Question

If $f(x)$ is a polynomial satisfying $ f(x) =\frac {1} {2} \begin{vmatrix} f{(x)} & f (\frac {1} {x}) - f(x) \\[0.3em] 1 & (\frac{1} {x}) \\[0.3em] \end{vmatrix} $ 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 $f\left(x\right) = a_{0}x^{n} + a_{1}x^{n-1} + .... + a_{n-1} x + a_{n} \left(a_{0} \ne0\right)$
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 $ [$\because \, a_0 \neq 0]$
$a_{0}a_{n-1}+a_{n}a_{1} = a_{1} \Rightarrow a_{0}a_{n-1} = 0 \Rightarrow a_{n-1} = 0,$
Similarly $a_{n-2} = 0, ........, a_{1} =0$
Again $a^{2} _{0} +a^{2} _{n} + a^{2} _{1} +a^{2} _{n-1} + ...... = 2a_{n}$
$\Rightarrow a^{2} _{0} + a^{2} _{n} = 2a_{n} \Rightarrow a^{2} _{0} = 2 - 1 = 1 \Rightarrow a_{0 } = \pm 1$
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.