AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Let the function f (x) is given by
f(x)=a0xn+a1xn−1+a2xn−2 +.....+an−1x+an
The given relation is
f(x).(

1

x

)=f(x)+f(

1

x

)
Substitute f(x) in the given equation
(a0xn+a1xn−1+....+an) (a0xn+a1xn−1+....+an)
Compare the coefficient of xn,
a0an=a0⇒an=1
Compare the coefficient of xn−1
a0an−1+ana1=a1
a0an−1=0
So,
an−1=an−2=.....=a1=0
a0=±1
f(x)=1±xn
f(4)=1±4n=257
4n=256
n=4
Therefore, f(3)=1+34=82