ACJC 2023 JC1 Promos 9649 (Solutions)

ACJC 2023 FM Promotional Exam Solution 1           5 5 4 32 23 4 53 2 4 253 2 2 53 cos 5 Re cos 5 i sin 5 Re cos isin Re cos 5i cos sin 10cos sin 10i cos sin 5cos sin cos 10 cos sin 5cos sin cos 10cos 1 cos 5cos 1 cos 16cos 20cos 5cos                                  Let cosx   53 53 16cos 20cos 5cos 1 0 16cos 20cos 5cos 1 cos 5 1 50 , 2 π,4 π,               2π 4π 6π 8π0, , , ,5555 2π 4π 6π 8πcos 0,cos ,cos ,cos ,cos5555 2π 4π 6π 8π1, cos ,cos ,cos ,cos5555 x     However, since  cos cos 2 π  , we have 2π 8πcos cos55 and 4π 6πcos cos55 which are repeated. 2 Let nP be the statement   d πes i n 3 2es i n 3d3 n xn x n nx xx   for all positive integers n. To prove 1P is true: LHS:          d es i n 3 3 ec o s 3 es i n 3d e3 c o s 3 s i n 3 xx x x x xxx x x   RHS:       ππ π2e sin 3 2e sin cos 3 cos sin 333 3 312e cos 3 sin 322 es i n 3 3 c o s 3 xx x x x xx xx xx            Since LHS = RHS, 1P is true.

Assume kP is true for some positive integer k. Try to prove 1kP is true.    1 1 1 1 d sin 3d dd sin 3dd d π2es i n 3d3 ππ23 e c o s 3 e s i n 333 3 π 1 π2e c o s 3 s i n 323 2 3 ππ π2es i n c o s 3 c o s33 k x k k x k kx kx x kx kx exx exxx k xx kk xx kk xx k x                                       1 1 πsin 333 ππ2e s i n 3 33 1 π2e s i n 3 3 kx kx k x k x k x                   Since 1P is true and 1kkPP  , by Mathematical Induction, nP is true for all positive integers n. 3(i) 5 π1e iz   Let iez     i2 1 πi5 π 3πii iπ55 π 3πii55 ee 52 1 π 21 π 5 e, e , e e, e , 1 k k k z            3(ii)     2 2 5 2 5 2 11p1 11 1 1 1 zz z zz z zz z zz         

The solutions of  p0z  are the solutions of 5 10z  which are not solutions of 10z Therefore, the solutions of  p0z  are π 3πii55e, ez   . 3(iii)  2 2 2 2 2 11p1 11 21 1 zz z zz zzzz ww                 2 10 15 2 ww w    3(iv) If πi 5ez then πi 51 ez   so π2cos 5w . Similarly, if 3πi 5ez  then 3πi 51 ez   so 3π2cos 5w . Since ππ0 52 , πcos 05  so π 15 π 152cos cos52 54   and 3π 15cos 54  4(i)      22 22 2 22 2 2 22 2 22 2 2 222 2 222 2 2 3( 1) 4 12 3( cos 1) 4 sin 12 3 cos 6 cos 3 4 1 cos 12 4c o s 6 c o s 9 6c o s 9 4c o s 4c o s 3cos 9cos 9 4 cos 4 cos 4c o s 9 4 cos 9cos3cos 4c o s 4c o s xy rr rr r rr rr r r                                  2 22 22 2 3cos 6 4 cos 4 cos 63 c o s 63 c o s or rejected 04c o s 4c o s 63 c o s 4c o s r rr r r              

4(ii) 4(iii) To find points of intersec