Set 2 Paper 1. Set 2 Paper 1. 1 Pearson Education Asia Limited Section A(1) (4) ( m. 1M m

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1 Set Paper Set Paper Section A() 5 5 ( m n ) m n. ( m ) m 6 5 ( ) m 6 n m 6 n. (a) 5.8 (c) The required probability () () 5 +. (a) m 5m n m m (m 5n ) 5. (a) 6. (a) m m (m 5n ) (m 5n ) ( m 5m n m )(m 5n ) m 5n ( m )( m )(m 5n ) x y x y ( x y) x y x y x y () x y If the value of y is increased by, then the value of x is increased by. ( x ) x and x x 7 8 x and x x and x 8 x x 8 7. Let x and y be the numbers of candies owned by Alice and Ben respectively. x y 8...() + x ( 5%) y...() By substituting () into (), we have ( 5%) y y 8.65y 8 y By substituting y = into (), we have x 8 x 78 The required difference (a) ABE DCE AE = DE and BE = CE (corr. sides, s) AED and BEC are isosceles triangles. + BEC is an isosceles triangle. BCE CBE 5 (base s, isos. ) ACD ( in semi-circle) CDE ( 5) 5 8 ( sum of ) CDE. (a) AE 6 5 DC Let AG = x. ( 6) (7.5.5) x 6x 55 x 8 AG 8 Total surface area of the prism ABCDEFGHIJ [ 6 ( ) 8] Section A() (a) Mean $ 56 and median $ $ () Pearson Education Asia Limited 7

2 Solution Guide and Maring Scheme (i) Let $x be the mean of the amounts spent on the three days. 856 x 5 8 x 67 The mean of the amounts spent on the three days is $67. (ii) Total amount spent on the other two days $(67) $ If the median of the amounts spent on these eleven days is $55.5, then the amount for one of these two days must be $55.5. The amount spent on the remaining day $( 55.5) $5.5 $5 It is impossible that the median of the amounts spent on these eleven days is the same as that found in (a).. (a) Consider AEB and BFC. AB = BC (property of square) AE = BF (given) ABE = BCF = (property of square) AEB BFC (RHS) Maring Scheme: Case Any correct proof with correct reasons. Case Any correct proof without reasons. (i) AEB BFC (by (a)) BAE = CBF (corr. s, s) AEB = 8 ABE BAE ( sum of ) = 8 BAE = BAE BGE = 8 CBF AEB ( sum of ) = 8 BAE ( BAE) = BGE is a right angle. (ii) AB = BC = In ABG, BG AB AG 6 (Pyth. theorem). (a) Let f ( x) x ( x 5), where and are non-zero constants. () By the remainder theorem, we have f ( ) ( ) ( 5) f () () ( 5) ()...() From (), we have () By substituting () into (), we have 7( ) By substituting = into (), we have ( ) f ( x) x f ( x) x x ( x )(x ( x 5) x 5 x 5 6x 5) Consider the equation x 6x 5. (6) () 8 The equation x 6x 5 has no real roots. The equation f ( x) has only one real root. Louis claim is agreed.. (a) (i) By substituting x = and y = into x y 8x my n, we have () () () 8() m () n n By substituting x = and y = into x y 8x my, we have () ( ) 8() m( ) (ii) The equation of the circle C is x y 8x 6y. m 8 6 Coordinates of S, (, 8) m 6 (i) A, S and P are collinear. (ii) A, S and P are collinear. AP is a diameter of the circle. The perpendicular bisector of AP passes through S. Pearson Education Asia Limited 7

3 Set Paper 8 ( ) Slope of AS The required equation is: y ( 8) ( x ) x y 6 (). (a) (i) The figure below shows the front view of Solid X. With the notations in the figure, h 5 r (7) 7 (corr. sides, ~ s) r 5 Volume of Solid X (7) 75 (ii) Volume of Solid Y Volume of water overflowed from the tan [( 75 5 ) ( 85%)] + (7 5) 78.78m.m Florence s claim is correct. Section B 5. The slope and the intercept on the horizontal axis of the graph are and respectively. (6) () log7 y log x ( ) log 7 y log x log y log x log y log x...() 8 Tae logarithms to the base on both sides of the expression y Ax, we have log y log ( Ax ) log y log A log log y log x log A...() Compare the coefficients of () and (), we have and log A 8 x A 7 8 y 7x () 6!! 6. The required probability + ( 5)! 8 7. (a) Note that the highest mars in the test is mars. Let µ and be the mean and the standard deviation of the distribution respectively....() ( ) () + () : (7 ) (.5 ) The mean of the distribution is 7 mars. The median = 75 and the mean = 7 The mean is smaller than the median. At least half of the students in the class whose results are greater than 7 mars. i.e. The standard scores of at least half of the students in the class are positive. Felix s claim is agreed. () () () Pearson Education Asia Limited 7

