HCI Paper D Sec 3 Higher Math Solutions

1 Paper D Sec 3 Higher Paper Solutions for students Qn Solutions 1(ai)          44 4 2 2 4 2 2 22 2 2 2 2 222 2 2 2 2 4 4 4 4 24 22 2 2 2 2 xy x x y y x y x y x y x y xy x xy y x xy y                 1(aii)        86 4 42 2 2 2 2 52 5 4 2 25 2 25 2 2 2 25 2 25 2 2 2 533 733 13 41 733                  Hence, the distinct prime factors are 13, 41 and 733. 2 Marks if students leave the answer as 533 and 733. 1(bi) 1 cos 1x   1(bii) 66 66 66 6 6 6 6 66 66 1 tan sec 1 sec tan 1From (i), 0 cos 1 sec 1 cos tan 0 sec tan 1 It can only be possible that sec 1 and ta n 0. i.e. cos 1 and tan 0 cos 1 and tan 0 360 , 180 ,0 ,180 ,360 xx xx xx x x x x xx xx xx x                         Alternatively,     66 2 33 2 2 3 3 2 32 2 2 2 1 tan sec Let tan . 1 1 (apply tan 1 sec ) 1 3 3 1 2 3 3 0 3 2 3 3 0 0 or 2 3 3 0 (no solution) tan 0 360 , 180 ,0 ,180 ,360 xx yx y y x x y y y y y y y y y y y y y x x                               

2 2(a)           22 24 8 4 3 8 4 3 2 6 8 2 12 8 2 12 2 6 6 2 6 2 2 6 6 2 6 2 24                 Alternatively, students may try squaring the expression first,            2 2 24 8 4 3 8 4 3 24 8 4 3 2 8 4 3 8 4 3 8 4 3 24 16 2 64 16 3 24 16 8 24 24 8 4 3 8 4 3 24                     2(bi) Transformation I: scaling of the graph along the x-axis by factor of 2 (stretching). Transformation II: scaling of the graph along the y-axis by factor of 1 3 (compression). Transformation III: translation of the graph to the negative direction of x- axis by 1 unit. Alternatively, Transformation I: translation of the graph to the negative direction of x-axis by 1 2 unit. Transformation II: scaling of the graph along the y-axis by factor of 1 3 (compression). Transformation III: scaling/stretching of the graph along the x-axis by factor of 2. Alternatively, Transformation I: translation of the graph to the negative direction of x-axis by 1 unit. Transformation II: translation of the graph to the positive direction of y-axis by 3 unit. Transformation III: scaling of the graph along the y-axis by factor of 1 12 (compression). 2(bii) The translation of the graph does not change the area enclosed by the graph and x-axis (1 mark) but the scaling does. The area will increase and decrease

3 proportionately with stretching and compression respectively.(1 mark) Hence, the area enclosed = 24 1 82 units3 3 9   3(a)     22 4 2 5 0 1 4 1y x y y x        The parabola has vertex (1, 1). Since the x-coordinate of the point is always equal to its distance from the fixed point (focus point) Q, we can infer that the directrix is x = 0 . Hence, the focus point Q has coordinates (2, 1). 3(bi)                                 2 22 2 1 1 or 212 2 1012 2 1 2 012 2 012 12 012 Multiply both sides with 1 2 , 1 2 1 2 0 2 or 1 1 or 2 x xxx x xx x x x xx xx xx xx xx xx x x x x x x x                        3(bii)          2 2e 1 e 1 1 2e Multiply both numerator and denominator by e , 2e 1 1 e 2 Let e , we have 2 1 12 From (i), 2 or 1 1 or 2 i.e. e 2 or 1 e 1 or e 2 Since x xx x x xx x x x x e y y yy y y y                   e is always positive, 0 e 1 or e 2 0 or ln 2 x xx xx       

4 4(a)    2 2 4 2 322 2 4 24Let . 2 3 2 2 3 0 1 3 0 241 or 3 (N.A. since 0) 2 24 12 24 12 2 4 2 6 xx xx xy x y y yy yy xyy x x x x x xx x                        Note: Minus 1 mark if student did not reject 22 7 while using other methods such as squaring the equation. 4(b)            22 22 sin cos sin cos tan cot cos sin cos sin sin cos sin cos cos cos sin sin sin cos sin cos sin cos sin cos sin cos sin cos sin cos sin cos sin cos sin cos 11 sin cos RHS x x x x xx x x x x x x xx x x x x x x xx x x x x xx xx xx xx x x x x xx             Alternatively,

