kedah trial stpm 2012-mathst paper 2(q&a)

7/31/2019 Kedah Trial Stpm 2012-Mathst Paper 2(Q&A)

1/13

S T P M K E D A H 2 0 1 2 - M T P A P E R 2 C H U / S M K K

1 P r o v e t h a t ( ) ( ) A A A A 3 s i n

4 1

6 0 s i n 6 0 s i n s i n = + . [ 4 m ]

2 G i v e n A B

= ( 2 k 1 ) i i i i j j j j a n d A C

= 6 i i i i + k j j j j a r e t w o v e c t o r s l i e o n a s a m e

p l a n e . F i n d t h e v a l u e o f k

( a ) i f t h e p o i n t s A , B a n d C a r e c o l l i n e a r ,

( b ) i f B A C = 9 0 o . [ 4 m ]

3 T h e d i a g r a m s h o w s a c i r c l e . P Q i s a t a n g e n t t o t h e c i

s t r a i g h t l i n e P R c u t s t h e c i r c l e a t S . S h o w t h a t

( a ) i f R Q S = Q P S , t h e n t h e l e n g t h o f c h o r d Q R i s t w i c e to f t h e c i r c l e ,

( b ) i f R Q S = 3 0 o , t h e n t h e l e n g t h o f c h o r d R S i s e q u a l t o t h c i r c l e .

[ 3 m ]

[ 3 m ]

4 F i n d t h e g e n e r a l s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n2 5 0 .

d y x y y

d x + =

D e d u c e t h a t t h e p a r t i c u l a r s o l u t i o n f o r w h i c h y = 1 w he x p r e s s e d i n t h e f o r m a n d s k e t c h t h e c u r v e .

S k e t c h , i n s e p a r a t e d i a g r a m , t h e s o l u t i o n c u r v e w h i c

p o i n t ( 1 , 0 ) .

[ 3 m ]

[ 2 m ]

[ 3 m ]

5 I n a p a r a l l e l o g r a m A B C D , X i s t h e m i d p o i n t o f A B a nd i a g o n a l A C a t P .

G i v e n t h a t A B = a , a , a , a , A D = = = = b , b , b , b , A P = = = = A C a n d D P = D X , e x p r e s s A P i n t e r m s o f ( a ) , a a a a a n d b b b b . ( b ) , a a a a a n d b b b b . H e n c e , d e d u c e t h a t P i s a p o i n t o f t r i s e c t i o n o f b o t h A

[ 2 m ] [ 3 m ] [ 5 m ]

6 E x p r e s s ( c o s x + 1 ) 2 + ( s i n x + 3 ) 2 i n t h e f o r m a + b c o s ( x ) , w h e r e a >

0 , b > 0 a n d 0 < < 1 2

. H e n c e ,

( a ) f i n d t h e s e t o f v a l u e s o f k s u c h t h a t t h e f o l l o w i n g e

( c o s x + 1 ) 2 + ( s i n x + 3 ) 2 = k 2 ,

[ 4 m ]

[ 3 m ]


2/13

( b ) s o l v e , f o r 0 < x < 2 , t h e e q u a t i o n

( c o s x + 1 ) 2 + ( s i n x + 3 ) 2 = 5 + 2 2 ,

( c ) f i n d t h e v a l u e s o f p a n d q s u c h t h a t 2 ( c o s x + 1 ) 2 + ( s i n x + 3 ) 2

l i e s i n

t h e i n t e r v a l ( p , q ) a n d t h e c o r r e s p o n d i n g v a l u e s o f x .

