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 intersection, let 2 2 63 c o s 22 c o s4c o s 63 c o s 22 c o s 04c o s          Using GC, 1.7489  or 4.5343  Sub. 1.7489  into equation of 2C :  2 2cos 1.7489 1.6457OP  Area of triangle OPQ     2 2 1 () s i n2 1 1.6457 sin 4.5343 1.74892 0.472 units (3 s.f.) OP OQ QOP   5(i) Auxiliary equation: 2 23 4 0mm  Roots are πi 63i2 em    Hence solutions are 1 2 4 y x C1 C2 P Q O

  ππii66 ππii66 2e 2e 2e e ππ ππ2 cos isin cos isin66 66 ππ2c o s i s i n 66 π2c o s s i n 6 nn n nn n n n n xX Y XY nn nnXY nnXY XY nnAB                                      π 6     Where A XY ,  iB XY 5(ii) Method 1      6 6 6 6 6 6 π 6 π2c o s s i n 66 ππ2c o s π sin π66 ππ2c o s s i n 66 ππ22 c o s s i n 66 64 n n n n n n nnxA B nnAB nnAB nnAB x                                    Method 2       65 4 43 32 43 2 32 21 1 32 1 21 2 1 21 23 4 2323 4 4 23 4 12 16 3 16 1 2 23 4 1 6323 4 1 6 23 4 24 3 144 32 3 64 24 3 2 3 4 144 96 3 64 144 96 3 144 nn n nn nn nn n nn nn n n nn n n nn n n n nn n xx x xx xx xx x x xx x x x xx x x xx x x x xx x                                  21 96 3 64 64 nn n xx x    5(iii) When 0n  : 3 2A 1n : 33 1 121 22 2 2 BB     6(a) 43 i 5z  represents a circle centred at (4, 3) and has a radius of 5 units.

15 i 55 izz   represents a perpendicular bisector of the line segment joining the points (1 , 5 ) and (5,5) . Cartesian equation of circle is   22 43 2 5xy  ---(1) Cartesian equation of perpendicular bisector is 2x  ---(2) Sub. (2) into (1):   22 24 3 2 5 y  2 61 2 0yy 63 6 4 ( 1 ) ( 1 2 ) 32 12y       The complex numbers are 23 2 1 i and 22 1 3 i . = = Re Im O 5

6(b)(i) From diagram, greatest value of 82 iwA B   22 5 (8 4) (2 ( 3)) 5 41 5 units (or 11.4 units) AC     6(b)(ii) From diagram,  32 1 3 62 1tan 0.61655 rad22     = = Re Im O 5 = = Re Im O 5 21 3

22(4 0) ( 3 3) 52CD     units 5sin 0.76616 rad 52 CF CD    1 3tan arg 3i2 w     i.e.  1.75 arg 3i 0.617 (3 s.f.)w    7(i) Amount of salt in the beaker after 10 g of salt is added is 1 10nu   . After it is mixed and k ml is removed, there is  1 100 10100 n k u    Adding k ml of pure water does not cha nge the amount of salt. Hence,  1 100 10100 nn kuu   . 7(ii)  1 1 11 0100 11 0100 10 nn n kuu kk u           Let 1 100 kp               1 2 2 2 2 3 32 2 21 0 1 1001 10 1 100 10011 10 100 1000 1 100 10 10 10 10 1 10 10 1 10 1 10 1 110 1 1110 11 10 1 1 10 1 1 nn n n n n nn n n n k n k up u p pp u p p pu p p pp u p p p pu p p p pu p p p p pp p kk k kk k                                   7(iii) Method 1 As n , nuL and 1nuL 

  0.9 10 0.1 9 90 LL L L    Method 2 90 1 0.9 n nu   As n , 0.9 0 n  90nu  7(iv)    90 9 90 1 0.9 90 9 90 0.9 9 0.9 0.1 ln 0.1 21.8ln 0.9 n n n n u n       Smallest n is 22. 7(v) 1000 10 1 1 100 n n ku k            As n , 10 100 n k 1000 10nu k 1000 10 50 1000 60 k k   Hence the range of values is 50 1003 k . 8(i) [Diagram] Light ray coming from Y hits mirror at A, reflects and intersects y-axis at S. Let R be at  0, r and RAY  . Since RA is a radius of the circle, sin k r  . RA is perpendicular to the tangent at A, so RAS RAY   By alternate angles, SRA RAY   . Therefore, RAS is an isosceles triangle with RSA S . By drawing the perpendicular from S to RA, we see that  cos 2 r SR  so 2cos rSR  . Therefore, the required distance OS is

2 2 2 11 2cos 11 21 s i n 11 21 OS OR RS r r r k r           8(ii)  2 22 22 22 22 yr r x yr r x yr r x yr r x          Because the part of the cross-section in the neighbourhood of the origin is where y r .  1 22 2 1 2 2 2 2 2 2 1 1 2 2 yr r x xrr r xrr r x r            8(iii) Compare the equation 2 2 xy r with 2 4x ay 42 2 rara  The focus is at 0, 0, 2 ra  . 8(iv) Using the answer from (i), 1 2 2 2 2 2 2 2 111 2