n e a r e s t d e g r e e ) �

( i i i ) A r e a o f ∆ A B C

1

=

2

  

A B

A C

⎛ 2 2 ⎞ 5 0 ) ( 3 ) 1 − ⎜ ⎜ 5 ⎟ ⎟ ⎝ ⎠

(

2 1

= 5

M

1

1 7

2 ( 3 )

2 3

=

s i n ∠ B A C 2

1

=

  

5

3 4 2 = 8 . 7 5 ( 3 s i g . f i g . )

A 1

O R   

i

j

k

3

− 5

4

2

− 2

− 1

   

( i i i ) A B × A C =

= ( 5 + 8 ) i – ( – 3 – 8 ) j j + ( – 6 + 1 0 ) k = 1 3 i + 1 1 jj + 4 k A r e a o f ∆ A B C

= = =

1 2 1 2

  

   

A B × A C 1 3

2

+ 1 1 + 4 2

2

M

1

3 0 6

2 = 8 . 7 5 ( 3 s i g . f i g . )

A 1

A v e c t o r e q u a t i o n o f t h e l i n e p a s s i n g t h r o u g h A a a n d B i s g i v e n b y    

M 1 r = ( 2 i + 3 jj – k ) + λ ( A B ) = ( 2 i + 3 jj – k ) + λ ( 3 i – 5 jj + 4 k ) = ( 2 + 3 λ ) i + ( 3 – 5 λ ) j j + ( – 1 + 4 λ ) k p = ( 2 + 3 t ) i + ( 3 – 5 t ) j j + ( – 1 + 4 t ) k A 1 T h i s h a s t h e f o r m g i v e n f o r t h e p o s i t i o n v e c t o r o f P . T h e r e f o r e , f o r a l l v a l u e s o f t , P l i e s o n t h e l i n e


2 0 1 2 T R IA L S T P M

B A H A R U


S M J K

J IT S IN

,P E N A N G

t h r o u g h A a n d B . O R p = ( 2 + 3 t ) i + ( 3 – 5 t ) j j + ( – 1 + 4 t ) k p = 2 i + 3 jj – k + t ( 3 i – 5 jj + 4 k ) . . . ( 1 ) s i n c e p = a + t A B o r s in c e p s a t i A a n d T h e re A a n d

O A s fie s B fo fo re B .

= 2 i + th e v e r a ll v f o r a ll

3 jj – k a c to r e q u a lu e s o f v a lu e s

n d A B = ( 3 i – 5 jj + 4 k ) , a t i o n o f t h e l i n e p a s s e s t h r o u g h t . o f t , P l i e s o n t h e l i n e t h r o u g h

F o r O P t o b e p e r p e n d i c u l a r t o A B ,   

  

O P ⋅ A B = 0 [ ( 2 + 3 t ) i + ( 3 – 5 t ) j j + ( – 1 + 4 t ) k ] ⋅ [ 3 i – 5 jj + 4 k ] = 6 + 9 t t – – 1 5 + 2 5 t – – 4 + 1 6 t = = – 1 3 + 5 0 t t = = t = = ∴ p = ( 2 + 0 . 7 8 ) i + ( 3 – 1 . 3 ) j j + ( – 1 + 1 . 0 4 ) k = 2 . 7 8 i + 1 . 7 jj + 0 . 0 4 k

8 .

D •

M

1

0 0 A 1 0 M 1 0 . 2 6 A 1

1 5

• X

1 B • 1 O •

•

1 A

2

• C

⎛ 3 ⎞ ⎛ 1 ⎞    ⎜ ⎟ ⎜ ⎟ G i v e n O A = 2 ,   O B = 2 ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 3 ⎠    

   

A X

   

   

      

= A C + C X = 2 O A + 2 O B = 2 ( O A + O B )

M

1

A 1


2 0 1 2 T R IA L S T P M

B A H A R U


+ 1 ⎞ ⎟ + 2 ⎟ = 2 + 3 ⎠⎟

⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎜ ⎟ ⎜ ⎟ ⎜ = 2 [ ⎜ 2 ⎟ + ⎜ 2 ⎟ ] = 2 ⎜ 2 ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎝ ⎝ 1 ⎠ ⎝ 3 ⎠ ⎛     ⇒ A X  i s  p a r a l l e l  t o ⎜⎜ ⎜ ⎝

