First slide

Functions (XII)

Question

A polynomial $f (x)$ satisfies the condition $f (x) . f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ and $f (10) = 1001,$ then the value of $f (20)$ must be

Moderate

Solution

$f (x) = x^{n} + 1$
$∵ f (10) = 10^{n} + 1 = 1001$ (given)
$∴ 10^{n} = 1000 = 10^{3}$
$\Rightarrow n = 3,$
Then $f (x) = x^{3} + 1$
$∴ f (20) = {(20)}^{3} + 1$
$= 8001$

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