If `f(x)` is a polynomial satisfying `f(x)f(1/x)=f(x)+f(1/x)` and `f(3)=28` then `f(4)=`

Options

A. 63
B. 65
C. 66
D. 27

Answer

B

Explanation

Let
`f(x)=a_(0)x^(n)+a_(1)x^(n-1)+a_(2)x^(n-1)+…..+a_(n-1)x+a_(n) `
Then,
`f(x)f((1)/(x))=f(x)+f((1)/(x))`
`implies (a_(0)x^(n)+a_(1)x^(n-1)+…a_(n-1)x+a_(n))xx ((a_(0))/(x^(n))+(a_(1))/(x^(n-1))+…..+(a_(n-1))/(x)+a_(n))`
`=(a_(0)x^(n)+a_(1)x^(n-1)+.....+a_(n-1)x+a_(n))+((a_(0))/(x^(n))+(a_(1))/(x^(n-1))+.....+(a_(n-1))/(x)+a_(n))`
On comparing the coefficient of like powers of x on both sides , we get
`a_(0) ne pm1 , a_(n)=1 and a_(1)=a_(2)=........=a_(n-1)=0`
`:. f(x)=x^(n)+1 or , f(x)-=x^(n)+1`
If `f(x)=-x^(n)+1`, then
`f(x)=-3^(n)+1 ne 28" " [ :' f(3)=28]`
So, `f(x)=x^(n)+1`
`implies f(3)=3^(n)+1 implies 28=3^(n)+1 implies n=3 `
`:. f(x)=x^(3)+1`
`implies F(4)=4^(3)+1=65`.