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KULIAH 1 - KONSEP PENDUDUK TEMA 1: PENDUDUK DAN SUMBER KULIAH 1 - KONSEP PENDUDUK 1.1. Alam Sekitar Fizikal & Manusia [Terbahagi kepada Tiga Fahaman] A. Determinisme Alam Sekitar · Pandangan determ...
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2 0 1 2 T R IA L S T P M
1 . T h e
(i) F (ii) S o
B A H A R U
M A T H E M A T IC S T S M K S e c t i o n A [ 4 5 m a r k s ] A n s w e r a ll q u e s tio n s in th is s e c tio f u n c t i o n s f a n d g a r e d e f i n e d a s f o l l o w s f : x ↦ x 2 – 2 x , x ∈ ℜ g : x ↦ 2 x + 3 x ∈ ℜ in d th e ra n g e o f f a n d s ta te , w ith a re a s o n , w h e th e r f h a s a n h o w th a t t h e e q u a tio n g f ( x ) = 0 h a s n o r e a l s o l u t io n a n d s ta f g f(x )
S T P A U L S E R E M B A N
N .S .
n
in v e rs e t e t h e d o m a i n
[ 3 ] [4 ]
6 + 7 x 2 . L e t f ( x ) = ( 2 − x ) ( 1 + x 2 )
(i) E x p re s s f(x ) in p a rtia l fra c tio n s [4 ] ( i i ) S h o w t h a t , w h e n x i s s u f f i c i e n t l y s m a l l f o r x 4 a n d h i g h e r p o w e r s t o b e n e g l e c t e d , 1 1 5 f ( x ) = 3 + 5 x − x 2 − xx 3 [ 4 ] 2 4 3 . F i n d t h e i n v e r s e o f m a t r i x A b y u s i n g e l e m e n t a r y r o w ⎛ − 3 − 1 6 ⎞
⎜
o p e r a t i o n s
⎟ − 4 ⎟ ⎜ ⎟ ⎝ − 5 − 2 1 1 ⎠
A = ⎜ 2
[ 4 ]
1
H e n c e , s o l v e t h e s i m u l t a n e o u s e q u a t i o n s 2 x + y – 4 z = 2 -3 x – y + 6 z = -2 -5 x - 2 y + 1 1 z = -5 4 . T h e c o m p l e x n u m b e r 1 + i 3 i s d e n o t e d b y z . ( i ) E x p r e s s z i n t h e f o r m r ( c o s θ + i i s s i n θ ) w h e r e
[3 ]
r > 0 a n d − π <
θ
≤ π . H e n c e , o r
o t h e r w i s e , f i n d t h e m o d u l u s a n d a r g u m e n t o f z 3 ( i i ) F i n d t h e v a l u e s o f a l l t h e c u b e r o o t s o f 4 + 4 i 3 i n t h e f o r m
[ 3 ] r ( c o s θ + i i s s i n
5 . ( i ) F i n d t h e C a r t e s i a n e q u a t i o n o f t h e e l l i p s e w i t h t h e p a r a m e t r i c e q u a t i o n s x = 3 c o s θ - 3 , y = s i n θ + 2 (ii) D e te rm in e th e c e n tre , v e rtic e s a n d fo c i a n d s k e tc h th e e llip s e 6 . T h x (i) (ii)
θ )
[ 4 ]
[2 ] [ 6 ]
e l i n e l h h a s e q u a t i o n r = jj + k + s ( i – 2 jj + k ) . T h e p l a n e p p h a s t h e e q u a t i o n + 2 y + 3 z = 5 S h o w t h a t t h e l i n e l l l i e s i n t h e p l a n e p p [ 2 ] A s e c o n d p l a n e i s p e r p e n d i c u l a r t o t h e p l a n e p p , p a r a l l e l t o t h e l i n e l a a n d c o n t a i n s t h e p o i n t w i t h t h e p o s i t i o n v e c t o r 2 i + j + 4 k . F i n d t h e e q u a t i o n o f t h i s p l a n e , g i v i n g y o u r a n s w e r in th e fo rm a x + b y + c z = d [6 ]
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 S e c t io A n s w e r a n y o n
IC S T S M K S T P A U L S E R E M B A N n B [ 1 5 m a r k s ] e q u e s t i o n i n t h i s s e c t i o n
7 . 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 s i t i o n v e c t o r s o f t h e p o i n t s A
a n d B a r e g i v e n b y
O A = 2 i + 3 j j – k a n d j + 2 k O B = 4 i – 3 j ( a ) U s e a s c a la r p ro d u c t to f in d th e a n g le A O B , c o rr e c t to th e n e a r e s t d e g re e
(b ) (i) T h e p o in t C o f AA B a n ( i i ) I f p p > 0 , v e rtic e s . (iii)A s tra ig h t o f th e s tra 8 . ( a ) T h e p o l y n fa c to rs (x (i) F in d th e (ii) F in d th e
N .S .
[ 4 ]
i s s u c h t h a t O C = 6 j j + p p k w h e r e p i s a c o n s t a n t . G i v e n t h a t t h e l e n g t h s
d AA C a r e e q u a l , f i n d t h e p o s s i b l e v a l u e s o f p p fin d th e a re a o f th e p a ra lle lo g ra m A B C D w h e re A , B , C a n d D
[ 3 ] a r e
[4 ] l i n e p a s s e s t h r o u g h C w h e r e p p < 0 a n d p a r a l l e l t o A B . F i n d t h e e q u a t i o n ig h t lin e in c a rte s ia n fo rm [4 ]
o m ia l p ( x -1 ) a n d (x v a lu e s o f re m a in d e
) = + a a r w
4 x 4 + a x 3 + b x 2 – x + 2 w h e r e a a n d b a r e r e a l c o n s t a n t s h a s 2 ) n d b a n d h e n c e fa c to ris e p (x ) c o m p le te ly [ 7 ] h e n p (x ) is d iv id e d b y (4 x + 1 ) [3 ]
( b ) F i n d t h e s e t o f v a l u e s o f x f o r w h i c h 1 3 < x − 1 x + 2