2018 H3_Solution

Suggested Solution Remarks 1i             1 1 1F0 1 11 1 1 1 1 1 11 ...2 2 3 1 11 1 n n r n r rr rr nn n                   As n , 1 01n  and so  F 0 1n  . 1ii (a)                 1 1 1 2 2F 11 11 1 1 1 1 11 ...2 2 3 1 2 2 2 2 2 2 ...1 1 1 2 1 1 11 1212 11 123 11 n n r x r r rx rx nn x x x nx nx n nx n nx                                    1ii (b) Since  1lim 0 1n n  and   2, if 02lim 1 0, if 0n x nx x   ,   1, if 0lim F 3, if 0 nn xx x   . There’s a need to distinguish the special case when 0x . 2i Let u a x .         0 0 0 0 f d f d f d f d a a a a a x x u u u u x x        2ii Since f is symmetrical about 1 2xa ,    ff x a x for 0 xa . -------------------- (*) Since f is continuous and x is also continuous , the function given by  fxx is also continuous on  0,a .                             00 0 00 00 00 f d f d by i f d by * f d f d 2 f d f d f d f d 2 aa a aa aa aa x x x a x a x x a x x x a x x x x x x x x a x x ax x x x x            

Alternative Since f is symmetrical about 1 2xa ,    11ff 22x a x a    for 11 22a x a   . Let 1 2v x a . Note that        g : f f g 22 aav v v v v v         , which shows that g is an odd function on ,22 aa .                 2 0 2 22 22 2 2 0 f d f d 22 f d f d2 2 2 0 f d since g is odd22 fd2 a a a aa aa a a a aax x x v v v a a av v v v v aa vv a xx                    2iii Since sin x and 2 1 1 cos x are continuous on  0, π , so is their product   2 sinh: 1 cos xx x  . We can also show that h is symmetrical about π 2x :        22 sin π sinh πh 1 cos π 1 cos x xxx xx       Applying the result in (ii):         ππ π1 22 000 2 11 sin π sin πd d tan cos221 cos 1 cos π π π π πtan 1 tan 12 2 4 4 4 x x x x x xxx                  3 A triangle with sides of lengths p, q and r exists iff p q r , q r p and r p q . 3i WLOG, let 0c b a   . Note that 1 1 1 c b ac b a c b a       . It suffices to show that 1 1 1 a b c a b c   . Indeed, 1 1 1 1 1 1 a b a b a b c a b c c c c           .

3ii WLOG, let 0c b a   . Note that c b a c b a     . It suffices to show that a b c . Indeed, we have   2 2a b a b ab a b c       which implies that a b c . 3iii By (ii), if a triangle with sides of lengths  a b c a ,  b c a b and  c a b c exists, then so does a triangle with side of lengths  a b c a ,  b c a b and  c a b c . By symmetry, it suffices to show that      a b c a b c a b c a b c        .              2 2 2 2 2 2 22 2 0