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 t