Q41 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

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
  • Correct answer
    C
    82
  • D
    244

Solution

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, $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ $$ \begin{aligned} \Rightarrow\left(a_0 x^n+a_1 x^{n-1}+\ldots+a_n\right) & \left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\ldots+a_n\right) \end{aligned} $$ 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 $$ \begin{aligned} & \Rightarrow a_0 a_{n-1}+a_n a_1=a_1 \\ & \Rightarrow \quad a_0 a_{n-1}+a_1=a_1 \\ & {\left[\text { as } a_n=1\right]} \\ & \Rightarrow \quad a_0 a_{n-1}=0 \\ & \Rightarrow \quad a_{n-1}=0 \quad\left[\text { as } a_0 \neq 0\right] \\ & \end{aligned} $$ Similarly, $\quad a_{n-1}=a_{n-2}=\ldots=a_1=0$ and $\quad a_0= \pm 1$ $$ \begin{aligned} & \therefore \quad f(x)=1 \pm x^n \\ & f(4)=1 \pm 4^n=257 \Rightarrow 4^n=256 \\ & \Rightarrow \quad 4^n=256 \quad \Rightarrow \quad 4=4 \\ & f(x)=1+x^4 \\ & \end{aligned} $$ So, $f(3)=1+3^4=82$

Hence, the correct answer is option: 'C'.

Relevant Questions