Tuyển tập đề thi Olympic tToán sinh viên quốc tế

Phudinhgioihan Diendantoanhoc.net
TUYỂN TẬP ĐỀ THI OLYMPIC
TOÁN SINH VIÊN QUỐC TẾ
International Mathematics Competition for
University Students
1994-2013
Mục lục
IMC 1994 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
IMC 1995 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
IMC 1996 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
IMC 1997 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
IMC 1997 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
IMC 1998 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
IMC 1998 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
IMC 1999 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
IMC 1999 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
IMC 2000 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
IMC 2000 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
IMC 2001 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
IMC 2001 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
IMC 2002 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
IMC 2002 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
IMC 2003 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
IMC 2003 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
IMC 2004 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
IMC 2004 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
IMC 2005 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
IMC 2005 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
IMC 2006 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
IMC 2006 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
IMC 2007 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
IMC 2007 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
IMC 2008 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
IMC 2008 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
IMC 2009 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
IMC 2009 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
IMC 2010 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
IMC 2010 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
IMC 2011 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
IMC 2011 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
IMC 2012 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
IMC 2012 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
IMC 2013 ngày 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
IMC 2013 ngày 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
1
International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1994
1PROBLEMS AND SOLUTIONS
First day — July 29, 1994
Problem 1. (13 points)
a) Let A be a n × n, n ≥ 2, symmetric, invertible matrix with real
positive elements. Show that zn ≤ n
2 − 2n, where zn is the number of zero
elements in A−1.
b) How many zero elements are there in the inverse of the n× n matrix
A =


1 1 1 1 . . . 1
1 2 2 2 . . . 2
1 2 1 1 . . . 1
1 2 1 2 . . . 2
. . . . . . . . . . . . . . . . . . . .
1 2 1 2 . . . . . .


?
Solution. Denote by aij and bij the elements of A and A
−1, respectively.
Then for k 6= m we have
n∑
i=0
akibim = 0 and from the positivity of aij we
conclude that at least one of {bim : i = 1, 2, . . . , n} is positive and at least
one is negative. Hence we have at least two non-zero elements in every
column of A−1. This proves part a). For part b) all bij are zero except
b1,1 = 2, bn,n = (−1)
n, bi,i+1 = bi+1,i = (−1)
i for i = 1, 2, . . . , n− 1.
Problem 2. (13 points)
Let f ∈ C1(a, b), lim
x→a+
f(x) = +∞, lim
x→b−
f(x) = −∞ and
f ′(x)+ f2(x) ≥ −1 for x ∈ (a, b). Prove that b−a ≥ pi and give an example
where b− a = pi.
Solution. From the inequality we get
d
dx
(arctg f(x) + x) =
f ′(x)
1 + f2(x)
+ 1 ≥ 0
for x ∈ (a, b). Thus arctg f(x)+x is non-decreasing in the interval and using
the limits we get
pi
2
+ a ≤ −
pi
2
+ b. Hence b− a ≥ pi. One has equality for
f(x) = cotg x, a = 0, b = pi.
Problem 3. (13 points)
2Given a set S of 2n − 1, n ∈ N, different irrational numbers. Prove
that there are n different elements x1, x2, . . . , xn ∈ S such that for all non-
negative rational numbers a1, a2, . . . , an with a1 + a2 + · · ·+ an > 0 we have
that a1x1 + a2x2 + · · ·+ anxn is an irrational number.
Solution. Let I be the set of irrational numbers, Q – the set of rational
numbers, Q+ = Q∩ [0,∞). We work by induction. For n = 1 the statement
is trivial. Let it be true for n − 1. We start to prove it for n. From the
induction argument there are n − 1 different elements x1, x2, . . . , xn−1 ∈ S
such that
(1)
a1x1 + a2x2 + · · · + an−1xn−1 ∈ I
for all a1, a2, . . . , an ∈ Q
+ with a1 + a2 + · · ·+ an−1 > 0.
Denote the other elements of S by xn, xn+1, . . . , x2n−1. Assume the state-
ment is not true for n. Then for k = 0, 1, . . . , n − 1 there are rk ∈ Q such
that
(2)
n−1∑
i=1
bikxi + ckxn+k = rk for some bik, ck ∈ Q
+,
n−1∑
i=1
bik + ck > 0.
