RVHS 2023 T4 TImed Practice Session 3 (Solutionto self mark

1 2023 Term 4 Timed Practice (Structured Remedial Session 3) Question 1. (N14/2/4(b)(ii)) It is given that 3iw=− . Without using a calculator, find the three smallest positive whole number values of n for which * nw w is a real number. [4] Solution Learning Points and self mark If a complex number is real, then: arg 0, , 2 , 3 ,...* nw w    =    ( ) ( )arg arg * 0, , 2 , 3 ,...n w w   − =    ( ) ( ) ( ) ( ) arg arg 0, , 2 , 3 ,... arg arg 0, , 2 , 3 ,... n w w n w w       − − =    + =    1 1arg tan 63 w − =− =−  0, , 2 , 3 ,...66 1 0, 6, 12, 18,... n n       − + − =          − − =    5, 11, 17n= 1-mark for writing out the arguments for purely real complex number. 1-mark for using result that ( ) ( )arg arg arg ** nw n w ww  =− , or any appropriate method. 1-mark for argument of w. 1-mark for all answers of n.

2 Question 2. (N12/I/6, parts and modified) The complex number z is given by 1 i 3z=− . Find the smallest positive integer n such that 1000nz  . State the modulus and argument of nz when n takes this value. [5] Solution Learning Points and self mark use 1 i 3z=− 1 i 3 2−= 1000nz  1000 n z  1 i 3 1000 2 1000 9.97 n n n −   Smallest positive value of n is 10. 1-mark for getting modulus of z. 1-mark for using result that nnzz = , or any appropriate method. 1-mark for correct answer. 1010 10 2 1024zz = = = 10 10 10arg 10arg 10 33 2 3 zz   = = − =−  = 1-mark for getting 1024. 1-mark for getting 2 3  .

3 Question 3. (N12/I/11, parts and modified) A curve C has parametric equations sinx =− , 1 cosy =− where 02  . (i) Sketch C, showing clearly the features of the curve at the points where 0 and 2= . [2] (ii) Without using the calculator, find the exact area of the region bounded by C and the x-axis. [5] Solution Learning Points and self mark (i) 1-mark for shape. 1-mark for both end-points written in coordinates form. (ii) ( )( ) 2 0 2 0 d 1 cos 1 cos d A y x      = = − −   2 2 0 1 2cos cos d    = − + ( ) 2 0 11 2cos 1 cos 2 d2    = − + + 1-mark for expressing the area in terms of integral involving the parameter  . 1-mark for expansion. 1-mark for using appropriate trigonometry identity that allows integration to take place.

4 ( )   2 0 1 sin 22sin 22 12 2 02       = − + +   = + − 3= 1-mark for correctly integrated expression. 1-mark for answer.