ACSI 2022 HL AA Prelim Paper 3 Solutions for students

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 1 PRELIMINARY EXAMINATION 2022 YEAR 6 IB DIPLOMA PROGRAMME MATHEMATICS : ANALYSIS AND APPROACHES HIGHER LEVEL Paper 3 Qn Solution 1. (a) Hence is a root of the above equation Comments: Generally well done. (b) so is a root of Any root Any integer power of  can be reduced exclusively to integer power of k only. Since there are 10 values of k , this leads to 10 distinct complex number arguments. Comments: poorly done. Many did not comprehend the question.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 2 Qn Solution (c) Using the Argand diagram, it can be observed that there is a total of 6 equally spaced roots and thus 5  = , Comments: Some draw argand diagram with wrong angles and lack of symmetry. Some did not label correctly according to what is required in the question. (d) Let sin b = , sin 2 c = , cos d = and cos2 e = Considering the symmetry of the Argand Diagram about the Imaginary axes, it can be observed that sin 4 b = , sin 2 c = , cos4 d =− and cos3 e =− Therefore sin 4 sin= , cos4 cos=− and sin3 sin 2= . Hence ( ) 2 2 2 2 2 2 sin 3 sin 4tan 3 tan 4 cos3 cos 4 sin 2 sin cos 2 cos 2sin cos sin cos2cos 1 2sin 2cos 1 2 tan 2 sec 2 tan tan1 tan tan tan 2                = =−− = −−− = − = − = − =

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 3 Qn Solution Alternative Solution for proving tan3 tan 4 tan tan 2   = Using the Argand Diagram Let gradient 1tan m = then by symmetry, 1tan 4 m =− Similarly, let gradient 2tan 2 m = then 2tan 3 m =− ( )( )12 12 tan 3 tan 4 tan tan 2 mm mm   = − − = = Comments: Surprising number did not utilize the argand diagram effectively and gave long winded workings to arrive at same result. Some misread the question and miss out the need to prove the trigo identities and lost marks then. Students generally quite weak in trigonometry. A few just used GDC to give approximate value and not the exact value needed for proof. (e) DE = 1− Or using 1 CD AB ED ED ED = = =  Since AB = 1 (alternative form of solution accepted) Comments: Generally well done.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 4 Qn Solution (f) 1 1=− (resulting from property of Golden Rectangle given in question or from the fact that DE can be expressed in the two forms above) 2 10 −− = Using quadratic formula, ( )1 1 4 1 2 15 2  − −= = Since  is the ratio of lengths, we reject the negative answer and therefore 15 2 += (proven) Comments: Common careless mistake in not rejecting the negative root. (g) Using compound angle formula, double angle formula and appropriate simple trigonometric identities, Starting from sin3 sin 2= , ( ) ( ) 22 sin 3 sin 2 sin 2 2sin cos sin 2 cos cos 2 sin 2cos sin 2sin cos 2cos 1 sin 2cos sin 0                  = += += + − − = Since sin 0  ( ) 222cos 2cos 1 2cos 0  + − − = and rearranging, 24cos 2cos 1 0− − = (shown) Comments: Some careless mistakes in proof or students. Many divided by the sin function which is not a good practice. Some missed out showing that sin 0  and hence lost a mark here.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 5 Qn Solution (h) Using quadratic formula (or otherwise) 2 4 4(4)( 1)cos 8 2 2 5 8 15 4   − −= = = As 5  = is an acute angle 15cos 42 +== , since 15 2 += . Comments: Some did not see the connection between this and the preivous part, did not use quadratic to simiplify and obtain the answer. A number of students again missed out showing that the negative angle is rejected and lost marks here. 2. (a) 2 2 2 2 2 2 2 2 20 20 dy udx du uydx d y du uydx dx dy uydx d y dy ydx dx = =− + = =− + + − = +−= Comments: Students must recognize that the solution of 2nd order differential equations is not in the syllabus- and therefore they cannot try to solve it using 1st order Differential Equations method. The question is teaching you how to find solutions to 2nd order DE. Students must learn how to show a proof correctly and state clearly what you set out to prove. (b) ( ) 2 2 2 2 2 0, rx rx rx rx ye dy redx dy redx e r r = = = + − = Since 0rxe  , 2 20rr+ − = Comments: Do not simply cancel off erx Need to state 0rxe  or erx>0.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 6 Qn Solution (c) ( )( ) 12 2 1 0 , 2 or 1 rr Therfore r r + − = =− = Comments: Students should just simply factorise and not apply quadratic eqn formula or even sketch the graph. (d) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 4 2 4 2 2 (4 4 ) (1 1 2) 0 xx xx xx x x x x x x xx y Ae Be dy Ae Bedx dy Ae Bedx d y dy ydx dx Ae Be Ae Be Ae Be A A e Be − − − − − − − =+ =− + =+ +− = + + − + − + = − + + − = Comments: Need to state clearly what you set out to prove. Some students write the equation throughout . (e) ( ) ( ) 2 2 2 2 2 20 20 2 1 0, 0 10 1 kx kx kx kx kx d y dy ydx dx k e ke e e k k e k k − + = − + = − + =  −= = Need to write 0kxe  (f) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 ( 2 ) 2 ( 2 ) 2 2 2 2 2 0 x x x x x x x x x x x y A Bx e dy Be A Bx e A B Bx edx dy A B Bx e Be A B Bx edx d y dy ydx dx A B Bx e A B Bx e A Bx e A B Bx A B Bx A Bx e =+ = + + = + + = + + + = + + −+ = + + − + + + + = + + − − − + + = Comments: Need to show clearly what you set out to prove.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 7 Qn Solution (g) ( ) 12 2 (1 2 ) 2 kx x r x r x xx A Bx e xe Ae Be e e − + +=++ Comments: Need to show clearly the shape of the graph, there is no straight line. Label clearly the intercepts, min point, state the equation of the horizontal asymptote y=0 (not x=0) .

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6/ 2022 Prelim Exam / Paper 3 / Solutions 8 Qn Solution (h) Graphically, there seems to an oblique asymptote As 1, 2x y x→ → + ( ) 2 3 (1 2 )lim 2 (1 2 )lim 2 12 02 1 2 x xxx xx xe ee x e x x −→ −→ + + += + += + =+ 1 2yx=+ Alternative Solution using long division 3 3 3 3 3 3 3 1 2 2 2 1 2 1 1 1+ 2 1 x 2 1x11 2lim 2 2 2 x x x x x x xx x ex x xe xe e e e xx e − − − − − − −→ + ++ + − −−  −−  + + = + +  Comments: Students need to show understanding of oblique asymptote, as x tends to infinity, y tends towards x+1/2. (i) ( ) 2 2 2 2 4 4 0, 4 4 1 0 2 1 0 21 1 2 d y dy ydx dx mm m m m − + = − + = −= = = Therefore the general solution is ( ) 1 2 x y A Bx e=+ Comments: This is quite poorly done- most students do not seem to understand the intend of the question, that is, teaching you how to solve a second order differential equation from the earlier parts- very few got it correct.