Also
(3)
n−1∑
k=0
dkxn+k = R for some dk ∈ Q
+,
n−1∑
k=0
dk > 0, R ∈ Q.
If in (2) ck = 0 then (2) contradicts (1). Thus ck 6= 0 and without loss of
generality one may take ck = 1. In (2) also
n−1∑
i=1
bik > 0 in view of xn+k ∈ I.
Replacing (2) in (3) we get
n−1∑
k=0
dk
(
−
n−1∑
i=1
bikxi + rk
)
= R or
n−1∑
i=1
(
n−1∑
k=0
dkbik
)
xi ∈ Q,
which contradicts (1) because of the conditions on b′s and d′s.
Problem 4. (18 points)
Let α ∈ R \ {0} and suppose that F and G are linear maps (operators)
from Rn into Rn satisfying F ◦G−G ◦ F = αF .
a) Show that for all k ∈ N one has F k ◦G−G ◦ F k = αkF k.
b) Show that there exists k ≥ 1 such that F k = 0.
3Solution. For a) using the assumptions we have
F k ◦G−G ◦ F k =
k∑
i=1
(
F k−i+1 ◦G ◦ F i−1 − F k−i ◦G ◦ F i
)
=
=
k∑
i=1
F k−i ◦ (F ◦G−G ◦ F ) ◦ F i−1 =
=
k∑
i=1
F k−i ◦ αF ◦ F i−1 = αkF k.
b) Consider the linear operator L(F ) = F ◦G−G◦F acting over all n×n
matrices F . It may have at most n2 different eigenvalues. Assuming that
F k 6= 0 for every k we get that L has infinitely many different eigenvalues
αk in view of a) – a contradiction.
Problem 5. (18 points)
a) Let f ∈ C[0, b], g ∈ C(R) and let g be periodic with period b. Prove
that
∫ b
0
f(x)g(nx)dx has a limit as n →∞ and
lim
n→∞
∫ b
0
f(x)g(nx)dx =
1
b
∫ b
0
f(x)dx ·
∫ b
0
g(x)dx.
b) Find
lim
n→∞
∫ pi
0
sinx
1 + 3cos 2nx
dx.
Solution. Set ‖g‖1 =
∫ b
0
|g(x)|dx and
ω(f, t) = sup {|f(x)− f(y)| : x, y ∈ [0, b], |x− y| ≤ t} .
In view of the uniform continuity of f we have ω(f, t) → 0 as t → 0. Using
the periodicity of g we get
∫ b
0
f(x)g(nx)dx =
n∑
k=1
∫ bk/n
b(k−1)/n
f(x)g(nx)dx
=
n∑
k=1
f(bk/n)
∫ bk/n
b(k−1)/n
g(nx)dx +
n∑
k=1
∫ bk/n
b(k−1)/n
{f(x)− f(bk/n)}g(nx)dx
=
1
n
n∑
k=1
f(bk/n)
∫ b
0
g(x)dx + O(ω(f, b/n)‖g‖1)
4=
1
b
n∑
k=1
∫ bk/n
b(k−1)/n
f(x)dx
∫ b
0
g(x)dx
+
1
b
n∑
k=1
(
b
n
f(bk/n)−
∫ bk/n
b(k−1)/n
f(x)dx
)∫ b
0
g(x)dx + O(ω(f, b/n)‖g‖1)
=
1
b
∫ b
0
f(x)dx
∫ b
0
g(x)dx + O(ω(f, b/n)‖g‖1).
This proves a). For b) we set b = pi, f(x) = sinx, g(x) = (1 + 3cos 2x)−1.
From a) and
∫ pi
0
sinxdx = 2,
∫ pi
0
(1 + 3cos 2x)−1dx =
pi
2
we get
lim
n→∞
∫ pi
0
sinx
1 + 3cos 2nx
dx = 1.
