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 ba ab ba ab ab ii () APA B      pa ba 0 0 AP  is parallel to vector AB  (Note : and AP AB     00 ) Since line l that passes through point A and is parallel to vector AB  ,  ,:APl   ra b a  iii 4aa b b Since 4 ab is a scalar, a is a scalar multiple of b,

6   Q Solution a and b are parallel vectors. Since , the angle between a and b, is either 0 or 180 , cos 1  . Given 4aa b b , 4| | | |c o s aa b b | | 4 | || || cos || | aa b b 2 2 | | 4 | || | | cos |, | cos | 1 Since | | 0 1|| , || 4 1 since 02     aa b a ab bb Alternative method: Given 4aa b b ,   2 2 4 4 Since 0 as 1, 4 1 since 02 //          ab ab abbb ab ab b ab ab b bb 6(a) (i)          y x O y = 3 x = 0

7   (a) (ii) fyx f'yx  02,  02, 3x 3x 0x 0x y = 3 y = 0 (b)  21 13yx        21 213yx          2 2 1 24 13 1 293 yx yx           21 293yx                 O y x x C’: Scaling parallel to the x-axis by a scale factor ½: Replace x with 2x B’: Translation of 4 units in the negative x-direction: Replace x with x + 4 A’: Reflection about the x-axis: Replace y with y

8   7(i) 21 5 233 xy xx   Intersection with axes: 0, 1 3   and 1 2 ,0  Asymptotes: x = 3, y = 2 For ln 1yx , Intersection with axes: When y = 0, x = 0. 0,0 Asymptote: x = 1 From the graph, the points of intersection are (₋1.33 , 0.844) and (0.195 , 0.217). Hence, solving 21 3 ln 1x xx   , from the graph, x = ₋1.33 (to 3 sf) or 0.195 (to 3 sf)  (ii) Solving 21 3 ln 1 xx x   , from the graph in (i), we have 1.33 or 0.195 1x x     (iii) 21 ln2 x xx   Let y = x + 1 y x O x = 1 x = 3 y = 2             (0 ,0)

9       21 1 ln 1 113 21 ln 13 x xx y yy       1.33 or 0.195 1 1 1.33 or 0.195 1 1 2.33 or 0.805 0 yy xx xx             8(i) 1k  This is because there is no image for x = 1 under f. (or, f1 is undefined) (ii) 3Let , for , 1 1xyx x x       3 1 13 3 3 13 3 1 xy x yx x xy y x xy x y x yy yx y                Since  1 3f, 1 yxy y     1 3f 1 xx x    Since   1ff x x , ,1xx   ,   21ff f x xx (iii)

10      gg f g gf,0 , 3 ,D1 R 1 R         Alternative method: Using fg to find range. 9 (i) 23,x ty t --- (1) 2dd 1 3 3 ddd 22 d yy t txxt t t    The equation of the tangent at the point with parameter t is  3 3 33 3 2 2 3 2 22 3 ( ) 22 33 23 0 ( P r o v e d ) . yt t xt yt t x t yt t xt yt x t         (ii) A cubic equation has at most 3 real roots. Given that (a , b) is a fixed point, the equation 323 0 ba t t  is a cubic equation in terms of t. Hence there are at most 3 real values of t for a fixed value of x and y and therefore at most 3 tangents can pass through the fixed point (a , b). (iii) When t = 2, 323 0 268 0 34 0 ( 2 ) yt x t yx yx          Since the tangent at P meets the curve again at Q   23,kk , substituting equation (1) into (2):