2024 Yr 6 RI H2 Math Timed Practice (Soln) (updated 11 Jul)

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments RAFFLES INSTITUTION 2024 Year 6 H2 Mathematics Timed Practice Questions and Solutions with comments   1 Consider the second order differential equation 2 2 dd dd xx abtt  , where a and b are non-zero constants and d d x b ta . (a) By substituting d d xy t , show that the above second order differential equation can be written as d d y ba yt  . [2] (b) Find y in terms of t and hence find x in terms of t. [5] (a) [2] Since d d xy t , then 2 2 dd d d dd d d yx x tt t t   .  2 2 dd dd xx abtt   becomes  d d y ay bt  . Rearranging, we have d d y ba yt  .  Similar question: Tut 8C Qn 7(i)(a) (b) [5] d d 1d d 1, since 0dd y ba yt yx ba ybaba yt t      Integrating with respect to t,  1 d1 d 1 ln , where ln e ee e , where e , 0 1 e at ac ac at at ac at ytba y ba y tc ca ba y a ta c ba y ba y ba y A A A yb Aa                        Since d d xy t ,  2 d1 ed 1 e, w h e r e e at at at x bAta Axb t B Baa bAtBaa             There are a few did not start off correctly and wrote: dyb a y t The more common mistakes  Not having the modulus after integrating 1 ba y  Not able to use the appropriate method to “drop” the modulus for ba y  Not writing down the constant c and d after completing the integration. Some wrote the answer as e atxK t C B   . Note that B and C are arbitrary constants (in this case, B can be any real value and C can be any non-zero value). However, K is not an arbitrary constant as bK a . So, in this instance, it is better to write the answer as e atbxt C Ba   Similar question: Tut 8C Qn 3, 7(i)(b), 10(i)

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments 2 Referred to an origin O, the position vectors of three points A, B and C are 325ij k , 42ij k and 2+ 4 ij k respectively. (a) Determine the value of  for which points O, A, B and C are coplanar. [2] (b) Given that OD OA OB       , find the value of  for which OD   is perpendicular to the y – axis. [2] (c) Hence, find the exact area of triangle OAD. [2] (a) [2] Method 1 A vector normal to plane OABC 31 24 52 16 11 14 OA OB               So, 16 16 11 2 11 0 14 4 14 16 22 56 0 39 8 OC                              Method 2 Given the points are coplanar, 31 22 4 45 2 st         , where s and t are constants. 3st  --- (1) 22 4 st  --- (2) 45 2 st --- (3) Using GC to solve the above 3 eqns, we get 39 5 9, , 84 8st   . This part is well done. (b) [2] 31 24 52 OD OA OB                  A direction vector of the y-axis is 0 1 0    . If OD   is perpendicular to the y – axis, This part is well done.

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments 0 10 0 30 124 1 0 252 0 OD                        (c) [2] 31 3 . 5 11 24 022 52 4 OD OA OB                          Area of OAD   1 2 33 . 5 1 202 54 8 1 5.52 7 1 64 30.25 492 1 5734 OA OD                   This part is well done.

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments 3 (a) Write down constants A and B such that for all values of x, 1( 2 4 ).x Ax B   [1] (b) By expressing 2 45x x in completed square form, find 2 1 d45 x xxx    . [4] (c) Hence, without using a calculator, show that 2 1 2 2 1 1dl n t a n45 2 2 x px qxx q        , where p and q are exact real constants to be determined. [4] (a) [1] From 1( 2 4 )x Ax B   , Compare coefficients of x: 121 2AA    Compare constants: 41 1AB B    Hence A = 1 2 , B = 1. This part is well done. (b) [4]     22 22 2 2 21 1 24 11 2 dd , f r o m 45 45 12 4 1 dd24 5 4 5 11ln 4 5 d2 21 1 ln 4 5 tan 2 ,2 xx xxxx xx x x xxx xx x xx x x xx c                   (a)   since 2 45 0xx  for all real values of x. (c) [4] Note that (1 ) , i f 1 ,1 1, if 1. xxx xx           2 22 12 2221 1 11 22 2 21 11 11 1 2 2 1 d45 11 d d45 45 1 ln ( 4 5) tan 22 1 ln ( 4 5) tan 22 11 1 1ln2 ln17 ln1 ln222 2 2 tan 1 tan 4 tan 0 x xxx xx xxxx xx xx x xx x                                          1 11 11 1 tan 1 1= ln 2 ln17 2 tan 1 tan 4 2 11 7= ln 2 tan 1 tan 4 24 11 7ln tan 4 , where 17, 4.22 4 pq                 This part is not well-done. Not many are able to split the limits correctly. Similar question: Assignment 8B Qn 4

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments 4 The functions f and g are defined by 21 4 2f : , , , 32 3 sing: , , 0 . xxx x x xxx x x      (a) Find 1f( ) x and state its domain. [3] (b) Explain why  2024f xx for 2 3x . [1] (c) On the same diagram, sketch the graphs of fyx and   2024fyx , giving the equations of any asymptotes. [3] (d) Explain why the composite function gf exists. [1] (e) Find the range of gf . [2] (a) [3] Let 21 4 32 xy x   .   1 3221 4 32 2 1 4 32 2 1 4 21 4 32 21 4f 32 yx x xy y x xy x y yx y xx x         1 ff 2DR , 3    . Quite a number of students did not realise that the graph of 21 4 32 xy x   has the horizontal asymptote 2 3y . Thus, not able to give the correct range of f (which is also the domain of the inverse function). (b) [1] Since 1f( ) f ( )xx  for 2 3x , then 2f() xx . Hence,      2024 2022 2 2022 2020 2 2020 2 ff f f ( ), from above =f f f( ) f xx x x x xx      Similar question: Tut 3 Qn 9(iii), Chap 3 Example 11(b) (c) [3] Note that 2024 2 f 3fDD,  . So, we only drew the graph for 2 3x Many also drew the graph of f in the 3 rd quadrant. Note that domain of f is given to be 2 3 , . O  2024fyx

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments (d) [1] Since  2 fg 3R, 0 , D   , so the composite function gf exists. (e) [2]   fg ffg f 22D , R , R 0.217, 0.928 (3 sf)33             Not many students are able to give the correct range.

______________________  2024 Yr 6 H2 Math Timed Practice Solution with Comments   5 The curve C is defined by the pa rametric equations 2x t and 3 2yt t , where 11 t . (a) Sketch the graph of C. [2] (b) Without using a calculator, find the equation of the normal to the curve C at the point where 1 2t . [4] (c) Find the exact area bounded by the curve C, the positive y-axis and the normal found in part (b). [4] (a) [2] Some students used the default window setting and only obtain the graph in the first quadrant. There is a need to change min 1t  and max 1t  as defined in the question. Coordinates of the end-points are to be stated on the graph. (b) [4] Method 1  2 d 2,d xx tt t    32 d 22 3 d yyt t t t   . 2d2 3 d2 y t x t   When 119, , 248txy , 1 423d1 1 d1 4 y x  Gradient of normal = 4 11 Equation of normal is 94 1 81 1 4 41 9 11 11 8 41 0 7 11 88 yx yx yx             Method 2 32 2 4 6 2 3 24 4 4 4yt t y t t t xx x    .  Differentiate with respect x, 2 2dd 4 8 324 8 3 .dd 2 yy x xyx xxx y       When 119, , 248txy ,  