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The domain of f(x)=sin^-1 5x is

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(0, $\frac{1}{5}$ )

$(\frac{- 1}{5}, \frac{1}{5})$

$[\frac{- 1}{5}, \frac{1}{5}]$

answer is B.

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Detailed Solution

Given f(x)=sin^-15x

we know that, domain of sin ^-1x is is [-1,1]

$∴$ $- 1 \leq 5 x \leq 1$ $\Rightarrow \frac{- 1}{5} \leq x \leq \frac{1}{5}$

$∴$ x $\in [\frac{- 1}{5}, \frac{1}{5}]$

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