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 veotor seaoe of real eol nomials of egree n or less. The transformation 43:T P P→ is efine b , ( )( ) ( ) ( ) ( ) 23f f 0 f(1) f 1 f 2 .T x x x x = + + − + Show that T is a linear transformation. [2] Fin a basis for the null seaoe of T. [7] Solution For ( ) ( ) 4f ,gx x P  ,  the “vectors” here are the polynomials and not x! ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )( ) 23 2 3 2 3 23 23 f g f + g 0 f + g (1) f + g 1 f + g 2 f 0 f(1) f 1 f 2 g 0 g(1) g 1 g 2 fg f f 0 f (1) f 1 f 2 is a linear transformati = f 0 f(1) f 1 f 2 f T x x x x x x x x x x T x T x T x x x x x x x Tx T       + = + + − + = + + − + + + + − + =+ = + + − +  + + − +  = on. ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 23 43 f f 0 f(1) f 1 f 2 0 f 0 f( 0 1) f 1 f 2 0 f since f For the null space of : P 0 f 1 0 0 f 2 =16 8 4 2 0 16 8 4 2 f 0 0 f 1 0 T x x x x x ax bx cx dx e x e a b c d T a b c d a b c d e a b c d a b c d e a b c d e = + + −    + + + = − = − + − + += = = − = = = + + + + = + + =  − + − ++ = + =  + + + = = + + + = = To show that it is a linear transformation, we need to show: 1. ( ) ( )( )T f g xx+ ( )( ) ( )( )T f T gxx=+ 2. ( )( ) ( )( )T f T f xx = All workings regardless how simple it is must be explicitly shown For null space, We need to find the set of polynomials of degree 4 such that ( )( )T f 0 x = i.e. ( ) 4 3 2T0ax bx cx dx e+ + + + =

5 Augmented matrix: 1 1 1 1 0 1 0 0 0.5 0 1 1 1 1 0 0 1 0 1 0 using GC 16 8 4 2 0 0 0 1 0.5 0 −       − − →          ( ) ( ) 4 3 2 4 3 2 0.5 0.5 f 0.5 0.5 2 2 a b c d x x x x x x x x x          =  =− =−  = = − − + = − − +   4 3 2Basis for null space of 2 2T x x x x = − − + (express answer in polynomial)

6 5 The half-line L with equation 8y mx=+ , 0x , 0m , is tangential to the locus 4z = at point Q represented by complex number 4z . (a) Sketch the locus 4z = and L on a single Argand diagram. [2] (b) Find the exact value of m. [2] (c) Sketch the locus of ( )4 5arg 6zz − =− on the same Argand diagram in part (a). [2] (d) The complex number iz x y=+ satisfies the following relations: 8y mx+ , 0x , ( )Im 0zz − , 4z  , ( )4 5 arg63 zz−  − − . Shade the region R that contains z. Hence find the least ( )arg 4iz+ , leaving your answer in radians, correct to 4 decimal places. [3] Solution (a) (b) Method 1 4cos 83 =  = Line L

7 Therefore tan 33m =− =− Method 2 Solve 2 2 2 4xy+= -------(1) and 8y mx=+ ----------(2) Sub (2) into (1): ( ) 22 8 16x mx+ + = 2 2 2 16 64 16x m x mx+ + + = ( ) 22 1 16 48 0m x mx+ + + = For tangent, 2 40b ac−= ( ) ( )( ) 2 216 4 1 48 0mm− + = 264 192 0m −= 3m= Since 0m , 3m=− (c) (d) Least ( ) 1 8arg 4i tan 4 3 z − + =   0.7137= Line L Region R

8 6 The polar equations of two curves are given by 1 : 1 cosCr =+ and ( )2 : 3 1 cosCr =− for 02  . (a) Sketch 1C and 2C on the same diagram, indicating clearly the symmetries and the exact polar coordinates of the point(s) of intersection of the curves. [4] (b) Find the exact polar coordinates of the P on C2, where 0,  and P is the furthest away from the x-axis. [4] (c) Find in exact form, the shortest distance between the points P and A, where A is the point of intersection between 1C and 2C in the first quadrant. [2] (d) Express the equation of C1 in parametric form using the given  as the parameter. Hence find the area of the curved surface generated when the segment OA on C1 is rotated 2 radians about the x-axis. [4] 6 Solution (a) ( )1 cos 3 1 cos 1cos 2 5 or 33    + = − = = 3At , 1 cos3 3 2r = = + =  5 5 3At , 1 cos3 3 2r = = + =  The eolar ooor inates of the eoints of interseotions are 0 = ( )3 1 cosr =− 1 cosr =+

9 3 3 5, and , 2 3 2 3            (b) ( ) ( ) ( )( ) ( ) ( ) ( )( ) 22 2 3 1 cos sin 3 1 cos sin d 3 sin sin cos (1 cos )d 3 sin cos cos 3 1 2cos cos 3 2cos 1 cos 1 r yr y              =− = = − =  + −  = + − = − + =− + − ( )( )dAt 0 3 2cos 1 cos 1 0d y  = − + − = 1cos or cos 12 24 or (rejected) 033     =− = = = = 22At , 3 1 cos 4.533 r 