The domain of f(x) = frac{1}{{sqrt {{{log }_{ | vedclass

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Question

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive: $\log_{\frac{\pi}{4}}(\sin^{-1} x) - 1 > 0$

Vedclass · Accepted Answer

$\left( 0, \frac{1}{\sqrt{2}} \right)$

Vedclass · Answer

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive: $\log_{\frac{\pi}{4}}(\sin^{-1} x) - 1 > 0$ $\Rightarrow \log_{\frac{\pi}{4}}(\sin^{-1} x) > 1$ Since the base $\frac{\pi}{4} $\sin^{-1} x Also,the argument of the logarithm must be positive: $\sin^{-1} x > 0 \Rightarrow x > 0$ Additionally,for $\sin^{-1} x$ to be defined,$x \in [-1, 1]$. Combining these conditions: $0 Taking the sine of all parts: $\sin(0) $0 Thus,the domain is $x \in \left( 0, \frac{1}{\sqrt{2}} \right)$.

The domain of $f(x) = \frac{1}{{\sqrt {{{\log }_{\frac{\pi }{4}}}({{\sin }^{ - 1}}x) - 1} }}$ is:

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