Problem 6. (25 points)
Let f ∈ C2[0, N ] and |f ′(x)| 0 for every x ∈ [0, N ]. Let
0 ≤ m0 < m1 < · · · < mk ≤ N be integers such that ni = f(mi) are also
integers for i = 0, 1, . . . , k. Denote bi = ni − ni−1 and ai = mi − mi−1 for
i = 1, 2, . . . , k.
a) Prove that
−1 <
b1
a1
<
b2
a2
< · · · <
bk
ak
< 1.
b) Prove that for every choice of A > 1 there are no more than N/A
indices j such that aj > A.
c) Prove that k ≤ 3N 2/3 (i.e. there are no more than 3N 2/3 integer
points on the curve y = f(x), x ∈ [0, N ]).
Solution. a) For i = 1, 2, . . . , k we have
bi = f(mi)− f(mi−1) = (mi −mi−1)f
′(xi)
for some xi ∈ (mi−1,mi). Hence
bi
ai
= f ′(xi) and so −1 <
bi
ai
< 1. From the
convexity of f we have that f ′ is increasing and
bi
ai
= f ′(xi) < f
′(xi+1) =
bi+1
ai+1
because of xi < mi < xi+1.
5b) Set SA = {j ∈ {0, 1, . . . , k} : aj > A}. Then
N ≥ mk −m0 =
k∑
i=1
ai ≥
∑
j∈SA
aj > A|SA|
and hence |SA| < N/A.
c) All different fractions in (−1, 1) with denominators less or equal A are
no more 2A2. Using b) we get k < N/A + 2A2. Put A = N 1/3 in the above
estimate and get k < 3N 2/3.
Second day — July 30, 1994
Problem 1. (14 points)
Let f ∈ C1[a, b], f(a) = 0 and suppose that λ ∈ R, λ > 0, is such that
|f ′(x)| ≤ λ|f(x)|
for all x ∈ [a, b]. Is it true that f(x) = 0 for all x ∈ [a, b]?
Solution. Assume that there is y ∈ (a, b] such that f(y) 6= 0. Without
loss of generality we have f(y) > 0. In view of the continuity of f there exists
c ∈ [a, y) such that f(c) = 0 and f(x) > 0 for x ∈ (c, y]. For x ∈ (c, y] we
have |f ′(x)| ≤ λf(x). This implies that the function g(x) = ln f(x)− λx is
not increasing in (c, y] because of g ′(x) =
f ′(x)
f(x)
−λ ≤ 0. Thus ln f(x)−λx ≥
ln f(y)− λy and f(x) ≥ eλx−λyf(y) for x ∈ (c, y]. Thus
0 = f(c) = f(c + 0) ≥ eλc−λyf(y) > 0
— a contradiction. Hence one has f(x) = 0 for all x ∈ [a, b].
Problem 2. (14 points)
Let f : R2 → R be given by f(x, y) = (x2 − y2)e−x
2
−y2 .
a) Prove that f attains its minimum and its maximum.
b) Determine all points (x, y) such that
∂f
∂x
(x, y) =
∂f
∂y
(x, y) = 0 and
determine for which of them f has global or local minimum or maximum.
Solution. We have f(1, 0) = e−1, f(0, 1) = −e−1 and te−t ≤ 2e−2 for
t ≥ 2. Therefore |f(x, y)| ≤ (x2 + y2)e−x
2
−y2 ≤ 2e−2 < e−1 for (x, y) /∈
M = {(u, v) : u2 + v2 ≤ 2} and f cannot attain its minimum and its
6maximum outside M . Part a) follows from the compactness of M and the
continuity of f . Let (x, y) be a point from part b). From
∂f
∂x
(x, y) =
2x(1 − x2 + y2)e−x
2
−y2 we get
(1) x(1− x2 + y2) = 0.
Similarly
(2) y(1 + x2 − y2) = 0.
All solutions (x, y) of the system (1), (2) are (0, 0), (0, 1), (0,−1), (1, 0)
and (−1, 0). One has f(1, 0) = f(−1, 0) = e−1 and f has global maximum
at the points (1, 0) and (−1, 0). One has f(0, 1) = f(0,−1) = −e−1 and
f has global minimum at the points (0, 1) and (0,−1). The point (0, 0)
is not an extrema point because of f(x, 0) = x2e−x
2
> 0 if x 6= 0 and
f(y, 0) = −y2e−y
2
< 0 if y 6= 0.
