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 21ca are nonzero vectors, then 21ca is parallel to b and so 21 ca b , where  is a nonzero constant. 21cb a Note that 0  corresponds to Case 1. Thus we can say that 21cb a , where  is a constant. (Shown) [Note that a and b are nonzero vectors and since ab , ab 0 . And so, c is a nonzero vector.] Common mistakes  21 21 21 ab a b (refer to Chap 4B notes page 15 property 4: () ( ) ( )    ab a b a b )  () 2 1 () ( 2 1 ) bc ab b c a (refer to Chap 4B notes page 15 property 3:    () () ()ba bc b ac )  ab ba (should be  ab ba ) (c) bc cb is the length of projection of c onto b. 2 (2 1 ) () 2 1 () 21(0)         bc b b a bb ba b   Since 5bc , 5  . Refer to Chap 4B notes page 13: The length of projection of a onto b is given by .ab , where ˆb is a unit vector in the direction of b . In this question, the unit vector is b. So, length projection b.c is the length of projection of c onto b, and not b onto c. Refer to Chap 4B notes page 9 property 3.

Page 4 of 19  4 (a) Write 1 (1 )rr  in partial fractions. [1] (b) Using your answer to part (a), find 2 1 1n r rr  . [2] Hence find (i) 2 2 1 1n rn rr  , [2] (ii) 111 23 34 45  [2] 4(a)  11 1 11rr r r  This part is well-done. (b) 2 11 11 1 1 11 12 11 23 11 34 . . . 11 1 11 1 11 1 nn rr rr r r nn nn n               The part is well-done. In fact, this is Example 5 in Chap 6B notes.

Page 5 of 19  (i)       2 2 1 2 22 11 1 11 1111 21 1 11 12 1 21 1 12 1 12 1 n rn nn rr rr rr rr nn nn nn nn n nn                       Quite a number of scripts did this part by using method of differences. However, the question states “Hence”, so should make use of the previous part answer to do part (i). Common mistakes  22 1 22 2 111 11 1nn n rn r r rr rr rr      (refer to Chap 6B notes page 4 property 2: 1 11 mm k rr r rk r r uu u           )  22 22 2 11 1 11 1nn n n rn rn n rrr rr rr       (no such rule) (ii)   2 1 111 23 34 45 1 1 11 11 2 11lim 1 12 1 2 r r n rr rr n                  The presentation for this part is not well-done. Refer to Chap 6B notes Example 8(iii) on page 12 for the correct presentation. Quite a number of scripts did not realized that this represents an infinite sum and wrote 2 111 1 23 34 45 1 n r rr    

Page 6 of 19  5 In this question you may use expansions from the List of Formulae (MF26). (a) Find the first four non-zero term s of the Maclaurin series of e1 c o s 3x x  , in ascending powers of x. [4] (b) It is given that the first three terms of th is series are equal to the first three terms in the series expansion, in ascending powers of x, of 21 cxab x  . Find the values of a, b and c. [4] 5(a)    223 23 2 23 23 e1 c o s 3 311 1 2! 3! 2! 912 26 2 91 922 1 23 2 72 522 26 x x xxxx xx xx xx x xx x                                    Alternative method: Let e1 c o s 3xy x e1 c o s 3xy x dee3 s i n 3d xx y yxx  2 2 2 2 dd de eee9 c o s 3dd d dde2 e e 9 c o s 3dd x xxx xx x yyy yxxx x yy yxx x       32 32 ddde3 e3 e e 2 7 s i n 3ddd xxx x yy y yxxxx   When 23 23 dd d0, 2, 2, 7, 25dd d yyyxy xx x      ,  23 23 72 5e1 c o s 3 2 2 2! 3! 72 522 26 x xx x x xx x             Easier to use standard series expansions of e x and cos3 x from MF26. There are various way to perform the repeated differentiation, e.g. use product rule on the RHS.

Page 7 of 19  (b) 2 1 12 2 22 2 2 23 1 1 1 1 1 cxab x bax c x a bbx xc xaa a bbxc xaa a                    Comparing with the series expansion in (a), 11 2 2aa   2 2 122 2 b baa     22 33 77 7 1 1 222 2 2 bb ccaa        Use the expansion of (1 ) nx . Factor out a and remember to apply power of –1 to it.

Page 8 of 19  6 (a) Let xya , where a is a positive constant. Show that d ln .d xy aax  [2] (b) A curve has equation  132 2 .xyy  Find the equation of the normal to the curve at the point 0, 1 . Give your answer in the form yA x B , where A and B are constants in the exact form. [6] 6(a) ln lnxya yxa  . Differentiate with respect to x: 1d d ln ln ln (shown).dd xyy ay a a ayx x    Alternatively, ln lnee xx ax aya  .  lnd e ln ln (shown).d xa xy aaax  Generally most of the students are able to show the result. (b)  132 2xyy  Differentiate with respect to x: 1dd 3 3 ln 3 2 ln 2 0dd xx yyy yx x   1 d3 l n 3 d3 2 l n 2 x xy yy x    . At 0, 1 ,     11 l n 3dl n