EJC 9758 2023 Prelim P1 Solution

2023 EJC Prelim Paper 1 Solutions Q1 Sub x = 0, 2ABC++= 22d e 2e 2ed xx xy AB Cx −= +− Sub x = 0, 22 3ABC+−= − 2 22 2 d e 4e 4ed xx xy AB Cx −= ++ Sub x = 0, 4 4 11ABC++= Solving simultaneously, 1, 1, 2A BC= −== . So particular solution is 22f( ) e e 2exx xyx −= = −+ + . Q2 (a) ( ) ( )1 11 1 2 3 3 2 2 222 32 2 3 22 2 22 2 218 183 3 3 3 3 333 nnn n nn n n nnn n n n n n nnnuSS + ++ − − − − − − −−− −= −= −− − =−= −= =  2 1 1 3 2 23 2 3 3 n nn n n n u u − − − − = = Since 1 n n u u − is a constant independent of n, the sequence is a geometric progression. (b) As n→∞, 218 18 183 n nS = −→  (Or using G.P. sum to infinity formula 2 3 6 181S∞ = =− ) Let the common difference be d. ( )9 2 4 8 182 88 4 3 2 d d d  − + = −+ = =

Q3 (a) (1 ) (1 ) OB OAOP λλ λλ +−= +−      0 2 22 1 (1 ) 1 1 2 11 λ λλ λ −       = +− =      − − −−    10 (1 ) 3 (1 ) 1 1 2(1 ) 0 2 22 OC OBOQ λ λλ λλ λλλ λ    +−    = = +− =+   +−     −−         32 2 31 PQ OQ OP λ λ λ − =−=  −       (b) PX OX OP= −      2 22 11 5/2 1 λ λ −     = −−    − −−   2 2 3/2 λ λ  = − − Since P, Q and X are collinear, PX k PQ=     for some ,0kk∈≠ 2 2 32 22 3/2 3 1 2 (3 2) (1) 2 2 (2) 3 (3 1) (3)2 1 From (2), 1 Sub into (1), 2 3 2 k k k k k k λλ λ λλ λλ λ λλ λ λλλ −     −=    −−   = − −−− − = −−−−− − = − −− =− = − −= − 2 2 2 3 20 1 or 2 (NA 0 1)2 Checking, 13 Sub into (3), 3 12 2 k λλ λλ λ λ λλλ λ + −= = = − << = − −+ =−  2 3 20 1 or 2 (NA 0 1)2 1 2 λ λλ λ λ + −= = = − << ∴= 

Q4 (a) ( ) ( )( ) 2 22 2 2 2f 1 2 11 11 11 11 13 11 24 11 35 11 46 11 31 11 2 11 11 31 1 21 nn rr n r n r r r rr rr nn nn nn nn = = = = = − = −+ = − −+ = − +− +− +− + +− −− +− − +− −+ =−− + ∑∑ ∑ ∑  (b) ( ) 31 11 2 2 n r rr − = +∑ Replace r with r – 1, ( )( ) 13 1 1 11 3 2 12 2 1 12 2 1 rn r n r rr r −= − −= = = − −+ = − ∑ ∑ 3 11 22 22 22 11 n rr rr= =  = −  −− ∑∑ 31 1 311 2 3 3 1 2 11 12nn   =−− −−−  +   23 1 1 132 3 3 1nn= −− +

Q5 (a) ( ) 1sin 3 cos d sin 4 sin 2 d2 1 cos 4 cos 2 24 2 1 cos 4 2cos 28 x xx x xx xx c x xc = + −= −+ = −++ ∫∫ (b) ( ) ( ) 2 22 e cos3 d 13cos3 e sin 3 e d22 x xx xx x xx= + ∫ ∫ ( ) ( )( ) 2 221 3e 3cos3 e sin 3 cos3 e d2 22 2 x xxx x xx =+−   ∫ ( ) ( ) ( ) 22 21 39cos3 e sin 3 e cos3 e d2 44 xx xx x xx=+− ∫ Moving the integral over, we have ( ) ( ) ( ) 2 2213 1 3cos3 e d cos3 e sin 3 e4 24 x xxx x x xc =++∫ ( ) ( ) ( ) ( ) 2 22 1 2 1 41 3cos3 e d cos3 e sin 3 e13 2 4 1 e 2cos3 3sin 313 x xx x xx x x c x xc = ++ = ++ ∫ Q6 (a) f( ) f( ) f( 4 )33 scale translateyyyx x x=   → =    → = + So the equation of C is 3f ( 4)yx= + . Since we know f(7) 9=− , substituting 3x= we get 27y=− . Hence the corresponding point is (3, 27)− . Additionally, we are given that f (7) 10′ = . So differentiating the equation for C we get 3f ( 4)yx′′= + . Again, substituting 3x= we get 30y′= i.e. the gradient at this point is 30. (b) The corresponding point is 1 9(7, )− . Differentiate 1 f( )y x= :

