)+f(x)]⋅⋅⋅⋅⋅⋅⋅(i) Let f(x)=a0xn+a1xn−1+...+an−1x+an (a0≠0) Putting this value of f(x) in Eq. (i), f(x)=
1
2
[f(x)⋅f(
1
x
)−f(
1
x
)+f(x)] ⇒f(x)⋅f(
1
x
)=f(
1
x
)+f(x) Hence, f(x)=xn+1 or −xn+1 Now, f(2)=2n+1=33 ⇒2n=32 ⇒2n=25 ⇒n=5 And f(2)=−2n+1=33 ⇒−2n=33−1=32=25 which is not possible. ∴f(x)=xn+1=x5+1 So, f(3)=35+1=243+1=244
