Set 5 Paper 1. Set 5 Paper 1. 1 Pearson Education Asia Limited Let x and y be the original numbers of apples and oranges respectively.

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1 Set 5 Paper Set 5 Paper Section A().. (a). (a) m n m n ( mn ) m n 6 5 m n 8 m n m 8 n 5 p pq p( q) 5 5 p pq 5 p q (b) q > 0 and the value of q increases. 5 The value of the denominator of q The value of p decreases. increases. () () y y y 6 y y y (b) is the greatest even number smaller than. The greatest value of y is.. The total number of taxi drivers = = Mean = Median = the th datum the th datum ( ) Standarddeviation () 8(.5) 7(.5) (.5) (.5) = (cor.to sig.fig.) () 5. Let x and y be the original numbers of apples and oranges respectively. x 8 y 7 + x y 6 By solving, we have 8 y y 6 7 y 8 7 y (a) 8 x (56) 6 7 The total number of apples and oranges left (6) (566) 00 AOB (b) AOB is a right-angled triangle. 0 0 units 00 units The perimeter of AOB () OA OB (Pyth. theorem) (00 00) units.56 units.units(cor.to sig.fig.) Alternative Solution OA = OB O OBA (base s, isos. ) O ( sum of ) 5 OA coso 0 cos5 0 units The perimeter of AOB (000 ) units.56 units.units(cor.to sig.fig.) Pearson Education Asia Limited 07

2 Solution Guide and Marking Scheme (c) AOB is a right-angled isosceles triangle. AOB has axis of reflectional symmetry. () 7. (a) The maximum absolute error 0. kg 0.05kg The least possible weight of a mega size pack of coffee bean ( ) kg 0.95kg (b) The maximum possible total weight of 8 smaller packs of coffee bean ( ) g 909g 0.909kg 0.95kg The claim is disagreed. 8. (a) f ( x) x ax a By remainder theorem, the required remainder f ( a) ( a) ( a)( a) a a (b) f(x) is divisible by x + a. By the converse of factor theorem, f ( a) 0 a 0 a f ( x) x ( ) x ( ) = x x By solvingf ( x) 0, wehave x( x x x 0 ) 0 x( x )( x ) 0 x 0, or 9. (a) Let be the angle of the sector. (6) ( ) (6) The angle of the sector is 0. (b) The area of the sector 0 (6 ) 60 Section A() 0. (a) oordinates of ( 8) G, 0 (, 0) (b) (i) The equation of Γ is : ( x ) y 7 x y 8x 0 (ii) Γ and are concentric circles. (iii) Radius of the circle () 8 0 ( 9) 5 The shortest distance from A to Γ radiusof Γ radiusof 7 5. (a) Driving speed of Paul 90 km/h 60km/h The required distance 0 60 km 60 0 km (b) Let x hours be the time needed by Henry to catch up with Paul after taking rest. From the graph, after Paul met Henry at 9:5, Henry continues his rest for 5 minutes before he drives again. Since they meet again, the distance they travelled are the same, i.e. 60 x 0x + x x x Henry will meet Paul hours, i.e. 5 minutes, after his rest. They will meet at 0:5. () () Pearson Education Asia Limited 07

3 Set 5 Paper (c) Time spent for Henry to reach town from town B 90 5 hour 0 hour 8.5 minutes Henry will reach town at about 0::0. Henry reaches town earlier than Paul with less than 0 minutes. Henry s claim is correct. (). (a) The volume of the circular cone (6) (6) 68 Let r be the base radius of the frustum. r r The required volume 68 ( )(66) 97 (b) Volume of the hemisphere () 5 Volume of water filled in the frustum (8 5 ) 96 () Let h be the height of water inside the frustum. ( )(8) 8 + ( h 8) ( )(8) 96 8 h h 8 5 h The depth of water in the vessel (). (a) The median of the distribution is.5. The total number of employees owning or credit cards is equal to that of to 5 cards. i.e. a b 5 a b 5 < b < a < and a and b must be integers a or b 7 a 0 b 6 () + (b) The mode of the original distribution is and the mode is increased by after five more employees are joined. All these five employees own credit cards. (i) ase : a =, b = 7 The standard deviation ase : a = 0, b = 6 The standard deviation.5885 The least possible standard deviation of the numbers of credit cards owned by the employees in the company is. (cor. to sig. fig.). (ii) ase : a =, b = 7 The inter-quartile range.5.5 ase : a = 0, b = 6 The inter-quartile range () The greatest possible inter-quartile range of the numbers of credit cards owned by the employees in the company is. (). (a) Let f ( x) kx kx, where k and k are non-zero constants. f ( ) 5 5 k k () f ( 7) 5 5 9k 7k () Pearson Education Asia Limited 07

