ACSI 2020 Y3EXP FYE AMath P2 Solutions

Confidential – for internal circulation only 2020 Y3 Express AMath Final Exam P2 Worked Solutions 1 Length of sides 3 22 3 22 += − 322 322 322 322 ++= × −+ ( )3 22 cm= + Perimeter of square ( )12 128 cm= + 2 (i) at the start of the experiment, 0x= . ( ) 040 1.4 40y= = (ii) ( )1000 40 1.4 x= ( )25 1.4 x= lg 25 lg1.4x= lg25 lg1.4x= = 10 hrs (to the nearest hour) 3 When 1x= , ( ) ( ) ( ) ( ) 32 1 3 1 2 1 16 0 0 3 c+ − + −=+ 6c= When 2x= , ( ) ( ) ( ) ( ) 232 23 2 16 2 1 1 2x x x b x x ax x c x+ − +− − −= −+ + ( ) ( ) ( ) ( ) ( ) 32 2 2 32 22 1 6 0 2 64 a+ − + −= + 2a= When 0x= , ( )16 4 6 2b+= 1b=−

2 4 (i) LHS 2 3 sec 1 tan tan θ θθ −= + ( ) 2 2 tan tan 1 tan θ θθ = + 2 tan sec θ θ= 2sin cos.cos 1 θθ θ= sin cosθθ= (ii) 2 2 3 sec 1 2costan tan θ θθθ − =+ 2sin cos 2cosθθ θ = 2sin cos 2cos 0θθ θ −= ( )cos sin 2cos 0θθ θ −= cos 0θ = (N.A) sin 2cosθ θθ−= tan 2θ = Ref ∠ of θ =63.4o 63.4 243.4ooorθ = 5 (a) 0.9 cos 30 02 ox− −= cos 30 0.92 ox −= , 30 30 602 o oo x− ≤− ≤ Ref ∠ of 130 cos (0.9)2 ox −−= = 25.842o 30 25.842 25.8422 oo ox or−= − 8.3 111.7oox or= (b) 2 32cos 0secx co x+=

3 22cos 3sin 0xx+= ( ) 22 1 sin 3sin 0xx−+= 22 2sin 3sin 0xx− += 22sin 3sin 2 0xx− −= ( )( )2sin 1 sin 2 0xx+ −= ( )1sin sin 2 . .2x or x N A= −= Ref ∠ = 6 radπ 7 11 66x rad or radππ= 6a) For ( ) ( ) 223 4 2 0px px− + − +> , D < 0 and a > 0 ( ) ( )( ) 2 4 42 3 2 0pp−−− < 2 16 0pp+< ( )16 0pp +< 16 0 p−<< Since a > 0, 23 0 p−> 2 3p< Answer: 16 0 p−<< (b) 24 32 7x x kx− += −− ( ) 24 3 90xk x+ − += For line not to intersect curve, 0D< . ( ) ( )( ) 2 3 44 9 0k−− < 9 15k−< < 14k = 7 (a) ( ) ( )0,5 10, 0P and R (b) Since Q lies on the perpendicular bisector, it is equidistant from P and R.

4 ( ) ( ) ( ) ( ) 22 22 7 0 5 7 10 0qq− +− = − +− 2249 10 25 9qq q+− += + 6.5q= Alternatively, Mid-point of PR ( )5, 2.5 Gradient perpendicular to PR = 1 50 0 10 − − − = 2 Equation of perpendicular bisector: ( )2.5 2 5yx−= − 2 7.5yx= − ( )2 7 7.5 6.5q= −= (C) PQ =QR since perpendicular bisector passes through Q. Mid-point of QS = 36.573 2,22  − +    ( )5, 2.5= Mid-point of PR = Mid-point of QS (opposite sides are parallel) PQRS is a rhombus. Alternatively, ( ) ( ) 22 2057 0 6.5 5 2PQ= −+ −= ( ) ( ) 22 2057 10 6.5 0 2QR=− +−= ( ) 2 2 3 20510 3 0 22QS = −++ =  ( ) 2 2 3 20530 5 22PS = − +− − = Since the four sides are equal, it is a rhombus.

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