Domain of f(x)=sin−1[2x2−3]+log2[log12(x2−5x+5)] where [t] is greatest integer function.

f(x)=sin−1[2x2−3]+log2[log12(x2−5x+5)]Take angle of Sin−1 as T1 which lies between T1:−1≤[2x2−3]≤1⇒−1≤2x2−3 0(x−(5−52))(x−(5+52))>0 T3: log12(x2−5x+5)>0⇒x2−5x−5<1⇒x2−5x+4<0⇒T3:x∈(1,4)T1∩T2∩T3=(1,5−52)

Domain of f(x)=sin−1[2x2−3]+log2[log12(x2−5x+5)] where [t] is greatest integer function.

f(x)=sin−1[2x2−3]+log2[log12(x2−5x+5)]Take angle of Sin−1 as T1 which lies between T1:−1≤[2x2−3]≤1⇒−1≤2x2−3 0(x−(5−52))(x−(5+52))>0 T3: log12(x2−5x+5)>0⇒x2−5x−5<1⇒x2−5x+4<0⇒T3:x∈(1,4)T1∩T2∩T3=(1,5−52)

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Domain of $f (x) = \sin^{- 1} [2 x^{2} - 3] + \log_{2} [\log_{\frac{1}{2}} (x^{2} - 5 x + 5)]$ where $[t]$ is greatest integer function.

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$[- \sqrt{\frac{5}{2}}, \frac{5 - \sqrt{5}}{2}]$

$[\frac{5 - \sqrt{5}}{2}, \frac{5 + \sqrt{5}}{2}]$

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

$[1, \frac{5 + \sqrt{5}}{2}]$

answer is C.

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Detailed Solution

$f (x) = \sin^{- 1} [2 x^{2} - 3] + \log_{2} [\log_{\frac{1}{2}} (x^{2} - 5 x + 5)]$
Take angle of $S i n^{- 1}$ as $T_{1}$ which lies between
$\begin{array}{l} T_{1} : - 1 \leq [2 x^{2} - 3] \leq 1 \Rightarrow - 1 \leq 2 x^{2} - 3 < 2 \\ \Rightarrow 2 < 2 x^{2} < 5 \Rightarrow 1 < x^{2} < \frac{5}{2} \\ T_{1} : x \in (- \frac{5}{2}, - 1) \cup (1, \frac{5}{2}) \end{array}$
Similarly
$\begin{array}{l} T_{2} : x^{2} - 5 x + 5 > 0 \\ (x - (\frac{5 - \sqrt{5}}{2})) (x - (\frac{5 + \sqrt{5}}{2})) > 0 \end{array}$
$\begin{array}{l} T_{3} : \log_{\frac{1}{2}} (x^{2} - 5 x + 5) > 0 \Rightarrow x^{2} - 5 x - 5 < 1 \\ \Rightarrow x^{2} - 5 x + 4 < 0 \Rightarrow T_{3} : x \in (1, 4) \\ T_{1} \cap T_{2} \cap T_{3} = (1, \frac{5 - \sqrt{5}}{2}) \end{array}$