SPM Additional Mathematics 2017, Paper 2 (Question 10 & 11)


Question 11:
Diagram 6 shows a curve y = 2x2 – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.
(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180o about the y-axis, the volume generated is 65π unit3.
Find the value of h.

Solution: 
(a)
y=2 x 2 18 dy dx =4x Gradient of straight line PQ=4 4x=4 x=1 When x=1,  y=2 ( 1 ) 2 18=16 Coordinates of K=( 1,16 ).


(b)
At x-axis, y=0 2 x 2 18=0 2 x 2 =18 x 2 =9 x=±3 The curve cuts the x-axis at ( 3,0 ) and ( 3,0 ). Area of shaded region = Area of triangleArea under the curve = 1 2 ( 51 )( 16 ) 1 3 ydx =32 1 3 ( 2 x 2 18 )dx =32| [ 2 x 3 3 18x ] 1 3 | =32| ( 2 ( 3 ) 3 3 18( 3 ) )( 2 ( 1 ) 3 3 18( 1 ) ) | =32| ( 1854 2 3 +18 ) | =32| 18 2 3 | =3218 2 3 =13 1 3  units 2


(c)
Volume generated=65π π h 0 x 2 dy =65π π h 0 ( y 2 +9 )dx =65π y=2 x 2 18 x 2 = y 2 +9 [ y 2 4 +9y ] h 0 =65 0( h 2 4 +9h )=65 h 2 4 9h=65 h 2 +36h+260=0 ( h+10 )( h+26 )=0 h=10   or   h=26 ( rejected ) Thus, h=10


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