${(3 + x)}^{p / q}$ can be expanded in ascending powers of $x$ if

- 1 < x < 1

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x > 3

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- 3 < x < 3

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x < - 3

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Solution

The correct option is C $- 3 < x < 3$
$(x + 3)^{p / q}$
$= 3^{p / q} (\frac{x}{3} + 1)^{p / q}$
This can be expanded in the powers of x only when
$| \frac{x}{3} | < 1$
$- 1 < \frac{x}{3} < 1$
$- 3 < x < 3$

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