[ 3 m ]

[ 5 m ]

7 T h e r e s u l t o f a v e t e r i n a r y s u r v e y o n 1 2 5 r a b b i t s , e a cp a r t i c u l a r d i s e a s e i s s u m m a r i z e i n t h e f o l l o w i n g t a b l

D i s e a s e d N o t D i s e a s e d T o t a l F e m a l e 2 5 3 5 6 0 M a l e 2 0 4 5 6 5 T o t a l 4 5 8 0 1 2 5

T h e c a s e h i s t o r y o f e a c h r a b b i t w a s k e p t o n a s e p t h r e e d i f f e r e n t r e c o r d c a r d s a r e s e l e c t e d a t r a n d o m

c a l c u l a t e t h e p r o b a b i l i t y t h a t

( a ) a l l t h r e e r e c o r d c a r d s r e l a t e t o r a b b i t s w i t h t h e ( b ) o n e r e c o r d c a r d r e l a t e s t o a f e m a l e r a b b i t w i t h

m a l e r a b b i t w i t h t h e d i s e a s e a n d o n e t o a f e m a l e r a

t h e d i s e a s e .

[ 3 m ]

[ 3 m ]

8 T h e p r o b a b i l i t y t h a t a s h o o t e r s t r i k e s a t a r g e t i n o n

p r o b a b i l i t y t h a t n o t l e s s t h a n t h r e e o f f i v e s h o t s f i r e d ,

A f t e r t h e s h o o t e r h a s c o m p l e t e d a o n e m o n t h s h

p r o b a b i l i t y t h a t h e s t r i k e s t h e t a r g e t i n o n e s h o t i s 0 .

a p p r o x i m a t i o n , f i n d t h e p r o b a b i l i t y t h a t i n a s a m p l e o

s h o t s s t r i k e t h e t a r g e t .

[ 3 m ]

[ 4 m ]

9 T h r e e r e d b a l l s a n d t h r e e w h i t e b a l l s a r e p l a c e d i n a

b y o n e , a t r a n d o m a n d w i t h o u t r e p l a c e m e n t . T h e r a

n u m b e r o f w h i t e b a l l s d r a w n b e f o r e t h e f i r s t r e d b a l l i

( a ) S h o w t h a t P ( X = 1 ) = 1 0 3 , a n d f i n d t h e r e s t o f t h e p r o b a b i l i t y d

f u n c t i o n o f X .

( b ) F i n d E ( X ) a n d s h o w t h a t V a r ( X ) = 8 0 6 3

.

[ 3 m ]

[ 4 m ]


3/13

1 0 T h e t a b l e s h o w s t h e d u r a t i o n s o f 6 0 j o u r n e y s o n t ht h e v a r i a t i o n s i n j o u r n e y t i m e s b e i n g c a u s e d b y v a r y

C a l c u l a t e , t o t h e n e a r e s t m i n u t e , e s t i m a t e s o f t h e d e v i a t i o n f o r t h e d u r a t i o n o f t h e j o u r n e y s .

W h e n t h e t i m e s f o r 4 0 o t h e r j o u r n e y s w e r e t a k e n , i t wa n d s t a n d a r d d e v i a t i o n f o r t h e t i m e s o f t h e s e 4 0 j o u

a n d 1 8 m i n , r e s p e c t i v e l y . F i n d a l s o , t o t h e n e a r e s t mm e a n a n d s t a n d a r d d e v i a t i o n f o r t h e d u r a t i o n o f a l l 1

T i m e T i m e T i m e T i m e o f o f o f o f j o u r n e y j o u r n e y j o u r n e y j o u r n e y i n i n i n i n h o u r s h o u r s h o u r s h o u r s

N u m b e r N u m b e r N u m b e r N u m b e r o f o f o f o f j o u r n e y s j o u r n e y s j o u r n e y s j o u r n e y s

4 . 6 4 . 8 2 4 . 8 5 . 0 7 5 . 0 5 . 2 1 6 5 . 2 5 . 4 2 1 5 . 4 5 . 6 1 2 5 . 6 5 . 8 2

[ 4 m ]

[ 5 m ]

1 1 T h e c o n t i n u o u s r a n d o m v a r i a b l e X i s t h e d i s t a n c e ,

k i l o m e t e r s , t h a t a p a r t i c u l a r c a r w i l l t r a v e l o n a f u l l t

t h a t

= , 1

, 2 4 1 0

, 0

) ( 2 x x x X P a a

w h e r e a i s a c o n s t a n t . S h o w t h a t a = 1 . ( a ) F i n d t h e p r o b a b i l i t y d e n s i t y f u n c t i o n o f X .

