a-

k(0)= 2×0 -5 = -5

k(1)= 2×1 -5 = -3

b-

k(-4)= 2×(-4) -5= -13

k(1/4)= 2×(1/4) -5= -9/2

c-

on a k(x) =0

alors 2x-5 = 0

d'où 2x=5

Cad x=5/2

S={5/2}

d-

on a k(x) = 1

2x-5=1

2x= 1+5

x= 6

x=6/2

x=3

donc k(3)=1

on a k(x)= 15

2x-5=15

2x=20

x=20/2

x=10

donc k(10)= 15