Correction - Pierre

f f(x) = 1 + 2x+|x|
ln(x)
f Df
f Dfg(x) =
2(ln x)2+ ln x−1
g(x)
f
(x7→ ln x)x∈R∗
+(x7→ ln x)x= 1 1
ln x
x∈R∗
+\ {1}. Df=R∗ +\ {1}
∀x∈R∗
+\ {1} |x|=x
f(x) = 1 + 2x+x
ln(x).
fDfDf∀x∈Df
f0(x) = 2 +
1×ln x−x×1
x
(ln x)2
=2(ln x)2+ ln x−1
(ln x)2.
∀x∈Df,(ln x)2>0f0g(x) = 2(ln x)2+ ln x−1
X= ln x X ∈]−∞ ; +∞[ 2X2+X−1
∆=12−4×2×(−1) = 9 >0
X1,2=−1±3
4
X1=−1X2=1
22X2+X−1
X
2X2+
X−1
−∞ −11
2+∞
+0−0+
X= ln x x =eXX=−1x=e−1X=1
2x=e1/2
lim
X→−∞ eX= 0 limX→+∞eX= +∞
g(x)
x
g(x)
0e−1e1
2+∞
+0−0+
f x = 1 f

x
g(x)
f0(x) =
g(x)
(ln x)2
f(x)
0e−1e1
2+∞
+0− − 0+
+0− − 0+
1 + e−1 1 + e−1
1+4e1/2 1+4e1/2
x+10
x+ 1 −3x+ 3 x2+ 1 ≤3.(E)
(E)P(x)
Q(x)≤0P Q
1P P
(E)
x+10
x+ 1 −3x+ 3 x2+ 1 ≤3
⇐⇒ x+10
x+ 1 −3x+ 3 x2+ 1 −3≤0
⇐⇒ x(x+ 1)(x2+ 1) + 10(x2+ 1) −(3x+ 3)(x+ 1) −3(x+ 1)(x2+ 1)
(x+ 1)(x2+ 1) ≤0
⇐⇒ x(x3+x+x2+ 1) + 10x2+ 10 −3x2−3x−3x−3−3x3−3x−3x2−3
(x+ 1)(x2+ 1) ≤0
⇐⇒ x4−2x3+ 5x2−8x+ 4
(x+ 1)(x2+ 1) ≤0,
(E)P(x)
Q(x)≤0P(x) = x4−2x3+ 5x2−8x+ 4
Q(x)=(x+ 1)(x2+ 1)

P(1) = 1 −2+5−8 + 4 = 0 1 P P (x−1)
x4−2x3+ 5x2−8x+ 4 x−1
−(x4−x3)x3−x2+ 4x−4
−x3+ 5x2−8x+ 4
−(−x3+x2)
4x2−8x+ 4
−(4x2−4x)
−4x+ 4
−(−4x+ 4)
0
x
P(x) = (x−1)(x3−x2+ 4x−4) = (x−1)Q2(x).
P Q2(1) = 0 Q2
(x−1)
1
Q2Q32
∀x∈R, Q2(x)=(x−1)Q3(x).
Q32a, b, c x
Q3(x) = ax2+bx +c
x3−x2+ 4x−4=(x−1)(ax2+bx +c)
⇐⇒ x3−x2+ 4x−4=(x−1)(ax2+bx +c)
⇐⇒ x3−x2+ 4x−4 = ax3+bx2+cx −ax2−bx −c
⇐⇒ x3−x2+ 4x−4 = ax3+ (b−a)x2+ (c−b)x−c
a= 1
−a+b=−1
−b+c= 4
−c=−4
a= 1 b= 0 c= 4 R(x) = x2+ 4
Q(x) = (x−1)(x2+4) P(x)=(x−1)2(x2+ 4)
x2+ 4 2
∆=02−4× ×1×4 = −16 <0
P(x) = (x−1)2(x2+ 4)
P(x)
Q(x)P(x) = (x−

1)2(x2+ 4) Q(x) = (x+ 1)(x2+ 1) P Q
(x+ 1) (x2+ 1)
x
P(x)
Q(x)
P(x)
Q(x)
−∞ −11+∞
+ + 0+
−0+ +
−+0+
(E)S= ]−∞ ;−1[ ∪ {1}
R
bxc+1
2=|x|.
x n ∈Zx∈[n;n+ 1[ bxc=n
n+1
2=|x|.
n+ 1 ≤0
n+ 1 ≤0n≤ −1n+1
2≤ −1 2<0|x| ≥ 0
n+ 1 >0n≥0n∈Z
n+1
2=x,
x∈[n;n+ 1[ x=n+1
2
∀n∈Nx=n+1
2
bxc+1
2=n+1 2+1 2=n+1 2
|x|=|n+1
2|=n+1 2,
x=n+1
2n∈N
S=n+1
2|n∈N