Problem 3. (14 points)
Let f be a real-valued function with n + 1 derivatives at each point of
R. Show that for each pair of real numbers a, b, a < b, such that
ln
(
f(b) + f ′(b) + · · ·+ f (n)(b)
f(a) + f ′(a) + · · ·+ f (n)(a)
)
= b− a
there is a number c in the open interval (a, b) for which
f (n+1)(c) = f(c).
Note that ln denotes the natural logarithm.
Solution. Set g(x) =
(
f(x) + f ′(x) + · · ·+ f (n)(x)
)
e−x. From the
assumption one get g(a) = g(b). Then there exists c ∈ (a, b) such that
g′(c) = 0. Replacing in the last equality g ′(x) =
(
f (n+1)(x)− f(x)
)
e−x we
finish the proof.
Problem 4. (18 points)
Let A be a n× n diagonal matrix with characteristic polynomial
(x− c1)
d1(x− c2)
d2 . . . (x− ck)
dk ,
where c1, c2, . . . , ck are distinct (which means that c1 appears d1 times on the
diagonal, c2 appears d2 times on the diagonal, etc. and d1+d2+· · ·+dk = n).
7Let V be the space of all n×n matrices B such that AB = BA. Prove that
the dimension of V is
d21 + d
2
2 + · · ·+ d
2
k.
Solution. Set A = (aij)
n
i,j=1, B = (bij)
n
i,j=1, AB = (xij)
n
i,j=1 and
BA = (yij)
n
i,j=1. Then xij = aiibij and yij = ajjbij . Thus AB = BA is
equivalent to (aii − ajj)bij = 0 for i, j = 1, 2, . . . , n. Therefore bij = 0 if
aii 6= ajj and bij may be arbitrary if aii = ajj. The number of indices (i, j)
for which aii = ajj = cm for some m = 1, 2, . . . , k is d
2
m. This gives the
desired result.
Problem 5. (18 points)
Let x1, x2, . . . , xk be vectors of m-dimensional Euclidian space, such that
x1+x2+ · · ·+xk = 0. Show that there exists a permutation pi of the integers
{1, 2, . . . , k} such that
∥∥∥∥∥
n∑
i=1
xpi(i)
∥∥∥∥∥ ≤
(
k∑
i=1
‖xi‖
2
)1/2
for each n = 1, 2, . . . , k.
Note that ‖ · ‖ denotes the Euclidian norm.
Solution. We define pi inductively. Set pi(1) = 1. Assume pi is defined
for i = 1, 2, . . . , n and also
(1)
∥∥∥∥∥
n∑
i=1
xpi(i)
∥∥∥∥∥
2
≤
n∑
i=1
‖xpi(i)‖
2.
Note (1) is true for n = 1. We choose pi(n + 1) in a way that (1) is fulfilled
with n + 1 instead of n. Set y =
n∑
i=1
xpi(i) and A = {1, 2, . . . , k} \ {pi(i) : i =
1, 2, . . . , n}. Assume that (y, xr) > 0 for all r ∈ A. Then
(
y,
∑
r∈A
xr
)
> 0
and in view of y +
∑
r∈A
xr = 0 one gets −(y, y) > 0, which is impossible.
Therefore there is r ∈ A such that
(2) (y, xr) ≤ 0.
Put pi(n + 1) = r. Then using (2) and (1) we have
∥∥∥∥∥
n+1∑
i=1
xpi(i)
∥∥∥∥∥
2
= ‖y + xr‖
2 = ‖y‖2 + 2(y, xr) + ‖xr‖
2 ≤ ‖y‖2 + ‖xr‖
2 ≤
8≤
n∑
i=1
‖xpi(i)‖
2 + ‖xr‖
2 =
n+1∑
i=1
‖xpi(i)‖
2,
which verifies (1) for n + 1. Thus we define pi for every n = 1, 2, . . . , k.
Finally from (1) we get∥∥∥∥∥
n∑
i=1
xpi(i)
∥∥∥∥∥
2
≤
n∑
i=1
‖xpi(i)‖
2 ≤
k∑
i=1
‖xi‖
2.
