Set 6 Paper 1. Set 6 Paper 1. 1 Pearson Education Asia Limited Section A(1) (Pyth. Theorem) (b) 24units Area of OPQ. a b (4)

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1 Set Paper Set Paper Section A().. a b a b 8 ( a b) a b ( 8) a b a b a b k k k h k. The weight of Sam 5kg( %) 5kg The weight of Benny 5kg( %). 5. (a).85kg 5kg Benny is the heaviest one among them, his claim is agreed. () 5 x x 5 5 x 8x 5 x x x x x The solutions of the compound inequality are POQ ( ) 9 OPQ is a right-angled triangle. () () x. () (b) OP units units Area of OPQ sq.units sq.units (Pyth. Theorem). (a) By considering the height of the solid, we have h r 55...() By considering the volumes of the cylinder and the hemisphere, we have r h r h 8r...( ) By solving () and (), we have h and r 5. (b) Total surface area of the solid ( ) 85. (a) Let f ( x) kx kx, where k, k are non-zero constants. f 8 (b) k k k k k () k () 8...() f () k k...( ) Solving () and (), we have k 5 and k x f ( x) x 5x 5x x f ( x) 5 5 5x 5 (x )( x 5) x or x 5 The coordinates of the intersections between the graphs of y = f (x) and y = 5 are, 5 and (5, 5). Pearson Education Asia Limited

2 Solution Guide and Marking Scheme 8. Join AE. DBE = CDB = 8 (alt. s, BE // CD) EAD = DBE = 8 (s in the same segment) EAB = 9 ( in semi-circle) BAD (a) In ABD, ABD = BAD = (base s, isos. ) ABF 8 In ABF, AFE BAF ABF (ext. of ) 9 r 9 s 8 5 t 8 9 (b) The required probability Section A(). (a) Let (x, y) be the coordinates of P. ( x 9) ( y 5) ( x 9) ( y 5) x y 8x 5y x y x y 9 ( x 5) ( y ) ( x 5) ( y ) x y 5 x y 8 The equation of is x y + 8 =. Alternative Solution Note that is the perpendicular bisector of points A and B. The mid-point of AB 9 5 5, (, ) The slope of AB The slope of The equation of is y ( x ) x y 8 (b) Putting y = into x y + 8 =, we have x () 8 x The coordinates of E are (, ). Putting x = into x y + 8 =, we have () y 8 () y The coordinates of F are (, ). EOF 9 EF is a diameter of the circle. (converse of in semi-circle) The radius of C EF ( ) ( ) The area of the circle = () =.595 < 5 The claim is incorrect.. (a) In ACE and ACF, CAE = CAF = 5 (property of square) AC = AC (common side) ACB =ACD = 5 (property of square) BCE = DCF (given) ACE ACB BCE 5 BCE 5 DCF ACF ACE ACF (ASA) () Marking Scheme: Case Any correct proof with correct reasons. Case Any correct proof without reasons. Pearson Education Asia Limited

3 Set Paper (b) (i) AE = AE = x EB = AB AE = a x x( a x) Area of AEC ACE ACF AE = AF (corr. sides, s) i.e. AE = AF Area of AEC = area of AFC Area of quadrilateral AECF x( a x) x( a x) (ii) Area of the quadrilateral AECF = If a and x are integral with x(a x) =, then x =, a x = and hence a = + = ; x =, a x = and hence a = + = 5; x =, a x = and hence a = + = 5; x =, a x = and hence a = + =. a a a 5 The possible values are,, x x x a 5 and. + x. (a) (i) The inter-quartile range 59 9 The mean 5. (ii) Number of students who pass the test = The percentage of students passing the test % 55% (b) (i) The maximum possible median score The minimum possible median score 5 (ii) Let x marks be the average score of SB class. 5. x 5. x 5 It is impossible that the average score of SB class is 5. (). (a) With the notation in the figure, 9 h 8 (Pyth. theorem) Consider the corresponding sides of similar triangles, a (8 8) 9 8 a b (8 8) 9 8 b 8 The area of the wet surface ( 8) (b) The remaining volume of the vessel m m 9 Mary s claim is agreed.. (a) By comparing the coefficient of c c ( a b)( ) a b f a b a b( ) a b x, we have f ( ) [( ) a( ) b][( ) ( ) ] Pearson Education Asia Limited

