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 argument. (b) Sketch an Argand diagram showing the points 1P , 2P and 3P where 1P , 2P and 3P represent the complex numbers 1z , 2z and 3z respectively. [2] 2b This follows from (a). Do note that you are required to indicate the modulus in addition to the argument i.e. distance from the origin. Also, ensure that the relative positioning of the 3 points is correct. 1 Im 1P O Re 5 6  3  3P 2P 1 2 2

2024 VJC Prelim Paper 2 Solutions (c) Find the area of triangle 12OP P . [2] 2c Area of triangle 12OP P ( )( )1 1 2 sin2 6 2 6 5sin 6 0.5   = + +  = = For those who have done (a) and (b) correctly, this was well done and students were able to apply the correct formula ( )1 sin2 ab c . (d) Find the smallest positive integer n for which ( ) * 2 n z is purely imaginary. [2] 2d ( ) 5πi* 6 2 5 πi 6 2e 2e n n n n z − −  =  = For ( ) * 2 n z to be purely imaginary, 5 π π π 3π 3π 5π 5π, , , , , ,...6 2 2 2 2 2 2 n− = − − − Smallest positive integer 3n= Most students were able to translate “purely imaginary” to the argument being an odd multiple of π 2 . So, either write it in a general form e.g. ( ) π21 2k+ or ( ) π21 2k− , or since the question is asking for the smallest n , you can also list out the first few (negative) odd multiples of π 2 and check which value of n works.

2024 VJC Prelim Paper 2 Solutions 3 The line 1l has equation ( )3 4 5 2 = − − + − −r i j k i j k , where  is a real parameter. The point A has position vector 2−+i j k . (a) The plane p contains the line 1l and the point A. Find a cartesian equation of the plane p. [3] 3a 3 1 1 4 2 , ; 2 5 1 1 r OA             = − + −  = −          −−      Vector parallel to p 1 3 2 1 2 4 2 2 1 1 5 6 3 −−               = − − − = =              −        Normal vector 1 1 5 1 2 2 3 1 1 −           =  − =          −      Equation of p 5 1 5 5 2 2 2 2 2 1 1 1 1 rr                = −  =                      Cartesian equation of p: 5 2 2x y z+ + = Make sure you copy the vector correctly and not make any silly mistakes at the start The origin O may not be in plane p, hence you cannot assume that OA is a vector in the plane (as it turns out, O is not in the plane). Also, always check that the normal of the plane you have obtained after cross product is correct. A quick way to check is via dot product: 51 2 1 ... 0 13 −        = =          and 51 2 2 ... 0 11         − = =      −    . (since the normal must be perpendicular to the 2 vectors used in the cross product.) Finally, if your normal is 10 4 2    , it is a good idea to reduce it to 5 2 1    first before finding the cartesian equation of the plane (so that the equation of the plane can be in the “reduced” form).

2024 VJC Prelim Paper 2 Solutions (b) Find the position vector of the point 'A , the reflection of the point A in the line 1l . [4] 3b Let F be the foot of the perpendicular from A to the line 1l , and B be the point on 1l with position vector 3 4 5−−i j k Let 1 3 2 2 4 2 1 5 6 BA −           = − − − =          −      2 1 1 1 112 2 2 2 2 666 1 1 1 BF −               = − − =− −               − − −        3 1 1 4 2 2 0 5 1 3 OF OB BF            = + = − − − =          − − −      ' '22 OA OAOF OA OF OA+=  = − 11 1 ' 2 0 2 2 137 OA        = − − =      −−     Most students were able to apply the method of finding the projection vector or finding the foot of the perpendicular but made several errors, e.g. For 2 2 6 −   , you will have to use 2 2 6 −   and not 1 1 3 −   which is only half of BA (contrast this with the idea of “reducing” the normal). Almost all students were familiar with applying the midpoint theorem which is good (c) The plane q is such that q is parallel to p and passes through the point with position vector 3−+jk . Find a cartesian equation of q and the exact shortest distance between p and q. [3] 3c 52x y z k+ + = Sub 3−+jk into the equation: 5(0) 2( 3) 1 5 kk+ − + =  =− Cartesian equation of q: 5 2 5x y z+ + =− Exact shortest distance between p and q 0 1 5 13 2 2 25 4 11 1 1 15 112 3001 7 30            = − − −       ++           −       = −           = Alternative Exact shortest distance between p and q 2 2 2 5 2 7 305 2 1 +== ++ Always draw a simple diagram (if you need) to help you to determine the points to be used in the planes