ACSI 2019 Y3EXP FYE AMath P1 Solutions

Anglo - Chinese School (Independent) FINAL EXAMINATION 2019 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 1 Friday 4 October 2019 1 hour 30 minutes SOLUTIONS ONLY

2 Answer all the questions. 1 23 11 6 0xx− +≤ (3 2)( 3) 0xx− −≤ 2 33 x≤≤ 2 22 1xk x+= − 22 (2 ) ( 2) 0x kx k− +− +−= For line not to meet curve, D<0: ( ) ( )( ) 2 2 42 2 0kk− −− −< ( )( )2 2 80kk− −− < ( )( )2 60kk− +< 62 k−< < Least integer 5k =− . 3

3 4 (i) 2 12 log 1 log log 3 km m m m −+ + 2 12 log 1 log log 3 km m m m −+ + = 120 2 23 −− + = 3 10 (ii) 2 2 5 25 log 4 log log log m m m m × × . 2 2 5 25 log 4 log log log m m m m × × = 2lg2 2lg 2lg5 lg 1lg lg2 lg5 lg2 mm m m ××× = 16 5 (a) Line: 12 2xy y= −+ 2 41xy y= −+ 1124 x xy = −+ 1124 x xy += Gradient = 4, Intercept = 2 (b) (i) ( ) 227ln 7 414yx −−−= − − 2ln 3 5yx= − 235xye −=

4 5 ii) As x approaches zero, 5ye −= . 6 (i) 13x−= or 13x−= − 4x= or 2x=− (ii) (iii) 13xx+−= 22 2 1 2 ( 1) 9x x x xx+ − ++ − = 222 2 82 2x x xx−= +− 222 2 82 2x x xx−= +− or 222 2 82 2x x xx− = −− + 24 4 80xx− −= No solution 2 20xx−−= ( 2)( 1) 0xx− += 2x= or 1x=− Alternatively, 13xx+−= 31xx=−− 31xx=−− or 13xx= −− 13xx−=− 13xx−=+ 13xx−=− or 1 3( . )x x NA−=− 13xx−=+ (N.A) or 13xx−= −− 21x or= −

5 (iv) For 31 ≤−+ xx , 12 x−≤ ≤ 7 (i) PQ = ( ) ( ) 22 9 3 11 5−+− = 62 units (ii) Centre = midpoint = 3 9 5 11,22 ++  = ( )6,8 Radius = 32 units Equation of circle is ( ) ( ) 22 6 8 18xy− +− = (iii) ( )9311 911 5yx −−= − − − 9 11yx= −++ Equation of tangent is 20yx+= (iv) When 0y= , 20x= ----- ( )20, 0R= Area of triangle PQR = 1 20 3 9 20 2 05 0 11 = 66 units² 8 (i) ( )22log log 2 3xx+ += ( ) 2 2log 2 3xx+= ( ) 2 28xx+= 2 2 80xx+ −= ( )( )2 40xx− += 2x= or 4x=− (NA)

6 (ii) 4 16 126log 2 log 2 x += 22log 4 6 12log 16x += 2 6 48x+= 21x= (iii) 39 3log 4 2 log (2 5) log ( 1) 1xx x− −− += 33 3log 4 log (2 5) log ( 1) 1xx x− −− += ( )( ) 3 4log 125 1 x xx =−+ ( )( ) 4 325 1 x xx =−+ 26 9 15 4xx x−−= 26 13 15 0xx− −= ( )( )36 5 0xx− += 3x= or 5 6x=− (NA) 9 (a) 11 2αβ+= − 15 2αβ = ------- (1) 2αβ αβ + =− ------ (2) (1) in (2): ( ) 5 22αβ += − New equation is 25 4 20xx+ += (b) (i) Let α and 1 α be the roots of the equation. 1 92 k kα α − = 92 kk−= 3k =

7 (ii) 212 1 k kα α −+= OR 23 18 9 6xx x