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−   . Therefore 5bd= Sub 5bd= into (4): 7 5 7 3 11 5 11 4 b d d dr b d d d ++= = =++ Need to explain the reason for 0d  . It is not just about getting 3 4r = For students who obtain 4 3r = , they need to realize that 11 r−   . (ii) ( )1 0.0041 1 1 nar aa r r r − − − − −  ( )1 0.004 0.0041 1 1 1 nara a a r r r r −   −  −    − − − −    Note that 0a and 11 r−   , then 01 a r − Divide throughout by 1 a r− , ( )0.004 1 1 0.004nr−  − −  0.004 0.004nr−   30.004 0.0044 n −    Note that 3 0 0.0044 n   − for all n. 3 0.0044 n   …….(*)

River Valley High School 5 ( )ln 0.004 3ln 4 n  Note that 3ln 04   . 19.19n Hence least n is 20. Alternatively, 0.004nS S S− ( )1 0.0041 1 1 naraa r r r − − − − −  (Since nSS  as a and r are positive.) 3 0.0044 n   …….(*) ( )ln 0.004 3ln 4 n  Note that 3ln 04   . 19.19n Hence least n is 20. Note: For those who starts with 0.004nS S S − ( )1 0.0041 1 1 nar aa r r r − − − − −  This is INCORRECT because nSS  as a and r are positive.) Students who work with equality, ( )1 0.0041 1 1 naraa r r r − −= − − −  have to use a table of values to justify the minimum value of n is 20. Many candidates are unaware 3ln 04   and make erroneous manipulation. Many set up the incorrect relationship ( )1 0.0041 1 1 nar aa r r r − − − − −  It will end up with 3 0.0044 n  − which is a valid inequality for all n. Candidates should realize that the starting relationship is incorrect, when they arrive at 30.004 4 n −  instead, many went on to make incorrect manipulation from this point onwards.

River Valley High School 6 4 Solution [7] Inequality (i) ( ) ( )( ) ( ) 2 2 22 2 2 6 1 101 2 6 1 1 1 01 2 6 1 1 01 36 0....(*)1 2 01 xx xx x x x x x x x x x x x xx x xx x ++ − + − + + − + − − + + − + − + − + − + − 2 or 0 1xx −   Comments: 1. Other than some careless algebraic mistakes made by some students, this part of the question is generally well done. 2. Several students solved the inequality in (*) of the workings by considering cases. Even though this is commendable and extremely useful for tougher order thinking questions, they should consider going for efficient method (number line) especially for this standard question. Common Mistakes: 1. Should not cross- multiply the denominator ( )1 x− ; can only be done if it is definitely positive which we are uncertain till x is solved. 2. Some students do not know how to combine into a single fraction with the correct denominator, giving the following: ( ) ( )( ) ( )( ) ( )( ) 2 2 2 6 1 101 112 6 1 01 1 1 1 xx xx xxxx x x x x ++ − + − −+++ − − + − + (ii) 22 6 11 1 xxx x +++ − Replace x with x Comments: 1. Majority of students were able to see that this type of question requires the use of substitution. Unfortunately, many were 2− 0 1 1 + – + –

River Valley High School 7 2 (No Solution) or 0 1 Taking intersection -1 1 x x x  −   not able to solve it properly subsequently. Common Mistakes: 1. Students made mistakes in solving 2x − , writing out solution such as: (a) 0x (b) 22 x−   Students needs to learn that for special inequality expression such as: (a) for 2x − ; then there’s no solution since x is never lesser than a negative value for all real values of x. Other examples are such as: (b) for 2x − ; then x . (c) for 2xe − ; then x , etc. 2. Similarly, when solving 01 x , students could consider writing it as: 0 and 1 and 1 1 Finding intersection, 11 xx xx x   −   −  

River Valley High School 8 5 Solution [9] Summation (i) 4 2 1 2 3 ( 1)( 2) 1 2 r r r r r r r + = + −+ + + + Comments: Generally, well done. Though several students used a rather long approach to solve for this 1-mark question. Do consider using substituting special values of r at the stage where: ( )( ) ( ) ( ) 42 1 2 2 1 r A r r Br r Cr r += + + + + + + so that the constants A, B and C can be solved faster, as opposed to comparing coefficients (which is generally more time- consuming). (ii) ( )( )2 2 42 12 1 2 3 12 1 2 3 2 3 4 1 2 3 3 4 5 1 2 3 4 5 6 1 2 3 21 1 2 3 11 1 2 3 12 1 2 1 3 2 3 2 3 3 1 1 2 3 2 n r n r r r r r r r r n n n n n n n n n n n n = = + ++ = + − ++ = + − + + − + + − + + + −−− + + −−+ + + − ++  = + + − + −+ + + =− 13 12nn −++ Comments: Most students knew that this is a quest