Paper 1

1 2018 H2 Mathematics Paper 1 (9758/01) – Suggested Solutions 1(i) Use Quotient Rule,   2 2 1 ln 1 lnd d d1 1 lnd yx x xxy x xx y xxx     Alternatively,   2 2 1 ln By Product Rule: d 1 1 1 lnd d1 1 lnd yx x y xx x x x y xxx      (ii) From (i), 22 22 2 d 1 ln 1 ln d ln 1 d d y x x x x x x xy x x x     Integrating both sides with respect to x: ee 2211 e 1 1 e 1 e ln 1 ddd d 1 1 1 ln 1 ln1 1 ln e 1e 21 e xy xxx x x yx yx x x          

2 (Otherwise method– Not advised) Using integration by parts, 2211 1 21 1 ln 1 1d ln d 1 1 1ln ln1 d1 11 11 1 21 e ee e e x x x xx x x exex ex ee e                                  2(i) Given  3 .... 1 and 2 7 ....(2) intersect,y y xx   Substitute (1) into (2) :    2 3 27 2 7 3 0 2 1 3 0 1 or 32 xx xx xx xx            2 3 27 2 7 3 0 2 1 3 0 1 or 32 xx xx xx xx          The x-coordinates of A and B are 1 or 32xx 2 d1ln d du 1 1 d vux xx vx x x   

3 (ii) Volume required          33 2 120.5 0.5 2 3 2 0.5 3 2 20.5 3 3 2 3 0.5 d 32 7 d 92 7 d 27 9 23 125 π units6 y dx y x xx x xx x x x                            3(i)               2 2 22 3 2 2 3 3 d 2 6 .... 1d Given ..... 2 Differentiate (2) with respect to : d du 2 ..... 3dd Substitute 3 into 1 : du 2 2 6d du 2 2 6d du 6 ..... 4d 6f ( ) yxy x y ux x y x uxxx x x ux uxx x ux uxx xx x x                0.5,6  3,1 y x

4 (ii) To solve (1), we solve (4) Integrate (4) with respect to x : 2 22 2 2 3 , where is an arbitrary constant 3 , where is an arbitrary constant 3 When 1, 2 , 1 3 u C Cx y CCxx y Cx x y C yx          4(i)          2 2 2 2 2 2 2 2 2 2 12 3 2 , 2 or 22 3 2 12 3 2 , 2 2 2 3 2 2 1For 2 or , 2 2 3 2 2 2 4 4 0 4 4 4 2 4 1322 1For 2 , 2 2 3 2 2 2 3 2 2 2 2 0 0 or 1 1 3 or 1 3 or 1 or 0 x x x x xx x x x x x x xx x x x xx x x x x x x x x xx xx x                                                      

5 (ii) 2 Solving 2 3 2 2 ,x x x    using the graph above: 1 3 1 or 0 1 3xx       y 2 = 2 3 2y x x   1 3,0  1 3,0 2yx  1,0  0.75,3.125  0, 2 x

6 5 f ( ) xax xb   2ff( ) = g( ) = , f ( )x x x x x                      2 2 2 2 2 2 22 ( ) ( ) ( ) 1 11 0 01 xa axb xxa bxb x a a x b xx a b x b x a a x b x x a bx x b x b a bx x a x ax bx bx ab b x x x ax a bx ab b x x x a abx b                                         By comparing coefficients : 2 : 1 0 1 bx b     Thus, 1b . For 2ff( ) = g( ) = , f ( )x x x x x  , f