The domain of the function f(x) = sin^{-1}[2x | vedclass

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(C) For the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$ to be defined:$1.$ For $\sin^{-1}[2x^2 - 3]$,the argument must sa

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$\left(1, \frac{5-\sqrt{5}}{2}\right)$

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(C) For the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$ to be defined:$1.$ For $\sin^{-1}[2x^2 - 3]$,the argument must satisfy $-1 \leq [2x^2 - 3] \leq 1$. Since the range of $\sin^{-1}$ is $[-1, 1]$,we require $-1 \leq [2x^2 - 3] Actually,for $\sin^{-1}(y)$,$y \in [-1, 1]$. Thus,$-1 \leq [2x^2 - 3] \leq 1$. This implies $-1 \leq 2x^2 - 3 $2.$ For $\log_2(\log_{1/2}(x^2 - 5x + 5))$,we need $\log_{1/2}(x^2 - 5x + 5) > 0$. Since the base $1/2 $3.$ Solving $x^2 - 5x + 5 > 0$: Roots are $\frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$. So $x \in (-\infty, \frac{5-\sqrt{5}}{2}) \cup (\frac{5+\sqrt{5}}{2}, \infty)$.$4.$ Solving $x^2 - 5x + 5 Intersection of all conditions:From $1$: $x \in [1, \sqrt{2.5}) \cup (-\sqrt{2.5}, -1]$.From $2, 3, 4$: $x \in (1, \frac{5-\sqrt{5}}{2})$.Combining these,the domain is $(1, \frac{5-\sqrt{5}}{2})$.

The domain of the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$,where $[t]$ is the greatest integer function,is:

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