SPM Additional Mathematics 2018, Paper 2 (Question 5 & 6)


Question 5:
Mathematics Society of SMK Mulia organized a competition to design a logo for the society.


Diagram 3 shows the circular logo designed by Adrian. The three blue coloured regions are congruent. It is given that the perimeter of the blue coloured region is 20π cm.
[Use π = 3.142]
Find
(a) the radius, in cm, of the logo to the nearest integer,
(b) the area, in cm2, of the yellow coloured region.

Solution:
(a)
6 arcs =20π 6rθ=20π 6r[ 60 o × π 180 o 3 ]=20π 2πr=20π r=10 cm

(b)

Area of yellow coloured region =3[ area of triangle OAB ]6[ area of segment ] =3[ 1 2 absinC ]6[ 1 2 r 2 ( θsinθ ) ] =3[ 1 2 ( 10 )( 10 )sin 120 o ]6[ 1 2 ( 10 ) 2 ( θsinθ ) ] =3( 43.3013 )6[ 50( 1.0473sin1.0473 ) ] change to rad mode θ= 60 o × 3.142 180 o =1.0473 =129.90396( 9.0612 ) =129.903954.3672 =75.54  cm 2



Question 6:
Diagram 4 shows the front view of a part of a roller coaster track in a miniature park.

The curve part of the track of the roller coaster is represented by an equation y= 1 64 x 3 3 16 x 2 , with point A as the region.
Find the shortest vertical distance, in m, from the track to ground level.

Solution:
y= 1 64 x 3 3 16 x 2  …………… ( 1 ) dy dx =3( 1 64 ) x 2 2( 3 16 ) x 1     = 3 64 x 2 3 8 x At turning point,  dy dx =0 3 64 x 2 3 8 x=0 x( 3 64 x 3 8 )=0 x=0  or 3 64 x 3 8 =0 3 64 x= 3 8 x= 3 8 × 64 3 x=8 Substitute values of x into equation (1): When x=0, y= 1 64 ( 0 ) 3 3 16 ( 0 ) 2 y=0 When x=8, y= 1 64 ( 8 ) 3 3 16 ( 8 ) 2 y=4 Thus, turning points : ( 0, 0 ) and ( 8,4 )

dy dx = 3 64 x 2 3 8 x d 2 y d x 2 =2( 3 64 )x 3 8       = 3 32 x 3 8 When x=0, d 2 y d x 2 = 3 32 ( 0 ) 3 8       = 3 8 ( <0 ) ( 0, 0 ) is maximum point. When x=8, d 2 y d x 2 = 3 32 ( 8 ) 3 8       = 3 8 ( >0 ) ( 8,4 ) is minimum point. Shortest vertical distance between track  and ground level is at the minimum point. Shortest vertical distance =54 =1 m


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