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 differentiati