ACSI 2021 Y3EXP FYE AMath P2 Solutions

FINAL EXAMINATION 2021 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 2 SOLUTIONS 1. Required Height ( ) ( ) ( ) 2 16 6 2 4 8 3 2 422 22 2 ++ = = − − ( )28 32 22 22 22 + += × −+ ( )2 1 68 2626 42 +++ = − 22 14 2 cm= + 2. ( ) ( ) ( ) 2228 5 32 1 3x Ax x B x C x+ −≡ − + − − − 2 228 5 9 12 4 2 3 3x Ax x x Bx Bx B C x+ −≡− + + − +− + ( ) ( ) ( ) 228 54 2 9 9 3x Ax B x B x B C+ −≡ + − + + +− Comparing corresponding coefficients: 2x : 48 4BB+= ∴= x : ( )2 4 9 17AA= − ×+ ∴= − 0x : 5943 6 CC−=+− ∴ = 3a. LHS: ( ) 22 2 2 1 sin1 sin(sec tan ) cos cos θθθθ θθ −−−= =  ( ) 2 2 1 sin 1 sin θ θ −= − ( ) ( )( ) 2 1 sin 1 sin 1 sin θ θθ −= −+ 1 sin 1 sin θ θ −= + = RHS (proven)

2 3b. 2(sec tan ) 3θθ−= for 0 360oo θ≤≤ . 1 sin 31 sin θ θ −⇒=+ 1 sin 3 3sinθθ−= + 4sin 2θ =− 1sin 2θ =− 1 1sin 302α −= =  210 ,330θ∴=  4a. 10 sin 9 2 cosec xx+= − for 0 360o x≤≤  210 sin 9 sinx x+= − 210sin 9sin 2 0xx+ += ( )( )5sin 2 2sin 1 0xx+ += 2sin 5x=− 1or sin 2x=− 1 2sin 23.5785α −= =  1 1sin 302α −= =  203.6 , 210 , 336.4 , 330x∴=   4b. sin 2 cos 244 ππθθ += +   for 924 44 π ππθ≤ +≤ . tan 2 1 4 πθ += 1tan 1 rad4 πα −= = 592 ,, 44 4 4 ππ π πθ∴+= 0, , 2 πθπ∴=      

3 4c. 1 cos 2 sin 2cos 2 2sin 12 xx xx−= −⇒ =+ ∴ 4 solutions 5a. 5b. ( )0.0004279 20000 100Me − = 0.0192g= 5c. 0.000427950 100 te−= 0.00042791 2 te−= 1ln 0.00042792 t=− 1 2ln 0.0004279t∴= − 1619.88 1620 ∴Half-life of radium is 1620 years. x y t M 100

4 6a. 23yx= --------(1) 3.y px= − --------(2) Solving (1) & (2) simultaneously: 233x px= − or 233x px>− 23 30x px− += 23 30x px− +> For curve to be above the line: ( ) ( )( ) 2 43 3 0p−− < 2 36 0p −< ( )( )6 60pp+ −< 66 p∴− < < 6b. ( )( )12 2 13 2 023 x xx−<< ⇒ + − < 26 20xx−−< Comparing corresponding terms: 6, 1pq∴= = − 7a. ( )21 1 06 2 PRm −−= =−− ⇒Equation for PR: 1 22yx= −+ 2QSm⇒= ⇒Equation for QS: ( ) ( )12 1yx−− = − 23yx∴= − 7b. Coordinate of M 1 22yx= −+ ------(1) 23yx= − -------(2) 1 22 32 xx− += − 5 52 x= 2, 1xy∴= = ( )2,1M∴ Coordinate of P Using midpoint M of PR: 61 2; 122 xy+− = = 2, 3xy∴= − = ( )2,3P − Coordinate of S Both diagonals intersect at M: 1 6 2; 1 1 3xy+ = − −+ = −+ 3, 3xy∴= = ( )3, 3S∴

5 8a. Table of values Linearizing the given equation: 2 2 3 3 3 ; y ax bx y ax bx y ax bx Y mX c ma cb = +− += + + = + = + ⇒≡ ≡ b. From the graph:  23 7Gradient: 4.0 4.05.5 1.5ma − ⇒∴−   3 - intercept: c 1.0 1.0y bx + ⇒∴ c. ( ) 2 20 0ax b x+− = 2 3 20 3 20 3 3 20 ax bx x yx y x + −= − ⇒= − +⇒= ⇒ Insert the horizontal line 3 20y x + = , or 20Y = then read off the corresponding value of x from the point of intersection. x 1 2 3 4 5 3y x + 5 9 13 17 21

6 8a. Graph of 3y x + against x _________________________________________________________________________________________ 2 4 6 8 10 12 14 16 18 20 0 1 2 3 4 5 6 7 22