Correction DS No 7 1 ES2 Exercice 1 1. f 2. f Exercice 2 Exercice 3 1.

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f[−5; 5] f(x) = x3+ 3x2−9x+ 2
f′(x) = 3x2+ 6x−9
f′(x)ax2+bx +c a = 3 b= 6 c=−9
∆ = b2−4ac = 62−4×3×(−9) = 144 >0√144 = 12
x1=−b−√∆
2a=−6−12 2×3=−3x2=−b+√∆ 2a=−6 + 12 2×3= 1
a= 3 >0
x−5−3 1 5
f′(x) + −+
f
−3
29
@@@R
−3
157
f[−5; 5] −3x=−5x= 1
f[−5; 5] 157 x= 5
0y=f′(0) ×(x−0) + f(0)
f′(0) = −9f(0) = 2
y=−9x+ 2
−1y=f′(−1) ×(x−(−1)) + f(−1)
f′(−1) = −12 f(−1) = 13
y=−12(x+ 1) + 13
y=−12x−12 + 13
y=−12x+ 1
X10 0,25
p(X= 6) = 10
6×0,256×(1 −0,25)10−6= 210 ×0,256×0,754
p(X61) = p(X= 0) + p(X= 1) = 10
0×0,250×(1 −0,25)10−0+10 1×0,251×(1 −0,25)10−1
p(X61) = 0,7510 + 10 ×0,25 ×0,759
E(X) = n×p= 10 ×0,25 = 2,5
X35 0,34
35
0= 1 35 7= 6724520
p(X= 5) ≈0,00569
p(X67) ≈0,05692
p(X>20) = 1 −p(X619) ≈1−0,99582612 ≈0,00417

n= 20
S p = 0,817 X
n= 20 p= 0,817
p(X=k) = 20
k×0,817k×0,18320−k06k620
p(X= 10) = 20
10 ×0,81710 ×0,18310 = 184756 ×0,81710×0,18310 ≈0,001
p(X64) ≈3,6×10−9≈0
p(X>1) = 1 −pX>1= 1 −p(X < 1) = 1 −p(X= 0) X
p(X= 0) = 20
0×0,8170×0,18320 = 0,18320
p(X>1) = 1 −0,18320 ≈1
n= 100
S p = 0,03 X
n= 100 p= 0,03
p(X= 2) = 100
2×0,032×0,9798 ≈0,23
p(X= 4) = 100
4×0,034×0,9796 ≈0,17
p(X>3) = 1 −pX>3= 1 −p(X < 3) = 1 −p(X62) ≈1−0,41977508 ≈0,58
E(X) = n×p= 100 ×0,03 = 3
P(X63) ≈0,65
0,65
n= 15
S p = 0,004 X
n= 15 p= 0,004
p(X= 2) = 15
2×0,0042×(1 −0,004)15−2= 105 ×0,0042×0,99613 ≈0,002
p(X>1) = 1 −p(X= 0) = 1 −15
0×0,0040×(1 −0,004)15−0= 1 −0,99615 ≈0,058
E(X) = n×p= 15 ×0,004 = 0,06