RVHS 2022 JC1 H2 MA Promo Std Soln

River Valley High School 1 2022 JC1 H2 MA Promo 1 Solution [5] System of linear equations Sub (1, 2): 2.....(1)abc+ + = Sub (-1, 0): 0 0....(2) abc abc − + + = − − = 2 d 2d ya bxxx=− When 0.5,x= 2 (0.5) 1 (0.25) ab − =− 4 1....(3)ba− =− Using GC, 1, 3, 2a b c= = =− 21 32yx x = + − Generally well done. Some students make mistakes in arithmetic manipulation when sub values into the differential equation, and couldn’t get eqn (3).

River Valley High School 2 2 Solution [6] Transformation (i) f ( ) f (2 )y x y x= → = ( ) ( )2 ,0 ,0aa → 0, 0,22 bb   →       A few mis -read the questions: - it is not about writing down whether it is possible or not - only required to write down the coordinates of the axial intercepts. Generally well done. (ii) f ( ) f ( 1)y x y x= → = + ( ) ( )2 ,0 2 1,0aa →− Generally well done. (iii) f ( ) f ( 1) f (2 1)y x y x y x= → = + → = + ( ) ( ) 12 ,0 2 1,0 ,0 2a a a → − → −  Need to revise on the correct sequence of transformation required to effect the desired change. (iv) f ( ) 2f ( ) 1y x y x= → = − ( ) ( )0, 0, 0, 12 b bb → → − Generally well done. (v) 1f ( )yx −= ( ) ( )2 ,0 0, 2aa → 0, ,022 bb   →       Many are unaware if the point ( ),xy lies on th e curve of f ( )yx= , then the point ( ),yx lies on the curve of 1f ( )yx −= . Some students mistook 1f ( )yx −= as 1 f ( )y x= Note that 11 f ( )f ( ) xx − =

River Valley High School 3 3 Solution [7] AP GP (i) Let b and d be the first term and common difference of the AP respectively. ( ) 1 11 1nar b d− =+ ( )72nar b d=+ ( ) 1 43nar b d+ =+ Equation ( ) ( )12− gives 1 4nnar ar d− −= 1 4 nnar ar d − − = Equation ( ) ( )23− gives 1 3nnar ar d +−= 1 3 nnar ar d +− = Hence 11 34 n n n nar ar ar ar+−−− = ( ) ( ) 1143 n n n nar ar ar ar+−− = − 114 7 3 0n n nar ar ar+−− + = ( ) 12 4 7 3 0nar r r− − + = ( )( )4 3 1 0rr− − = since a and r are non-zero. 3 4r = or 1r= Reject 1r= since the GP a converging GP where 11 r−   . Therefore 3 4r = Alternatively ( ) 1 11 1nar b d− =+ ( )72nar b d=+ ( ) 1 43nar b d+ =+ (2)  (1) : 1 7 11 n n ar b d ar b d− += +  7 11 bdr bd += + ---(4) Many candidates are unable to set up (1), (2) and (3) and work with it meaningfully. Need to explain the reason for 1r

River Valley High School 4 (3)  (2) : 1 4 7 n n ar b d ar b d + += +  4 7 bdr bd += + ---(5) 7 11 bd bd + + 4 7 bd bd += + ( ) ( )( ) 2 7 4 11b d b d b d+ = + + 2 2 2 214 49 15 44b bd d b bd d+ + = + + 250bd d−= ( )50d b d−= 0d = or 5bd= Reject 0d = since 01dr=  = but GP a converging GP where 11 r−   . Th