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 