RVHS 23 Prelim H2MA P1 Std Soln

River Valley High School 1 2023 H2MA Prelim Paper 1 Markers Comments 1 Solution [6] Inequality P1 Q1 (i) ( ) ( )( ) 2 2 2 4 4 1 012 21 01 2 1 xx xx x xx −+ +− − −+ Many students did not factorize the denominator correctly, missing out the negative sign to ( )( )1 2 1xx−+ , which resulted in errors to solutions Method 1: ( ) ( )( ) ( ) ( )( ) 2 2 21 01 2 1 Since 2 1 0 for all real values of , 1 2 1 0 x xx xx xx − −+ − − +  1 or 12xx −  Some students still incorrectly wrote ( ) 2 2 1 0 x− or stated always positive. Method 2: ( ) ( )( ) 2 21 01 2 1 x xx − −+ 1 or 12xx −  (ii) ( ) ( )( ) ( ) ( ) 21 21 2 2 21 01 2 2 2 2 1 0 1 2 2 2 x xx x xx + + − +− −  +− Replace with 2 ,xx Most students could identify the correct replacement to use. However, most students missed out  in the question and thus missed out the solution 1x=− . Many students also did not give the reason for rejection correctly i.e. 2x > 0 for all real x 1 2x=− 1x= 1 2x= + + − − o  

River Valley High School 2 112 OR 2 1 OR 2 22 (rejected, since 0 1 2 is positive for all real values of .) x x x x xx x −  =  =−

River Valley High School 3 2 Solution [6] Complex Numbers P1 Q2 (i) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 22 22 22 11 * 1 i 2 i1 1 i * 1 i 2 1i 1 i 1 i 2 1i 1 2i 1 i 11 1 21 1 1 2 2 1 1 2 1 4 2 2 2 1 2 4 4 1 z z a a a a a a a a a a a a a a a a a − =+ +− =++ −+ =−+ −+ =−+ −+ = ++ − + + = + − + = + − + = + 2 4 3 0 1 or 3 (rejected since 2) aa aa − + = = Some students incorrectly applied properties of modulus. Common errors include: • ( ) ( ) 2 i 1 1 1aa+ − = − − • 11zz− = − • ( ) 11 * 1 i 2 z z − =+ ( ) 11 * 1 i 2 z z − = + It is necessary to state both roots for a before rejecting the one which doesn’t satisfy the criteria. (ii) Consider 1arg * n z zw −  ( ) ( ) ( ) 1arg * arg 1 arg * arg arg i arg 1 i 4 2 4 4 2 zn zw n z z w n n n      −=  = − − − = − − −  = − − −  = If 1 * n z zw −  is purely imaginary and negative, then: Most students attempted to find 1arg * n z zw −  using properties of argument, however only some were successful. Note: It is incorr ect to assume w = 1 + i just because arg(w) = 4  Many students were not able to identify correctly the possible values of the arg ument for purely imaginary and negative.

River Valley High School 4 1 3 7 11arg , , ,...* 2 2 2 n z zw   − = or 3 22 k + , k + Thus, 3 7 11, , , ...2 2 2 2 3, 7, 11, ... n n    = = 3 smallest positive values of 3,