Set Paper Set Paper. B. C. B. C. C 6. D 7. A. D. A. A. C. C. B. B. C 6. C 7. C. A. B. D. B. D. A. A. B 6. B 7. D. D. C. A. C. D. D. A. D 6. D 7. A. A. C. C. B. D. B. D. A Section A. B ( 7) 7 ( ) 7 ( ) 7 7. C a(b ) b 7 6ab a b 7 6ab b a 7 b(6a ) a 7 a 7 b 6a. B ( ) ( ) [ ( )][ ( )] ( )( ). C ( p)( ) ( q) ( r) p p q r Compare the coefficients of both sides, we have p q and p r p : q : p : r : p p p : q 6 : p : r 6 : p : q : r 6: : q r Alternative Solution Substitute = into the equation, we have ( q) ( r) q r r q r : q : Substitute = into the equation, we have p r p r p : r : p : r 6: and r : q : p : q : r 6: :. C f() is divisible by +. f ( ) ( ) k( ) 6 6 6 k k f ( ) ( ) ( )( ) 6 6. D The quadratic equation + k = has no real roots. i.e. ( ) ()( k) k k is a possible value of k. The answer is D. 7. A m n 6...() m n 7...( ) () : 6m + n =...() () : 6m + n =...() () (): 7n n 6 By substituting n = 6 into (), we have m (6) 6 m m n 6 Pearson Education Asia Limited 7
Solution Guide and Marking Scheme. D The graph opens upwards. a > I is true. The equation of the ais of symmetry is b c The ais of symmetry cuts the -ais at (, ). b c b c II is true. The y-intercept of the graph is negative. a( b)( c) abc III is true. The answer is D.. A and 6 6 6 6 The solution is.. A Let be the cost of a doll. Profit per cent ( %) () ( %) % On the whole, Betty gained %.. C The required interest 7 $ % $ (cor.to thenearest dollar). C : y : y y : z : i.e. y z z y % y z y y z y 6 ( y) : ( z) :. B y k y, where k Let and y be the original values of and y respectively. New value of = ( + %) =. k k New value of y y.. B. C Percentage change in y y y % y % y is decreased by %. h cm Let h cm be the height of the base of the prism. h ( 6) (Pyth. theorem) h 6 Total surface area of the prism ()( 6) ( 6 ) cm cm Pearson Education Asia Limited 7
Set Paper 6. C AEG ~ CDG (AAA) Area of AEG AE Area of CDG CD Area of CDG cm 6cm Consider ADG and AEG. AEG ~ CDG EG DG AE CD (corr. sides, ~ s) ADG and AEG have the same height. Area of AEG EG Area of ADG DG Area of ADG cm cm Area of BAC area of DCA (6 ) cm 6cm BEF ~ BAC (AAA) Area of BEF EB Area of BAC AB Area of BEF 6cm 6cm Area of EFCG (6 6) cm cm 7. C AFB 6 ABF = BAD= (property of square) BAF 6 ( sum of ) In ABG and ADE, AB = AD (property of square) BG = DE (given) ABG = ADE = (property of square) ABG ADE (SAS) DAE = BAG = (corr. s, s) EAF. A For a regular n-sided polygon, ( n ) its interior angle ( sum of polygon) n its eterior angle 6 (sum of et. s of polygon) n From the question, we have ( n ) 6 6 n n n() 6 6 n(6) n() 7 n. B Draw a perpendicular AJ from A to GH and join AH. AJ FG DE BC 7 JH GH FE CD BA 6 6 AH AJ JH (Pyth. theorem) 7 6. D ACB ( in semi-circle) In ABC, AC coscab AB CAB.6 Let O be the centre of the semi-circle. In AOC, OA = OC = OB = 6 cm (radii) OAC OCA (base s, isos. ) COB OCAOAC (et. of ) CAB 6.77 Height of AOC 6sin6.77cm.67cm Pearson Education Asia Limited 7
Solution Guide and Marking Scheme Area of the shaded region 6.77 6.67 (6) cm 6.7cm (cor.to thenearest.cm ). B With the notations in the figure, O is the centre of the circle and join AO, OD. 7 Refle AOD 6 (s at a pt.) 7 6 6 AED refle AOD( at centre twice at ce ) Alternative Solution Join AD.. A CA tan B AB 7 AB tanc AC 7 7 tan B : tanc : 7 7 76 76: tan ( ) sin(6 ) sin tan sin sin sin cos sin cos sin sin sin cos sin. A With the notations in the figure, Let EAD = k. Then ADE = k and AED ( 7 ) k (arcs prop. to s at ce ) k In ADE, k k k ( sum of ) 6k k AED CAB CAY BAY 6. D AB CA 7 BC ABC is a right angle triangle. (converse of Pyth. theorem) Pearson Education Asia Limited 7
Set Paper In ABC, AB tanbca AC 6 km km BCA 6.6 XCA = CAY = (alt. s, XC // AY) XCB 6.6 67 (cor.to thenearest integer) The bearing of B from C is N67 E.. B Coordinates of the image (, ( )) (, ) 6. B The slopes of L and L are negative. and a c a c L is steeper than L and their slopes are negative. Slopeof L slopeof L c a a c ( a and c ) I is true. -intercept of L > -intercept of L b d b d II is not true. y-intercept of L = y-intercept of L b d a c ad bc III is true. The answer is B. 7. D A(, ) + y 6 + y + = C(, ) y y 6 y y ( ) Centre, (, ) Radius Diameter of the circle.67766 I is true. Shortest distance between A and the circle ( ) [ ( )] Shortest distance between B and the circle (6 ) [ ( )] The shortest distance between A and the circle is less than the shortest distance between B and the circle. II is true. Let and be the inclinations of AC and BC respectively. ( ) Slope of AC tan. ( ) Slope of BC 6 tan ACB. 6.6 ACB is an obtuse angle. III is true. The answer is D. B(6, ) Pearson Education Asia Limited 7
