2026 RI H2 Math Year 6 Preliminary Exam Paper 2 (Solutions with comments)

RAFFLES INSTITUTION H2 Mathematics (9758) 2026 Year 6 Page 1 of 24 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments 1 (a) The complex number w is such that 2 * 6 2 3 iw ww   . Find exactly the possible values of ,w giving your answer in the form ix y where x and y are real numbers. [3] (b) The complex numbers found in part (a) are represented by points A and D in an Argand diagram as shown below. Points B and C represent complex numbers b and c respectively. Given that ABCD is a rectangle with2AD AB , find complex numbers b and ,c giving your answers in the form iu v , where u and v are real numbers. [3] (c) By finding arg( ),b find the exact value of 5tan12  . [2] Solution Comments (a) [3] 2 * 6 2 3 iw ww   Let iw x y  , then      2 2 2 2 2 2 i i i 6 2 3 i 2i 6 2 3 i 2 2i 6 2 3 i x y x y x y x xy y x y x xy                2 2Comparing real part: 2 6 3 3 Comparing imaginary part: 2 2 3 1 3 i or 3 i x x x xy y w w              Most students handled this part well. C B Re O A Im D

2026 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments _____________________________________________________________________________________ Page 2 of 24 (b) [3] From the Argand diagram, point A represents 3 i and point D represents 3 i  . Let the midpoint of BC be E, then E is represented by the complex number i 3 i 1 3i   . Hence,   3 i 1 3i ( 3 1) ( 3 1)i 3 i 1 3i ( 3 1) ( 3 1)i b c                    This part challenges many students. Useful ideas include multiplying a complex number by i and the geometric effects of adding/subtracting complex numbers. Note that we rotate OA  about O by 2  anticlockwise to obtain OE . Also, OB OA OE       and OC OD OE       (c) [2] Since , . 4OA AB BOA    Also, 1 1arg( ) tan 63a        . Hence, arg( ) arg( ) 6 4 5 12 b a BOA          5 Im( ) 3 1 3 1 3 2 3 1tan 2 312 Re( ) 3 13 1 3 1 b b              A key element of this part is explaining why the exact value of argb is 5 12  , enabling us to determine the exact value of 5tan12  from the real and imaginary parts of b found in part (b). Thus, numerical verification using the GC alone is insufficient.

2026 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments _____________________________________________________________________________________ Page 3 of 24 2 (a) Find 2 2 3 d .2 3 x xx x    [3] (b) Using the substitution tan ,2 xu show that 2 1 2 d d2 cos 3x ux u   . Hence find the exact value of 2 0 1 d .2 cos xx   [6] Solution Comments (a) [3] 2 2 3 d2 3 3 1 5 1 d4 3 4 1 3 5ln 3 ln 14 4 x xx x xx x x x c                       OR     2 2 2 22 2 2 2 3 d2 3 2 2 1+ d2 3 2 3 2 2 1+ d2 3 1 4 1 21ln 2 3 ln 2(2) 1 2 1 1ln 2 3 ln 4 3 x xx x x xx x x x x xx x x xx x c x xx x c x                               Generally well done, except that some missed out the modulus sign for lnor arbitrary constant, c in the final line of the answer.       2 3 3 1 3 1 2 3 1 3 5When 1: 4 3When 3 : 4 x A B x x x x x A x B x x B x A                

2026 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments _____________________________________________________________________________________ Page 4 of 24 (b) [6] Method 1:    2 2 2 2 2 2 2 2 1 d2 cos 1 d2 2cos 1 1 d1 2cos 1 2 d1 11 2 1 2 d1 2 2 d (shown)3 x x xx x x uu u uu uu                         Method 2:    2 2 2 2 2 2 22 1 1 1 2 1 d2 cos 1 2 d1 12 1 1 2 d1 2 d (shown)3 u u u xx uu u u uu uu                    2 0 1 20 1 1 0 1 1 d2 cos 2 d3 12 tan3 3 2 1tan3 3 2 3 6 93 3 3 xx uu u                        For the “show” part, most were able to differentiate the given substitution and made use of trigo identities to complete the proof. However, do note that as this is a “show” question, clear working is required to earn full credit. This last part was generally well done, although some did not evaluate 1 1tan 3  which is a special angle. Since tan2 xu When , tan 12 4 When 0, 0 x u x u       tan2 xu   2 2 2 2 2 2 2 1cos2 1 cos 2cos 12 2 11 2 1 1 1 1 x u xx u u u u u             Let tan2 xu   2 2 2 2 d 1 secd 2 2 1 1 tan2 2 1 12 2d d1 u x x x u x u u          

2026 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments _____________________________________________________________________________________ Page 5 of 24 3 The function f is given by f : e , x ax a x    , where a is a positive constant. It is given that the range of f is ,k , where 0k . (a) Find the range of values of a. [2] (b) Show that 1f does not exist. [2] (c) If the domain of f is restricted to ,m , state the largest value of m in terms of a for which 1f exists. [1] (d) Using your answer found in part (c), find 1f in similar form. [3] The function g is given by g : ln 1 , .xx x a a      (e) Show that the composite function gf exists and find the range of gf in terms of a. [3] Solution Comments (a) [2] Since 0 for all , e 1 e 1 e 1 x a x a x a x a x a a k             Since 0, 1 0 1 k a a     The objective of this question is to find the range of values of a given that the maximum y-value is positive. Students may either use inequalities or a graphical approach to determine k in terms of a. Note that this question is not about solving x. (b) [2] Method 1: Since f 1 f 1 ea a a     , f is not one-one. Hence, 1f does not exist. Note that method 1 is more time efficient than method 2!

2026 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments _____________________________________________________________________________________ Page 6 of 24 Method 2: Since the horizontal line 2y cuts the graph of  fy x at more than one point, f is not one-one. Hence, 1f does not exist. (c) [1] Largest m a . (d) [3]     Let e . For , e e ln ln x a x a a x y a x a y a a y a x a y x a a y                   1f : ln , 1x a a x x a     It would be worthwhile to resolve the modulus first by considering the domain of f, before making x the subject. Do note that a function cannot take on 2 y-values for 1 x-value – hence the rule should not include a sign. Students should also be mindful to present the answer in similar form as required by the question. (e) [3] f g f g R , 1 , D , Since R D , gf exists. a a      Students should state both Rf and Dg in interval notation before concluding that f gR D . It is also important for students to show clearly that the upper bound of Rf is indeed smaller than that of Dg.