Question

Fie functia f:R->R,f(x)=a$\sqrt{2x}$+b$\sqrt{5}$ ;a,b∈Q
Demonstrati ca daca punctul A(√2,√5)∈Gf,atunci f este functie constanta.

Answer 1

f(√2)=√5
f(√2)=a*√(2*√2)+b*√5=√5=.>
a=0  si  b=1
f(x)=0*√2x+√5=√5=constanta