Metas, objetivos e indicadores de seguridad vial

n

k

en

_ k)! + (k +

1) n

k

n 1

�

(k + l) !

(n

- k)! =

�

(k + I) ! +

�

k!(n - k) ! =

n

k I n n! n

( 1 1 ) 1

n

(

n

)

=

L

(k + I)! +

n

!

L

k ! (n - k ) ! =

L ki -

(k + I ) !

+

n!

L

k

=

k=O k=O k=O k=O

Since

1 2n

= 1 -

---

(n

+ -.

+ I) ! n! 1. 1m --- = 1m

1

1. - =2n

0

,

n-+oo (n

+ I) !

n-+oo n!

196

5. MATHEMATICAL ANALYSIS

it follows that

r

k(n - k)! + (k + 1)

n�� (k + 1)!(n

_

k)! = 1.

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No.

2(1975),

pp.

52,

Problem

2281)

14.

Note that lim XZ

= 1.

z-+O _z>O Indeed, !�o ¥

l�(-z)

lim XZ = lim ez1nz = e "'>O z

=

_{e z>O}

= 1,

",-+0 ", -+0

z>O ",>0

after applying L'Hospital rule. Then

xz+1

lim _{z_o} --=

1.

x

z>o

Let

e

>

O.

There is an integer

nee)

>

0

such that for any integer

n � nee),

we have

or

�+1

1 - e < k < 1 + e, k = 1,2, ... , n. _n

2

Summing up from

k = 1

to

k = n

and using algebraic manipulations yields

n (�) �+1

L n2

1 - e <

k=

1 n k < 1 + e, n � n (e),

L n

2 k=1

- - - e - - + -_{2 2 n n} 1 1 (

1 e)

n (-_nk

2

) �+1 < - + - e+ - + -1 1 2 2 n n ' (

1 e)

k= l for any integer

n

�

nee).

Therefore

n (

k) �+1

₁

lim

-

= -.

n-+oo t; n

2

2

(Dorin Andrica)

15. (

i

)

We have

1 +x+ x + · .. +x + " ' =

2

n

1

--

,

I - x

Ixl <

1. Hence so 5.2. SOLUTIONS

(1 + x + x2 + ... )dx = 1 -x'

00

1

q

I: _{nqn =}

In

q

_

1

'

q

>

1.

n=1

(

ii

)

For any

Ixl < 1

we have Hence so

1 + x4 + x8 + .. . + x4n + ... = 1 - x4

1o 1/q (1 + x4 + x8 + .. . )dx = 1

0

1/q

_dx - x '

00

1

1 1

(

1)

1 1

� (4n + l)q4n+l

=

- q + 4

ln

1 - q2 + '2

arctg

q

'

197

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No.

2(1977),

pp.

70,

Problem

3091;

Gazeta Matematica

(

GM-B

)

, No.

1(1981),

pp.

40,

Problem

18608)

16.

The number () has the decimal representation

() = 0.0 .. . 010 .. . 010 .. . 01 .. . 10 ... 01 '-.;--' '-.;--'

'-.;--'

kl k2 kn

where

k1, k

2

'

_{. . . ,}

k

_n are the number of zeros between two consecutive ones.

Because

we have

kl < k2 < k3 < .. . < kn <

. .

.

, hence () does not have a periodical decimal representation.

It follows that () is irrational, as claimed.

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No.

2(1981),

pp.

73,

Problem

4661)

17.

Note that 00

1

An = 1 + L (k! )

n >

1

k=2 and that 00

1

00 1

An = -1 + L (k!)n < -1 + L k! = -1 + e < 2.

k=O k=O

198

5.

MATHEMATICAL ANALYSIS

Assume by way of contradiction that there are positive integers

n,p, q

with

q :f

1 such that P.

_q

=

An.

Then

1 1

pqn-l(q

- 1

)

1 =

A

+

(q

+ l

)n

+

(q

+ l

)n(q

+

2)n

+ .. .

<

1

1

< A +

_(q

₊

_l)n

+

_(q

_{+ l}

_)n(q

_{+ l}

_)n

+ . . . = 1

(

1 1

)

1 = A +

_(q

1 + + + . . . = A +

( ) ,

+ 1

) n

(q

+ 1

) n (q

+ 1

) 2n

q

+ 1

n

-1

for some integer

A � O.

It follows that

1 1

B =

_(q

₊

_{l)n (q}

+ _{+ l}

_)n(q

₊

_2)n

+

.. ·< 1 _'

so A + B is not integer, which is false.

Therefore

An

is irrational for all n

�

1.

(Dorin Andrica)

18.

1

)

Using the limit

lim � = 1

n-+oo

and taking the limits in both sides of the equality, we obtain

so

k

= 8 .

