Question

If $f(x)$ is a polynomial satisfying $f(x)=\frac{1}{2}\left|\begin{array}{cc}f(x)& f(\frac{1}{x})-f(x)\\ 1& (\frac{1}{x})\end{array}\right|$ f(2) = 17, then the value of f (5) is

• A. 126
• B. 626
• C. -124
• D. 624

Answer 1

Answer: (B)

$f\left(x\right)=\frac{1}{2}\left[f\left(x\right).f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)-f\left(x\right)\right]$ $\Rightarrow f\left(x\right).f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ ...(1) Let $f\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+....+a_{n-1}x+a_n\left(a_0\ne 0\right)$
Putting in (1), we get
$\left(a_0{x}^n+a_1{x}^{n-1}+....+a_{n-1}x+a_n\right)$
$\left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+.....+\frac{a_{n-1}}{x}+a_n\right)$
= $a_0{x}^n+a_1{x}^{n-1}+....+a_{n-1}x+a_n+\frac{a_0}{x_n}+\frac{a_1}{x^{n-1}}+....+\frac{a_{n-1}}{x}+a_n$
Comparing the various co-efficients, we get
$a_0{a}_n=a_0\Rightarrow a_n=1$ [$\because a_0\ne 0]$
$a_0{a}_{n-1}+a_n{a}_1=a_1\Rightarrow a_0{a}_{n-1}=0\Rightarrow a_{n-1}=0,$
Similarly $a_{n-2}=0,........,a_1=0$
Again $a_0^2+a_n^2+a_1^2+a_{n-1}^2+......=2a_n$
$\Rightarrow a_0^2+a_n^2=2a_n\Rightarrow a_0^2=2-1=1\Rightarrow a_0=\pm 1$
Hence $f(x)=x^n+1$ or $-x^n+1$
Now $f(2)=2^n+1$ = 17 $\Rightarrow 2^n=16=2^4\Rightarrow n$ = 4
Clearly, $f(x)=-x^n+1$ is not possible.
$[\because f(2)=-2^n+1=17\Rightarrow -2^n$ = 16 not possible]
$\therefore f(x)=x^4+1\therefore f(5)=5^4$ + 1 = 625 + 1 = 626.