Problem 6. (22 points)
Find lim
N→∞
ln2 N
N
N−2∑
k=2
1
ln k · ln(N − k)
. Note that ln denotes the natural
logarithm.
Solution. Obviously
(1) AN =
ln2 N
N
N−2∑
k=2
1
ln k · ln(N − k)
≥
ln2 N
N
·
N − 3
ln2 N
= 1−
3
N
.
Take M , 2 ≤ M < N/2. Then using that
1
ln k · ln(N − k)
is decreasing in
[2, N/2] and the symmetry with respect to N/2 one get
AN =
ln2 N
N


M∑
k=2
+
N−M−1∑
k=M+1
+
N−2∑
k=N−M

 1ln k · ln(N − k) ≤
≤
ln2 N
N
{
2
M − 1
ln 2 · ln(N − 2)
+
N − 2M − 1
lnM · ln(N −M)
}
≤
≤
2
ln 2
·
M lnN
N
+
(
1−
2M
N
)
lnN
lnM
+ O
(
1
lnN
)
.
Choose M =
[
N
ln2 N
]
+ 1 to get
(2) AN ≤
(
1−
2
N ln2 N
)
lnN
lnN − 2 ln lnN
+O
(
1
lnN
)
≤ 1+O
(
ln lnN
lnN
)
.
Estimates (1) and (2) give
lim
N→∞
ln2 N
N
N−2∑
k=2
1
ln k · ln(N − k)
= 1.
International Competition in Mathematics for
Universtiy Students
in
Plovdiv, Bulgaria
1995
1PROBLEMS AND SOLUTIONS
First day
Problem 1. (10 points)
Let X be a nonsingular matrix with columns X1, X2, . . . , Xn. Let Y be a
matrix with columns X2, X3, . . . , Xn, 0. Show that the matrices A = Y X
−1
and B = X−1Y have rank n− 1 and have only 0’s for eigenvalues.
Solution. Let J = (aij) be the n× n matrix where aij = 1 if i = j + 1
and aij = 0 otherwise. The rank of J is n − 1 and its only eigenvalues are
0′s. Moreover Y = XJ and A = Y X−1 = XJX−1, B = X−1Y = J . It
follows that both A and B have rank n− 1 with only 0′s for eigenvalues.
Problem 2. (15 points)
Let f be a continuous function on [0, 1] such that for every x ∈ [0, 1] we
have
∫ 1
x
f(t)dt ≥ 1− x
2
2
. Show that
∫ 1
0
f2(t)dt ≥ 1
3
.
Solution. From the inequality
0 ≤
∫ 1
0
(f(x)− x)2 dx =
∫ 1
0
f2(x)dx− 2
∫ 1
0
xf(x)dx +
∫ 1
0
x2dx
we get ∫ 1
0
f2(x)dx ≥ 2
∫ 1
0
xf(x)dx−
∫ 1
0
x2dx = 2
∫ 1
0
xf(x)dx− 1
3
.
From the hypotheses we have
∫ 1
0
∫ 1
x
f(t)dtdx ≥
∫ 1
0
1− x2
2
dx or
∫ 1
0
tf(t)dt ≥
1
3
. This completes the proof.
Problem 3. (15 points)
Let f be twice continuously differentiable on (0,+∞) such that
lim
x→0+
f ′(x) = −∞ and lim
x→0+
f ′′(x) = +∞. Show that
lim
x→0+
f(x)
f ′(x)
= 0.
2Solution. Since f ′ tends to −∞ and f ′′ tends to +∞ as x tends to
0+, there exists an interval (0, r) such that f ′(x) 0 for all
x ∈ (0, r). Hence f is decreasing and f ′ is increasing on (0, r). By the mean
value theorem for every 0 < x < x0 < r we obtain
f(x)− f(x0) = f ′(ξ)(x− x0) > 0,
for some ξ ∈ (x, x0). Taking into account that f ′ is increasing, f ′(x) <
f ′(ξ) < 0, we get
x− x0 < f
′(ξ)
f ′(x)
(x− x0) = f(x)− f(x0)
f ′(x)
< 0.
Taking limits as x tends to 0+ we obtain
−x0 ≤ lim inf
x→0+
f(x)
f ′(x)
≤ lim sup
x→0+
f(x)
f ′(x)
≤ 0.
