SJI 2024 Year 6 HL MAA Prelim Examination Paper 3 (MS)

Year 6 HL MAA Preliminary Examination 2024 Paper 3 (Mark Scheme) Qn Suggested Solutions Marks 1 Complex polynomials, conics, calculus [Max mark: 24] (a)(i) Solve 32( ) 2 0p z z z= + − = Method 1: GDC Method 2: Analytical ( ) ( )( ) ( ) 2 2 (1) 1 1 2 0, 1 is a factor. ( ) 1 2 2 by inspection/long division etc ( ) 0 2 2 4(1)(2)1 or 2 1 i pz p z z z z pz zz = + − =  − = − + + = −  − = = =−  Since 1 2 3arg arg argz z z−     , 1 2 3 1 i, 1, 1 i. z z z =− − = =− + (M1) OR (M1) A1 A1 A1 (a)(ii) (b) (c)(ii) (a)(ii) A1 (label P, Q, R in the correct order) (b) A1 (label A, B, C) (c)(ii) A1 (label H, K) (c)(i) ( ) 32 2 ( ) 2 '( ) 3 2 Solving '( ) 0, 3 2 0 20 or 3 p z z z p z z z pz zz zz = + − =+ = +=  = =− M1 A1 A1 (d) Coordinates of the midpoints: ( ) 11A 1,0 , B 0, ,C 0,22    −−        (M1)A1 (or labelled in diagram)

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 2 (e)(i) Since ( ) 2, ,0 3hc =− and ( ) ( ), 0,0kc = , the centroid ( ) 1 2 1, 0 ,0 ,02 3 3mc     = − + = −        . Therefore, the equation of the ellipse is 2 2 22 1 3 1 x y ab  + += ……………. (*) Subst ( )A 1,0− into (*), we have ( ) 2 2 2 2 3 0 1 n 4 show 9 a a − += = OR The ellipse’s semi-major axis, 2AM 3a== , where M denotes the centroid. 2 4 9a= . A1 (seen anywhere) M1 (subst ( )1,0− ) A1 AG OR M1A1 ( 2AM 3a== ) AG (ii) Method 1: Using a 2nd set of coordinates Subst 10, 2   into (*), we have 22 2 22 11 1 1 132 1 1 4 4 4 3 9 bbb          + =  + =  = Method 2: Using 2 2 2−=a b m Subst 2 41, 93am== from above, we have 2 2 2 2 4 1 1 9 3 3 b a m=− = − = By any method, or a more tedious sequence of substitutions: Solving, 22 41, 93ab== . (There is no need to find ,ab .) Hence, the equation of the ellipse is 2 2 1 3 141 93 x y  + += (accept equivalent forms) M1 (subst another coordinates A1 OR M1 (using 2 2 2a b m−= ) A1 A1

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 3 (Sketch of the ellipse – not required but always good to check answer, time permitting) (f) By functions transformation, observe that the vertices are formed by a vertical stretch of factor 2. Original ellipse: 2 2 1 3 141 93 x y  + += For a vertical stretch of factor 2, Replacing y by 1 2 y : 22 11 32 141 93 xy   +       += That is: 2 2 1 3 144 93 x y  + += Simplifying: 2 2143 33xy + + = (shown) (M1) Attempt to do a vertical stretch A1 AG (Sketch confirms the vertical stretch of factor 2, time permitting)

Year 6 Mathematics: Analysis