( b ) S h o w t h a t E ( X ) = 3

1 3 .

( c ) T h r e e i n d e p e n d e n t o b s e r v a t i o n s o f X a r e t a k e n .

t w o o f t h e o b s e r v a t i o n s a r e l e s s t h a n 4 . 5 a n d o n e i s g

[ 2 m ]

[ 3 m ]

[ 3 m ]

1 2 A f a c t o r y p r o d u c e s b o t h m e t a l r o d s a n d t u b e s . T h e

o f a m e t a l t u b e i s d i s t r i b u t e d N ( 2 . 2 , 0 . 0 0 0 9 ) . T h e d i a

r o d i s d i s t r i b u t e d N ( 2 . 1 5 , 0 . 0 0 0 4 ) .

( a ) F i n d t h e p e r c e n t a g e o f t u b e s w i t h i n t e r n a l d i a m e( b ) F i n d t h e p e r c e n t a g e o f r o d s w i t h d i a m e t e r g r e a tt h a n 2 . 2 c m . ( c ) A r o d a n d a t u b e a r e c h o s e n a t r a n d o m . F i n d t h e

w i l l n o t p a s s t h r o u g h t h e t u b e .

( d ) T h r e e p a c k e t s , e a c h c o n t a i n i n g a r o d a n d a t u b e

F i n d t h e p r o b a b i l i t y t h a t t w o p a c k e t s c o n t a i n r o d s t h

r e s p e c t i v e t u b e s a n d t h e o t h e r w i l l n o t .

[ 3 m ]

[ 3 m ]

[ 5 m ]

[ 3 m ]

x

x

x

5

, 5 4

, 4


4/13

S T P M K E D A H 2 0 1 2 - M T P A P E R 1 - M A R K I N G S C H E M E

1 . 2 x = 3 y

2 x = 3 1 x

x l n 2 = ( 1 x ) l n 3

x ( l n 2 + l n 3 ) = l n 3

x = l n 3 l n 6

.

2 .

3 .

2 s e c = d d x

,

t a n 0 = 0 , t a n 4

= 1

( 1 + x 2 ) 2 = ( 1 + t a n 2 ) 2 = s e c 4

( ) =

+ 4

0

2

4

1

0 2 2 s e c

s e c

1

1

1

d x x

d

= 4 0 2 c o s

d

( ) ( ) + = +

1

0

4

0 2 2

2 c o s 1 2 1

1

1

d d x x

= 1 2

4

0

1 s i n 2

2

+

B 1

M 1 A 1

A 1

B 1

M 1

M 1 A 1

M 1

A 1

A 1

M 1

B 1


5/13

= 1 8

[ + 2 ] .

4 .

5 . c o s 3 = x y a x

c o s 3 = y x a x

d x d y

x x y 3 2 ) ) ( 3 ( + = a x a s i n

d x

d y x y x 3 2 3 + = a x a s i n

( ) ( ) ( ) ( ) a x a x d x d y

d x y d

x x y d x d y

x c o s 3 2 3 2 2 2 2

3 2 = + +

+

a x a d x d y

x d x

y d x x y

d x d y

x c o s 3 6 3 2 2 2 2

3 2 = + + +

a x a d x y d

x x y d x d y

x c o s 6 6 2

2

2 3 2

= + +

S u b i n t o

3 2 2

2 3 2 6 6 y x a

d x y d

x x y d x d y

x = + +

0 6 6 3 2 2

2 3 2 = + + + y x a

d x

y d x x y

d x

d y x

D 1 : x y =

D 1 : s h a p e o f 2 = x y

D 1 : p o i n t s ( 2 , 0 ) , ( 0 , 2

T h e s o l u t i o n s e t i s } { < < x x x , 4 1 : B 1

A 1

1

2

1 2

M 1 A 1

M 1 A 1

M 1


6/13

( ) 0 6 6 2 2 2 2

2 =

+ + + y x a y

d x d y

x d x

y d x x

( ) 0 6 6 2 2 2 2

2 = + + + y x a d x d y

x d x

y d x

6 .

