- Since the value of
*f*(*x*,*y*) is unchanged when we swap*x*with*y*,2Also*f*(*x*+*y*) d*ydx*=*f*(*x*+*y*) d*ydx*.*f*(*x*+*y*) d*y*=*f*(*z*) d*z*=*f*(*z*) d*z**f*(*z*) =*f*(1 +*z*) for all*z*. Since*f*(*z*) d*z*= 1999, we conclude that*f*(*x*+*y*) d*ydx*= 1999/2. - For
a = 1,
b = 0 and
*x*= 1, we have*f*(1)*f*(0) =*f*(1). Therefore*f*(0) = 1. By differentiationaholds for all*f'*(a*x*)*f*(b*x*) + b*f*(a*x*)*f'*(b*x*) =*f'*(*x*)*x*, and for all a,b satisfying a^{2}+ b^{2}= 1. Hence (a + b)*f'*(0) =*f'*(0) holds. By taking a = b = 1/, we wee that*f'*(0) = 0. Set*c*=*f''*(0). By Taylor's theorem,*f*(*y*) = 1 +*cy*^{2}/2 + e(*y*^{2}), where lim_{y - > 0}e(*y*^{2})/*y*^{2}= 0. By taking a = b = 1/ again, we see that*f*(*x*/)^{2}=*f*(*x*) for all*x*. By repetition, for every positive integer*m*,*f*(*x*) = (*f*(2^{-m/2}*x*))^{2m}.*x*, and d > 0. There is a positive integer*N*such that for all*m**N*, 2^{-m}(|*c*| + d)*x*^{2}< 1, and(1 + 2Now let^{-m - 1}(*c*- d)*x*^{2})^{2m}*f*(*x*)(1 + 2^{-m - 1}(*c*+ d)*x*^{2})^{2m}.*m*- > . We obtain*e*^{(c - d)x2/2}*f*(*x*)*e*^{(c + d)/x2}.*f*(*x*) =*e*^{cx2/2}. Using the condition*f*(1) = 2, we conclude that*f*(*x*) = 2^{x2}. - Note that any eigenvalue of
*A*_{n}has absolute value at most*M*, because the sum of the absolute values of the entries in any row of*A*_{n}is at most*M*. We may assume that*M*> 1. By considering the characteristic polynomial, we see that the product of nonzero eigenvalues of*A*_{n}is a nonzero integer. Write*d*=*e*_{n}(d). Then we have*M*^{n}d^{d}> 1. This can be written as*e*_{n}(d)/*n*< ln(*M*)(ln(1/d)). - The points inside the box which are distance at least 1 from
all of the sides form a rectangular box with sides 1,2,3, which has
volume 6. The volume of the original box is 60. The points outside
the box which are distance at most 1 from one of the sides have volume
3 X 4 + 3 X 4 + 3 X 5 + 3 X 5 + 4 X 5 + 4 X 5 = 94

plus the points at the corners, which form eight 1/8th spheres of radius 1, plus the points which form 12 1/4th cylinders whose heights are 3,4,5. It follows that the volume required is

60 - 6 + 94 + 4p/3 + 12p = 148 + 40p/3. - By differentiating
*f*(*f*(*x*)) =*x*, we obtain*f'*(*f*(*x*))*f'*(*x*) = 1. Since*f*is continuous,*f'*(*x*) can never cross zero. This means that either*f'*(*x*) > 0 for all*x*of*f'*(*x*) < 0 for all*x*. If*f'*(*x*) > 0 for all*x*, then*x*>*y*implies*f*(*x*) >*f*(*y*), and we get a contradiction by considering*f*(*f*(*a*)) =*a*. We deduce that*f*is monotonically decreasing, and since*f*is bounded below by 0, we see that lim_{x - > }*f*(*x*) exists and is some nonnegative number, which we shall call*L*. If*L*> 0, then we obtain a contradiction by considering*f*(*f*(*L*/2)) =*L*/2. The result follows.**Remark**The condition*f*(*a*)*a*is required, otherwise*f*(*x*) =*x*would be a solution. - (i)
- Obviously
*n*> 4. Next,*n*5 because 4 divides 3 + 5. Also*n*6 because 3 divides 6 and*n*7 because 7 divides 3 + 4. Finally*n*8 since 4 divides 8, and*n*9 since 3 divides 9. On the other hand*n*= 10 because 3 does not divide 4, 10 and 14. Furthermore 4 does not divide 3, 10 and 13, and 10 does not divide 3, 4 and 7. - (ii)
- Suppose
{3, 4, 10,
*m*} is contained in a set which has property**ND**. Then 3 should not divide*m*, so*m*is not of the form 3*k*. Also 3 should not divide 10 +*m*, so*m*is not of the form 3*k*+ 2. Furthermore 33 should not divide*m*+ 4 + 10, so*m*is not of the form 3*k*+ 1. Here*k*denotes some integer. If*s*has property**ND**and contains 3,4 10 and*m*, then*m*cannot be of the form 3*k*, 3*k*+ 1, 3*k*+ 2. This is impossible and the statement is proven.