If $f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16}) (1 + x^{32})$ then $f^{1} (x)$ at $x = 0$ is

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answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{(1 - x) (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16}) (1 + x^{32})}{1 - x} \\ = \frac{(1 - x^{2}) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16}) (1 + x^{32})}{1 - x} \\ = \frac{(1 - x^{4}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16}) (1 + x^{32})}{1 - x} \\ f (x) = \frac{1 - x^{64}}{1 - x} \\ f^{1} (x) = \frac{(1 - x) \frac{d}{d x} (1 - x^{64}) - (1 - x^{64}) \frac{d}{d x} (1 - x)}{{(1 - x)}^{2}} \\ = \frac{(1 - x) (- 64 x^{63}) - (1 - x^{64}) (- 1)}{{(1 - x)}^{2}} \\ = \frac{- 64 x^{63} + 64 x^{64} + 1 - x^{64}}{{(1 - x)}^{2}} \\ = \frac{63 x^{64} + 1 - 64 x^{63}}{{(1 - x)}^{2}} \\ {[f^{1} (x)]}_{a t x = 0} = \frac{1}{{(1 - 0)}^{2}} = 1 \end{array}$