2021 HCI H3 Math Prelim P1 Solutions

Q1 Solutions Comments (i)                  2 2 2 2 2 22 2 2 11 1 1 1 1 1 1 1 1 k k k k k k k k k NN N N N N N N NN NN NN NN                                          2 2 22 1 22 22 00 22 22 00 2 2 0 22 11 1 11 11 1 1 1 1 1 1 2 2 1 202 k k kk iikk k i k i ii iikk ik i k i ii ik i ki i kk NN NN N N N N kk N N N Nii kk N N N Nii k NNi kk N N N                                                                              242 2 2 22 2 4 2 1 2 2 14 2 1 , if is e ven 2 2 1 2 10 2 4 2 1 , if is odd1 k k k k k k k NN k Nkk k k kN N N N N k N N kk                                     Many students tried to show this using MI. This is a valid method, but either requires binomial expansion similar to the solution here or uses       11 1111 kk kkkk x x x x x x x x            which requires 2 predecessors and therefore 2 base cases. For k even, 2 k is an integer, hence       22 2 4 2 2 2 2 2 1 2 10 2 4 21 k k k k k k kN N N N N k Nk                      is an integer since N and k r   are also integers

For k odd, 1 2 k is an integer, hence       22 2 4 2 1 2 2 2 2 1 2 10 2 4 21 1 k k k k k k kN N N N N k NNk                         is an integer since N and k r   are also integers. Therefore     2 2 11 1 k kNN NN     is an integer for all positive integer k. (ii) Let     2 2 11 1 k kI N N NN      , where I is the integer from part (i). Since   2 1 1 k NN is positive and   2 1 1 1 1 221 k kkNNN     for 2N  ,   2 21 1k I N N I     . Therefore the integer closest to   2 1 k NN is I. Some students neglected to show that what the closest integer is. Alternatively, students should at least mention that the closest integer is AT MOST   2 1 k NN away. Notice that for any real number 1 2x , we have 2 2 2 2 22 121 2 1 12 1 1,2 1 14 5 4 x x x xx xx x x x x                   Most students were able to show this part of the inequality.

Hence for 52 4N  , we have 2121 2N N N    2 11 11 2 2 NN N      2 11 2 21 k k N NN      Hence   2 1 k NN differs from I by less than 12 2 k N    .

Q2 Solutions Comments (i)           23 23 1 1 1 1 1 ax bx cx a b c x ac bc ab x abcx ac bc ab x abcx                 Therefore, ,q ac bc ab   r abc and 0.abc   (ii)       23 223 23 2 2 3 4 5 ln 1 2 2 qx rx qx rx qx rx qqx rx x qrx            Some students did not use the Maclaurin expansion in MF26, and instead went through differentiation. (iii)         23ln 1 ln 1 ln 1 ln 1qx rx ax bx cx        Coefficient of         1 1 1 11 1 1 1 n n n n n n nn n abcxT n n n           where . n n n n abcT n  Most students were able to use the correct logarithm law. They should directly look for the general term in the expansion (given in MF26), instead of generalizing from the coefficients of 23, , ,...x x x (iv)       21 22 31 33 51 55 1 1 1 T q T q T r T r T qr T qr                Hence       2 2 2 3 3 3 2 2 2 3 3 3 23 5 5 5 5 6 23 5 a b c a b c a b c a b c TT qr T abc              The question states that the results in (ii) and (iii) should be used. Students need to follow this instruction strictly.

Q3 Solutions Comments Using :ux     0 a 1 b 0x x x   Using e:xu     e a e b e 0x x x xx   , which implies    1 a b 0xx   since e 0.x  Subtracting the two equations, we obtain    1 1 b 0 xx   . Therefore, and  a. 1 xx x  (i) From 1d d uy ux , we have 22 22 222 2 2 2 2 2 2 2 d 1 d 1 d d d d 1 d 1 d 1 d 1 d 1 d d d 1 d 1 1 d 1 d 1 d 1 d 1 d d 1 0d 1 d 1 y u u x u x u x u u u x u u x u x u x x u x x u x u u x x u x x u x u ux x x x                                 Many students wrongly wrote 2 22 d 1 d 1 d d d d y u u x u x u x   (without the power 2 in the last term). (ii) Using (1), the general solution is e.xu Ax B Therefore,  1 d 1 e.de x x uy A Bu x Ax B    When 0, 2,xy 2 2 . AB B B A B BA    Hence,  1 1 e e.ee x x xxy A AAx A x    Some students found the values of d d u x and/or d d y x when 0, 2.xy They are not required in the question. The approach should be to first find the general solution to (2), with the help of the general solution of (1) given in the question, and then substitute the initial condition to find the particular solution.   1b 1x x 

Q4 Solutions Comments For convenience, we will draw all diagrams using a net diagram with sides Top, Bottom, East, South, West, North: T E S W N B Symmetry for this question is difficult. H2 syllabus deals with rotational symmetry. There is a need to consider other axis of rotation here. (i) 3 colours. Since a vertex shares 3 faces adjacent to each other, they must be all different in colour. The only way a cube can be coloured in 3 colours is when opposite sides are coloured in the same colour. Hence, there are 3 7 35C  such cubes. In 3 colours, this is the only possible result up to rotation: R B Y B Y R There are 35 such cubes. This part was generally well done. (ii) Since 6 colours are used, we have 6 7 7C  ways to choose the colours used. With an uncoloured cube, there are too many rotational symmetry in 3D. Hence, we start by colouring in 1 side. If red is available, we will paint red. (o.w. use orange) Since the cube will end with 1 coloured side and 5 uncoloured, there is 6 61 way to colour it in. Rotate the cube to have the coloured side bottom. R This reduces the symmetry to 2D, similar to a square table question. Since only the top side is unique, there’s 5 colour choice, and the remaining 4 sides is a square rotational symmetry. Alternative method: 7 6 6 1 4 6!C C   7 6C for choosing the colours, 6! for arranging 6 colours, 6 1C to choose the bottom side (locking it in to reduce 3D to 2D), 4 for rotational symmetry.

Number of ways 04!5 47 21 (iii) If 4 colours are used, 2 colours must be used twice and painted opposite each other. There are 2 7 21C  ways these colour can be chosen and painted. The remaining 2 colours have 2 5 10C  choices and can be painted in 2! 2 ways. These are the same: (hence 2! 2 ) R R Y G Y B Y B Y G R R Number of ways 21 2! 2 21010  Most students have some idea of tackling this question, but may have missed out some order of rotational symmetries or considered identical cases. If 5 colours are used, 1 colour must be used twice and painted opposite each other. There are 7 ways this colour can be chosen and painted. The remaining 4 colours have 4 6 15C  choices and can be painted in 4! 234  ways. ( 2 when the cube is flipped upside down) Eg. (similar when top/bottom flipped) R R Y G B I I B G Y R R Number of ways 3157 3 15 3-colours: 35 4-colours: 210 5-colours: 315 6-colours: 210 Total no. of ways 35 210 315 210 770    

Q5 Solutions Comments (i)                 2233 2 2 22222 2222 d 33 d 24 34 b a b a b a b ab abaxx b a b abaxx b a b ab a b a b a            Since 0,ba we have                   222 22 22 22 43 43 4 4 3 4 4 3 b ab a