(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.