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,