4 Solution Guide and Marking Scheme By solving () and (), we have k and k 0 5 f ( x) 5x 0x For f ( x) 0, we have 5x 0x 0 x 6x 8 0 ( x )( x ) 0 x or x (b) The y-coordinates of A and B are 0. From (a), we have x or x. The coordinates of A and B are (, 0) and (, 0) respectively. For (c, 0), solving f ( x) 0, we have 5x 0x 0 x( x 6) 0 x 6 or x 0 (rejected) oordinates of (6, 0) The area of the quadrilateral O ( 6 )0 sq.units 60sq.units Section B x 5. y mn... () By substituting (0, 5) into (), we have 5 mn 0 m 5 By substituting (, 5) and m = 5 into (), we have 5 5n n 7 n x y 5( ) x y 5 y x log 5 6. Note that the numbers of dots in the patterns form an arithmetic sequence. The total number of dots in the first m patterns [5 ( m )] m [5 (m )] m(m 7) () () () + m m(m 7) 88 7m ( m 96)(m 8) 0 96 m 8 The greatest value of m is 9. () 7. (a) The coordinates of the centre of are (5, 5) and the radius of is 5. The equation of is [ x ( 5)] ( y 5) 5 ( x 5) ( y 5) 5 () (b) The equation of L is y x k. Substitute y x k into the equation of, ( x 5) ( x k 5) 5 x kx ( k 5) 0 Let M be the mid-point of. The x-coordinate of M k k The y-coordinate of M k k k k k The coordinates of M are,. Let X be the centre of the circle. AM = BM XM (line joining centre to mid-pt. of chord chord) The distance between the centre of and XM k k 5 5 k 5 (0 k) Pearson Education Asia Limited 07

5 Set 5 Paper 8. (a) P (at least red balls are drawn) = P ( red balls) + P ( red balls) (b) P (at most one red ball is drawn) = P (all the balls drawn are white) + P (one red ball is drawn) P(all the balls drawn are white) 6 0 () 6 P (all the balls drawn are white at most one red ball is drawn) (c) P ( red balls are drawn) The expected points Peter will be awarded () Peter s claim is agreed. () 9. (a) (i) Let M be the mid-point of B. = A and BM = M AM B (prop. of isos. ) BM cos 6 cos (cor.to sig.fig.) (ii) Alternative Solution onsider. A = AB 76 By the sine formula, sinab sinba B sin76 sin( ) (cor.to sig.fig.) 55 A B 0 The area of the paper card ( )(6)sin76 ( ) sin Note that ( )(6)sin76 is a constant and ( ) sin varies as sin. Also note that the area of the paper card is the greatest when A B 90. When increases from 55 to 90, the area of the paper card increases. When increases from 90 to 0, the area of the paper card decreases. (b) (i) AD is the angle bisector of and A. D DB and AD B (prop. of isos. ) 8 i.e. D DB By the cosine formula, DB D B cosbd ( DB)( D) 6 cosbd ()() BD (cor.to sig.fig.) (ii) Note that AD and BD are the height and the base of the pyramid D respectively. AD DB (Pyth. theorem) 5 Pearson Education Asia Limited 07

6 Solution Guide and Marking Scheme Area of BD ( BD )( D)sinBD ()()sin Volume of pyramid D area of BD AD (cor.to sig.fig.) In BAN, sinban BAN (cor.to sig.fig.) The angle between and the plane AD is.. (7) (iii) Let N be the projection of B on the plane AD. ADB = 90 N lies on D. i.e. The required angle is BAN. In D, sinbd BD sin In BAN, sinban BAN (cor.to sig.fig.) The angle between and the plane AD is.. Alternative Solution Area of AD ( D)( AD) ()( ) Let N be the projection of B on the plane AD. area of AD volumeof pyramid D Pearson Education Asia Limited 07