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 23tan y x     To show   , we can show 2  .       2 00 2 00 2 00 2 20 00 2 00 22 000 00 2 0 2t a n 211t a n 23 23 43 3 43 43 ta 4 n2 23 43 4 xx yy yx y yx yx xxy xy x                      

    00 0 0 00 0 0 2 00 22 0 00 0 00 2 2 0 0 00 2 0 23 23tan tan 11t a nt a n 22 3 23 23 22 43 4 tan yy y y xx x x yx x xy y xy xx xy x                                  Since   tan 2 tan 2      over the interval  0, π , we have 2   . Hence the angles made by the line segments OP and OP with the tangent T is the same.

Error! Reference source not found. Paper 1 6 Solution (a) 1 12 2 22 n n nn XX   Substitute 2n nnvX : 1 1 22 nnvv  Let 1 2 n nvA B   , where A, B are constants Substituting 1,ovv : 0 1 6 15 2 vA B vA B    Solving, 2, 4AB . 124n nv   12 2 1 2 22 nn n nn n Xv X    (b) (i) As nul  as n , 2 2 22 5 24 5 024 45 024 ( 024 2 5)( 1) ll l ll l ll l l ll l            5 OR 1ll   . (b) (ii) Using GC, When 2.01a , the sequence increases towards the limit 5l . When 1.99a , the sequence decreases towards the limit 1l . (b) (iii) 22 1 41 3 ( 2 ) 92 2( 2) 2( 2) nn n n nn uu uu uu      Note 2(2 ) 9 0nu   If 2nu  , 1120 , 20 2nn n uuu     

If 2nu  , 1120 , 20 2nn n uuu      When 2.01 2a  , all terms in the sequence will be less th an –2, hence the se quence conv erges to 52   . When 1.99 2a  , all terms in the sequence will be greater than than –2, hence the sequence converges to 12   .

Error! Reference source not found. Paper 1 7 Solution (a) 11 2 12(, ,31) nn n n nn n xx k x x xkx k xn n          (b) 12(1 ) 0nn nxkx k x    Auxiliary equation: 2 (1 ) 0kk   2 2 1( 1 ) 2 1( ) 4 1 2 kkk kk        OR 1k  General solution: () n nx AB k , where A, B are constants When 1n , 1 8 ----- (1)xA B k When 2n , 2 2 23 ----- (2)xA B k  (2) – (1): 15 (1 )B kk  82 3 1 5 811 kA kk    11581 , 1 n nxk n k       (c) When 1k , 1 as as n n nnkx     When 01 k , 1 0 as 8 as 15 1 n n nnkx k        If 1k , we would expect the spread of the virus to escalate and affect the whole population. If 01 k , we would expect the spread of the virus to stop, and the total number of Omega variant cases in Singapore to approach approximately 158 1k  . (d) Any of the following or equivalent:  k is unlikely to be constant for an extended period of time in a virus outbreak  k may decrease if a lot of the population has gotten the virus (eg. if there is herd immunity)

 nx cannot grow indefinitely due to population size limits  If k is not an integer, nx takes non-integer values which is not feasible in real life. (e) Consider 1 for some , 2nnxx n n   1k : 112 fo a 3 ll ,rmm m mxx x x m m    For 2n : 21 15xx For 3n : 1 2 2 23 1 1 15 nn n n nn xx x x xx xx        1 for all , 15 ,2 nnnnx x   (constant indep of n) Therefore nx is an arithmetic progression with a common difference of 15. (f) When 5n , 4k  : 1 4 5 35 ( 4 ) , 5 323 1283 n nxn x x     When 6n , 1k and nx increases as an arithmetic progression with common difference 1283 323 960 . 1283 ( 5)960, 5n nxn    1012 0083 ( 0 5 9.0802 14.08 ) 02 59 6 0 n n n     Alert Orange will be triggered on the 15th week of the outbreak.