SAJC H2 Promo 2023 (Solutions)

[Turn Over 2023 SAJC H2 Math Promo Solutions Q Solution 1 Let 32 , a 1nSa n b n c n d   When 1, 5 (1) When 2, 8 4 2 20 (2) When 3, 27 9 3 57 (3) When 4, 64 16 4 128 (4) n abcd na b c d na b c d na b c d                           Using GC to solve (1), (2), (3) and (4), 2, 1, 4, 0ab cd   . 3224nSn nn   2(a)          1 2 21 2 22 21 22 sin 2d d 214 s i n 2 8 12 14 21 4 s i n 14 82 14 x x x xx x x xx x x x                   2(b) 24 3e 4xy  ---(1) Differentiate with respect to x: 4 4 d21 2 ed d 6e (1)d x x yy x yy x    Differentiate (1) with respect to x:   22 4 2 4 2 2 22 2 2 dd 24edd 83 e 8 4 , from (1) 83 2 dd 8 32 (Shown)dd x x yy xx y y y y yyy xx        where 32k 

2   Q Solution 3(a)           1 2 11 61 66 161 2 1 6 162 12 13 212 n n r nn rr ur r rr nnn n n nn n nn n                      (b) (i) Let   1 12 1 2 AB C rr r r r r    By cover-rule,   11 0102 2A     1 111 2B      11 22 12C     1 12rr r = 11 1 2 1 2( 2)rr r 

3   Q Solution (b) (ii)       1 1 1 1 12 1 12 11 1 2 1 2( 2) 11 1 222 ( 3 ) 11 1 432 ( 4 ) 11 1 642 ( 5 ) 11 1 8 5 2(6) 11 1 22 1 2 ( ) 111 21 2 ( 1 ) 11 1 21 2 ( 2 ) N N r N r N r S rr r rr r rr r NN N NN N NN N                                          11 1 1 42 ( 1 ) 12 ( 2 ) 11 1 42 1 2 2 11 1 42 1 2 11 1 42 1 2 NN N NN NN NN                                     Since   1 012NN  ,  for N  , 1 4 NS  .

4   Q Solution 4(i)   1 1 1 1 ln3 e e e e e 1, since is an arithmetic progression3 n n n n n n u n u n uu uu n w w u                 Since 1 1 3 n n w w   is a constant (independent of n), the sequence of terms given by ,nwn  is a geometric progression with a common ratio of 1 3 .  (ii) 1 1 3 n n w w   is the common ratio of the geometric progression (given). 1 3r  1r  ---(*) Hence, 1 r r w    converges. (iii) ln 1 3 1e 3w    ln 3 ln 3 ln 3 ln 3 ln 3 ln 3 l 1 1 n3 1 l 3 1 1n 1 1e 0.0051e 1e 1e 1e 10 . 0 0 51e 1e e0 . 0 0 5 1 0.0053 n n n n w ww ww                      n 1 3 n   4 0.0123 > 0.005 5 0.0041 < 0.005 6 0.0014 < 0.005 Smallest possible value of n = 5

5   Q Solution 5i By Ratio Theorem, 2 3OM  ab  1Area of triangle 2 114| 2 |23 1 | 2 | ----- (*)6 1 |2 |6 1 ||3 || 1 2 OBM OM OB             ab b ab bb ab 0 ab ab Alternative solution:  1Area of triangle 2 114| 2 |23 1 |2 |6 1 |2 |6 1 ||3 1 || s i n c e || ||3 || 1 2 OBM OB OM                 ba b babb ba 0 b