2024 Y6 Timed Prac Rev Paper 2 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6 2024 Year 6 Timed Practice Revision Practice Paper 2 Solutions Source: 2020 Year 6 Term 3 Timed Practice Section A: Pure Mathematics (60 marks) 1 The complex number z is given by ixy , where x and y are real numbers. (i) Express y in terms of x if  2arg 2z  . [2] (ii) State the values of x and y if  Re 0z  and 2z  . [1] Using these values of x and y, find the smallest positive integer n for which * n z z is a negative real number. [2] (i) [2] i 22 i ( 2 ) e e zr zr       2 2   2,kk    ,4 kk      tan 1 y x    yx ,0x Alternative Given izxy  2arg 2 32a rg or 22 3arg or is in the 2nd or 4th quadrant44 , 0 z z zz yx x           Represent 2z on an argand diagram! (ii) [1] 2, 2xy 

2      (iii) [2] 2i2z  arg 4z     i* 1 i e e n n nn zr rz r      i1 e n     1n     2,k k     (1 ) ( 21 ) ,4 1 21 ,4 nk k n kk              83 ,nk k   The smallest positive integer n is 3. Alternative 0, ( 1) , 3 , 5 ...4nn    The smallest positive integer n is 3. For * n z z to be a negative real number means that * arg n z z   2 (a) By writing  2 2 1rr  in partial fractions, find an expression for  2 2 2 . 1 n r rr  [ 3 ] (b) A geometric series has first term a and common ratio r, where a and r are non- zero and 1r  . The 3rd and 9th terms of the series are 448 and 7 respectively. Given also that the sum of the first n terms is 1197, find the values of a, r and n. [ 4 ]   (a) [3] Let   2 2 111 A BC rr rrr      2( 1 ) ( 1 ) ( 1 ) ( 1 )Ar r B rr C rr       Let 0, 2 Let 1, 1 Let 1, 1 rA rC rB         2 22 1 1 111 rr rrr        2 22 21 2 1 111 nn rr rr rrr       To apply MOD write in “order” 121 11rr r  .

3      121 123 121 234 121 345 121 456 121 321 nnn       12 1 21 121 11 nn n nn n       2 2 21 1 1 211 n r nnrr   (b) [4] 28 448, 7ar ar   6 1 64r 1 , 2r  1792a Sum of first n terms  1792 1 11971 nr r      11792 1 21if , 1197 12 2 n r       1 1 11971( ) 2 2 1792 13 4 1 25 1 2 n n      0.586n (rej as n is an integer) 11792 1 21if , 1197 12 1 2 n r       1 3 11971( ) 2 2 1792 n  

4      11() 25 1 2 n  Note that n must be an odd integer, 11()25 1 2 n    11ln ln2 512n    19, 1792, 2na r      Alternative Observe that since every term is positive, and the first term 1792 sum of terms =1197,an 1 2r  otherwise the sum can only get larger f