Oraux-2015-solutions

Telechargé par ABDELLAH BEN ZANNOU
A=
1a a2a3
1/a 1a a2
1/a21/a 1a
1/a31/a21/a 1
A
A
f∈ C(R,R)
xR, f(x) + x
0
(xt)f(t)dt = 1
C1, C2, C3, C4A
C2=aC1, C3=a2C1, C4=a3C1
A
43 = 1 x+ay +a2z+a3t= 0
λ1=λ2=λ3= 0 λ4(A)=4 λ4= 4
(A)C1
A
f∈ C(R,R)xR, f(x) + x
0
(xt)f(t)dt = 1
xR, f(x)=1xx
0
f(t)dt +x
0
tf(t)dt
t7→ f(t)t7→ tf(t)x7→ x
0f(t)dt x 7→ x
0tf(t)dt
C1RfC1
xR, f(x) = xf(x)x
0
f(t)dt +xf(x)
xR, f(x) = x
0
f(t)dt
fC1fC2
xR, f′′(x) = f(x)
fRy′′ +y= 0
A B xR, f(x) = Acos x+Bsin x
xR, A cos x+Bsin x+x
0
(xt)(Acos t+Bsin t)dt = 1
⇒ ∀xR, A cos x+Bsin x+xx
0
(Acos t+Bsin t)dt x
0
t(Acos t+Bsin t)dt = 1
xR, A cos x+Bsin x+x(Asin xBcos x+B)[t(Asin tBcos t)]x
0+x
0
(Asin tBcos t)dt = 1
⇒ ∀xR, A cos x+Bsin x+Ax sin xBx cos x+Bx (Ax sin xBx cos x)+[Acos t+Bsin t]x
0= 1
⇒ ∀xR, A cos x+Bsin x+Bx + (Acos x+ABsin x) = 1
⇒ ∀xR, Bx +A= 1
x A = 1 B= 0
xR, f(x) = cos(x)
A= 1 B= 0
f= cos
f
C2
A=
33 2
1 5 2
1 3 0
M3(R)
R R2=A R
RM3(R)A=R2
f:RR
(x, y)R2, f x+y
2=1
2(f(x) + f(y)) R
fRf(0) = f(1) = 0
f
f f
f(2x) = 2f(x)
f
fR
A=
33 2
1 5 2
1 3 0
χA(x) =
3x3 2
1 5 x2
1 3 x
=
2x3 2
2x5x2
2x3x
χA(x) = (2 x)
13 2
1 5 x2
1 3 x
= (2 x)
13 2
0 8 x4
0 6 2x
= (2 x)[(x8)(x+ 2) + 24]
χA(x) = (2 x)(x26x+ 8) = (2 x)(x2)(x4)
χA(X) = (X2)2(X4)
E4(A)
dim(E2(A)) = 3 (A2I3)
A2I3=
13 2
1 3 2
1 3 2
dim(E2(A)) = 3 1 = 2 = (2)
χA(X) = (X2)2(X4) R[X]
AM3(R)
PGL3(R)A=P.
200
020
004
.P 1
R=P.
2 0 0
02 0
0 0 2
.P 1R2=P.
200
020
004
.P 1=A
A Q(X) =
λSp(A)
(Xλ) = (X2)(X4)
A
RM3(R)R2=A
Q(A) = (A2I3)(A4I3)=0=(R22I3)(R24I3)
T(X) = (X22)(X24) = (X2)(X+2)(X2)(X+ 2)
RR[X]R
fRR
(x, y)R2, f x+y
2=1
2(f(x) + f(y)) f(0) = f(1) = 0
x y =x
xR, f xx
2=f(0) = 0 = 1
2(f(x) + f(x))
xR, f(x) = f(x)f
x y = 2 + x
xR, f(x+ 2) f(x) = f(x+ 2) + f(x) = 2f(x+2)+(x)
2= 2f(1) = 0
xR, f(x+ 2) = f(x)f
f[0,2]
MR+,xR,|f(x)|6M
fxR,|f(x)|6M f
x y = 0
xR, f x+ 0
2=1
2(f(x) + f(0)

=0
)xR, f(x) = 2fx
2
xxR, f(2x) = 2f(x)
xR, f(4x) = 2f(2x) = 4f(x)
xR,kN, f(2kx) = 2kf(x)
f x0Rf(x0)̸= 0 kN, f(2kx0)=2kf(x0)
k+±∞
f x0f(x0)
f
fRR
(x, y)R2, f x+y
2=1
2(f(x) + f(y))
f(0) = f(1) = 0
x7→ ax +bR
xR, g(x) = f(x)(ax +b)gR
a b g g(0) = g(1) = 0
g(0) = 0 f(0) b= 0 b=f(0)
g(1) = 0 f(1) ab= 0 a=f(1) b=f(1) f(0)
g:x7→ f(x)(f(1) f(0))
  
a
xf(0)

