Question

If $f(x)$ is a polynomial function satisfying $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(4)=257$, then $f(3)=$

• A. 28
• B. 65
• C. 82
• D. 244

Answer 1

Answer: (C)

Let
$f(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n} $
Then, $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$
$\Rightarrow \, \left(a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}\right) $
$\left(\frac{a_{0}}{x^{n}}+\frac{a_{1}}{x^{n-1}}+\ldots+a_{n}\right)$
On comparing the coefficient of $x^{n}$, we have
$a_{0} a_{n}=a_{0} $
$\Rightarrow \, a_{n}=1$
Comparing the coefficient of $x^{n-1}$, we have
$\Rightarrow \, a_{0} a_{n-1}+a_{n} a_{1}=a_{1}$
$\Rightarrow \, a_{0} a_{n-1}+a_{1}=a_{1}$
[ as $ a_{n}=1]$
$\Rightarrow \, a_{0} a_{n-1}=0$
$\Rightarrow \, a_{n-1}=0$
[ as $ a_{0} \neq 0]$
Similarly, $ a_{n-1}=a_{n-2}=\ldots=a_{1}=0$
and $a_{0}=\pm\, 1$
$\therefore \, f(x)=1 \pm x^{n}$
$f(4)=1 \pm 4^{n}=257 $
$\Rightarrow \,4^{n}=256$
$\Rightarrow \, 4^{n}=256$
$ \Rightarrow \, 4=4$
$f(x)=1+x^{4}$
So, $f(3)=1+3^{4}=82$