Devoir Surveillé MPSI : Analyse, Algèbre, Trigonométrie

2015/2016
f(x)=(x+ 2) e1
x
f f(x)=(x+ 2) e1
x
n∈N∗lim
x→+∞
ex
xn= +∞lim
x→−∞ xnex= 0
lim
x→0
ex−1
x= 1
Dff
f
f
lim
t→0−
f0(t) (G)f
(G)f
xy=yx
N2(E)x∈Ny∈N
(E)
xy=yx
0< x < y
f f(x) = ln(x)
xe= 2.7 10−1
ln ab=bln(a)a > 0b∈R
f
(x, y) (E)f(x) = f(y)
(x, y) (E)x∈]1, e[y∈[e, +∞[
(E)
zn= (z−1)n= 1
z∈C
z= 1
z−1= 1
(z, n)∈C×N∗
(z)n= 1
(z−1)n= 1

2015/2016
t∈Ra= sin(t) + cos(t)
sin(t) cos(t)a
sin3(t) + cos3(t)a
sin5(t) + cos5(t)a
n∈N∗z∈C(z+i)2n+1 −(z−i)2n+1 = 0
2z4−4z3+ (1 −4i)z2−(6 −8i)z−3+4i= 0
z
2z4−4z3+ (1 −4i)z2−(6 −8i)z−3+4i= 0
n∈Nω=e2iπ
n
1 + ω+ω2+··· +ωn−1= 0
n−1
X
k=0
cos 2kπ
n= 0
n−1
X
k=0 ωk−1
2n
p∈[[0, n]]
n−1
X
k=0
ωk p n p = 0 p=n
A > 0φ∀x∈R,cos(x) + √3 sin(x) = Acos (x−φ)
cos(x)−sin(x)
R
f(x) = cos(x) + √3 sin(x) + √2
Rcos(x)−sin(x)−√2
2= 0
[0,2π]cos(x) + √3 sin(x) + √2
cos(x)−sin(x)−√2
2

2015/2016
f(x)=(x+ 2) e1
x
f f(x)=(x+ 2) e1
x
n∈N∗lim
x→+∞
ex
xn= +∞t=−xlim
x→−∞ xnex= 0
lim
x→0
ex−1
x= 1
Df=R∗
+∞lim
x→+∞f(x) = +∞lim x→+∞
f(x)
x= 1 lim
x→+∞(f(x)−x)=3
f(x)−x=xe1
x−1+ 2e1 x
f Y =X+ 3
−∞ f
f Y =X+ 3
0+lim
x→0+f(x) = +∞X= 0
0−lim
x→0−
f(x) = 0
fR∗R∗
∀x∈R∗f0(x) = (x+ 1)(x−2)
x2e1
x
fR∗R∗
f0(x) = (x+ 1)(x−2)
x2e1
x
x
f0(x)
f(x)
−∞ −10 2 +∞
+0− − 0+
−∞−∞
1
e
1
e
0
+∞
4√e4√e
+∞+∞
t=1
x0−
t−∞
lim
t→−∞ ttet= 0 lim
t→0−
f0(t) = 0
f0−
xy=yx
(E)x∈Ny∈Nxy=yx0< x < y
f f(x) = ln(x)
xe= 2.7 10−1

2015/2016
f(x) = ln(x)
xfR∗
+∀x∈R∗ +f0(x) = 1−ln(x)
x2f
Y= 0
X= 0
x
f0(x)
f(x)
0e+∞
+0−
−∞−∞
1
e
1
e
00
(x, y) (E)xy=yxyln(x) = xln(y)
x y f(x) = f(y)
(x, y) (E)f(x) = f(y)
f]0, e[ [e, +∞[f(x) = f(y)x<y
x∈]0, e[y∈[e, +∞[f(x) = f(y)>0 ]0, e[f
]1, e[x∈]1, e[y∈[e, +∞[
]1, e[ (x, y)x= 2 y > e
ln(y)
y=ln(2) 2f[e, +∞[0,1 e
[e, +∞[fln(2)
24f(4) = ln(2) 2
(x, y) (E) (x, y) = (2,4) (2,4)
zn= (z−1)n= 1
(S)z∈C
z= 1
z−1= 1
z(S)
z∈U z z =eit t∈[0,2π[z−1= 1 ⇐⇒ 2 sin t
2t=π 3
t=5π
3z=−j z =−j2
(S)
(S)−j, −j2
(S2) (z, n)∈C×N∗
(z)n= 1
(z−1)n= 1
(z, n) (S2)
z=z−1= 1 z=−j−j2(S2) (−j)n= 1 = −j2n
n n
z−j−j2n(z, n)
(S2)
(S){(−j, 6p) ; p∈N}[−j2,6p;p∈N
t∈Ra= sin(t) + cos(t)

2015/2016
a2= sin2(t) + 2 sin(t)×cos(t) + cos2(t) = 1 + 2 sin(t) cos(t)sin(t) cos(t) = a2−1
2
a3= sin3(t)+3 sin2(t) cos(t)+3 sin(t) cos2(t)+cos3(t) = sin3(t) + cos3(t)+3 sin(t) cos(t) (sin(t) + cos(t))
sin3(t) + cos3(t) = a3−3aa2−1
2sin3(t) + cos3(t) = a3−a2 2
sin5(t)+cos5(t)a5sin(t) cos(t) sin3(t)+cos3(t)
sin5(t) + cos5(t) = a5−a4
4
n∈N∗(E) (z+i)2n+1 −(z−i)2n+1 = 0
i z 6=i(E)
z+i
z−i2n+1
= 1 z+i
z−i
z(E)⇐⇒ ∃k∈[[1,2n]] 
z+i
z−i=e2ikπ
2n+1 ⇐⇒ ∃k∈[[1,2n]] 2zsin kπ
2n+ 1e
ikπ
2n+1 = 2 cos kπ
2n+ 1e
ikπ
2n+1
z(E)⇐⇒ ∃k∈[[1,2n]] z= cotan kπ
2n+ 1
2n
2z4−4z3+ (1 −4i)z2−(6 −8i)z−3+4i= 0
z
2z4−4z3+ (1 −4i)z2−(6 −8i)z−3+4i= 0
x
x⇐⇒ 2x4−4x3+ (1 −4i)x2−(6 −8i)x−3+4i= 0 ⇐⇒ 2x4−4x3+x2−6x−3+i−4x2+ 8x+ 4= 0
⇐⇒ 
2x4−4x3+x2−6x−3=0
−4x2+ 8x+ 4 = 0 ⇐⇒ 
2x2+ 3x2−2x−1= 0
x2−2x−1=0 ⇐⇒ x∈n1 + √2,1−√2o
1 + √2 1 −√2
z⇐⇒ 2x2+ 3 −4ix2−2x−1= 0
S=(1 + √2,1−√2,√2
2+i√2,−√2 2−i√2)
n∈Nω=e2iπ
n
n−1 + ω+ω2+··· +ωn−1= 0
n−1
X
k=0
cos 2kπ
n= 0
ωk−1
2= 4 sin2kπ
n= 2 −2 cos 2kπ nn−1
X
k=0 ωk−1
2= 2n
p n ωp= 1
n−1
X
k=0
ωk p =np n ωp
n−n