A Level 2015 P2 Solutions for 9780_CWY_2018

2015 H2 Maths Paper 2 (edited for new H2 Syllabus 9758) Qn Suggested Solutions Section A (Pure Math) 1(i) Max height is when d 0d h t  . 11 16 010 2 1 162 32 m h h h    (ii) Separating variables and integrating w.r.t. t,  1 21 1dd 10116 2 2 ht h     112 16 22 10h t c    114 16 2 10h t c    When t = 0, h = 0: 16c 140 16 1602th   When the tree is 16m, 140 16 (16) 1602 46.9 years t t     2(i) Let the angle between L and the x-axis be . cos  = 21 3 . 0 60 | 49 1                   = 2 7   = 73.4 Acute angle between L and the x-axis = 73.4 (ii) 12 : 2 3 , 46 L                      r . Let B be a point on L such that 33PB  In general, 1 2 1 2 1 2 1 2 f '( ) d f '( )[ f ( )] d f ( ) [ f ( )] 2[ f ( )] x x x a x x ax ax C a x C         In particular, 1()( ) d ( )( 1) n n a bxa bx x C bn     P B B’ 33 L N

and let N be the point on L closest to P. 1 2 2 2 3 5 for some 4 6 6 12 73 26 PB OB OP                                            2 2 2( 1 2 ) ( 7 3 ) (2 6 )PB           = 254 70 49 254 70 49 33    Square both sides and solve for  , 2 2 2 54 70 49 33 0 49 70 21 0 7 10 3 0 (7 3)( 1) 0 31 or 7                    1 2 3 2 1 3 = 1 4 6 10 OB                                or 1 2 13 312 3 = 5774 6 46 OB                                 The points are (3, 1, −10) and 1 (13, 5, 46)7  . Hence, as N is the midpoint of BB’, 3 13 17 1 1 1 1 5 = 12 7 7 10 46 58 ON                                (iii) Normal to the new plane: 1 2 36 7 3 = 2 2 6 11                              36 1 36 . 2 2 2 4 11 4 11                                 r Cartesian equation: 36 2 11 4x y z   .

3(ai) 2 1f ( ) 1x x  , x > 1 Any horizontal line y = k intersects the graph of f ( )yx at most once.  f is a one-one function and so its inverse exists. (ii) 2 1f ( ) , 11xx x Let 2 1 1y x  Make x the subject: 2(1 ) 1yx  2 2 1 1 1 y yx yx y yx y    As 1x , 11 1yx yy    1 1f ( ) 1 , , 0x x x x      1 ( ,0).ffDR    (b) Let 2 2g( ) 1 xyx x   . 2 2 (1 ) 2 (2 ) 0 (1) y x x yx x y         The range of g corresponds to the values of y for which (1) has real root x i.e. D  0. 2 2 1 4 (2 ) 0 1 8 4 0 yy yy       Solve 21 8 4 0yy   : 8 64 4(4) 1132(4) 2y    −6 −4 −2 2 4 6 −4 −3 −2 −1 1 2 3 4 x y x = 1 y = 0

33R : 1 or 1 22 g y y y       4(bi)     2 2 4 8 3 2 1 2 3 2 2 3 2 1 AB r r r r A r B r         Let 33: 2 2 1 122r B