b
R
g(0) = g(1) = 0
xR, f(x) = ax +b f
R
an=+
0
sin t
n
t(1 + t2)dt
(an)
uMn(R)
MMn(R), u(M) = 1
3(2MtM)
u u
(u) det(u)
nNt7→ sin t
n
t(1 + t2)]0,+[
sin t
n
t(1 + t2)
t0
t
n
t(1 + t2)=1
n
1
1 + t2
t0
1
nlim
n+
sin t
n
t(1 + t2)=1
n
t7→ sin t
n
t(1 + t2)[0,+[
1
n[0, a], a > 0
uR,|sin u|6|u| ∀nN,tR,
sin t
n
t(1 + t2)
6t/n
t(1 + t2)=1
n(1 + t2)
nNt7→ 1
n(1+t2)]0,+[ [0,+[
t7→ sin t
n
t(1 + t2)]0,+[
annN
nN,|an|=+
0
sin t
n
t(1 + t2)dt
6+
0
sin t
n
t(1 + t2)
dt 6+
0
dt
n(1 + t2)=π
2×1
n
lim
n+an= 0
• ∀nN, an=+
0
sin t
n
t(1 + t2)dt =1
n+
0
sin t
n
t
n×1
1 + t2dt
nN,t]0,+[, gn(t) = sin t
n
t
n×1
1 + t2
lim
u0
sin u
u= 1 t]0,+[,lim
n+gn(t) = 1
1 + t2
|sin u|6|u|
nN,t]0,+[,|gn(t)|61
1+t2t7→ 1
1+t2]0,+[
[0,+[
lim
n++
0
gn(t)dt =+
0
1
1 + t2dt = [ t]+
0=π
2
an=1
n+
0
sin t
n
t
n×1
1 + t2dt =1
n+
0
gn(t)dt
an
t0
π
2n
λRMMn(R)− {0}
u(M) = λ.M 2MtM= 3λM
⇒ −tM= (3λ2)M
M= (2 3λ)tM
=M= (2 3λ)(2 3λ)M
=(2 3λ)2= 1 M̸= 0
=23λ= 1 2 3λ=1
=λ= 1 λ=1
3(u)1,1
3
• ∀MMn(R), u(M) = M2MtM= 3MtM=M
u
Mn(R)E1(u) = An(R)n(n1)
2
• ∀MMn(R), u(M) = 1
3M2MtM=MtM=M
1
3u
Mn(R)E1
3(u) = Sn(R)n(n+1)
2
Sn(R)An(R)Mn(R)
uMn(R)u
Mn(R)Sn(R)An(R)u
n(n1)
2= dim(E1(u)) n(n+1)
2= dim(E1
3(u)) 1
3
(u) = n(n1)
2+1
3
n(n+ 1)
2=3n23n+n2+n
6=n2n
3
det(u) = 1n(n1)
2×1
3n(n+1)
2
=1
3n(n+1)
2
aR
f:t7→ 1
t+a
fR+R+
f(t)dt
u=et
A∈ Mn(R) 2A3+ 3A2+ 6AIn= 0
A
AMn(C)
det(A)>0
D={(x, y)R2, x2+y261}f
(x, y)∈ D, f(x, y) = x33x(1 + y2)
fD
t7→ t+aR+xR, x >1
f:t7→ 1
t+aR+
[0, b]
f(t) = 1
t+a=2
et+et+ea+ea62
et= 2et
t7→ 2etR+f
J=+
0
dt
t+a=+
0
2etdt
e2t+ 2 a et+ 1
u=etJ=+
0
2du
u2+ 2u a + 1
J=+
0
2du
(u+a)22a+ 1 =+
0
2du
(u+a)22a=+
0
2du
(u+a+a)(u+aa)
=+
0
2du
(u+ea)(u+ea)
2
(u+ea)(u+ea)=(u+ea)(u+ea)
(u+ea)(u+ea)×2
eaea=1
a1
u+ea1
u+ea
+
0
1
u+e±adu J
x
0
1
u+e±adu
x > 0, Jx=x
0
dt
t+a=1
ax
0
1
u+eadu x
0
1
u+eadu
Jx=1
aln u+ea
u+eax
0
=1
aln x+ea
x+ealn e2a
x+,x+ea
x+ea1 ln x+ea
x+ea0
J= lim
x+Jx=ln(e2a)
a=2a
a
A∈ Mn(R) 2A3+ 3A2+ 6AIn= 0
A(2A2+ 3A+ 6In) = InA A1= 2A2+ 3A+ 6In
P(X) = 2X3+ 3X2+ 6X1A
n P
xR, P (x) = 6x2+ 6x+ 6 = 6(x2+x+ 1) >0
PR−∞ +
α P (0) = 1P α > 0
P(X)
P(X)β β
α, β β P (X)
P(X)C[X]
AMn(C)
C(A)⊂ {α, β, β}
AMn(C)
1 / 88 100%

Oraux-2015-solutions

Telechargé par ABDELLAH BEN ZANNOU
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