ACSI 2017 Y3EXP FYE AMath P1 Solutions

Year 3 Express Additional Mathematics Paper 1 Solutions 2017 Final Examinations 1a. Given 32() 2 2f x x ax x=+ ++ ( 1) 1 2 2fa∴ − = −+ −+ 11 a−= −+ 0a∴= 1b. 3 22xx++ ( )( ) 21 31x xx= + −+ − ( ) 2Quotient is 3xx∴ −+ 2 Given 12 111 RR R= + 2 3 11 3 31 1R + ⇒= − − 33 13 3 31 +− += − 342 2 −= 32= − 2 3 1 2 R∴= − 3 33 12 22 += × −+ 2 32R∴= +

2 3 (i) 3 x 1 42 x= 5 4 3 2x = 4 53 1.383...2x ∴= = and 3 2.17...1.383y= = Point of intersection: (1.38,2.17) (ii) x y 1 42yx= 3y x= (1.38, 2.17) O

3 4i. Using Y mX c= + 3Y Xc= + Passing through (1,5): ( )( )5 31 c= + 2c∴= 32YX= + Given pxxq y−= xy qy px−= yyq p x−= ∴ yyq p x= + Gradient: 3 3mq⇒ = ∴= -intercept: 2 2y cp⇒ = ∴= 4ii. From part (i): 3 2 yy x = + or 32YX= + When 20 2 yxy x−= ⇒ = Using ( )32 2y= + 8y∴=

4 5 (a) ( )( )61 8xx− + ≥− 2 066 8 xxx ≥+ −− + 2 5 14 0x x− ++≥ 2 5 14 0x x−−≤ ( )( )7 20xx− +≤ 27 x∴− ≤ ≤ (b) 2 84 5 xmmx = −− ( ) 2 4 5 80mx x m−+ −= For line to meet curve: ( ) ( ) 2 4 45 80mm− − −≥ 216 20 32 0mm− +≥ 220 32 16 0mm− + +≥ 25 8 40mm− + +≥ 25 8 40mm− −≤ ( )( )5 2 20mm+ −≤ 2 25 m∴− ≤ ≤

5 6 (a) 8 16xxee −= − 2 8 16 0xxee − += ( ) 2 40xe −= 4xe = ln 4x⇒= (to 3 sig fig)1.39x∴= (b) Working with Base 3 ( ) ( ) 2 93 9 12 log 2 log 4 log 3xx − −= ( ) ( ) 2 33log 2 log 4 2xx − −= 2 3 2log 2 4 x x  = − 2 22 34 x x∴= − 22 9 36 0xx⇒ +−= ( )( ) ( ) 29 9 4 2 36 22x −± − − = 2.55 or 7.05x∴= − Alternatively: Working with Base 9 ( ) ( ) 22 992 log 2 log 4 2xx − −= ( ) ( ) 2 992 log 2 2 log 4 2xx − −= ( ) ( ) 2 99log 2 log 4 1xx − −= 2 9 2log 1 4 x x  = − 22 94 x x∴= − 22 9 36 0xx⇒ +−= ( )( ) ( ) 29 9 4 2 36 22x −± − − = 2.55 or 7.05x∴= − *

6 7 (i) Given equation of C1 : 22 2 6 60xy xy+ − − += ( )1 26 1 ; 3 centre of : 1,322ab C−−= = = = ∴−− 22 11 3 6 2 radius of 2 unitsrC= +−= ∴ = Alternatively, 2 22 2 1 6 3 619xx yy− ++ − + = −++ ( ) ( ) 22 21 32xy−+− = ( )1centre of : 1,3C∴ 1radius of 2 unitsC∴= Alternatively: (ii) Given ( 4, 11)A and ( 8, 7 )B on 2C : Let centre of 2 =(4, )Cb 11 7 1148 ABmm ⊥ −= = −∴ =− Using AO BO= : ( )4 8 11 7Midpoint of , 6,9 22AB ++= = AO BO= Equation of perpendicular bisector of :AB∴ ( ) ( ) ( ) ( ) 2 2 22 4 4 11 8 4 7 bb−+−=−+ − ( )91 6yx−= − Taking sq on both sides:( ) ( ) 22 11 16 7bb− =+− 3 (1)yx⇒ = + −−−−−−− Diff of 2 squares: ( ) ( ) 22 11 7 16bb− −− = 2Centre of also lies on: C 4 (2)x= −−−−−−− ( )( )11 7 11 7 16bb bb−+− −−+ = Sub (2) into (1): 437y=+= ( )( )18 2 4 16b−= 7b∴= ( )2Centre of is 4,7C∴ ( )2Centre of 4,7C∴= ( ) ( ) 2 2 2Radius of = 4 4 11 7 4 unitsC∴ −+− = ( ) ( ) 2 2 2Radius of = 4 4 11 7 4 unitsC∴ −+− = ( ) ( ) 22 2 2Equation of : 4 7 4Cx y∴ − +− = ( ) ( ) 22 2 2Equation of : 4 7 4Cx y∴ − +− = (iii) Distance between the centres of 1 C and 2C = ( ) ( ) 2 24 1 7 3 5 units−+− = Since 12 245rr+ =+> 12 and intersects. CC∴

7 8 Given the function ( ) 3sin 2 2fx x = − , (i) Period = 180 Amplitude = 3 (ii) For 0 180 0 2 360xx°≤ ≤ ° ⇒ °≤ ≤ ° 3sin 2 2 0x−= 2sin 2 3x⇒= 1 2sin 41.813α −= =  2 41.81 , 138.19x∴=  20.9 , 69.1x∴=  (iii) sketch the graph of ( ) 3sin 2 2fx x = − for 0 180x°≤ ≤ ° , showing the x – and y – intercepts clearly. [Correct shape of graph, period, amplitude] [Correct x, y – intercepts, and turning points] ( ) 3sin 2 2fx x = − y x

8 9 (ai) Given ( ) 2() 2 9 2f xxx= −+ 32() 2 9 2fx x x⇒ =−+ . ( ) ( ) 32111 1 222 2let : ( ) 2 9 2 0xf= = − += * ( )2 1 is a factorx∴− (aii) By synthetic division: ( )( ) 21 2() 2 8 4fx x x x=− −− OR ( )( ) 221 42x xx= − −− 21 2( ) 0 0 or 2 8 4 0fx x x x=⇒ −= − −= 1 or 2x∴= ( )( ) ( ) 28 8 42 4 2622x ±− − = = ± 1 or 4.45 or 0.4492x∴= − (aiii) 222sin (sin 4) cos 1θθ θ −+ = − 222sin (sin 4) 1 sin 1θθ θ − +− = − 32 22sin 8sin 1 sin 1 0θθ θ− +− += 322sin 9sin 2 0θθ⇒ − += (NA)sin sin sin1,2 6 ,2 62 θθ θ⇒= = = −+ 1 1 2sin 30α −= =  1 0.449sinα −= 30 ,150θ∴=  26.67=  206.7 , 333.3θ∴=  (b) 13 5sec cos5 13θθ= ⇒= (i) 12sin 13θ =− (ii) ( ) 12tan 180 tan 5θθ−= − =