[ ] 2 d f( ) d f( ) yx x x ′=− We are given that f (7) 9=− and f (7) 10′ = . So At 7,x= ( )7 2 1 d (10) 10 8d 9x y x = = −= − − Q7 (a) When x = 0, 32320tt += . ( ) 2 320tt += , 0t = or 2 3t =− When t = 0, 20 11y= += . When 2 3t =− , 3 2 2 2 59213 3 27y  = −− + − +=   . When 1t =− , 1x=− and 4y= . When 1t = , 5x= and 2y= . (b) The curve is a circle with centre (0, 1) and radius k . (c) Substitute 2332xt t= + and 3 221yt t= −+ + into 2C ,

( ) ( ) 23 322 2 32 2 1 11 6tt tt+ +−+− =+ ( ) ( ) 66 54 549 12 4 4 4 16 0t tt ttt+++ − +− = 65410 8 8 16 0ttt+ + −= Using GC, 0.90823t = (reject root smaller than 1− and non-real roots) Substituting, 3.8973x= and 1.9006y= . The coordinates of the point of intersection are (3.90,1.90). (d) There can be 0, 1 or 2 points of intersection. Q8 (a) (b) Any 2 points with the same y-value, e.g. any 2 of (0,0) , (π,0) , (2π,0) (c) From graph in part (a), ( ]fR 4,5= . Since ( ] [ ]fgR 4,5 0, 2 D π= ⊆= , the composite function gf exists. 4.5 x 0 y

(d) Since ( ]fR 4,5= , we consider the graph of g( )yx= or sinyx= restricted to 4 5x< ≤ . When 4 5x< ≤ , g( ) 0.71 57x≤ <−− , so the range of gf is [ )1, 0.757−− . Q9 a ( )( ) ( ) ( ) 1 2 2 2 22 2 2 22 4 2 2 2 4 1 1 1 13 1 221 ...2 2! 31 ...28 ax ax ax ax aa xx − = +− −  −−   =+− − + − + = ++ + The expansion is valid for 22 1ax−< 22 1ax < since 22 0a x ≥ 2 2 1x a< since a is positive From graph, 11 xaa−<< b Since 1 2 1 d cos 1 x xC x −−= + − ⌠ ⌡ , substituting 1a= into the expansion obtained in (a) and integrating, 2 14 3cos 1 d 28 xx xx− ≈− + +⌠⌡ x 0 y

51 3 3cos 6 40 xxcx x− ≈−− − for some constant c Since 1 π 2cos 0− = , substituting 0, 2xc π= = . 3 15 π3cos ...2 6 40 xxx−∴ = −− + c 11 35 122 00 3cos d d 0.65760 (5 )2 6 40 xxx x x x sfπ− ≈ −− − =∫∫ d By GC, 1 12 0 cos d 0.65757 (5 )x x sf− =∫ Estimate in (c) is accurate to 4sf but not to 5sf. To improve estimate, we can include higher-order terms in the Maclaurin series expansion of 1cos x− Q10 (a) (cos isin ) (cos isin ) ((cos cos ) i(sin sin )) 2cos cos i 2sin cos22 22 2 cos cos isin22 2 zw rr r r r αα ββ αβ αβ αβ αβ αβ αβ αβ αβ αβ + =+++ = +++ +− +−= +⋅ −+ += + 2 cos 2zw r αβ−+= arg( ) 2zw αβ++= Alternative ii ii i 22 2 i 2 ee ee e e 2cos 2 2 cos cos isin22 2 zw rr r r r αβ αβ αβ αβ αβ αβ α βα α β β +− −  −   +  + = +  = + −= ⋅  − = + +  + 2 cos 2zw r αβ−+=

arg( ) 2zw αβ++= (b) 152(2)cos 4cos 2 32 12 12 6zw ππ π+= − = =  15arg( ) 2 12 12 4zw πππ+= + = (c) 2 2zv w= = 5ππarg( ) 2arg( ) arg( 1 π 41) 22 2v zw ×−= −= = − (d) Since πarg( ) 4zw+= and πarg( ) 4v =− , angle POQ is a right angle. Thus a rea of triangle OPQ 21 2 3 nits2 uz wv= += . Alternatively, Area of triangle OPQ 2 1 ππ(2 3)(2)sin2 44 2 2 π u 3 ni 2 t3 s sin −=    − =   =  P Q

Q11 (a) ( )( ) ( )( ) ( ) ( ) 0 0 0 2 2 0 2 Volume d12 d12 d2 2 ln( 1) 2 ln( 2) 2 ln( 1) 2 ln( 2) 212 ln 2 222l 2 3 2 3 22 13 6 n 2 ln26 6 2 k k k k k kxxx xxx xx kx kx kk kk kk k kk kx kx k kx x x k k k k π π π π π π π  = + −−  ++  = ++ = + = +− ++ = +− ++   += +  +   + −++  − + −  = + ∫ ∫ ∫ 2 ()6 shownk +  (b) 0 2d 3 k kx x−⌠ ⌡ ( ) 0 16d 3 3 k kx x= −−−⌠ ⌡ ( ) 2 0 3 6 3 2 3 k k x −= −    ( ) ( ) 3 32 2 24 3 k k = −−  ( ) 3 2 3 3 2 241k = −  (c) in π0π 3A kk = = − out 3 2 ππ 3A kkk −= = 