⎛ 4 ⎞ ⎜ ⎟ ⎜ 4 ⎟ = 8 ⎜ 4 ⎟ ⎝ ⎠

S M J K

⎛ 1 ⎞ ⎜ ⎟ ⎜ 1 ⎟ ⎜ 1 ⎟ ⎝ ⎠

J IT S IN

,P E N A N G

A 1

1 ⎞

⎟

1 ⎟ . ( S h o w n ) 1 ⎠⎟ B 1

⎛ 3 ⎜ ( i ) E q u a t i o n o f A X : r = ⎜ 2 ⎜ 1 ⎝               C D = O D − O C = 2 O B − ⎛ 2 ⎞ ⎛ ⎜ ⎟ ⎜ = ⎜ 4 ⎟ − ⎜ ⎜ 6 ⎟ ⎜ ⎝ ⎠ ⎝

⎞ ⎛ 1 ⎞ ⎟ + λ ⎜ 1 ⎟ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ ⎠ ⎝ 1 ⎠     3 O A 9 ⎞ ⎛ − 7 ⎞ ⎟ ⎜ ⎟ 6 = − 2 ⎟ ⎜ ⎟ ⎟ ⎜ 3 ⎠ ⎝ 3 ⎠⎟ ⎛ 2 ⎞ ⎛ − 7 ⎞ ⎜ ⎟ ⎜ ⎟ ∴ E q u a t i o n o f C D : : r = ⎜ 4 ⎟ + µ ⎜ − 2 ⎟ ⎜ 6 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 3 ⎠

( ii) A t p o in t o f in te r s e c tio n 3 + λ = 2 − 7 µ ⇒ λ + 2 + λ = 4 − 2 µ ⇒ λ + 1 + λ = 6 + 3 µ ⇒ λ − ( 1 ) – ( 2 ) : 5 µ = − 3 ⇒

⎛ 3 + λ ⎞ ⎜ ⎟ , 2 + λ = ⎜ ⎟ ⎜ 1 + λ ⎟ ⎝ ⎠ 7 µ = − 1 … 2 µ = 2 … 3 µ = 5 …

µ = −

M

⎛ 2 − 7 µ ⎞ ⎜ ⎟ 4 − 2 µ ⎜ ⎟ ⎜ 6 + 3 µ ⎟ ⎝ ⎠ … … …

1

A 1

M

1

M

1

(1 ) (2 ) (3 )

3 5

⎛

3 ⎞ 1 6 F r o m ( 1 ) , λ = − 1 − 7 ⎜ − ⎟ = 5 ⎝ ⎠ 5 1 6 C h e c k ( 3 ) , L H S = λ − 3 µ = − 3 5

⎛

3 ⎞ − ⎜ ⎟ = 5 = R H S ⎝ 5 ⎠


2 0 1 2 T R IA L S T P M

B A H A R U


S M J K

⎛ 3 1 ⎞ ⎜ 5 ⎟ ⎜ ⎟ 2 6 ⎟ ∴ p o s i t i o n v e c t o r o f p o i n t o f i n t e r s e c t i o n i s ⎜ ⎜ 5 ⎟ ⎜ ⎟ 2 1 ⎜⎜ ⎟ ⎟ ⎝ 5 ⎠ ⎛ 1 − 3 ⎜ ( i i i ) A B = O B − O A = 2 − 2 ⎜ ⎜ 3 − 1 ⎝   

  

   

⎛ − 2 ⎞ ⎛ − 1 ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ 0 2 0 = = ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎠ ⎝ 2 ⎠ ⎝ 1 ⎠ ⎞

⎛ − 1 ⎜ 0 ⎜        ⎜ 1 A B ⋅ A X ⎝ c o s ∠ B A X =         = 1 + 1 A B A X − 1 + 1

=

2

3

⎞ ⎟ ⎟ ⎟ ⎠

J IT S IN

,P E N A N G

A 1

⎛ 1 ⎞ ⎜ ⎟ ⋅ ⎜ 1 ⎟ ⎜ ⎟ ⎝ 1 ⎠ 1 + 1 + 1

M

1

M

1

A 1

= 0

∴ ∠ B A X = 9 0

�

   

i

   

j

k

( i v ) O C × O D = 9 6 2

=

3

4

6

3 4

6

6 i −

9

3 2

6

j +

2

9

6 k 4

= ( 3 6 – 1 2 ) i –– ( 5 4 – 6 ) j j + ( 3 6 – 1 2 ) k = 2 4 i –– 4 8 jj + 2 4 k a r e a o f t h e p a r a l l e l o g r a m O C X D    

2 4

1

M

1

   

= O C × O D

=

M

2

A 1

+ ( − 4 8 ) + 2 4 2

2

= 3 4 5 6 = 5 7 . 7 9


MathematicsT STPM Baharu PENANG 2012

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