2)

Using the limit

and the relation

k

times s times

lim

n (

� -1

)

= In a

n-+oo

n

( \Ial -

1

)

+ n

( \Ia2

-1

)

+ . . . +

n ( ifiik

-

1) =

we obt�in after taking limits: This implies

as desired.

(Dorin Andrica,

Romanian Mathematical Regional Contest "Grigore Moisil" ,

1999)

5.2.

SOLUTIONS

19.

Let

Pn

=

IT (

1 -

�).

Then

k=1 Xk+l

hence 1 1

Xn+2 Xn+l Xn+2

-

(n

+

l)Xn+l xn+l

Pn+l - Pn

=

(n

+

I)! -� = (n

+ 1

)

1 =

(n

+

1)

r It follows that Because 1 1

x2 x3

xn+l

--

= - + - + - + . . . + --- =

Pn+l PI

_{2! 3!}

_(n

+ 1

)1

x x2

Xn+l

= 1 + 11 + 2f + . . . +

(n

+

1)1'

(

X x2

xn+l )

lim 1 + - + - + · · · +

---

=ez

,

n-+oo

_{I! 2!}

_(n

+ 1

)

1

it follows that lim

n-+oo Pn+1

= e-x , as desired.

199

(Titu Andreescu

and

Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No. 1

(

1

977),

pp.

49,

Problem

2843)

so

20.

Interchanging

x

and

y

we obtain

Let Then

f(lnx

+

Alny)

= 9

(v'X)

+

g (.JY) , x,y

E

(0,00)

f(

ln

y

+

Alnx)

= 9

(v'X)

+ 9

(.JY) , X,y

> o.

a = ln x +

Alny

and b

= Iny

+

Alnx.

Ab-a

Aa.-b

x = e� and y = e�,

hence

f₍

a

)

=

f(b)

= ₉

)

+ ₉

)

, a, b E R

It follows that

f

is a constant and let

f (x) = C.

Then for

x

=

y

we have

g(.jX)

=

�,

so 9 is a constant and

g(x)

=

�.

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No.

1-2(1979),

pp.

51,

Problem

3827)

21.

Consider the function

F

:

I

-t lR,

200

5.

MATHEMATICAL ANALYSIS

Note that

F

is continuous and

hence there is

c

E

(a, b)

such that

F(c) = O.

It follows that

f(c)

= �

a - c b - c + m2

and the solution is complete.

(Dorin Andrica)

22.

Note that

g(x) = g(y)

implies that

g(g(x)) = g(g(y))

and hence

x = y

from the given equation. That is, 9 is injective. Since 9 is also continuous, 9 is either strictly increasing or strictly decreasing. Moreover, 9 cannot tend to a finite limit

L

as

x

----+ 00 ,

or else we'd have

g(g(x)) - ag(x) = bx,

with the left side bounded and the right side unbounded. Similarly, 9 cannot tend to a finite limit as

x

----+ -00. Together with

monotonicity, this yields that 9 is also surjective.

Pick

Xo

arbitrary, and define

Xn

for all

n

E

Z

recursively by

Xn+l = g(xn)

for

n

>

0, and

Xn-l = g-1 (xn)

for

n

<

O.

Let

rl =

+ 4b)j2

and

r2 =

+ 4b)j2

be the roots of

x2 - ax - b =

0, so that

rl

> 0 >

r2

and 1 >

Irll

>

Ir2

1. Then there exist

C

l,

C2

E lR such that

Xn

=

clr� + c2r�

for all

n

E

Z.

Suppose 9 is strictly increasing. If C2 f:. 0 for some choice of

xo,

then

Xn

is dominated by

r�

for

n

sufficiently negative. But taking

Xn

and

Xn+2

for

n

sufficiently negative of the right parity, we get 0 <

Xn

<

Xn+2

but

g(xn)

>

g(Xn+2),

contradiction. Thus

C2 = O.

Since

Xo = Cl

and

Xl

=

Clrl,

we have

g(x) = rlX

for all

x.

Analogously,

if 9 is strictly decreasing, then

C2

= 0 or else

Xn

is dominated by

r�

for

n

sufficiently

positive. But taking

Xn

and

Xn+2

for

n

sufficiently positive of the right parity, we get o <

Xn+2

<

Xn

but

g(Xn+2)

<

g(xn),

contradiction. Thus in that case,

g(x) = r2X

for all

x.

(Titu Andreescu,

The " William Lowell Putnam" Mathematical Competition, 2001, Problem B-5)

23.

Setting

x = y

= 0 yields

f(O)

=

O.

For

x = y

we obtain

f2(2x) = 4f2(X)

and

then

f(

2;C

) = 2f(x),

since

f(x) � O.

