EJC 2023 JC1 MYE 9649 (Solutions)

Error! Reference source not found. Paper 1 1 Solution Let nP be the proposition 122n n  where ,3nn  . When n = 3: 31 2 21 9 6 3    3P is true as 16 9 . Assume kP is true for some 3,kk  : 122k k  To prove 1kP is true:    21 2 22 2 2kk k       222212 1 11 0 2 a 2 3s kk k k kk k            22 12k k  , as 22 (1 )2 kk  Since 3P is true and kP is true  1kP is true, by Mathematical Induction, nP is true for all positive integers 3n .

2 Solution (a)      2 2222 22 99 cos sin cos cos sin cos cos sin cos 9 rr rr r             Using ,s ncos i yxr r   ,  22 22 9 22 9 xy x xx y y     (b) 1 sins cos 1 1cosinry x rr        22 22 2 2 22 1 1 9 222 2 1 9 8 22 xx x x xxx xx x x             Hence the points of intersections are  22 , 1 22 and  22 , 1 22 Otherwise: Using    22 cos sin cos 9rr r    and csin os 1rr   ,  2 2 2cos 1 9 cos 2 xrr       Hence the points of intersections are  22 , 1 22 and  22 , 1 22

Error! Reference source not found. Paper 1 3 Solution (a) 2f( ) 1x x x  2 2( 1) 3 2f x xx   3 3( 2) 4 3f x xx   4 4( 3) 5 4f x xx   Conjecture: (1 )f() (1 ) n nx nx nx n   (b) Let nP be the proposition (1 )f() (1 ) n nx nx nx n   where n  . When n = 1: LHS f ( ) 21RH 2 1S x x xx    1P is true. Assume kP is true for some k  : (1 )f() (1 ) k kx kx kx k   To prove 1kP is true: 1 1 (2 )(1 ):f ( ) (1 ) k k kx kPx kx k      1 (1 )f( ) f (1 ) 2( 1) 2 ( 1) (1 ) (2 )(1 ) (1 ) k kx kx kx k kx k k x k kx k kx k kx k             Since 1P is true and kP is true  1kP is true, by Mathematical Induction, nP is true for all n  . (c) For f()n x to be defined, 1x n n  .  Largest subset of the reals for f()n x to be defined for all n  is 1:, nxn nx    .

4 Solution (a) 2 sin 3r  is undefined for sin3 0  Then the range of values are or or or33 3 3 2ππ π 2ππ 0 π       . Hence the equation of the asymptotes are π 2π0, , a n d 33, π  . (b)

Error! Reference source not found. Paper 1 5 Solution (a) Since the coefficients of 2x and 2y must have different sign, 0A . (b) Since the equation of H is 22 24 3 4 0Ax y y   , the centre is 0, 3 and by symmetry, the other focus, O is 0, 2 3 . Comparing  2 2131 2yx A    to a standard equation of a hyperbola,  2 2 22 1yh x ab   , we have  2 2 2 2 13 1 1 1 b AeA a        (c)  222224 3 4 0 3 1 2 xyy x y      For curve H, the derivative will be  0 0 d d 23 xy x y   The angle that T makes with the horizontal,  1 0 0 tan 23 x y       The angle that OP makes with the horizontal, 1 0 0 tan y x     The angle that OP makes with the horizontal, 1 0 0