4 Solution Guide and Maring Scheme 8. (a) In ABD, AB BD sinadb sinbad 5 6 sinadb sin6 ADB ABD = 8 BAD ADB ( sum of ) = (cor. to sig. fig.) by cosine formula, CD BC BD ( BC)( BD) coscbd CD 6 ()(6)cos (cor. to sig. fig.) Note that the claim is true only when AD and CD lie on the same vertical plane. Suppose AD and CD lie on the same vertical plane. Let M and N be points on BD and CD respectively such that AM BD and AN CD. Then, MN BD. In ABD, AD BD sinabd sinbad AD 6 sin58.58 sin6 AD 5.7 In ABM, BM ABcosABM 5cos DM BD BM (6 7.8) BC CD sinbdc sincbd sinbdc sin78 BDC.688. (a) In MDN, DM DN cosmdn cos In ADN, DN cosadn AD In ACD, by cosine formula, AC AD CD (5.7)(6.6665) ( AD)( CD)cosADC AD and CD do not lie on the same vertical plane. ADC is not the angle between AD and the plane BCD. The claim is disagreed. () y x 6x x...() y...() By substituting () into (), we have x 6x x x ( ) x 6...(*) of (*) ( ) ()(6 ) 6 ( ) ( ) > The quadratic equation (*) has two distinct real roots. P and L intersect at two distinct points. Maring Scheme: Case Any correct proof with correct reasons. Case Any correct proof without reasons. Case Incomplete proof with any one correct step and one correct reason. (i) Note that and are the roots of (*). and 6 ( ) ( ) ( ) (6 ) () 6 6 Pearson Education Asia Limited 7

5 Set Paper (ii) Distance between A and B ( ) 6 (by (i)) ( ) The distance between A and B is at least. The claim is agreed.. (a) Note that A, I and J are collinear. I is the incentre of OAB. JAO = JAB J is the circumcentre of OAB. J is the centre of the circle passing through O, A and B. i.e. OJ = AJ = BJ JAO = JOA and JAB = JBA (base s, isos. ) JOA = JBA JA = JA (common side) JAO JAB (AAS) AO = AB (corr. sides, s) i.e. OAB is an isosceles triangle. Maring Scheme: Case Any correct proof with correct reasons. Case Any correct proof without reasons. Case Incomplete proof with any one correct step and one correct reason. Refer to the figure below. A O y B (i) Let (h, 7) be the coordinates of B. By (a), we have AB = AO. [ h ( 5)] (7) [ ( 5)] ( x h h 5 56 () ) h h h = 7 or 7 (rejected) OAB = OB is a diameter of the circle. (converse of in semi-circle) 7 7 Centre of the circle OAB, (5, 85) Radius of the circle OAB (5 ) (85 ) 85 The equation of C is ( x 5) ( y 85) 85. (or x y 7x 7y ) Alternative Solution Let (h, 7) be the coordinates of B. By (a), we have AB = AO. [ h ( 5)] (7) [ ( 5)] ( h h 5 56 ) h h h = 7 or 7 (rejected) Let the equation of C be x y Dx Ey F. C passes through O(, ). We have F. C passes through A( 5, ) and B(7, 7). We have ( 5) () D( 5) E() (7) (7) D(7) E(7) 5D E 6 7D 7E 8 By solving, we have D 7 and E 7 The equation of C is x y 7x 7y. (ii) Let the equation of L be y x c. By substituting y x c into x y x x 7x 7y, we have x ( x c) 7x 7( x c) cx c 7x 7x 7c x (c ) x c 7c...(*) Since y x c is the tangent to C, we have of (*) = (c ) ()( c 7c) c 6c 576 8c 6c c c 576 c 8 or 8(rejected) The equation of L is y x 8. The x-intercept of L = 8 < 85 The claim is disagreed. (8) 5 Pearson Education Asia Limited 7

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