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(