VJC 2024 H2 JC2 Math Prelim P1 Solutions

2024 VJC Prelim Paper 1 Solutions 1 (a) Express 2 33 8 22 15 x xx − ++− as a single algebraic fraction. Hence, without using a calculator, solve exactly the inequality 2 33 8 22 15 x xx − >−+− . [4] 1a ( ) ( ) ( )( ) 2 2 2 2 2 2 33 8 22 15 33 8 2 2 15 2 15 2 43 2 15 211 53 x xx xxx xx xx xx x xx − ++− −+ +− = +− −+= +− −+= +− Solve: ( ) ( )( ) 2 211 053 x xx −+ >+− Since ( ) 2 10x−≥ for all x∈  , ( ) 2 2 1 10x− +> for all x∈  . Hence, we solve ( )( )5 30xx+ −> . 3 or 5xx∴ > <− There is a need to explain cl early why the numerator is always positive. The better method is to complete the square and state that ( ) 2 10x−≥ , so ( ) 2 2 1 10x− +> . Some only described the discriminant to be negative or said that there are no real roots – both statements do NOT imply that the numerator is always positive. Note: 1 is NOT a critical value. (b) Using your answer to part (a), find the set of values of x for which 2 42 33 8e 2e 2e 15 x xx − >−+− . [2] 1b Replace x with 2e x : 22 2 e 3 or e 5 (rej. e 0 for all ) xx x x > <− >∈ ln 3: 2xx∴∈ >  Explain clearly why 2e5x <− is rejected. Many showed working up to ( )2 ln 5x<− to reject the statement which is incorrect. Firstly, ( )ln 5− is undefined. Secondly, IF the inequality were 2e5x >− , would you say that ( )2 ln 5x>− and reject to conclude that there are no solutions?!

2024 VJC Prelim Paper 1 Solutions 2 The sum of the first n terms of a sequence, ru is given by 1 1 ( 1) ! n r r nu n= = − +∑ . (a) Find nu in terms of n, for 2n≥ , expressing your answer as a single algebraic fraction. [2] 2a 1 11 2 111 ( 1) ! ! ( 1)( 1) ( 1) ! 1 ( 1) ! nn n rr rr u uu nn nn nn n nn nn n − = = = − −= − −− +  −+ − += +× −−= + ∑∑ Note that 1 n r r u = ∑ is usually denoted by nS . Here, 1n nnuSS −= − . (b) Show that 5 1 30 n r r u = <∑ , for all 5n≥ . [2] 2b 4 5 11 411 ( 1) ! 5! 1 30 ( 1) ! 1 , since 0 for all 530 ( 1) ! nn r rr r rr u uu n n n n n nn = = = = − = − −− +  = − + < >≥ + ∑∑∑ The explanation to why 1 30 ( 1)! n n− + is less than 1 30 is to correctly mention that 0( 1)! n n >+ . Most commonly seen answer was that 0( 1)! n n →+ . While it is true, it does not explain why 11 30 ( 1)! 30 n n−< + . (c) Explain why 1 r r u ∞ = ∑ is a convergent series. [1] 2c 111 1 ( 1) ! ( 1) ( 1) ! ( 1) ( 1) ! nn n n nn n n −= − = − + +××− +×− 1 ( )( ) 1As , ( 1) ( 1)! , 0,1 1! 1 0 1, which is a constant. n r r n nn nn u = →∞ + × − →∞ →+× − →−=∑ Hence 1 r r u ∞ = ∑ is a convergent series. Many gave incomplete explanations to say that as n→∞ , ( ) 1!n+ →∞ , so ( 1)! 0n n+ → . This completely disregards the fact that the numerator, n→∞ too and did not show complete understanding of why ( 1)! 0n n+ → .

2024 VJC Prelim Paper 1 Solutions 3 The functions f and g are defined by 6f: 3 axx x − −  for x∈  , 3x≠ , 9b≠ , g: e xx − f