5     22 22 2 2 2 2 sin cos sin cos tan cot sin cos cot sin cos tan cos sincos sin sin cos sin coscos sin cos sin cos sin sin cos sin cos 11 sin cos RHS x x x x xx x x x x x x xxxx xx xxxx xx x x x x xx xx                           Accept any other equivalent methods. 5(i)                                                     2 2 2 2 2 2 Let ( ) 2 6. Since 2 2 1 and 2 1 2 , from given conditions, by remainder theorem, 2 3 2 4 2 1 3 1 4 7 1 1 4 3 2 2 4 6 2 2 2 2 2 6 0 1 1 1 2 1 6 0 1 1 1 2 1 6 0 22 f x P x x x x x x x x x x x P P P P fP fP fP PP                                                                     2 2 2 2 6 0 By factor theorem, 2 , 1 , 1 and 2 are all factors of . x x x x fx          Hence   2 26P x x x   has both factors 2 2xx and 2 2xx . Alternatively,

6                          2 1 2 22 1 22 1 2 1 22 2 2 2 22 2 2 Let 2 ( ) 3 4. 26 2 ( ) 3 4 2 6 2 ( ) 2 2 ( ) 1 2 6 has factor 2. Similarly, let 2 ( ) 4. 26 2 ( ) 4 2 6 2 P x x x Q x x P x x x x x Q x x x x x x Q x x x x x Q x P x x x x x P x x x Q x x P x x x x x Q x x x x x x Q                                                             2 2 2 2 22 2 22 ( ) 2 2 ( ) 1 2 6 has factor 2. 2 6 has both factors 2 and 2 x x x x x Q x P x x x x x P x x x x x x x                      5(ii) From (i), since P(x) is a degree 4 polynomial and each of 2 2xx and 2 2xx has degree 2, and the leading term of   2 26P x x x   is x4,             2 2 2 2 2 2 22 2 2 42 2 6 2 2 2 2 2 6 2 2 6 4 2 2 P x x x x x x x P x x x x x x x x x x x x x x                           Note: Minus 1 mark if students never explain using the degree of polynomial/leading term. 6(a) 3 3 45 75 15 45 15 and 75 15 15 15 45 75 15 15 15 3 a b c cc ab cc ab cc ab cc ab              Alternatively,

7   Let 45 75 15 0 ln 45 ln 75 ln15 i.e. ln 45 ln 75 ln15 ln 45 ln 75, ln15 ln15 ln 45 ln 75 ln15 ln15 ln 45 75 ln15 3 a b c a b c k a b c cc ab cc ab              6(bi)   3 3 1fg 2 1 2 f and g are inverse functions. xxx      6(bii)                   1 2017 'fg ' 1 3 3 3 3 2 2 2 fg...fgfg g g f 21 2 1 0 When 1, 2 1 1 1 0. By factor theorem, 1 is a factor of 2 1. 1 2 2 1 0 1 or 2 2 1 0 2 4 2 1 0 xx xx xx xx xx x x x x x x x x x x                             No solution 1x   7(i) Gradient of OX =   01 101   Mid-point of OX: 11, 22  Equation of perpendicular bisector of OX: 1 1 1 2 1 2 1 yx yx        Gradient of OY = 07 107  

8 Mid-point of OY: 77, 22   Equation of perpendicular bisector of OY: 7 1 7 2 1 2 7 yx yx        Since the centre of the circle is the intersection of the two perpendicular bisectors of the chords, we have 1 7 yx yx    Solving the simultaneous linear equations,   3 and 4 Centre of the circle: 3, 4 xy  Radius of the circle = 223 4 5     22 22 equation of the circle: 3 4 25 or 6 8 0x y x y x y          7(ii) Greatest distance occurs when (2, 6), centre of the circle and the point on the circumference lie on a straight line. Hence, Greatest distance =     22 5 2 3 6 4 5 5      Note: Minus 1 mark if answer is left as 30 10 5 .

9 7(iii) From the sketch, one of the tangent lines ZP is parallel to y-axis. (only needed when