SPM Additional Mathematics 2019, Paper 2 (Question 4)


Question 4:
Diagram 2 shows the curve y = 4xx2 and tangent to the curve at point Q passes point P.

Diagram 2

(a) Show that h = 3. [4 marks]
(b) Calculate the area of the shaded region. [4 marks]


Solution:
(a)
y=4x x 2 dy dx =42x At point Q( h, 4h h 2 ) dy dx =42h Equation of tangent at Q y y 1 = dy dx ( x x 1 ) y( 4h h 2 )=( 42h )( xh )

At point P( 2, 5 ), x=2, y=5 54h+ h 2 =( 42h )( 2h ) 54h+ h 2 =84h4h+2 h 2 h 2 4h+3=0 ( h1 )( h3 )=0 h=1 ( rejected ), h=3

(b)
Area of shaded region = Area of trapeziumArea under the curve = 1 2 ( a+b )h 2 3 y dx = 1 2 ( 5+3 )1 2 3 4x x 2  dx =4 [ 4 x 2 2 x 3 3 ] 2 3 =4[ 189( 8 8 3 ) ] =49+( 8 8 3 ) = 1 3  unit 2

Leave a Comment