(1) If a=e+ 1, then

F(x)= radical sign (e x+x-e- 1)

F(y0)= radical sign (e y0+y0-e- 1)

e^y0+y0-e- 1>； =0

y0= 1

f( 1)=0

F(f( 1))=f(0)= radical sign (1-e- 1)= radical sign (-e) This is not true, so B and D are incorrect.

(2) If a = e (- 1)- 1

F(x)= radical sign (e x+x-e (- 1)+ 1).

e^x+x-e^(- 1)+ 1>； =0

e^y0+y0-e^(- 1)+ 1>； =0

y0=- 1

f(y0)=f(- 1)=0

F(f(- 1))=f(0) = radical sign (1-e (-1)+1) = radical sign (2-E (- 1).

Is established

(3) If a=0

Then f(x)= radical sign (e x+x)

F(y0)= radical sign (e y0+y0)

0 & lt=e^y0+y0<； =e+ 1

0 & lt= f(y0)& lt； = root sign (e+ 1)

F(f(y0))= radical sign (e f (y0)+f (y0)) = y0 (-1< = y0 <; = 1)

y0 & gt=0 f(y0)>= 1

F(f(y0)>= radical sign (e+ 1)> 1, so a cannot be 0.

So you can only choose a.