L e t ( ) ( ) 4 :

4 :

2

1

+ = =

x m y l

x m y l

i n t e r c e p t 2 5 5 4 = + y x a t p o i n t F ( x 1 , y 1 ) a n d G ( x 2 , y 2 ) .

( ) [ ]

( )

m m

x

m m x

m m x x

x m x

5 4 2 0 2 5

2 0 2 5 5 4

2 5 2 0 5 4

2 5 4 5 4

1

1

1 1

1 1

+ + =

+ = + = + = +

+ + = 4

5 4 2 0 2 5

1 m m

m y

m m y 5 4

9 1 +

=

+ + +

m m

m m

F 5 4

9 ,

5 4 2 0 2 5

( ) [ ]

( )

m m x

m m x

m m x x

x m x

5 4 2 0 2 5

2 0 2 5 5 4

2 5 2 0 5 4

2 5 4 5 4

2

2

2 2

2 2

+ =

+ = + = + = + +

+

+ = 4

5 4 2 0 2 5

2 m m

m y

m m

y 5 4

4 1 2 +

=

+ +

m

m

m

m G

5 4

4 1 ,

5 4

2 0 2 5

A 1

B 1

( a ) M 1

A 1

A 1

( l 1 o r l 2 )

A 1


7/13

8 5 4

4 1 5 4

9 5 4 2 0 2 5

5 4 2 0 2 5

2 2

=

+

+ +

+

+ +

m m

m m

m m

m m

2 2 2

8 5 4

3 2 5 4

4 0 =

+ +

+ m m

m m

6 4 2 5 4 0 1 6

2 6 2 4 2

2

= + + m m m

0 1 6 4 0 1 6 2 = + + m m

0 2 5 2 2 = + + m m

( ) ( ) 0 2 1 2 = + + m m

2 , 2 1 = m

7 . ( a ) | w | = 4 1 ,

a r g w = t a n 1 4 5

= 2 . 4 7 r a d . ( 3 s . f . )

( b ) z w

= 3 + 2 i 5 + 4 i

x 5 4 i 5 4 i

= 7 4 1

2 2 4 1

i

( c ) Z ( 3 , 2 ) , W ( 5 , 4 ) , Z * ( 3 , 2 ) , P ( x , y )

m i d - p t o f Z Z * = m i d - p t o f W P

( 3 , 0 ) = ( x 5 2

, y + 4 2

)

x = 1 1 , y = 4

p = 1 1 4 i .

8 .

( b )

M 1

A 1

M 1 A 1

M 1 A 1

A 1

A 1

M 1

B 1

5 l n 2 l n x y =

5 l n 2 1 =

d x d y

y

5 l n 2 y d x d y =

2 5 l n 5 2 x

d x

d y =

M 1

A 1

M 1


8/13

( ) 1

0

2 2 2 5 l n 5 d x x x

= ( )

1

0

2 2 5 l n 5 2 5 l n d x x x

= 1

0

2 2 2 5 l n 5 2 5 l n 5

d x x x x

= [ ] 1 0 2 2 5 2 5 l n 5 x x x

= 5 0 l n 5 2 4

H e n c e , m = 5 0 , n = 2 4

9 . ( a ) D o m a i n : } { 5 , : = x x x g

R a n g e : } { 0 , : = y y y g

( b ) ( ) 3 ) ( + = x x f g

( ) ( ) 3 2

5 3 2 + = + + x b x a

( ) ( ) [ ] 2 2

3 2

5 3 + = + + x b x a

1 2

= a

a = 2

0 2

5 = + b

b = 5

B 1 B 1

M 1

M 1 A 1

A 1 A 1

B 1

B 1

M 1

A 1

A 1


9/13

( c )

S i n c e t h e l i n e p a r a l l e l t o t h e x - a x i s i n t e r s e c t s t h e c u r v e y = f i s o n e - t o - o n e f u n c t i o n .

f ( x ) = 2 ( x + 3 ) 2 5

2 ( y + 3 ) 2 5 = x

2 ( y + 3 ) 2 = x + 5

( ) 2

5 3 2

+ = + x y

2 5

3 + = + x y

2

5 3

+ = + x y

2 5

3 + = x y

5 , 2

5 3 ) ( 1 + = x x x f

1 0 .

