ACJC_Solutions_Paper_2

1 Anglo-Chinese Junior College H2 Mathematics 9740 Qn Paper 2 Solution 1 (i) (ii) 2.5 3 4 3 0.5 7 2 0.5 5 , , , 2.5 4 3 0.5 3 7 0 1.5 5 0 ,: i j k ---(1) ---(2) ---(3) (1) (2), (3) 2 k kk kk Equating components k k k Solving Equations k                                     . 1 2. ,            2, 2, 2.5 (2) 2 7 2 3 2 8 2 2 2 6 7,8,6 r Intersection point is When k           2 Let  y f x         1sin 2 1 2 2 2 2 22 2 22 2 2 2 2 2 2 2 ln sin 2 diff. w.r.t. 12 14 1 4 2 diff. w.r.t. 4 1 4 2 14 4 1 4 2 1 4 1 4 4 2 2 1 4 4 4 (shown) xye yx x dy y dx x dyxy dx x x dy d y dy xdx dx dxx dy d y dyx x xdx dx dx d y dyx x y dxdx d y dyx x y dxdx                                3 2 2 2 3 2 2 32 2 32 diff. w.r.t. 1 4 8 4 4 4 1 4 12 8 x d y d y d y dy dyx x x dx dxdx dx dx d y d y dyxx dxdx dx         

2         01 ' 0 2 '' 0 4 ''' 0 16 f f f f       23 81 2 2 ... 3f x x x x     1sin 2 6 1 23 6 sin 2 6 12 2 1 4 1 1 8 11 2 2 4 4 3 4 21 3 xee x x x e                             3i 1 sin diff. w.r.t. cos xu x dx udu           2 2 2 2 2 21 2 2 1 sin 1 sin cos 2 2sin 1 2sin sin cos 1 sin cos cos cos 2 1 2 sin 2 1 42 2sin cos 1 42 1 12 sin 122 x x dx u u u du u u u u du u u du u du u du u uc uu uc x x x x c                               

3 ii a) b)   11 92 00 11 29 00 12 10 2 21 0 Area =2 2 2 =2 2 2 112 2 2 sin 12 2 2 10 2 122 10 4 19 52 R x x x dx x x dx x x x dx x xdx x x x xx x x x                                 211 229 00 Vol. of revolution formed when R is rotated completely about -axis 2 2 2 14.995 x x x x dx x x dx         4 (i) Method 1: 11 : 1 2 03 rline l                    Since (1,1,0) lies on 2p , 11 11 01 11 2 a a a                       Method 2: Since direction vector of line l is perpendicular to normal of 2p ,

4 4(ii) 11 20 31 1 2 3 0 2 a a a                        Method 3: 1 1 1 12 1 1 1 11 2 a a a a a                               Method 1: Normal method 1 1 : 1 2 1 01 : 0 1 11 01 0 1 2 11 11 1 3 r r p line AN                                                   0 1 1 110 1 1 331 1 4 ON                                 