We prove that

f(nx) = nf(x), n � 1.

Assume that

f(kx) = kf(x)

for all

k = 1,2, .. . , n.

We have

f2((n + l)x) - f2((n - l)x) = 4f(nx)f(x),

then

5.2.

SOLUTIONS

201

hence

f((n

+

l)x)

₌

(n + l)f(x),

as desired.

It follows that if p,

q

are positive integers then

qf (�)

=

f(P) = pf(l),

so

f (�)

=

�f(l)

and

f(r)

=

r f(l)

for any positive rational

r.

Setting

x

= 0 in the initial condition gives

then

f(y)

=

f( -y),

for all real

y,

hence

f(r)

=

Irlf(l),

for all rational numbers

r.

We prove that

f(x) = Ixlf(l)

for all real numbers

x.

Let

x

be an arbitrary real number and let

(rn)n>l _-

be a sequence of rational numbers with lim

_�oo rn = x.

Because and

f

is a continuous function, it follows that

hence

lim

f(rn)

= lim

Irnlf(l) = f (

lim

rn)

,

n�oo

n�oo

n�oo

f(x) = f(l)lxl·

Note that

a = f(l) �

0, therefore the desired functions are

f(x) = alxl

for some

a � O.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 90, Problem 3203)

24.

(i) Let

a

and

b

be fixed points for

f

and

g,

respectively. We have

a - f(a) �

0,

b

=

g(b) �

0

so

a � b.

Consider the function

<p

: lR ----+ lR,

<p(x) = f(x) + g(x) - x.

The function

<p

is continuous because functions

f

and 9 are continuous. Moreover,

202 5. MATHEMATICAL ANALYSIS

and

cp(b)

=

f(b)

+

g(b) - b

=

f(b)

::;

O.

By the Intermediate Value Theorem, there is an Xo E

(a, b)

such that

cp(xo)

=

0,

hence

(f

+

g)(xo)

=

Xo,

as desired.

(ii) Let 0: and

(3

be fixed points of the functions

cp

and

'l/J,

respectively. We have

o

::;

0:

=

cp ( 0:) ::; 1, (3

=

'l/J ((3)

�

1

so 0:

::; (3.

Consider the function

w :

lR -+ lR,

w(x)

=

cp(x)'l/J(x) - x.

The function

w

is continuous because functions

cp

and

'l/J

are continuous. Moreover,

w(o:)

=

cp(o:)'l/J(o:) - 0:

=

O:('l/J(o:) - 1) � 0,

w ((3)

=

cp((3)'l/J ((3) - (3

=

(3( cp((3) - 1) ::; O.

Likewise, there is an ,0 E

[0:, (3]

such that

w(,o)

=

0,

hence

(cp'l/J)(,o)

= ,0, as

desired.

(Titu Andreescu,

"Asupra unor functii cu punct fix" , Revista Matematica Timi§oara (RMT), No.

1(1977),

pp. 5-

10)

25.

We have

�� � (_cp(x)

+

cp (�)

+ . . . +

cp (�))

=

= lim

- <p(O)

+

1 <p(�) - <p(O)

+ . . .

+ 1 <p(�) - <P(O))

=

z-+o x - 0 2

� _

0

n �

- 0

2

n

=

cp'(O) (1

+

�

+ . . . +

�)

2

n '

since

cp(O)

=

0

and

cp

is differentiable at the origin.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No.

2(1977),

pp.

71,

Problem

3095;

Gazeta Matematica (GM-B) , No.

3(1979),

pp.

111,

Problem

17671)

so

26.

Consider the function

f : (0,00)

-+ lR,

f(x)

= In

ax

.

We have

x

!' (x)

=

1 -_x

In ax

2

!,(x)

=

0

if and only if

x

= �.

_a

e

It follows that I-" = - is the maximum point of the function I so I-" is the only a

point such that I

(

x

) � f(l-")

for all positive real numbers x. Hence

5.2. SOLUTIONS 203

for all

x

>

0,

as desired.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT) , No.

2(1978),

pp. 54, Problem

3

54

7)

27.

Consider the function

F : [a, b]

-+ lR,

F(x)

=

f(x)e->'f(z),

>. E R

Function

F

is differentiable, since I and I' are differentiable, and

F(a)

=

F(b).

By Rolle's theorem it follows that there is c E (a,

b)

such that F' (c) =

O.

On the other hand,

F' (x)

=

e->'f(z)

(I"

(x)

- >'(/' (X))2),

hence II/(c) - >'(/' (c))2 =

0,

as desired.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT) , No.

2(1981),

pp.

76,

Problem

4677)

28.

Let

g : [0,2]

-+

[0, �), g(x)

= arccos /(x) . Then

f(x)

=

cosg(x)

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