IJC_JC2_H2_Maths_2012_Solutions_Paper_2

1 IJC/2012/8864/BT/Marking Scheme Solutions for 2012 H2 Maths Prelim 2 Paper 2 1(i) 1(ii) 2(i) 1 4 1 22 2 2 1 d 1 4 u uu u ⋅ −   −    ∫ 1 4 1 22 2 2 2 1 d 1 4 u uuu u = ⋅ − −∫ 1 4 1 2 2 2 1 d 1 4 u u = − −∫ ( ) 1 1 4 1 2 2 1 sin 2 2 u−  = −   1 1 1 1 1 sin sin 2 2 2 − −       = − −             1 2 6 4 24 π π π   = − − =     ( )f 2 y x ′= − 1x = 1x = − O y x 4− 4 y x O ( )fy x = 1y = − 2x = 2x = − 2−

IJC/2012/9740/02/Prelim 2/Solutions 2 2(ii) Area = ( ) 4 22 2 1 1 1 2 2 d 2 4 2 4 x x x   +    − ∫ 21 unit 4 24 π= + 3(i) 1, , . n n n x l x l +→ ∞ → → ( ) ( ) 2 2 10 3 10 3 3 10 0 2 5 0 l l l l l l l l = − = − + − = − + = Solving, 2 or 5 (rejected) l l = = − 3(ii) ( ) 2 2 2 1 10 3 2 6 3 3( ) n n n n x l x x l x + − = − − = − = − 3(iii) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 1 1 1 1 1 Since 0 3( ) 0 0 0 (since 0) n n n n n n n n n n x l l x x l l x x l x l x l x l x l x l + + + + + + + < − > − = − − > − + > − > + > >

IJC/2012/9740/02/Prelim 2/Solutions 3 4 Volume of cone V 21 3 r h π= , where 2r θ= and 2 2 16 4 h π θ = − . ( ) 2 2 2 1 2 16 4 3V π θ π θ = − 4 2 2 4 4 4 3V π θ π θ = − 2 4 6 8 43V π π θ θ = − Differentiating with respect to θ, ( ) ( ) 1 2 4 6 2 3 5 2d 8 1 4 16 6 d 3 2 V π π θ θ π θ θ θ −   = − −       When d 0d V θ = , 2 3 5 16 6 0 π θ θ − = ( ) 3 2 2 16 6 0 θ π θ − = 2 2 16 6 π θ = 2 2 8 3 πθ = 2 3 2 2 28 8 8 43 3 3 V π π π π     = −         6 6 8 256 512 3 9 27 V π π π= − 48 256 3 27 V π= 4128 9 3 V π= 4128 3 27 V π= 128 , 27 p q = = θ 28 3 π − 28 3 π 28 3 π + gradient + ve 0 - ve Shape / _ \ Therefore V is a maximum when 28 3 πθ = (verified).

IJC/2012/9740/02/Prelim 2/Solutions 4 5 12 0 12 8 , 4 , 4 0 7 7 OB OD DB             = = =             −      /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Equation of line DB: 12 12 8 4 0 7 r λ         = +         −    12 4 5 OE     =       /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Let N be the foot of perpendicular from E to line DB . 12 12 12 8 4 , 4 4 7 5 7 k k ON k EN k k k +        = + = +         − − −     /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp . 0 12 12 4 4 . 4 0 5 7 7 144 16 16 35 49 0 51 209 1896 1 1468 209 357 dEN k k k k k k k ON =         + =         − − −     + + + + = = −     =    