CJC_9758_2023_Prelim_P2_Solutions_2

Page 1 of 31 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 2023 JC2 PRELIM EXAM PAPER 2 SOLUTION Q1 ( )( ) 2 2 2 2 2 2 2 2 2 25 22 25 202 2 5 2 4 02 65 0( 2) 65 0( 2) 15 0( 2) xx xx xx xx x x x x xx xx xx xx xx xx xx +− − +− −− + − − + − − + − − −+ − −− −  0 or 1 2 or 5x x x    . − + − + 0 1 2 + 5 x

Page 2 of 31 Solution of 2 2 25 22 xx xx +− − is the complementary of solution of 2 2 25 22 xx xx +− − So, replacing x with x , solution of 2 2 25 22 xx xx +− − will be 0 1 or 2 5xx    For 01 x , 1 1, 0xx−    For 25 x , 5 2 or 2 5xx−  −   Thus, range of values: 5 2 or 2 5xx−  −   or 1 1, 0xx−    x 5 −5 −2 2 y=2 y=5 y y=1 1 1

Page 3 of 31 Q2 (a) ( ) ( ) ( ) ( ) 2 2 e sin 1 d e cos e sin 1d e cos sin 1 d e sin cos e cos sind 2e cos where 2 shown x xx x xx x y x x y xxx xx y x x x xx xk − −− − −− − = + − = − + = − + = − − − − =− =− (b) ( ) ( ) ( ) 3 3 d 2e sin 2e cosd 2e sin cos xx x y xxx xx −− − =− − − − =+ ( ) ( ) ( ) ( ) When 0 : f 0 1 f ' 0 2 f '' 0 2 f ''' 0 2 x= =− = =− = 2 3 2 3 1 2 2 2 ... 2! 3! 1 2 ...3 xxyx xxx =− + −  +  + =− + − + +

Page 4 of 31 (c) ( ) ( ) ( ) ( ) ( ) 1 3 2 2 3 22 3 22 3 22 3 2 3 2 23 1 2 ...e sin 1 3 cos 2 21 ... 2! 1 2 ... 1 23 1 2 ... 1 1 2 ...3 1 2 ... 1 2 ...3 1 2 2 4 ...3 131 2 3 ... 3 x xxxxx x x xx x x xx x x xx x x xx x x x x x x − − − + − + ++− = −+ = − + − + + −  = − + − + + + − − +  = − + − + + + + =− − + + − + + =− + − + +

Page 5 of 31 Q3 (a) Method 1 ( ) ( ) 1 13 2 13 2 1 13 2 MN P P P P M P NP =+−− = − + When 0P= , 1 13M = When 13 2P= , 2 13N = ( ) ( ) 112 13 2 13 13 13 2 1 1 2 13 13 2 P P P P PP =+−− =+  − ( ) ( ) ( ) ( ) 1 2 d1 13 2d 26 11 d 1d13 2 26 1 1 2 1 d 1d13 13 2 26 1 2 1d d 1d13 2 2 1ln ln 13 2 2 1ln 13 2 2 e13 2 tC P PPt PtPP PtPP P P tPP P P t C P tCP P P + − =− =− += − =− − − = + =+− =− −          

Page 6 of 31 1 2 1 2 1 2 11 22 11 22 1 2 1 2 1 2 e13 2 ee13 2 e13 2 13 e 2 e 1 2 e 13 e 13 e 1 2 e 13 e2 tC t C t tt tt t t t P P P P P P P A AP P A A AP A AP A A + − =− =− =− =− +=  =   = + + When 0t= , 2P= , 132 12 2 4 13 92 2 9 A A AA A A = + += = =

Page 7 of 31 ( ) 1 2 1 2 213 9 2e2 9 26 shown 9e 4 t t P − − = +  = +

Page 8 of 31 Method 2 ( ) ( ) 2 22 1 '2 d1 13 2d 26 11 d 1d13 2 26 11 d 1d13 2 26 1 1 1d 1d2 2613 13 44 13 13 1 1 144ln 13 13132 262 444 13 12ln '2 13 2 e tC P PPt PtPP PtPP Pt P P tC P P tCP P P −+ =− =− =− −=    −−       −− − = + −+ − =− + − =          

Page 9 of 31 1 '2 1 '2 1 2 1 2 1 2 1 2 13 2 e 13 2 ee 13 2 e 13 e2 131e 2 13 2 4 e tC t C t t t t P P P P P P P AP PA P A A −+ − − − − − − = − = − = =   + −=  =   − When 0t= , 2P= ,