Solution Guide and Marking Scheme. D m m ( m)( m ) m m m m ( m )( m ) m or m. C Mean = 6 7 67 ( a) Inter-quartile range a. A Let {, +,, + 7} be Group A and {, +, 6, + } be Group B. 7 Mean for Group A 6 Mean for Group B I is true. Arrange the data in Group A in ascending order:,, +, + 7 Median for Group A Arrange the data in Group B in ascending order: 6,, +, + Median for Group B II is true. Standard deviation for Group A =.7 Standard deviation for Group B =. III is not true. The answer is A. Section B. C 6 ( ) ( )( ) 7 ( )( ) The L.C.M. of 6, and 7 is ( )( ) ( ).. D 6 6 76 6 (6 ) 6 6 (6 ) 6 6 6 6 6 6 6 BC. D a Sum of roots a ( i) ( i). A 6 a b Product of roots b ( i)( i) ( ) (i) ( ) a + b = + 7 The shaded region is the common region of y the system of inequalities. y 6 y By solving, the vertices of the shaded region are (, ), (6, ), (, ) and,. The value of 6y + for (, ) = The value of 6y + for (6, ) = The value of 6y + for (, ) = The value of 6y + for, = The least value of 6y + formed by D is. 6 Pearson Education Asia Limited 7
Set Paper. D Let r be the common ratio of the sequence. ar...() ar 7...( ) 7 ( ) () : r r r a or I may not be true. a ar r 7 a a r II must be true. a a a7... an.7 III must be true. The answer is D. 6. D Let T(n) be the general term of the sequence. T () S() k() () k T () S() S() [ k( ) ()] ( k ) (k ) ( k ) k Common difference = T () T () (k ) ( k ) k k First term = T() = + = The nth term T ( n) ( n )( ) n n 7 Alternative Solution Let T(n) be the general term of the sequence. T( n) S( n) S( n ) ( kn kn n) [ k( n ) ( n )] n [ k( n kn k Common difference = T ( n) T( n ) k k The nth term T( n) n ) ( n )] ( ) n ( ) n 7 7. A When =, y = p = I is true. The graphs of y = p and y = r are symmetric about the y-ais. p r p r pr II is true. From the graph, for the same positive value of, q is greater than p. q > p III is not true. The answer is A.. A ( cos sin cos ) cos cos cos (cos )(cos ) cos or or 6 or The equation sin cos has roots.. C Let f ( ) sin. From the figure, the period of the function is (6 ) 7. The function is obtained by enlarging the graph of y = sin along the -ais to time. i.e. y f 7 Pearson Education Asia Limited 7
Solution Guide and Marking Scheme For a sine function y = sin, y The range of y is. From the figure, the range of y = 6 =. The function is obtained by enlarging the graph of y = sin along the y-ais to times. i.e. y f ( ) The minimum value of the function y = sin should be and the minimum value shown in the graph is. The function is obtained by translating the graph upwards by units. i.e. y f The figure shows the graph of y sin.. C ABC is a right-angled triangle. AC cm (Pyth. theorem) cm AE AC cm cm In ADC, AD cosdac AC In ADE, by the cosine formula, DE AD cm AE ( AD)( AE)cosDAE ()() cm DE. B Join AB. cm ATB TAB TBA ( sum of ) TAB TAB ACD = TAB = ( in alt. segment) In ACD, ACD CAD ADB (et. of ) CAD CAD. D L // L Slopeof L slopeof L Let y = + h be the equation of L. y h...() y...( ) By substituting () into (), we have ( h) h h ( h ) ( h L is a tangent to C. of (*) i.e. ( h ) ()( h h h 6h h ) ) 6 h h h ( h )( h ) The equation of L is y y...(*) h or (rejected) Alternative Solution ( ) Coordinates of the centre of C =, (, ) Let L be the straight line parallel to L and passing through the centre of C. In TAB, TA = TB (tangent properties) TAB = TBA (base s, isos. ) Pearson Education Asia Limited 7
Set Paper L // L Slope of L = slope of L The equation of L is y ( ) y L // L Slope of L = slope of L y-intercept of L = y-intercept of L + (y-interept of L y-intercept of L ) The equation of L is y y. B P(passthefirst part passat least onepart) P(passthefirst part) P(passat least onepart) P(passthefirst part) P(both fail). D C C The required probability C. A Let be the standard deviation of the eamination. Peter s standard score = 7 Amy s standard score. 7. Pearson Education Asia Limited 7