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 ( )x must be a self-inverse function     1ff xx  Hence, 1f ( ) 1 xax x    6(i) Given that a  3b = 2a  c a  3b  2a  c = 0 a  (3b  2c) = 0 Either (3b  2c) = 0 or a = 0 or (3b  2c) is parallel to a As a0 , (3b  2c) is parallel to a.  (3b  2c)=  a for some  . (shown)

7 (ii) Given that a and c are unit vectors,                       22 2 22 2 22 2 22 2 2 1, 4 and angle between and 60 From (i), 3 2 for some 3 2 3 2 32 3 2 3 2 1 9 12 4 9 12 2 4 9 4 12 2 4 1 124 124 2 31                                      a c b b c b c a b c b c a a b c a b c b c b b b c c c bc 7(i)   22 22 2 2 2 2 41 ,2 2 8 ... 1 xy x xy x y x xy      Differentiating (1) with respect to x,       2 2 2 dd4 16 2 2 dd d 2 16 4 2d d2 shown ... 2d 2 16 yyx y x y x yxx y xy y x x yx y x y x xy y            (ii)   22 When 1, From 1 , 2 8 1 11 or 33 1At point 1, ,3 12d 17 9 .11d 54 2 1633 x yy y y x                     

8         1At point 1, , 3 12d 17 9 11d 54 2 1633 1 1 17Tangent of equation at point 1, : 1 ... 13 3 54 1 1 17Tangent of equation at point 1, : 1 ... 23 3 54 y x yx yx                                (1) (2), 2 17 17 113 54 54 2 17 13 27 1 17 17 1 1 1054 17 3 1 ,017 xx x x y N                  8(a) Given 1 1 2 2 , where tan , 1, 5 , 15,n n nu u A A is a cons t n u u            21 32 At 1, 2 15 2 5 5 At 2, 22 2 15 2 5 40 n u u A A A n u u A          

9 (ii)         1 2 3 2 1, 2 5 .....(1) 2, 4 2 15 .....(2) 3, 8 3 140 .......(3) Using GC, solving 1 , 2 and 3 , 15 , 5 , 52 n nu a bn c n u a b c n u a b c n u a b c abc                      (iii)               1 1 1 1 1 2 15 2 5 52 15 2 5 5 12 2 2 1 115 552 2 1 2 515 2 1 1 5 2 15 515 2 1 22 n r r n r r n n n r r r r n n n u r r nn n n n n nn                                   

10 9(i) 2 2 2 2 2 sin 2 , 2sin , 0 2 dd 2 2cos 2 4sin cosdd d 4sin cos d 2 2cos 2 4sin cos 2 2(2cos 1) 4sin cos 4(1 cos ) 4sin cos 4sin cos sin cot (shown) xy xy y x                                     Alternative, 22 sin 2 , 2sin , 0 2 dd 2 2cos 2 4sin cosdd d 4sin cos d 2 2cos 2 4sin cos 2(1 cos 2 ) xy xy y x                         2 2 4sin cos 2(2sin ) 4sin cos 4sin cos sin cot (shown)           