NYJC 9758 2024 Promo Solutions

Q1 Suggested Answers (a) 2 2 e xayb c x   Sub 21, 2e 1xy 22e2 e 1ab c  ---(1) After scaling, 2 2 22 4ee 2 x xaayb c b c xx        Sub 21, 8 exy   1 24e 8 eab c    ---(2) After translation 2 2 e1xayb c x    Sub 233, 2e 4xy ௔ ଷ ൅𝑏 𝑒 ଶ√ଷ ൅𝑐ൌ 2𝑒ଶ√ଷ െ 3---(3) Using GC, 3, 2, 4abc 

Q2 Suggested Answers (a) Method 1:  22 49 2 5xx x   Since  22 20 , 25 0xx   . Therefore, 2 49xx is always positive for all real values of x. Method 2:  22 44 4 1 92 0ba c    Since the discriminant <0 and that the coefficient of 2x is positive, therefore, 2 49xx is always positive for all real values of x. (b)   22 2 49 2 0 67 xx x xx     Since 2 49 0xx   2 2 2 0 67 x xx      2 2 2 0 32 x x       2 2 0 32 32 x xx     + + - + 2 3 2 3 2  2 or 3 2 3 2xx   

Q3 Suggested Answers (a) (b)  2 2 2 2 253 23 xxk xx      2 223xy k  2 25 00 33 25 99 106 9 106 3 k        106 3k Q4 Suggested Answers (a)  43 76 88 1Up q U p q      54 70 76 2Up q U p q      Solving equation (1) and (2): 32, 0.5pq (b) Method 1:    32 2 2 21 1 1 10 0 0 88 32 0.5 112 112 32 0.5 160 160 32 0.5 256 Up q U U U Up q U UU Up q U U U                   (5.37, 0.157) (-0.372, 1.59) y x

(b) Method 2:    32 1 0 0 0 1 2 11 22 111 222 111 222 32 32 32 32 32 32 88 32 32 32 256 UU U U U U          (c) Method 1: Using GC to get 64 Method 2: As 1,, tttC L C L    132 2 26 4 64 LL LL L    The readings decreases and converges to 64. Q5 Suggested Answers (a) 34 1 231 0 12 1 7 ab               Area of ΔAOB 1 2 1 1 102 17 1 11 0 0 2 8 92           ba 1 3902 units2

(b) Method 1: Perpendicular ht of tetrahedron = 0 13 11 3 0 11 100 289 1 10 1 10 17 1 390 39 7 OC                          Vol of tetrahedron OABC = 1 390 30 532 390      units3 Method 2 Let X be the foot of perpendicular from C to plane OAB. 1 10 o 3 1 1 , f r som e 17 OX             Equation of plane OAB: 1 10 0 17    r Since the point X also lies on the plane OAB, 31 11 0 0 11 7 3 10 100 17 289 0 390 30 1 13 1 10 17                               1 1 1 1 33 11 013 17 1 1013 1 11 7 CX OX OC                                  221perpendicular height, 1 10 1713 390 13 30 390 CX        

(c) Method 1 Perpendicular distance from B to line OA ˆ 390 941 195 5.287     a o b r Method 2 Area of OAB = 11 390 ( )22 OA h Thus, perpendicular distance from B to line OA = 390 195 14 7h  Q6 Suggested Answers (a) NEW 2 2nSn n   2 1 22 22 2 21 2 1 22 1 2 2 21 n nnn Sn n uSS nnn n nn nn n n         1 21 1 2 1 23 21 2 nnuu n n nn          Since 1 2nnuu  is a constant, it is an arithmetic series. 2d  21 12 2 1 22 21 3 nun an n an n a       (b) 6 5 5 12 3 8 3 8 1 32 1 2 a u ar r r        

 1 1 112 1 2 11 2 181 2 n n n n ar S r              1 12 11 2 8 aS r     0.001 18 1 8 0.0012 18 1 8 0.001 02 n n n SS               From GC, n 181 8 0 . 0 0 12 n      12 49.53 10  > 0 13 52.34 10  < 0 Therefore, the least value of n is 13.

Q7 Suggested Answers (a) fR) [0, g fgR D D    Thus gf does not exist. (b) Method 1: Consider the different scenarios: If 0  , g will have a turning point at x  , making g not an one-one function and 1g will not exist. Hence, 0  . Method 2: We need 2 0g' ( ) 1x x   for 1g to exist. Since 2 0, 0x  Also, when 0  , g xx is also an one-one function. Hence, 0  |2 0 |yx 0  0  0 

(c) 2 4 4g' ( ) 1 0 2 g( ) x x x xx x     From graph, least  = 2 since we need g to be one-one function. (or since 0,x  = 2)  2 2 2 12 4 4 4(1)(4) 2 16since 2, 2 1g(1 6 ) 2 x xy y x yyx yyx x xx x x          1 ggDR [ 4 , )   Q8 Suggested Answers (a)   2 22 2ln e ln e e e dd2e 1 2 e ( s h o w n )dd y yy yx y x x yyyy xx           . OR Differentiate w.r.t. x gives 2 d112 de ey yy xx  since  22 ln e e e yyx x    That is, 2d2e d yyy x  . Differentiate w.r.t. x gives

2 2 22 2 22 2 dd d22 e dd d dd d edd d y y yy yyy xx x yy yyy xx x         When 0x , 1y , d1 d2 e y x  , 2 22 d3 d4 e y x  . 22 22 113 1 311 2e 2! 4e 2e 8eyx x x x         . (b)  2 2 2 ln e ln e 1 e 1l n 1 e 11 e2 e 1 e 2e xx x xx xx                   1 2 1 2 2 2 222 22 2 2 ln e ln e 1 e 2e 11 1 221 2e 2 ! e2e 2e 131 2e 8e yx x xx xx xx xx                               which agrees with the expansion in (a). (c) 11 0 e 1ln ln e10 10         2 2 2 2 2 11 3 11 2e 10 10 8e 131 20e 800e 800e 40e 3 800e         