VJC 2023 JC1 Promos 9649 (Solutions)

1(a) 3 3 2 2 1 10 (1 ) ( 1 ) 0 Since is non-real, it satisfies 1 0. w w ww w ww w        2 1.ww   - GC is not allowed and given 2 marks, the intent is not to solve for w. (b)    22 22 2 2 2 32 (1 2 3 )(1 2 3 ) 3 LHS: 1 2( ) 1 2( ) 12 ( 1 ) 12 ( 1 ) 11 () 1 1( 1 )1 3 ww w w ww w w w w ww ww ww w                     Alternative  22 3 4 2 3 23 4 23 4 2 LHS = (1 2 3 ) (2 4 6 ) (3 6 9 ) 1 5 11 13 6 1 5 11 13 6 1 14 11 11 14 11( 1) 3 RHS ww w w w ww w ww ww ww w w w w ww                       - Make good use of result in (a) to simplify. - Note the efficient way to evaluate 4.w 2(a) 22 22 22 2 22 2 2 distance from ( , ) to (0,2 ) = ( 2 ) distance from ( , ) to line = (2 ) (2 )( ) (2 )( ) () () (2 ) (2 ) (2 ) ( s h o w n ) xy d x y d xy x d xy d x d xy d x d yd x d x xd x xd x dxd yd d dx                  From definition, d > x. |x – d| = d – x

(b)  3 3 23 33 2 32 3 22 3343 22 Volume = 2 d 2 d (2 )2 d 22 d( 2 ) d () ( 2 ) ( 2 ) d2 (2 ) (2 4 2+ )922 2 3 4 d d d d d d dd dd d d dd dd dd xy y xy y dy d yyd dy y y y dy d yd y yd yd y d dy d y ddd d d dd                                     34 4 3 3 3 2 43 8 cu. units3 dd ddd d      Alternative (Disc Method) 2( 2) ( 2) 2 ( 2)yd d dx yd d dx      Required volume = volume generated by upper curve Uy – volume generated by lower curve Ly  22 22 00 222 0 2 0 2 0 3/2 2 0 3 3 dd 2( 2 ) 2( 2 ) d 8( 2 ) d 182 ( 2 ) d2 (2 )4 3/2 8 03 8 cu. units3 dd uL d d d d yx yx dd d x dd d xx dd d x x dd d d x xd dd x d d                             Alternative: Expand 2 (2 )yy d ; the method shown is simpler to evaluate limits. By shell method, 2 2 2 2 (2 ) (2 ) (2 )2 (2 ) 22 Ax y y yd d dx yddx d dy dx d       

3(a)            22 22 3 2 2 2 122 2 21 23 21 23 4646 2 21 23 2 0 (2 1) & (2 3) 0 for 021 23 kk k kkk kk k k kkkk kk kk kkk               22 2 21 23 kk kk   (b) Let Pn be the proposition that 2 , .21 n n nun n    . Consider P1: LHS 1 2 given3u RHS  2(1) 2 21 1 3  Hence 1 2 3u  1P i s t r u e Assume that Pk is true for some positive integer k, 2 21 k k ku k   . To show Pk+1 is true, 1 22 23 k k ku k    LHS of Pk+1 1 1 22 23 22 2 23 21 22 22 2 22 since 23 23 21 23 22 23 k k k k k u k uk kk kk kk k k kk k k k k                     1 1 22 23 k k ku k