RI H2 Math Promo Paper 2023 (Soln)

RAFFLES INSTITUTION 2023 Year 5 H2 Mathematics Promotion Exam Questions and Solutions with comments     Page 1 of 19  1 The first 3 terms of a sequence are given by 1 1823u  , 2 200u  and 3 2023u  . Given that nu is a quadratic polynomial in n, find nu in terms of n. [4] 1 Let 2 nua n b n c  . 1 1823: 1823u abc   2 200 : 4 2 200ua b c   3 2023: 9 3 2023ua b c   Using GC, we have 1723, 6792, 6892.ab c  Therefore, 21723 6792 6892.nunn  Mostly well done. Some students mistook “quadratic polynomial” as “quartic polynomial” and were hence unable to solve the question. Students should take note that a sequence is not restricted to only AP or GP. It can just be a generic sequence.

Page 2 of 19  2 Do not use a calculator in answering this question. Solve the inequality 2 565 31 xxx x   . [4] Hence solve (a) 2 42 2 565 31 xxx x   , [2] (b) 2 5l n(ln ) 6 ln 5 13 l n xxx x   . [2] 2             2 2 565 0 31 551 0 31 13 1 150 31 53 4 1 1 031 53 4 031 xxx x xxx x xxx x xx x x xx x x                      45 3x or 1 03 x Most students were able to tackle this question successfully. Students are advised to check their work to avoid careless mistakes. (a) Replace x with 2x , 2 42 2 565 31 xxx x   2 45 3x or 21 03 x  No solution, since 2 0x  or 0x                                    0x Students are reminded that complex numbers cannot be ordered. Hence they should not appear in inequalities. (b) Replace x with ln x , 2 ln 5(ln ) 6ln 5 3ln 1 xxx x    45l n 3x or 1 ln 03 x 54 / 3 ee x  or 1/3e1 x  This part was well done. Students could see the replacement and understand that the ln function is an increasing one, hence there will be no change to the inequality signs.

Page 3 of 19  3 Vectors a and b are such that the magnitude of  a is 2 and b is a unit vector perpendicular to a. (a) Find the area of the parallelogram with adjacent sides formed by the vectors 3ab  and 54ab . [3] A vector c is such that 21 bc ab . (b) Show that 21cb a , where  is a constant. [2] (c) Give the geometrical meaning of b.c and find the possible values of  if 5b.c . [3] 3(a) Area of the parallelogram (3 ) ( 54 ) 5( ) 15( ) 4( ) 12( ) 19( ) 19 sin 90 38           ab ab aa ba ab bb ba ba Common mistakes  (3 ) ( 54 ).ab ab  (3 ) ( 54 )ab ab   (without the modulus)  5( ) 15( ) 4( ) 12( )    aa ab ab bb (note that ba ab ) (b) 21 () 2 1 () () 2 1 () (2 1 )         bc ab bc ab 0 bc ba 0 bc a 0 Case 1: If 21ca Note that 21ca satisfy the equation (2 1 ) bc a 0 Case 2: If 21ca Since b and 2