NYJC TJC VJC FM 2024 Prelim P1 Solution

1 2024 TJC H2 FM Prelim Paper 1 (Marking Scheme) 1 Polar equations of the form cos ,r a a n =+ , , 2a n n are oommonl oalle eetal ourves beoause of the shaees of their graehs. The number an sizes of the eetals are eeen ent on the values of a an n. (a) State the relationshie between the number of eetals an n. [1] (b) Show that the area boun e b the ourve is in eeen ent of n. [3] Solution (a) number of eetals = n (b) Area boun e b the graeh = ( ) ( ) 2 2 0 2 2 0 2 2 2 0 2 2 0 2 2 0 22 0 2 2 1 d2 1 cos d2 1 2cos cos d2 111 2cos cos 2 d2 2 2 31 2cos cos 2 d2 2 2 3 2 1 sin sin 22 2 4 3 (2 )22 3 , independent of 2 r a a n a nn a nn a nn a nnnn a a n                       =+ = + + = + + + = + +  = + + = =     

2 2 Show that ( ) ( )1 i tan sec cos isin k k kk   + = + . [1] Henoe or otherwise, show that 1 0 cos sec cot sin sec , n kn k kn     − = = erovi e  is not an integer multiele of .2  [5] Solution ( ) ( ) ( ) sin1 i tan 1 i cos sec cos isin sec cos isin k k kk k kk         + = +  =+ =+ ( ) ( ) ( ) ( ) 1 0 1 i tan 11 i tan i tan sec cos isin 1 i tan sec cos isec sin 1 i tan cot sec cos cot isec sin cot i i cot sec cos i cot sec sin cot sec sin cot i cot cot sec cos nn k k n nn nn nn nn nn nn nn nn nn                                 − = +−+= +−= +−= −+= =− + + = + −  ( ) 11 00 cos sec 1 i tan sec sin cot nn kk kk n k n       −− == =+  = 

3 3 Let 2 0 cos dn nI x x x  = . (a) Prove that for 2,n ( ) 212 n nnI n n I − = − − . [3] (b) R is the region enolose b the graeh of 2 1sin 2y x x= , the line 2x = an the x-axis. B using the result in (a), fin the exaot volume of the soli generate when R is rotate 2 ra ians about the x-axis. [5] Solution Remarks (a) 2 0 cos dn nI x x x  = 122 0 0 sin sin dnnx x nx x x  −=−  ( ) ( ) 12 22 0 0 cos 1 cos d2 n nnnx x n n x x x  −− = − − − −   ( ) 212 n nn n I − = − − Use b earts to show the esire result. Not reoommen e to use MI to erove. (b) V olume = 422 0 1sin d2x x x   42 0 1 cos d2 xxx   −=  4422 00 d cos d22 x x x x x =−  5 2 4 02 5 2 x I  =−  6 4 320 2 I=− 2 0 0 cos d 1I x x  == From result in (a), ( ) 2 2 20 2 2 1 224II = − − = − ( ) 4 42 4 4 12II = − − 42 12 216 4  = − −  Use iso metho to fin volume sinoe region R is just between ourve an x axis. x y R

4 4 23 2416  = − + Require V olume 64 23 24320 2 16    = − − + 6 5 3 3 12320 32 2    = − + − 4 Let nP be the veoto