Since this happens for all x0 ∈ (0, r) we deduce that lim
x→0+
f(x)
f ′(x)
exists and
lim
x→0+
f(x)
f ′(x)
= 0.
Problem 4. (15 points)
Let F : (1,∞) → R be the function defined by
F (x) :=
∫ x2
x
dt
ln t
.
Show that F is one-to-one (i.e. injective) and find the range (i.e. set of
values) of F .
Solution. From the definition we have
F ′(x) =
x− 1
lnx
, x > 1.
Therefore F ′(x) > 0 for x ∈ (1,∞). Thus F is strictly increasing and hence
one-to-one. Since
F (x) ≥ (x2 − x)min
{
1
ln t
: x ≤ t ≤ x2
}
=
x2 − x
lnx2
→∞
3as x →∞, it follows that the range of F is (F (1+),∞). In order to determine
F (1+) we substitute t = ev in the definition of F and we get
F (x) =
∫ 2 ln x
ln x
ev
v
dv.
Hence
F (x) < e2 lnx
∫ 2 ln x
ln x
1
v
dv = x2 ln 2
and similarly F (x) > x ln 2. Thus F (1+) = ln 2.
Problem 5. (20 points)
Let A and B be real n × n matrices. Assume that there exist n + 1
different real numbers t1, t2, . . . , tn+1 such that the matrices
Ci = A + tiB, i = 1, 2, . . . , n + 1,
are nilpotent (i.e. Cni = 0).
Show that both A and B are nilpotent.
Solution. We have that
(A + tB)n = An + tP1 + t
2P2 + · · ·+ tn−1Pn−1 + tnBn
for some matrices P1, P2, . . . , Pn−1 not depending on t.
Assume that a, p1, p2, . . . , pn−1, b are the (i, j)-th entries of the corre-
sponding matrices An, P1, P2, . . . , Pn−1, B
n. Then the polynomial
btn + pn−1t
n−1 + · · · + p2t2 + p1t + a
has at least n + 1 roots t1, t2, . . . , tn+1. Hence all its coefficients vanish.
Therefore An = 0, Bn = 0, Pi = 0; and A and B are nilpotent.
Problem 6. (25 points)
Let p > 1. Show that there exists a constant Kp > 0 such that for every
x, y ∈ R satisfying |x|p + |y|p = 2, we have
(x− y)2 ≤ Kp
(
4− (x + y)2
)
.
4Solution. Let 0 0 such
that
f(x, y) =
(x− y)2
4− (x + y)2 ≤ Kp,δ
for every (x, y) ∈ Dδ = {(x, y) : |x− y| ≥ δ, |x|p + |y|p = 2}.
Since Dδ is compact it is enough to show that f is continuous on Dδ.
For this we show that the denominator of f is different from zero. Assume
the contrary. Then |x + y| = 2, and
∣∣∣∣x + y2
∣∣∣∣p = 1. Since p > 1, the function
g(t) = |t|p is strictly convex, in other words
∣∣∣∣x + y2
∣∣∣∣p < |x|p + |y|p2 whenever
x 6= y. So for some (x, y) ∈ Dδ we have
∣∣∣∣x + y2
∣∣∣∣p < |x|p + |y|p2 = 1 =∣∣∣∣x + y2
∣∣∣∣p. We get a contradiction.
If x and y have different signs then (x, y) ∈ Dδ for all 0 < δ < 1 because
then |x−y| ≥ max{|x|, |y|} ≥ 1 > δ. So we may further assume without loss
of generality that x > 0, y > 0 and xp + yp = 2. Set x = 1 + t. Then
y = (2− xp)1/p=(2− (1 + t)p)1/p =
(
2− (1 + pt + p(p−1)
2
t2 + o(t2))
)1/p
=
(
1− pt− p(p− 1)
2
t2 + o(t2)
)1/p
= 1 +
1
p
(
−pt− p(p− 1)
2
t2 + o(t2)
)
+
1
2p
(
1
p
− 1
)
(−pt + o(t))2 + o(t2)
= 1− t− p− 1
2
t2 + o(t2)− p− 1
2
t2 + o(t2)
=