Set 2 Paper (a) (i) (ii) (b) The coordinates of R = ( 5, (a) Range = 8 2 (3) (b) New Mean. New variance
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1 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 by, then the value of x is increased by. M A (. (a r % M r r 0 A (b The required interest 0% $ M $ 6 (cor. to the nearest dollar A ( 6. (a (i (ii POQ OP OQ A PQ + units units M The required perimeter ( + + units 0 units A (b The coordinates of R (, 0 A (or any other possible answers ( 7. (a Range Mean Variance (.6 + (.6 + 6(.6 + 8(.6 + 6(6.6 + (7.6 + ( (b New Mean New variance (.7 + (.7 + 6(.7 + 8(.7 + 6(6.7 + (7.7 + ( Change in the variance of the distribution (a ABE DCE AE DE and BE CE (corr. sides, s AED and BEC are isosceles triangles. M A A A ( A + A (b BEC is an isosceles triangle. BCE CBE (base s, isos. M ACD 0 ( in semi-circle In BCD, CDE + ( ( sum of M CDE 0 A ( Pearson Education Asia Limited 0
2 Solution Guide and Maring Scheme. (a AE 0 6 DC 7.. Let AG x. (0 6 ( x 6x M x 8 AG 8 A (b Total surface area of the prism ABCDEFGHIJ [ 6 + ( ] M 0 A Section A( (a Mean $ 6 and median $ $. A + A ( (b (i Let $x be the mean of the amounts spent on the three days x M 8 + x 67 The mean of the amounts spent on the three days is $67. A (ii Total amount spent on the other two days $(67 00 $0 If the median of the amounts spent on these eleven days is $., then the amount for one of these two days must be $.. M The amount spent on the remaining day $(0. $. < $0 It is impossible that the median of the amounts spent on these eleven days is the same as that found in (a.. (a Let f ( x x + ( x +, where and are non-zero constants. By the remainder theorem, we have f ( ( + ( + ( ( + 7 f ( 0 ( ( A ( A ( A M + A By solving, we have and. f ( x x ( x + x x (b f ( x x x ( x (x + 6x + Consider the equation x + 6x + 0. Δ (6 (( 8 The equation x + 6x + 0 has no real roots. The equation f ( x 0 has only one real root. Louis claim is agreed. y (a (i x..y 6 0x 60 y 80x + 70 (ii The required charge $[80(. + 70] $70 (b (i correct shape of graph correct coordinates for (, 0 and (.7, 00 A A A ( A ( M (ii Let $S be monthly service charge of using. GB for Pacage B. S 0 00 M. S 0 00 S 0 > 70 Eric s claim is disagreed. A (. (a (i Γ is the angle bisector of L and L with a negative slope. A + A (ii Point of intersection of L and L is (,. A M M A A ( Pearson Education Asia Limited 0
3 Let (x, y be the coordinates of P. y tan(80 x y x x + y 6 0 (b Centre of C,, By substituting L.H.S. +, 6 M A ( into the equation of the locus of P, R.H.S. Centre of C does not lie on Γ. Γ does not divide C into two equal halves.. (a (i The figure below shows the front view of Solid X. M A ( With the notations in the figure, h M r (7 7 (corr. sides, ~ s M r Volume of Solid X π ( + π ( (7 M 7π A (ii Volume of Solid Y 6 7π M 0π A (6 (b Volume of water overflowed from the tan [( 0 7π + 0π 0 ( 8%] M (70π m.0 m Florence s claim is correct. A ( Section B. The slope and the intercept on the horizontal axis of the graph are and respectively. log7 y 0 [ log x ( ] M log 7 y log x + log y log x + A log y log x + ( 8 Tae logarithms to the base on both sides of the expression y Ax, we have 6. (a 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 A x 7 y 7x 8 A ( (b n n + n + n n + n n + n + n ( n ( n 0i i 0i + i i( 0 i( 0 + n n M A ( Pearson Education Asia Limited 0
4 Solution Guide and Maring Scheme i i 0 M A (a The required number of ways 7 C C M 0 A ( (b (i The required number of ways ( +!! M 70 A (ii The number of ways that no Chinese boos are next to each other!! M The number of ways that all Chinese boos are next to each other ( +!! M 76 The number of ways of all the English boos next to each other is the greatest. Vincent s claim is agreed. A ( ( (a ABC 08 M Join AC and AD ACB 6 M ACD Consider ABC, by the cosine formula, AC AB + BC ( AB( BC cos ABC + (( cos08 M AC Area of pentagon ABCDE ((sin08 + ((8.006sin 7 M.0 (cor. to sig. fig. A ( (b F is the mid-point of AB. AB DF M Consider the acute-angled triangle CPD, by the sine formula, ( CD PD sin CPD sin PCD sin PCD sin CPD PD M When P moves from F to A, PD increases and PCD decreases. sin CPD decreases, i.e. CPD decreases as P moves from F to A. A (. (a (i Let (, q be the coordinates of Q. H is the orthocentre of OPQ. QH OP q M q The coordinates of Q are (,. A (ii Let x + y + x + y + 0 be the equation of the circle, where, and are real constants. M By substituting (0, 0, (, and (, into the equation, we have ( By solving, we have 7, and 0. M The equation of the circle is x + y 7x + y 0.A ( (b (i x + y 7x + y 0 By solving, we have x + y m x + ( m x 7x + ( m x 0 x ( m + x + m + m 0...(* M L intersects the circle OPQ at two distinct points. Δ of (* > 0, i.e. [ ( m + ] (( m + m > 0 M m + 6m + 6 > 0 ( m + ( m 8 < m < 8 A (ii Let M(x, y be the mid-point of AB. (6 + x M When x, y 6. The coordinates of M are (,. PQ 7, PH ( + 8, QH ( + (, PM ( + ( and QM ( + ( 6 M Pearson Education Asia Limited 0
5 Consider PHQ, by the cosine formula, ( cos PHQ ( 8( PHQ > 0 Consider PMQ, by the cosine formula, ( + ( 6 7 cos PMQ ( ( 6 0 PMQ > 0 PHQ + PMQ > 80 P, H, Q and the mid-point of AB are not concyclic. M + A A (8 Pearson Education Asia Limited 0
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