ACSI 2022 Promo Paper 1 ans

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 1 FINAL EXAMINATION 2022 YEAR 5 IB DIPLOMA PROGRAMME MATHEMATICS : ANALYSIS AND APPROACHES HIGHER LEVEL Paper 1 Qn Solution Marker’s Comments 1. 9 12 3 27 0xx−  + = 23 12 3 27 0xx−  + = Let 3xy= 2 12 27 0yy− + = ( 9)( 3) 0yy− − = 93y or= 21x or= General well done with some careless mistakes.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 2 2. 33 2 9 9log 1 log lg lg 2lg 9 125 5xy= + = − + 2 9 9 2 125 81lg lg 2lg lg9 125 5 9 9 25 − + =  lg10 1== Therefore 3log 1 3xx=  = 331 log 1 log 0yy+ =  = Hence 1y= Many students made mistake when trying to simplify the log expression on the RHS. 3. (a) 32ln ln 3ln 2lnx x x x− = − = ln x 2ln ln 2ln lnx x x x− = − = ln x Since the common difference is the same, they form 3 consecutive terms in AP Students are required to show. But many simply find the difference and did not explain why are they finding the difference. A handful find the general terms when the question clearly state that there are only three terms.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 3 (b) 10 10 1 10ln (ln ln )2 k k x x x = =+ 5(11ln )x= 55ln 55 ln 1 x x = = xe= Some students used GP formula instead of AP. Did not realise that (a) and (b) are related. Some treat x as the dummy variable.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 4 4. (a) General term 26 6 12 3 6 ( ) ( ) 6 () rr rr px xr pxr −− −− = = Constant term: Power of x is 0 12 3 0r − = 4r= Term = 226 154 pp = Generally well done. Most errors are due to wrong computation of 6C4 or 6C2 (b) 215 60p = 22p or=− Generally well done. Most errors are due to carelessness. 5. (a) Let ( ) 32 99f x x x x= − − + ( )10f = Therefore, by factor theorem, 1x− is a factor of ( )fx . Let ( )( ) 3 2 2 9 9 1x x x x ax bx c− − + = − + + By inspection, 1a= and 9c=− Comparing coefficient of 2x : 1 ba− = − 1 1 1 0ba=− + =− + = ( )( ) ( )( )( ) 3 2 2 9 9 1 9 1 3 3x x x x x x x x− − + = − − = − − + Generally well done. (b) 32 99 02 x x x x − − + − ( )( )( )1 3 3 02 x x x x − − + − using answer from (a) 31 x−   and 23 x Many students could not work with equations with inequalities. This is an area for further work. –3 1 2 3

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 5 6. ( )( ) 2 11 2 3 2 2 1 2 xx x x x x +− =− − + − Let 2 1 2 3 2 2 1 2 x A B x x x x + =+− − + − ( ) ( )1 2 2 1x A x B x+ = − + + 2x= : 335 5BB=  = 1 2x=− : 1 5 1 2 2 5 AA=−  =− ( ) ( ) 2 1 3 1 2 3 2 5 2 5 2 1 x x x x x + =−− − − + Generally well done 7. (a) 2 1 1 1 3 2 4 2 3 1 1 10 kk −   − 2 2 1 31 RR RR − −⎯⎯⎯→ 2 1 1 1 3 0 2 4 3 0 0 9 3 kk −  − −− In order to have no solution, 2 90k −= and 30k− Therefore, 3k =− . Mostly well done and students can do manual ref. Some still need to wok on identifying no solution and infinite solutions from looking at the ref final matrix. (b) For infinite number of solutions, 3k = . Then we get 1 1 1 3 0 2 4 3 0 0 0 0 −  − . Let ,z = Then 32 4 3 2 2y z y + =−  =− − and 393 3 2 3 22x y z x   + − =  = − − − + = + Therefore, the general solution is 9 32x =+ , 3 22y =− − and z = Accept: 4.5 3 1.5 2 01 x y z             = − + −                Common mistakes due to conceptual error identified in 7a. Many wasted time doing ref again.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 6 8. (a) 23+ = −a b b a 22 23+ = −a b b a ( ) ( ) ( ) ( )2 2 3 3+  + = −  −a b a b b a b a 4 4 6 9 +  +  =  −  + a a a b b b b b a b a a 2 2 2 2 4 4 6 9+  + = −  +a a b b b a b a Since a and b are unit vector, then 1==ab . 10 5 10 + =ab 1 2=ab Failure to use shorter method by using the concept u.u =|u|^2. There are attempts to prove using 2 dimensional vectors which is not general. For IB students, if this were the solution path to take, the least they can do is work on 3 dimensional vectors. But this tedious method is not recommended. (b) 1 12cos 1 1 2 == 3  = (accept 60 ) Generally well done 9. (a) ( ) 2tan 2 abxy ab ++= − tan tan 2 1 tan tan 2 x y a b x y ab ++ =−− tan 2 1 tan 2 a y a b a y ab ++ =−− because ( )arctan tanx a x a=  = 222 2 tan tan 2 2 tan tana y a b ab y a b a y ab y+ − − = + − − ( ) ( ) 222 tan 1 1y a b a+ = + tan 2 by= Most have a good start but did not finish solving to the end.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 7 (b) 2tan 2 bxa b −== + Consider 12tan arctan arctanab     +        12 21 ab ab + = − 2 2 ba ab += − 2 42 2 2 22 bb b bb b −+ += − −+ 2 2 2 4 2 2 4 2 b b b b b b + + −= − − − 2 2 4 14 b b +=− =−+ Therefore 12tan arctan arctan 1ab     + =−        Then ( )12arctan arctan arctan 1ab    + = −       Since 1arctan a   and 2arctan b   are between 0 and  , 120 arctan arctanab     +         . Therefore 1 2 3arctan arctan 44ab    + = − =       Poorly done. Students should spend more time being familiar with inverse trigonometric functions. 10. (a)(i) Let ( ) ( )( )24f x a x x= + − . Substitute ( )0,16 : ( )( )16 2 4a=− 2a=− ( ) ( )( ) 22 2 4 2 4 16f x x x x x=− + − =− + + Surprisingly, many students were unable to find the correct expression for f.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 8 (ii) ( ) 22 4 16f x x x=− + + ( ) 22 2 16xx=− − + ( ) 2 2 1 1 16x=− − − + ( ) 2 2 1 18x=− − + Although the diagram is not drawn to scale, many students assume turning is (1,18) which happens to be correct this time round. They should not assume that the diagram is drawn to scale. All turning will usually be indicated in the question itself. (iii) Sequence of transformation: 1. Vertical stretch of scale factor 3 2 . 2. Reflection in x-axis. 3. Translation of 3 0   . Note: The order is not important for this question. Poorly attempted. For stretching, the factor must be positive (IB context). If it is negative, they should indicate by stating that there is a reflection. Many wrong words are used to describe the transformations. Words like shift and move are used instead of translation. Words like compression and enlargement are used instead of stretch. Should indicate vertical stretch or stretch parallel (the word “along” is not accepted) to the y-axis. (iv) 33, 2 ba +− Poor done due to the mistakes in (iii).

ACS (Independent) / Mathematics Department / Mathematics HL / Year 5 / 2022 Final Exam / Paper 1 / Solutions 9 (b)(i) A handful of students did not state the equation of the asymptotes. The curve must shows that it is asymptotic. Many students drawn the curve away from the asymptote. (ii) Poorly attempted. Note that the curve at the x-intercept should not be pointed. It should be a turning point at the x-intercept.

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