- Set
*r*=*x*^{2}+*y*^{2}. Then*f*(*x*,*y*) = ln(1 -*r*) - 1/(2*r*- (*x*+*y*)^{2}), so for given*r*, we see that*f*is maximized when*x*+*y*= 0. Therefore we need to maximize ln(1 -*r*) - 1/(2*r*) where 0 <*r*< 1. The derivative of this function is1/(which is positive when*r*- 1) + 1/(2*r*^{2}) = (2*r*^{2}+*r*- 1)/(2(*r*- 1)*r*^{2})*r*< 1/2, 0 when r=1/2, and negative when*r*> 1/2. It follows that the maximum value of this function occurs when*r*= 1/2 and we deduce that*M*= - 1 - ln 2. - We cut the cone along
*PV*and then open it out flat, so in the picture below*P*and*P*_{1}are the same point. We want to find*Q*on*VP*_{1}so that the length of*MQP*is minimal. To do this we reflect in*VP*_{1}so*P*_{2}is the image of*P*under this reflection, and then*MQP*_{2}will be a straight line and the problem is to find the length of*MP*_{2}.Since the radius of the base of the cone is 1, we see that the length from

*P*to*P*_{1}along the circular arc is 2p, hence the angle*PVP*_{1}is p/3 because*VP*= 6. We deduce that*PVP*_{2}= 2p/3, and since*VM*= 3 and*VP*_{2}=*VP*= 6, we conclude that*MP*_{2}= = 3.

- We calculate the volume of the region which is in the first octant
and above
{(
*x*,*y*, 0) |*x*__>__*y*}; this is 1/16 of the required volume. The volume is above*R*, where*R*is the region in the*xy*-plane and bounded by*y*= 0,*y*=*x*and*y*= , and below*z*= . This volume is*dzdydx*+*dzdydx*= d *ydx*+ d*ydx*= *x*d*x*+ (1 -*x*^{2}) d*x*= [- (1 - *x*^{2})^{3/2}/3]_{0}^{1/}+ [*x*-*x*^{3}/3]_{1/}^{1}= 1/3 - 1/(6) + 2/3 + 1/(6) - 1/ = 1 - 1/.

Therefore the required volume is 16 - 8. - We shall prove that
*AB*=*BC*. Using the cosine rule applied to the triangle*ABC*, we see that*BC*^{2}=*AB*^{2}+*AC*^{2}- 2(*AB*)(*AC*)cos 70. Therefore we need to prove*AC*= 2*AB*cos 70. By the sine rule applied to the triangle*APC*, we find that*AP*= 2sin 50, so we need to prove = 2(1 + 2sin 50)cos 70. However sin(50 + 70) + sin(50 - 70) = 2sin 50cos 70, sin 120 = /2 and sin(50 - 70) = - cos 70. The result follows.

- Since
S1/
*a*_{n}is a convergent series of positive terms, we see that given*M*> 0, there are only finitely many positive integers*n*such that*a*_{n}<*M*. Also rearranging a series with positive terms does not affect its convergence, hence we may assume that {*a*_{n}} is a monotonic increasing sequence. Then*b*_{2n + 1}__>__*b*_{2n}__>__*a*_{n}/2, so the terms of the sequence {1/*b*_{n}} are at most the corresponding terms of the sequence2/Since S1/*a*_{1}, 2/*a*_{1}, 2/*a*_{2}, 2/*a*_{2}, 2/*a*_{3}, 2/*a*_{3},...*a*_{n}is convergent, so is the sum of the above sequence and the result now follows from the comparison test for positive term series. - We shall assume the theory of writing permutations as a product of
disjoint cycles, though this is not necessary. Rule 1 corresponds to
the permutation (1 2 3 4 5 6 7 8 9 10) and Rule 2 corresponds to the
permutation (2 6)(3 4)(5 9)(7 8). Since
(2 6)(3 4)(5 9)(7 8)(1 2 3 4 5 6 7 8 9 10) = (1 6 8 5 2 4 9 10)(where we have written mappings on the left) has order 8, we see the position of the cats repeats once every 16 jumps. Now 10p.m. occurs after 900 jumps, hence the cats are in the same position then as after 4 jumps and we conclude that the white cat is on post 8 at 10p.m.