VJC 2024 H2 JC2 Math Prelim P2 Solutions

2024 VJC Prelim Paper 2 Solutions General Comments • The rubric of the paper states that non -exact numerical answers should be given correct to 3 significant figures. • The use of graphing calculators is encouraged, but in questions where a calculator is prohibited, you need to show sufficient working in answering that question. • Need to be aware that every step shown in a given answer question needs to maintain an appropriate level of accuracy. • Read the question carefully. Section A: Pure Mathematics (40 marks) 1 (a) Show that ( ) 22 222 11 11 211 xx xxx = + + − + − − . [1] 1a ( ) 22 2 2 2 2 22 22 22 2 1 1 1 1 1 1 1 11 (1 ) (1 ) 1 1 1 (shown) 2 xx x x x x xx xx xx x + + − + − − + + − + + −= + − − = + + − Mostly well done Remember that: ( )( ) 22a b a b a b+ − = − . (b) Hence use appropriate expansions from the List of Formulae (MF26) to find the first two non - zero terms in the series expansion of 22 2 11 x xx+ − − in ascending powers of x for 0x . [3] 1b ( ) ( ) ( ) ( ) 2 22 11 22 22 1 1 1 1 222 2 2 22 2 2 2 4 4 11 1 112 1 1 1 11 ... 1 ...2 2 2! 2 2 2! 11 2 ...24 11 8 x xx xx x x x x x x    −−       + − −  = + + −         = + + + + − + − +           = − + − Mostly well done Do note that repeated differentiation is not allowed here as question says “ use appropriate expansions from MF26”.

2024 VJC Prelim Paper 2 Solutions (c) State the set of values of x for which the series expansion is valid. [1] 1c 2 2 2 1, 0 1 10 xx x x   − (or draw the graph) ( )( )1 1 0xx+ −  11 x−    Set of values of : : 1 1, 0x x x x −    Learn how to solve inequalities properly and not “jump” the step. (d) It is given that the two terms found in part (b) are equal to the first two terms in the series expansion of ( )cos bax . Find the possible value(s) of the constants a and b. [2] 1d ( ) ( ) 2 4 2 1cos 1 1 28 1 and 2 428 1 and 22 b b ax ax x a b ab  −  − == = = Quite a few arithmetic errors spotted here, e.g. ( ) 2 2 22 b bax ax . Do practise well to gain fluency

2024 VJC Prelim Paper 2 Solutions 2 Do not use a calculator in answering this question. The complex numbers 1z , 2z and 3z are such that 2πi 31 ez =− , 2 3iz =− + and 1 3 2 zz z= . (a) Express each of 1z , 2z and 3z in the form ier  , where 0r and ππ −   . [3] 2a 2πi 31 2πiiπ 3 5πi 3 πi 3 e e .e e e z − =− = = = 1 2 1i π tan 3 πi π 6 5πi 6 3i 2e 2e 2e z − −  − =− + = = = 1 3 2 πi 3 5πi 6 π 5πi 36 7πi 6 5πi 6 e 2e 1 e2 1 e2 1 e2 zz z − −− − = = = = = Polar form for complex numbers was poorly performed in general. Do note that iπ1e−= (i.e. modulus = 1, argument = π . When in doubt, please plot the complex number as a point on an argand diagram – this will help you to better determine the correct modulus and argu