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 or x∈  , ln 3x≥ . The function f is self-inverse if 1 f () f ()xx − = for all values of x in the domain of f. It is given that f is self-inverse. (a) Find the value of a . [3] 3a 1 6Let 3 36 36 36f () axy x xy y ax yx ya xx xa − −= − −=− −= − −= − Given that 1f () f ()xx −= , 636 3 ax x x xa −− =−− . ∴ a = 3 Some wrote INCORRECTLY that 1 f () f () f ()x x xx − = ⇒= . It is important to know that 1 f( ) f ( ) ff( )x x xx − = ⇒= . (b) Find the exact value of 6 f( π) . [1] 3b 1 2 f () f () ff( ) f() xx xx xx −= = = 6 222f( π) f f f (π) π= = (c) Find the exact range of fg. [3] 3c Given 36 2f( ) 3 33 xxx xx −− = = −−  [ ) gf g g fg 1 15D ln3, R 0, R ,238   = ∞  → =  → =    The first method is to find gR and use it as the new domain of f. To do this, it is advisable to sketch the graphs of g and of f. y x ln 3 1/3 y x 2 15 8 O 1 3

2024 VJC Prelim Paper 1 Solutions Alternative ( ) 3e 6fg e3 x xx − − −= − , ln 3x≥ The alternative is to draw directly the graph of ( ) 3e 6fg e3 x xyx − − −= = − , and note that [ )fg gDD ln 3,= = ∞ . Some who did this method, left out the horizontal asymptote of the graph or incorrectly stated that it to be 3y = . BOTH methods should be learnt and internalised. 4 (a) Given that a , b and c are non-zero vectors such that ( ) ( )+×+=×ab ac bc , and ≠bc , find the relationship between a , b and c . [4] 4a ( ) ( ) ( ) ( ) ( ) 0 00 0 ab a c bc aa acbabc bc acba aa a cb ba ab + × += × ×+×+×+×=× +×+×= ×= × − = ×= −×                       Since 0a≠ and bc≠ , a is parallel to cb− . The notation of 0 is still NOT seen in most answers. Note that 0 and 0 are NOT to be used interchangeably. Some answers still show misunderstanding that ba× and ab× are equal but they are NOT. Note that ( ) 0a cb×−=   does not directly imply that a is parallel to cb− . One has to check and state/explain that neither a nor cb− is 0 . (b) It is g iven instead that a, b and c satisfy the equation abc 0++= with 2a = , 3b = and 4c = . Find the value of ab bc ca⋅+⋅+⋅ . [3] 4b ( ) ( ) ( ) ( ) 222 222 .0 . . . 2. 2. 2. 0 2. . . 0 2342 . . . 0 29... 2 abc abc aa bb cc ab bc ac a b c ab bc ac ab bc ac ab bc ac ++ ++ = + + +++= +++ ++ = +++ + + = ++= −                       

2024 VJC Prelim Paper 1 Solutions 5 It is given that ( ) 1f 5n nn −= where n is a positive integer. (a) By considering ( ) ( )f f1rr−+ , find an expression for 2 41 5 n r r r = − ∑ . [3] 5 a ( ) ( ) 1 1f f1 55 5 14 1 55 rr rr rrrr rr r − +− += − −− −= = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 41 f f1 5 f2 f3 f3 f4 f 1f f f1 f2 f 1 21 5 5 nn r rr n r rr nn nn n n = = − =  − + = − +− + + −− +−+ = −+ += − ∑∑  (b) Hence find an expression for 1 1 46 5 n r r r + = + ∑ . [3] 5 b 1 1 46 5 n r r r + = + ∑ Replace r by 1r− ( )11 11 11 2 11 22 1 1 4 16 42 55 41 3 55 11 125 522 3 15 5 1 5 11 2 3 1 20 20 55 rn n rr rr nn rr rr n n n n r r r n n −= + −+ −= = ++ = = + + −+ += −= +    −  + = −+   −   + = −−  ∑∑ ∑∑ When we use the replacement of variable method, we need to change the upper limit as well. Learn to recognise a sum of a GP, like in this case: 1 2 3 5 n r r + = ∑ is a sum of a GP with first term 2 3 5 and common ratio 1 5 .

2024 VJC Prelim Paper 1 Solutions 6 (a) Find the exact value of 3 12 21 2 2sin d 1 x x x − − ⌠ ⌡ . [3] 6 a ( ) 3 12 21 2 3 2 1 21 2 3 2 21 1 2 2 2 11 22 2 2sin d 1 12 sin d 1 sin 31sin sin22 ππ 36 π 12 x x x xx x x − − − −− − = ⋅ − =  = −    = −   = ∫ ∫ Be familiar with all the standard forms. Know how to check if the integrand is truly of the form ( ) ( ) f' f x x , or of the form ( ) ( )f' f n xx . Check through all standard forms before even thinking about integration by parts. Those who did it by parts spent much more time on this question. (b) Find the exact value of π 3 0 cos 2 dxx∫ . [3] 6 b πππ 334 π00 4 ππ 43 π0 4 cos 2 d cos 2 d cos 2 d sin 2 sin 2 22 1 31 2 42 31 4 x x xx xx xx = +−  = −   = −− = − ∫∫∫ Sketching a graph is good way to tell when the expression is positive or negative. If you cannot visualise clearly, sketch it! There is a need to split the integral into parts where the expression cos 2x is positive and negative within the interval π0, 3   .

2024 VJC Prelim Paper 1 Solutions (c) Find 22 1 d23 xx kx k−+ + ⌠⌡ , where k is a positive constant. [4] 6 c ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 22 2 22 2 2 22 1 d23 1 d 23 1 d 3 1 d 4 1 d 2 21 ln22 2 1 ln43 xx kx k x x kx k x xk k k x xk k x k xk k xk Ck k xk kx Ck kx −+ + = −− + = −−−+ = −− + = −− +−= +−− += +− ∫ ∫ ∫ ∫ ∫ Alternative (