Problem 1. Let a be a fixed real number. Prove that ax ≥ x + 1 for any x ∈ R if and only if a = e.
Solution : If a = e then we study the behavior of the function f (x) = ex − x − 1. The derivative f 0 (x) = ex − 1 has
negative sign on (−∞, 0) and positive sign on (0, ∞), so the function has a minimum at x = 0. But f (x) ≥ f (0)
can be writen ex ≥ x + 1.
Suppose now ax ≥ x + 1 for any x. Taking x = 0 we see a ≥ 2. Consider the function f (x) = ax − x − 1. The
ln ln a
which is a minimum for the function f . By hypothesis
derivative f 0 (x) = ax ln a − 1 has the root x0 = −
a
(ln ln a) ln a − a(ln a − 1)
(ln a − 1) ln a − a(ln a − 1)
f (x) ≥ 0 for all x ∈ R, therefore f (x0 ) ≥ 0. Bur f (x0 ) =
≤
≤
a ln a
a ln a
0. Consequently ln ln a = ln a − 1 = 0, so a = e.
Problem 2. Let a > 0 be a fixed real number. Prove that ax ≥ xa for all x > 0 if and only if a = e.
ln x
Solution : The inequality is equivalent with f (x) ≤ f (a), for any x > 0, where f (x) =
. Therefore x = a is a
x
maximum for f , but this function has a maximum only in x = e.
Problem 3. Consider the sequence defined by
limit
1+
1
n
n+xn
= e. Prove that xn is decreasing towards the
1
.
2
1
− x. We prove f is increasing on [1, ∞). It
ln(x + 1) − ln x
1
1
− ln(x + 1) + ln x ≥ 0.
− 1 ≥ 0 or equivalently g(x) = p
suffices to prove f 0 (x) =
2
x(x + 1) ln (1 +
x(x + 1)
! 1/x)
1
x + 1/2
Since g 0 (x) =
1− p
< 0, we have g(x) ≥ lim g(x) = 0.
x→∞
x(x + 1)
x(x + 1)
1
1
1
1
1
− = ,
For the limit of the sequence it suffices to prove that lim f (x) = , or with y = , lim
x→∞
y→0
2
x
ln(1 + y) y
2
which can be proved easily using l’Hopital’s rule or a Taylor expansion.
Solution : By hypothesis xn = f (n), where f (x) =
√
Problem 4. Solve the equation 2x + 2
1−x2
= 3.
Solution : Obviously, the solutions of the equation satisfy x ∈ [0, 1]. We prove x = 0√and x = 1 are the only
solutions
of the equation. The function
f : [0, 1] → R is increasing on the interval [0, 2/2]
√
√
√ and decreasing on
[ 2/2, 1]. The identity f (x) = f ( 1 − x2 ) makes it suffficient to prove f increasing on [0, 2/2]. The function
√
2x
2x
g(x) =
is decreasing on [0, 1], since g 0 (x) = 2 (x ln 2 − 1) < 0. Then f 0 (x) = x ln 2 g(x) − g( 1 − x2 ) > 0,
x√
x
√
since x < 1 − x2 on [0, 2/2].
Problem 5. Let f (x) be a positive-valued function over the reals such that f 0 (x) > f (x) for all x. For what k
must there exist N such that f (x) > ekx for x > N ? [P1994]
Solution : The condition can be written g 0 (x) > 0, where g(x) = ln f (x) − x and an equivalent form for f (x) > ekx
is
g(x) > (k − 1)x
(0.1)
k−1
For k ≥ 1, and g(x) =
x there is no N such that 0.1 is satisfied for x > N . If k < 1 then lim (g(x)−(k−1)x) =
x→∞
2
∞, so there is an N such that for x > N 0.1 holds.
