Question

If the polynomial satisfies $f\left(x\right)=\frac{1}{2}\left|\begin{array}{cc}f\left(x\right)& f\left(\frac{1}{x}\right)-f\left(x\right)\\ 1& f\left(\frac{1}{x}\right)\end{array}\right|$ and $f\left(2\right)=17$, then the value of $f\left(3\right)$ is

• A. $81$
• B. $27$
• C. $26$
• D. $82$

Answer 1

Answer: (D)

$82$

Given that $f\left(x\right)=\frac{1}{2}\left|\begin{array}{cc}f\left(x\right)& f\left(\frac{1}{x}\right)-f\left(x\right)\\ 1& f\left(\frac{1}{x}\right)\end{array}\right|$

$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$

$\Rightarrow$ $f\left(x\right)f\left(\frac{1}{x}\right)-f\left(x\right)=f\left(\frac{1}{x}\right)$

$\Rightarrow$ $f\left(x\right)=\frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right)-1}$ .. (i)

Also, $f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$

$\Rightarrow$ $f\left(x\right)f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)=f\left(x\right)$

$\Rightarrow$ $f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{f\left(x\right)-1}$ ..(ii)

On multiplying Eqs. (i) and (ii), we get,

$f\left(x\right)f\left(\frac{1}{x}\right)=\frac{f\left(\frac{1}{x}\right)f\left(x\right)}{\{f\left(\frac{1}{x}\right)-1\}\{f\left(x\right)-1\}}$

$\Rightarrow$ $(f\left(\frac{1}{x}\right)-1)(f\left(x\right)-1)=1$ ..(iii)

Since, $f\left(x\right)$ is polynomial function, so $\left(f\right(x)-1)$ and $f\left(\frac{1}{x}\right)-1$ are reciprocals of each other.

Also, $x$ and $\frac{1}{x}$ are reciprocals of each other.

Thus, Eq. (iii) can hold only when

$f\left(x\right)-1=\pm x^n$, where $n\in N$

$\therefore$ $f\left(x\right)=\pm x^n+1$ but $f\left(2\right)=17$

$\Rightarrow$ $\pm 2^n+1=17$ $\Rightarrow$ $2^n=16$

$\Rightarrow$ $2^n=2^4$ $(\because 2^n>0)$

$\Rightarrow$ $n=4$

So, $f\left(x\right)=x^4+1$

Hence, $f\left(3\right)=3^4+1=82$
Ans: 82