SPM Additional Mathematics 2018, Paper 2 (Question 7 – 9)


Question 7:
( a ) Prove sin( 3x+ π 6 )sin( 3x π 6 )=cos3x ( b ) Hence, ( i ) solve the equation sin( 3x 2 + π 6 )sin( 3x 2 π 6 )= 1 2  for 0x2π  and give your answer in the simplest fraction form in terms of π radian. ( ii ) sketch the graph of y=sin( 3x+ π 6 )sin( 3x π 6 ) 1 2  for 0xπ.

Solution:
( a ) Left hand side, sin( 3x+ π 6 )sin( 3x π 6 ) =[ sin3xcos π 6 +cos3xsin π 6 ][ sin3xcos π 6 cos3xsin π 6 ] =2[ cos3xsin π 6 ] =2[ cos3x( 1 2 ) ] =cos3x( right hand side )


( b )( i ) sin( 3x 2 + π 6 )sin( 3x 2 π 6 )= 1 2 ,0x2π cos 3x 2 = 1 2 3x 2 = π 3 ,( 2π π 3 ),( 2π+ π 3 ) 3x 2 = π 3 , 5π 3 , 7π 3 x= 2π 9 , 10π 9 , 14π 9


( b )( ii )  y=sin( 3x+ π 6 )sin( 3x π 6 ) 1 2  for 0xπ. y=cos3x 1 2




Question 8:
Diagram 5 shows triangles OAQ and OPB where point P lies on OA and point Q lies on OB. The straight lines AQ and PB intersect at point R.
It is given that  OA =18 x ˜ ,  OB =16 y ˜ , OP:PA=1:2, OQ:QB=3:1, PR =m PB  and  QR =n QA , where m and n are constants. ( a ) Express  OR  in terms of    ( i ) m,  x ˜  and  y ˜ ,    ( ii ) n,  x ˜  and  y ˜ , ( b ) Hence, find the value of m and of n. ( c ) Given | x ˜ |=2 units, | y ˜ |=1 unit and OA is perpendicular to OB calculate | PR |.

Solution
(a)(i)
OR = OP + PR  = 1 3 OA +m PB  = 1 3 ( 18 x ˜ )+m( PO + OB )  =6 x ˜ +m( 6 x ˜ +16 y ˜ )

(a)(ii)
OR = OQ + QR  = 3 4 OB +n QA  = 3 4 ( 16 y ˜ )+n( QO + OA )  =12 y ˜ +n( 12 y ˜ +18 x ˜ )  =( 1212n ) y ˜ +18n x ˜


(b)
6 x ˜ +m( 6 x ˜ +16 y ˜ )=( 1212n ) y ˜ +18n x ˜ 6 x ˜ 6m x ˜ +16m y ˜ =18n x ˜ +12 y ˜ 12n y ˜ by comparison; 66m=18n 1m=3n m=13n…………..( 1 ) 16m=1212n 4m=33n…………..( 2 ) Substitute (1) into (2), 4( 13n )=33n 412n=33n 9n=1 n= 1 9 Substitute n= 1 9  into (1), m=13( 1 9 ) m= 2 3


(c)
| x ˜ |=2| y ˜ |=1  PR = 2 3 PB  = 2 3 ( 6 x ˜ +16 y ˜ )  =4 x ˜ + 32 3 y ˜ | PR |= [ 4( 2 ) ] 2 + [ 32 3 ( 1 ) ] 2   = 1600 9   = 40 3  units



Question 9:
A study shows that the credit card balance of the customers is normally distributed as shown in Diagram 6.

(a)(i) Find the standard deviation.
(ii) If 30 customers are chosen at random, find the number of customers who have a credit card balance between RM1800 and RM3000.
(b) It is found that 25% of the customers have a credit card balance less than RM y.
Find the value of y.

Solution:
(a)(i)
μ=2870,x=3770 P( X>3770 )=15.87% P( Z> 37702870 σ )=0.1587 P( Z>1.0 )=0.1587 37702870 σ =1.0 σ=900


(a)(ii)
P( 1800<X<3000 ) =P( 18002870 900 <Z< 30002870 900 ) =P( 1.189<Z<0.144 ) =1P( Z1.189 )P( Z0.144 ) =10.11720.4427 =0.4401 Number of customers=0.4401×30   =14


(b)
μ=2870,x=y P( x<y )=25% P( Z< y2870 900 )=0.25 y2870 900 =0.674 y=2263.40


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