4 Solution Guide and Marking Scheme (b) Since f ( ) f, we have a b a b b 9 b Note that the coefficient of x in the expansion is + a. So, by comparing the coefficient of x, we have a a x (x x x )(x or For x x, x ( ) x f ( x) x ) () x ( ) ()( ) (by (a)) For x x, x Note that 8, ( ) are irrational numbers. The claim is agreed., and Section B 5. (a) The total postage cost saved $[.. ( 5%)..... ( 5%) ] 8. [ (.85) ] $.85 $85. ( 5%) $. (cor.to sig.fig.) (b) The total postage cost saved since the start of the plan $[( (.85)...]. $.85 $ $. $.5 The manager s claim is agreed.. (a) The required probability C5 C C C C C5 () () 9 9 (b) The required probability () () Alternative Solution Consider the of x x. ( ) ()( ), which isnot a perfect square So, x x does not have rational roots. Consider the of x x. ( ), which isnot a perfect square So, x x does not have rational roots. The claim is agreed.. (a) Join BD. BD 8 (Pyth. Theorem) Area of BCD s ( s )( s )( s ), where s.58 Alternative solution Height of BCD Area of BCD.58 Pearson Education Asia Limited

5 Set Paper 8. (a) (b) (i) Area of ABCD (cor.to sig.fig.) The volume of the tetrahedron is a maximum when the plane ABD is perpendicular to the horizontal plane. Let x be the altitude of ABD from A to BD x 8 x.8 Volume of tetrahedron ABCD (cor.to sig.fig.) (ii) When the angle between the plane ABD and the horizontal plane is, the height of tetrahedron ABCD.8 sin. The height of tetrahedron ABCD is halved when the angle between the plane ABD and the horizontal plane is, while the base area remains unchanged. The volume of tetrahedron ABCD is also halved. Alan s claim is correct. (k) ( )( k k k 8k ) Since the graph is opening downwards and does not cut the x-axis, the graph is completely below the x-axis. () () (b) f ( x) ( x kx k k ) k ( x kx k ( x k) ( k ) k ) k The coordinates of the vertex are (k, k ). () (c) When PQ is the shortest, P and Q are the vertices of the graphs of y = f(x) and y = g(x) respectively. The coordinates of P are (k, k ). The graph of y = g(x) can be obtained from reflecting the graph of y = f (x) along the x-axis and then translating upwards by units. The coordinates of Q are (k, k + 5). Since the x-coordinates of P and Q are the same, PQ is a vertical line. The equation of the perpendicular bisector of PQ is ( k y y ) ( k 5) 9. (a) Join BF and produce BH to meet AC at K. BFO ACB ( s in the same segment) H is the orthocentre of ABC. AOB BKC 9 In KBC, KBC 8 BKC KCB ( sum of ) 8 9 ACB HBO 9 BFO In BFO, FBO8 BOF BFO( sum of ) 9 BFO HBO BO is the common side and HOB FOB 9. BOF BOH (ASA) OF = OH (corr. sides, s), i.e. O is the mid-point of HF. () () (b) (i) Let x y kx k y k be the equation of the circle, where k, k and k are real constants. By substituting (, 8), (, ) and (, ) into the equation, we have 8 8k k ( ) k k k k By solving, we have k, k and k. The equation of the circle is x y x y. (or ( x ) ( y ) ) (ii) By substituting x = into the equation, we have y y ( y )( y 8) y or 8 5 Pearson Education Asia Limited

6 Solution Guide and Marking Scheme The coordinates of F are (, ). O is the mid-point of HF. OH = OF = The coordinates of H are (, ). (iii) E is the circumcentre of ABC. E is the centre of the circle, i.e. the coordinates of E are (, ). ER BC The coordinates of R = (, ) The coordinates of G () () (), (, ) 8 Slope of AG Slope of RG Slope of AG = slope of RG A, G and R are collinear. (9) Pearson Education Asia Limited

Set 2 Paper (a) (i) (ii) (b) The coordinates of R = ( 5, (a) Range = 8 2 (3) (b) New Mean. New variance

Set 2 Paper (a) (i) (ii) (b) The coordinates of R = ( 5, (a) Range = 8 2 (3) (b) New Mean. New variance Section A( ( + + + +. ( ( ( M M 7 A 7 (. (a.8 A (b 00 (c A A (. (a x + xy (x ( x + y A (b. (a x x xy + y (x (x ( x + y M ( x ( x A x x + y ( x + y x x + y x M ( x y A (b If the value of y is increased