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 and Approaches HL Preliminary Examination 2024/P3 4 (g) Method 1: Finding the gradient of the tangent Observe that ZY is a tangent to the ellipse at midpoint ( )0,1 . Gradient at ( )0,1 02 11 ( 1) ZYm −= = =− −− . Similarly, gradient at ( )0, 1− ( )02 11 ( 1) XYm −−= = = −− . Method 2: GDC graph using gradient of appropriate functions 2 2 2 143 33 41 333 xy yx  + + = = − +  From GDC, numerical derivative, ( ) ( ) 0,1 0, 1 d 1d d 1d y x y x − =− = M1 A1 A1 OR M1 A1 A1 OR M1 A1 A1

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 5 Method 3: Implicit Differentiation Differentiating implicitly w.r.t. x, 2 2d 1 d 43d 3 d 3 1d6 2 0 3d d 3 1 d3 xyxx yxy x y xxy     + + =          + + = =− + ( )0,1 d 3 1 01d 1 3 y x =− + =− , ( )0, 1 d 3 1 01d 1 3 y x − =− + =−  By whichever method: Since ( )( )XY ZX 1 1 1mm  = − =− , the tangents are perpendicular. OR XY ZY 1 XYO 45 1 ZYO 45 m m =   =  =−   =  Hence, XYZ 45 45 90 = + =  . OR M1 A1A1 A1 OR A1

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 6 2 DE, Euler’s and modified Euler’s method [Max mark: 31] (a) 00 0 0 00 0 0 2( ) 0 2( ) 0 1 d 2 d No need for ln as 0ln ln 2( ) ln 2( ) e e yx yx xx xx yxy yyy y x x y xxy y y yy − − =   − = −   = − = =  M1A1 M1 A1 Alternatively, ( ) 0 0 0 2 2 2 00 2 0 2 0 2( ) 0 1 d 2 d No need for ln as 0ln 2 e e e e At , , e e e x c c x x x x xx yxy yyy x c yA xy yA Ay yy + − − =   = +   = = = = = =  M1 A1 M1 A1 (b) 0 1 1 1 Apply Euler's method: f ( , ) with f ( , ) 2 , n n n n xxy y h x y x y y h n− − − −= + = = 11 1 2 2 0 (2 ) (1 2 ) (1 2 ) (1 2 ) n n n n n n y y h y hy hy hy −− − − =+ =+ =+ =+ 0 0i.e 1 2 (shown) n n xxyy n  −=+   M1 (apply Euler’s method with correct 11f ( , )nnxy−− ) A1 1(1 2 ) nhy −+ AG Alternatively, 1 0 0 0 2 2 1 1 0 0 0 0 0 0 0 (2 ) (1 2 ) (2 ) (1 2 ) 2 (1 2 ) (1 2 )(1 2 ) (1 2 ) (1 2 ) i.e 1 2 (shown) n n n n y y h y h y y y h y h y h h y h h y h y y h y xxyy n = + = + = + = + + + = + + = + =+  −=+   M1 A1 AG

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 7 (c) ln 1 ln ln 1 1 w w nvn n n + = + = 2 2 ln 1 0lim ln lim 1 0 1 1 lim 1 lim 1 nn n n w nv n w w n n n w w n w → → → → + =    −  += −  =   + = M1 M1 A1 A1 (d) lim ln lim e w nn v w v → → =  = Since 1 , i.e. lim 1 e nn w n wwv nn →    = + + =       i.e. 1 e as . n ww nn + → → (deduced) R1 AG (e) Let ( )02w x x=− . 02( )02( )lim 1 e n xx n xx n − → − + = Since 0 012 n n xxyy n  − =+   , hence, 02( ) 0lim e ( ) (shown)xx nn y y y x − → == A1 R1

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 8 (f) Apply the Modified Euler’s Method:  111 f ( , ) f ( , )2 nnn n n n hy y x y x y− −−= + + 0with f ( , ) 2 , xxx y y h n −==  11 11 1 1 1 1 2 2 1 21 1 1 2 12 (2 2 ) (1 ) (1 ) (1 ) 1 1 (2 2 ) (1 ) ( f ( , ) f ( , )2 2 1 ) (1 1 1 1 2 ) 1 nn n n n n nn nn nn n n n n nn nn n n n n n n n n hy y y y y h y hy h y h y hyy h y y y y h y h y h y h y h y x y x y y h h h yy hy h −− −− − − − − − − − −− −− −− − = + + = + + − = + += − = + + = + + −  = + + =+ + = − += − 2 12 0 1 1 1 1 n n n h yh hyy h −− += − += − i.e. 0 0 0 1 (shown) 1 n n xx nyy xx n − += − − M1 (with correct 111f ( , ) , f ( , 2) 2nn n n nn xy x yy y−− − = = ) A1 M1 AG (g) d 2 (*) d y yx =− for which 1y= when 0x= . When 00 0, 1xy== , the exact solution is 02( ) 0 e xxyy −= i.e. 2e xy= when 0.5, e.xy== M1 A1 (h) The graph of 2e xy= is concave upwards. OR 2 2 d 0 d y x  OR d d y x increases Hence the tangent drawn using Euler’s method gives an under-estimate of the true value of y. R1

Year 6 Mathematics: Analysis and Approaches HL Preliminary Examination 2024/P3 9 (i) If 0.1, then 5.hn== (i) Euler’s Method ( )5 5 0.5 0 11 2 1 1.2 2.488 3d.p. n n ny nn y  −    = + = +        == (ii) Modified Euler’s Method 0 0 0 1 . 1 n n xx nyy xx n − += − − ( ) 0 5 5 0.5 0 111 2 0.5 0 111 2 1.1 2.727 3d.p.0.9 nn nn nny y y nn y −   ++   =  =   −   −−    == M1 A1 A1 Modified Euler’s Method A1 (j) Euler’s Method 0.5 0 11 2 1 n n ny nn  −    = + = +     