2023 NYJC Promo (Soln)

1 of 19 2023 NYJC H2 Math Promo Solutions   Q1 Suggested Answers (a) (b) Using GC, the graphs in (a) intersect at 1.2028, 4.5396, 5.4576x For 2 115 1( 5 )xx  11 . 2 0x or 4.54 5 or 5 5.46xx             y  x         1.2028 4.5396 5.4576       y = 0

2 of 19 Q2 Suggested Answers (a)    2d ln 1d x xx  22 11 2 1 211 x x xx        2 22 11 11 x x xx x      2 1 1 x   (b) 1 2 2 22 22 21 ( 1 2 ) ( 2 ) ( 2 ) 212 ( 12 ) 1) 12 12 42 (1 2 ) (2 ) d 51 t 5 and ( 51 xx x xxx x xx xx x x x                  

3 of 19 Q3 Suggested Answers (a) O, A, C and B forms a parallelogram. Or O, A, C and B are vertices of a parallelogram. (b) OC ab     11 122 33 333OM OA OC       baab ab a        Area of triangle OAM 11 22 3 11 23 2 3 1 6 OA OM            baa bbaaa 0a ab    (b) Alternative method: Area of triangle OAC 11 22 11 22 1 2 OA OC          aa b aaab 0ab ab    Area of triangle OAM 1 area of triangle3 1 6 OAC  ab (c) The shortest distance from A to OB =  3  bab a b 53ab Area of triangle OAM 11 5 3 5366 6   ab   (c) Alternative method 1: Area of triangle OAM 1 6 ab = 1 6  area of parallelogram OACB   = 1 6 base OB  height of parallelogram 15 35366     Alternative method 2:

4 of 19 Let  be the angle between a and b Area of triangle OAM 11 OR (2 area of )66 11 1 sin = (2 5 3)66 2 13 5 3 5 35 66 6 AOB              ab ab a a  (d) Since ON : NM =3:1, 33 3 1 44 3 4 4ON OM      baa b     31 1 1 44 4 4 AN ON OA AB     ab a b a        Since AN   is parallel to AB  and A is a common point, A, N and B are collinear. AN:NB = 1:3

5 of 19 Q4 Suggested Answers (a) (1) Translate the curve by 1 unit in the negative x-direction (2) Reflect the curve in the y-axis (3) Scale the curve by factor 2 parallel to the y-axis (1) Reflect the curve in the y-axis (2) Translate the curve by 1 unit in the positive x-direction (3) Scale the curve by factor 2 parallel to the y-axis (1) Translate the curve by 4 units in the negative x-direction (2) Reflect the curve in the y-axis (3) Scale the curve by fact or 1/4 parallel to the x-axis (b)(i) (b)(ii)    y (0,2) x (0,0) (2,0) (1,0) ()    y (0,1) () (2,1) x O (1,1)

6 of 19 Q5 Suggested Answers (a) Greatest value of a = ½ Explanation (for students) Since  21 2g ( ) e x x   is an even function that is symmetrical about 1 2x , for g to be a 1-1 function, the largest 1 2D( 1 , )g  . Alternatively, use GC to find the minimum point of g ( )yx . (b) Given 0.a So, Note that 1 gggDD ( 1 , 0 )   (c)(i)   2211 22 2 hg 1h g () he 1 l ne 1 2 D( 1 , 0 ) xx xx            (c)(ii) “Hence” method: Let   1 3 2hgx    3 2hg ( ) , 1 0xx     2 2 1 3 2 311 22 11 2 2 1111 (reject since 1 0)22 22 1121hg 2 22 x x xx x x             (c)(ii) “Otherwise” method:                 x y O                 (0,0)

7 of 19 2 1Let hg( ) 1 , 1 0 2yx x x         2 1 12x y      1 11 1 (Reject 1 since 1 0)22 1hg 1 2 xy xy x xy y                1 1 3 2 3 2 1hg 1 2 11211hg 1 22 22 xx          

8 of 19 6 Suggested Answers (a)   2 2 22 f( 1 ) 2 f( ) f( 1 ) l n( 1 ) 2l n ( ) l n( 1 ) ln ln ln 1 11 11 rr rr r r rr r r rr                 (b) 2 22 1l n 1 f ( 1 )2 f ( )f ( 1 ) f( 1 ) 2 f( 2 ) f( 3 ) f (2) 2f (3) f (4) f( 3 ) 2 f( 4 ) f( 5 ) f (4) 2f (5) f (6) ... f( 3 ) 2 f( 2 ) f( 1 ) f ( 2) 2f ( 1) f ( ) f ( 1) 2f ( ) f ( 1) f ( 1 )2 f ( 2 )f ( 2 )f () NN rr rr rr NN N NN N NN N N                              2f ( ) f ( 1) ln1 ln 2 ln ln( 1) 1ln 2 NN NN N N        (c) 11 1 22 2 N NN   As N  , 1 02N  thus 11 1 22 2 N . Hence 2 2 1ln 1 ln 1 2 N r r      which is a constant/finite value. Hence the series converges. 2 2 1ln 1 ln 1 2r r           (d) Method 1:    222 2 21 2 22 22 11 1 ln 1 ln 1 ln 1 12 1ln ln ln ln 111ln ln ln ... 21 22 23 1 11 22 ( 2 ) 42 2(2 1) 2 1 rk k rr k kk k r rr k kk kk                                    

9 of 19 Method 2 (Change of index):     222 2 1 11 1 ln 1 111ln ln ln ... 21 22 23 1 2r kk k kr                     2 21 2 21 2 22 22 ln 1 ln 1 ln 1 ln 1 12 1ln ln ln ln Replace with 2 1 22 1 11 22 ( 2 ) 4 2(2 1) 2 21 rk rk k rr k rr k kr k r rr k k k k k                                    

10 of 19 Q7 Suggested Answers (a) d1 e2 ed2 ttx t  d e2 ed tty t  de 2 e 1d e2 e2 tt tt y x     At ln 4,t  ln 4 ln 4 ln 4 ln 4 de 2 e 7 1d5 e2 e2 y x     Gradient of normal at ln 4t is 5 7 (b) At ln 4,t  3 ,2x  9 2y . Equation of normal is 95 3 27 2yx     14 63 10 15yx   57 3 9xy (shown) (c) Sub 1 e2 e2 ttx  and e2 etty  into 57 3 9xy 15 e 2e 7 e 2e 392 tt tt      19e 8e 78 0tt   Let eta  819 78 0a a  219 78 8 0aa    41 9 2 0aa  4a or 2 19a ln 4t or 2ln 19t  (reject)