Corrigé Examen QCM corrigé mathématique générales(divers)
Q
1
.
On
doit
calculer
la
lim
ite
à
droite
et
à
gauche
de
f
(
x
)
=
x
2
−
1
x
2
−
1
en
effet
,
lim
x
→
1
+
f
(
x
)
=
0
et
lim
x
→
1
−
f
(
x
)
=
+
∞
donc
pas
de
lim
ite
Q
2
.
∀
n
>
0
,
x
n
=
2
x
n
−
1
donc
g
é
ometrique
de
raison
q
=
2
,
le
probl
é
me
se
pose
en
0
car
u
0
=
0
pour
tan
t
u
1
≠
2
u
0
.
la
r
é
ponse
exacte
on
peut
pas
conclure
Q
3
.
5
solutions
dans
ℝ
S
=
0
,
2
,
−
2
,
2
+
6
,
−
2
+
6
Q
4
.
On
sait
que
f
d
é
rivable
en
a
si
et
seulement
si
f
admet
un
DL
1
(
a
)
qui
vaut
:
f
x
=
f
a
+
x
−
a
f
′
a
+
ο
x
−
a
x
→
0
on
l
'
applique
cette
formule
on
obtient
8
f
a
f
′
a
.
Q
5
.
f
'
3
=
1
/
2
Q
6
.
∀
x
∊
ℝ
,
x
1
−
x
≤
1
/
4
donc
u
n
1
−
u
n
≤
1
/
4
⊛
u
n
+
1
−
u
n
1
−
u
n
≥
0
car
0
≤
u
n
≤
1
d
'
ou
la
croissance
de
u
n
.
d
'
apres
⊛
et
par
passage
au
lim
ite
on
aura
l
(
1
−
l
)
≤
1
/
4
l
(
1
−
l
)
≥
1
/
4
d
'
ou
l
(
1
−
l
)
=
1
/
4
⇔
l
=
1
/
2
Q
7
.
le
d
é
veloppement
lim
it
é
donne
u
n
=
−
1
n
α
/
2
1
+
1
n
α
/
2
−
1
/
2
=
−
1
n
α
/
2
1
−
1
2
n
α
/
2
+
o
1
n
α
=
−
1
n
α
/
2
−
1
2
n
3
α
/
2
+
o
1
n
3
α
/
2
la
serie
du
terme
−
1
n
α
/
2
verifie
la
condition
de
convergence
des
s
é
ries
altern
é
s
.
−
1
2
n
3
α
/
2
+
o
1
n
3
α
/
2
∼
−
1
2
n
3
α
/
2
CV
si
3
2
α
>
1
en
conclusion
∑
2
+
∞
−
1
n
−
1
n
+
n
α
CV
si
α
>
2
/
3
Q
8
.
l
'
equation
f
z
=
1
n
'
as
pas
de
solution
ou
encore
1
n
'
as
pas
d
'
antecedents
donc
la
premiere
et
la
deuxieme
affirmation
sont
fausses
.
la
troizieme
est
fausse
(
on
prend
f
2
i
=
3
∉
E
).
la
seule
affirmation
qui
est
correcte
est
f
E
=
i
ℝ
(
f
1
=
i
donc
vrai
puisque
1
∊
E
).
Q
9
.
Soit
B
une
partie
de
E
de
cardinal
k
.
le
nombre
de
partie
A
de
E
tq
A
⊂
B
est
card
P
B
=
2
k
en
plus
on
sait
qu
'
il
y
a
C
n
k
partie
de
B
à
k
elements
de
E
.
On
a
ainsi
card
A
,
B
∊
P
2
E
/
A
⊂
B
=
∑
k
=
0
n
2
k
C
n
k
=
3
n
CQFD
Q
10
.
on
note
X
=
k
por
l
'
apparition
de
la
boule
blanche
pour
la
premiere
fois
au
ki
é
me
tirage
.
P
X
=
1
=
2
n
,
P
X
=
2
=
n
−
2
n
×
2
n
−
1
,
P
X
=
3
=
n
−
2
n
×
n
−
3
n
−
1
×
2
n
−
2
...
P
X
=
k
=
2
n
−
k
n
n
−
1
Or
E
X
=
∑
k
=
1
n
k
.
P
X
=
k
=
2
n
n
−
1
∑
k
=
1
n
k
n
−
k
Or
on
sait
que
:
∑
k
=
1
n
k
=
n
n
+
1
2
et
∑
k
=
1
n
k
2
=
n
n
+
1
2
n
+
1
6
par
calcul
on
trouve
E
X
=
n
+
1
3
Q
11
.
S
=
7
k
+
3
,
7
k
+
5
,
k
∊
ℤ
Q
12
.
On
utilise
sin
x
−
x
≤
x
3
6
∑
k
=
0
n
sin
1
n
+
k
−
1
n
+
k
≤
∑
k
=
0
n
1
6
n
+
k
3
≤
∑
k
=
0
n
1
6
n
3
→
n
→
+
∞
0
car
1
6
n
3
ne
d
é
pend
pas
de
k
donc
notre
somme
CV
→
n
→
+
∞
∑
k
=
0
n
1
n
+
k
=
ln
2
(
Int
é
grale
de
Reimann
)
Q
13
.
lim
x
→
0
f
(
x
)
×
f
(
2
x
)
×
f
(
3
x
)
×
.....
×
f
((
n
−
1
)
x
)
x
n
−
1
=
lim
x
→
0
f
(
x
)
1
×
f
(
2
x
)
x
×
f
(
3
x
)
x
×
....
×
f
((
n
−
1
)
x
)
x
=
1
×
2
×
3
×
...
×
(
n
−
1
)
=
(
n
−
1
)!
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Q
14
.
on
s
é
que
5
4
≡
1
13
et
A
n
=
1
−
5
4
n
1
−
5
n
alors
1
−
5
n
A
n
=
1
−
5
n
4
=
0
13
donc
1
−
5
n
=
0
13
et
n
=
4
k
ce
qui
fait
A
n
=
0
13
ssi
n
≠
4
k
.
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