=

+

+

2 2 2

2 5 1 2 2 4

1 0 0

0 1 0 0 0 1

3 2 2

1 2 1 1 0 1

3 2 2

1 2 1 1 0 1

3 2 2

1 2 1 1 0 1

b a

=

+

+

2 2 2

2 5 1

2 2 4

0 0

0 0

0 0

3 2 2

2

0

9 1 0 1 0

4 6 5

4 2 1

b

b

b

a a a

a a a

a a

=

+ + + + + + + + + +

2 2 2

2 5 1 2 2 4

3 9 2 1 0 2 1 0

4 2 6 5 4 2 1

b a a a

a b a a a b a

B 1

S h a p e D 1

A l l c o r r e c t D 1

s i n c e 3 y

M 1

A 1

M 1

A 1

f ( x )

x - 3

- 5


10/13

6

1 5

= = +

a

a

1 1 4 1 = = + + b b a

( )

= + +

2 2 2

2 5 1

2 2 4

3 2 2

1 2 1

1 0 1 2 b I a P P P

=

6 0 0

0 6 0

0 0 6

( ) I b I a P P P 6 2 = + +

=

2 2 2

2 5 1

2 2 4

6 1 1 P

=

5

2

3

3 2 2

1 2 1

1 0 1

z

y

x

=

5

2

3

2 2 2

2 5 1

2 2 4

6 1

z

y

x

=

0 2 1

3

z y x

0 , 2 1

, 3 = = = z y x

A 1

A 1

A 1

B 1

B 1

A 1

M 1

B 1


11/13

1 1 .

( a )

( b )

C o n s i d e r ,

S i n c e

H e n c e

H e n c e ,

M 1

A 1

M 1 A 1

A 1

M 1 A 1


12/13

1 2 . ( a ) t h e a s y m p t o t e i s x = 5 4 .

( b ) f ( x ) = 4 x + 4 4 x + 5

.

= 4 ( 4 x + 1 ) ( x + 1 ) 4 x + 5

f ( x ) = 4 + 4 ( 1 ) ( 4 x + 5 ) 2 ( 4 )

= 4 ( 4 x + 7 ) ( 4 x + 3 ) ( 4 x + 5 ) 2

W h e n f ( x ) = 0 , ( x , y ) = ( 1 , 2 ) o r ( 1 4

, 1 8

+ l n 4 )

M 1

A 1

A 1

c )

0 1 1

6 1

1 3 1

1 2 1

4 2 3 4

= +

+

x x x x

B 1

M 1 r e f e r t o ( a )

A 1

s i g n o f f

( x ) :

s i g n o f f

( x ) :

1 1 4

x

3 4

+ +

5 4

+ +

B 1

M 1 f o r f a n d f

M 1 A 1 f o r s t a t i o n a r p o i n t s

M 1 d e t e r m i n e n a t u ro f s t a t i o n a r y p o i n t s

A 1 f o r f o r f


13/13

l o c a l m a x i m u m a t ( 1 , 2 )

l o c a l m i n i m u m a t ( 1 4 , 1 8 + l n 4 )

( c ) W h e n f ( x ) = 0 , x = 3 4

{ 7 4

i s n o t i n t h e g i v e n d o m a i n }

i n f l e x i o n a t ( 3 4

, 9 8

+ l n 2 )

( e ) m a x i m u m f o c c u r s a t x = 1

m a x . f i s 2 + l n 9

A 1

A 1

M 1 A 1

A 1

B 1

D 1 f o r s h a p e

D 1 f o r a s y m p t o t e

D 1 f o r p o i n t s i n ( b ) & ( c )

x O

y

( 1 , 2 + l n 9 )

( 1

4 ,

1

8 + l n 4 )

( 1 , 2 )

( 3

4 ,

9

8 + l n 2 )

4 5 = x

( d )

kedah trial stpm 2012-mathst paper 2(q&a)

Documents