CJC 9758 2023 Prelim P1 Solutions

Page 1 of 25 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 2023 JC2 PRELIM EXAM PAPER I SOLUTION Q1 Method : 2 1 (1) 2 4 24i (2) zw wz += − = + From (2): = 2 4 24izw −− Substitute into (1): 2(2 4 24i) 1 4 7 48i ww ww − − + = − − = ( ) 22 22 22 22 2 2 2 Let i 4( i) 8 48i+1 = (4 7) (4 48)i = Comparing Imaginary parts, 4 48 0 12 Comparing Real parts, 47 4 7 12 ** 4 7 144 15 56 95 0 19 or 515 w a b a b a b a b a b b b a a b aa aa aa aa =+ + − − + − + − + −= = − = + − = + − = + − − =  =− =

Page 2 of 25 From **, since 4 7 a positive real number, 19 19when , 4 7 4 7 015 15 19reject 15 a aa a −= =− − = − −    =− 5, 12, = 2(5 12i) 4 24i 6 5 12 , 6 a b z w i z  = = + − − =  = + = Method : 2 1 (1) 2 4 24i (2) zw wz += − = + 2 1 a positive real number Let and iz z x w a b+ =  = = + From (2): 2( ) 4 24ia bi x+ − = + Comparing Real and Imaginary parts, 24 2 24 12 ax bb −= =  = 22From (1): 2 1 (3) x a b+ = + Substitute 12 and 2 4 into (3):b x a= = −

Page 3 of 25 ( ) ( ) 22 22 2 2 22 2 2 2 4 1 12 4 7 12 ** 4 7 144 16 56 49 144 15 56 95 0 19 or 515 98 or 6 15 aa aa aa a a a aa aa xx − + = + − = + − = + − + = + − − =  =− =  =− = However 2 1 a positive real number, 98 98When , 2 1 2 1 015 15 98 19reject and 15 15 z xz xa += =− + = − +    =− =− 6, 5, 12 5 12 , 6 x a b w i z  = = =  = + =

Page 4 of 25 Q2 (a) When 1x n= , ( ) 2 1 11 n n y= + When 2x n= , ( ) 2 2 21 n n y= + … … When 1nx n −= , ( ) 2 1 11 n n n n y − − = + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 2 2 2 3112 3112 3112 1 1 1 1... 1 1 1 1 1 1 2 3 1... 1 1 1 1 11 2 3 ... 1 2 3 1 1 1 2 3 1 2 3 n n n n n n n n n n n n n n n A n n n n n n nnn n n n n n n nn n n n n − − −                = + + +       + + + +         −= + + + + + + + +  −= + + + + + + + +− = + + + + + ( ) ( ) ( ) 1 22 2 1 2 1... 1 1 shown n r n nn r n nr − =  − ++ +− = +

Page 5 of 25 (b) ( ) ( ) 1 0 11 2 0 112 0 2 2 2 lim d 1 1d 11 12 2 21 1 22 n xAx x x x x x → − = + =  +  +=   =−   

Page 6 of 25 Q3 (a) rq− is parallel to p ⟹ r q p r q p   −= =+ The point R lies on a line that passes through the point Q and is parallel to the vector p . (b) ( ) 0 0 r q p r p q p r p q p −  =  −  =  =  2 4 2 5 1 5 3 2 3 2 5 3 7 x y z x y z −                − =  −                      − + =− The point R lies on a plane that contains the point Q and is perpendicular to the vector p .

Page 7 of 25 Q4 (a) There is only one(positive) real root in the equation f ( ) 0.x = Since the equation has all real coefficients, then the two other roots must be a pair of complex conjugates. (b) 321 2i 2 7 16 0, 2( 11 2i) 7( 3 4i) 16(1 2i) 0 15 0 15 (shown) Since is a root of x x x c c x c c − − + + = − + − − − + − + = += = = − (c) Since all the coefficients are real, is another root1 2ix= + 32 . 2 7 16 0of x x x c− + + =    ( )    ( ) ( ) 32 2 2 7 16 15 0 (1 2i) (1 2i) 2 =0 ( 1) 2i) ( 1) 2i) 2 =0 2 5 2 =0 x x x x x x k x x x k x x x k − + − = − + − − − − + − − −  − + − Comparing the coefficient of constant term (or by long division), 5 15 3kk− =−  = Therefore, the last root is 3 2x= The roots are 21 2i, 1 i 3 , 2x x x −= = =+ (d) 322 7 16 0x x x c− + + = 1Replace with x w 32 23 1 1 12 7 16 15 0 2 7 16 15 0 w w w w w w      − + − =           − + − =

Page 8 of 25 3 ; 2 2 3 1 1 1Hence, the roots are 1 2i ; 1 2i 1 1 2 i1 2i 5 5 1 1 2 i1 2i 5 5 w w w w w w −= = = = = + =+ +− = − =

Page 9 of 25 Q5 (a) Method : The nth term, nu , is always one degree less than nS since 1n n nu S S −=− . If nS is quadratic, nu would be linear but it is not since there is no common difference between consecutive terms. Method : Proof by Contradiction Suppose 2 nS an bn c= + + 2 nS an bn c= + + 4 4 2 4 2 6 9 3 6 12 6 16 4 6 38 44 abc a b c a b c a b c + + =− + + =− − =− + + =− + = + + = + = Using G.C., no solution found. Hence nS cannot be a quadratic polynomial. (b) 32 nS an bn cn d= + + + 4 8 4 2 4 2 6 27 9 3 6 12 6 64 16 4 6 38 44 a b c d a b c d a b c d a b c d + + + =− + + + =− − =− + + + =− + = + + + = + = Using G.C., 2, 5, 1, 0a b c d= =− =− = 3225nS n n n= − −

Page 10 of 25 (c) ( ) ( ) ( ) ( ) ( ) ( ) 1 3232 3 2 3 2 2 3 2 3 2 2 2 5 2 1 5 1 1 2 5 2 3 3 1 5 2 1 1 2 5 2 11 15 6 6 16 6 n n nu S S n n n n n n n n n n n n n n n n n n n n n nn −=− = − − − − − − − − = − − − − + − − − + − + = − − − − + − = − + (d) ( ) ( ) ( ) ( ) ( ) 2 1 10 10 9 11 10 12 11 2 1 2 2 2 2 1 29 22 2 + + + + + 6 2 16 2 6 6 9 16 9 6 24 32 342 m nn n mm mm m uu uu uu uu uu uu uu mm mm − = −− − − =− − − − − =− = − + − − +  = − − 