Consider the function $g (x)$ defined as $g (x) (x^{2^{2008}} - 1) = (x + 1) (x^{2} + 1) (x^{4} + 1) . . . . . . (x^{2^{2008}} + 1)$ . The vaule of $g (2)$ equals

1

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2^{2008} - 1

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2^{2008}

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2

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Solution

The correct option is A $1$
$g (x) (x^{(2^{2008} - 1)} - 1) = (x + 1) (x^{2} + 1) (x^{4} + 1) . . . . . . (x^{2^{2007}} + 1) - 1$
Multiplying and dividing by (x-1) on RHS, we get
$g (x) [(x^{2008} - 1) - 1] = \frac{(x + 1) (x^{2} + 1) . . . (x^{2007} + 1) - (x - 1)}{x - 1}$
$g (x) [(x^{2008} - 1) - 1] = \frac{(x^{2} - 1) (x^{2} + 1) . . . (x^{2007} + 1) - (x - 1)}{x - 1}$
Upon simplifying we get.
$g (x) [(x^{2008} - 1) - 1] = \frac{(x^{2008} - 1) - (x - 1)}{x - 1}$
$g (x) = \frac{(x^{2008} - 1) - (x - 1)}{(x - 1) ((x^{2008} - 1) - 1)}$
Therefore $g (2) =$
$= \frac{(2^{2008} - 1) - (2 - 1)}{(2 - 1) ((2^{2008} - 1) - 1)}$
$= \frac{(2^{2008} - 2)}{2^{2008} - 2}$
$= 1$

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Similar questions

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