Problem 6. Let f be an infinitely differentiable real-valued function defined on the real numbers. If
1
n2
f
= 2
,
n = 1, 2, 3, . . . ,
n
n +1
compute the values of the derivatives f (k) (0), k = 1, 2, 3, . . . . [P1992]
1
we have g( n1 ) = 0, n = 1, 2, 3... By the theorem of Rolle there is a sequence
1 + x2
1
(1)
(1)
<
(*) and g 0 (xn ) = 0. As a consequence of (*) the sequence xn is decreasing
n
Solution : With g(x) = f (x) −
(1)
(xn )n such that
1
(1)
< xn
n+1
2
(2)
(1)
(2)
(1)
to 0, hence g 0 (0) = 0. Using again Rolle’s theorem ther is a sequence (xn )n such that xn+1 < xn < xn (**)
(2)
(2)
and g (2) (xn ) = 0. From (**), xn is decreasing to 0, and g (2) (0) = 0. By recurence we prove in this way that
1
= 1 − x2 + x4 − x6 + ... and consequently
g (k) (0) = 0, for any k. Thus f (k) (0) = φ(k) (0), where φ(x) =
1 + x2
f (2k+1) (0) = 0, f (2k) (0) = (−1)n (2n)!.
1
Problem 7. Let f be a function such that f (1) = 1 and f 0 (x) = 2
for all x ≥ 1. Show that lim f (x)
x→∞
x + f 2 (x)
π
exists and is less than 1 + .
4
Solution The function has positive derivative so is increasing. Then the limit lim f (x) exists and f 0 (x) ≤
x→∞
1
1
=
,
for
any
x
≥
1,
which
proves
that
the
function
g(x)
=
f
(x)
−
arctan x is decreasing. Hence
x2 + f (1)2
x2 + 1
π
π
π
=
g(x) ≤ g(1) = 1 − and consequently f (x) ≤ arctan x + 1 − . Therefore lim f (x) ≤ lim arctan x + 1 −
x→∞
x→∞
4
4
4
π
1+ .
4
Problem 8. Let f : (−2, 2) → R be a function of class C 2 such that f (0) = 0. Show that the sequence (un )
n
X
defined by un =
f ( nk2 ) is convergent and compute its limit.
k=1
k
k
k
− f (0) = 2 f 0 (ck ), and dk ∈ (0, ck ) such that f 0 (ck ) − f 0 (0) =
Solution : Let ck ∈ 0, 2 be such that f
n
n2
n
n
n
n
X
X
X
k 0
k
k2
00
ck f 00 (dk ). Then |un −
f
(0)|
=
|
c
f
(d
)|
≤
max |f 00 (x)|. Passing at limit with n → ∞ we
k
k
2
2
n
n
n4 x∈[0,1]
k=1
k=1
k=1
f 0 (0)
obtain lim un =
.
2
Problem 9. a) Solve in real numbers the equation 2x + 4x = 3x + 5x .
b) Solve in real numbers the equation 2x + 5x = 3x + 4x .
Solution : a) For x > 0, 3x + 5x > 2x + 4x , and for x < 0, 3x + 5x < 2x + 4x , so x = 0 is the only solution.
b) The mean value theorem for the function f (t) = tx shows
is c #∈ (2, 3) and d ∈ (4, 5) such that
" there
x−1
d
3x − 2x = xcx−1 and 5x − 4x = xdx−1 . The equation becomes x
− 1 = 0, and has the solutions x = 0
c
and x = 1.
2
2
Problem 10. Solve in real numbers the equation 5x + 5x = 4x + 6x .
2
2
Solution : Write the equation as 5x − 4x = 6x − 5x . Using the mean theorem for the function f (t) = tx ,
2
there is c ∈ (4, 5) such that 5x − 4x = xcx−1 . Similarly for the function g(t) = tx , there is d ∈ (5, 6) such that
2
2
2
2
6x − 5x = x2 dx −1 . The equation becomes xcx−1 = x2 dx −1 . One obvious solution is x = 0. Looking for non-zero
solutions, we see that necesarily x > 0.
1+x 1−x
d
Suppose 0 < x < 1. Then x =
> 1. Contradiction.
c
1+x 1−x
d
< 1. Therefore x = 0 and x = 1 are the only solutions.
For x > 1, we obtain the contradiction x =
c
Problem 11. Evaluate the limit lim n2
n→∞
1
1+
n+1
n+1
1
− 1+
n
n !
.
Problem 12. Let f (x) = (x2 − 1)n , and let Pn (x) be the n-th derivative of f (x). Prove that Pn (x) is a
polynomial of degree n with n real, distinct roots in (-1,1).