SOLUTION : (i) $\sqrt{\log _{1 / 3} \log _4\left([x]^2-5\right)}$
Domain
(i) $\quad \log _{1 / 3} \log _4\left([x]^2-5\right) \geq 0$ or $\log _4\left([x]^2-5\right) \leq 1$
or $[x]^2 \leq 9 \quad$ or $\quad x \in[-3,4)$
(ii) $\quad \log _4\left([x]^2-5\right)>0$
or $[x]^2-5>1$
or $\quad x \in(-\infty,-2) \cup[3, \infty)$
(iii) ${[x]^2-5>0} \\$
$x \in(-\infty,-2) \cup[3, \infty) \\$
$\text { Now }(i) \cap(i i) \cap(\text { iii }) \\$
$\Rightarrow \quad x \in[-3,2) \cup[3,4)$
(ii) $\quad f(x)=\frac{1}{[|x-1|]+[|12-x|]-11}$
Case-I $x>12$
$f(x)=\frac{1}{[x]-1+[x]-12-11} \quad \Rightarrow \quad f(x)=\frac{1}{2([x]-12)}$
Now for $f(x)$ to be defined $[x] \neq 12 \quad \Rightarrow \quad x \notin[12,13) \quad$ but $x>12$
$x \notin(12,13)$
Case-II $1 \leq x \leq 12$
$f(x)=$\frac{1}{[x]-1+12+[-x]-11}=\left\\{ll}$
$\frac{1}{[x]+(-1-[x])} \text { if } x \in I \\$
$\text { not defined } \text { if } x \in I$
$\Rightarrow \quad x \notin\{1,2,3,4,5,6,7,8,9,10,11,12\}$
Case-III
x<1
$f(x)=\frac{1}{1+[-x]+[-x]+12-11} \\$
$f(x)=\left\$
$\frac{1}{2(1-[x])} \text { if } x \in I \\$
$\frac{1}{-2[x]} \text { if } x \notin I$
$\{\quad \Rightarrow \quad x \notin(0,1) \quad \because \quad x<1\right}$
(iii) $\quad f(x)=(x+0.5)^{\log _{(0.5+x)} \frac{x^2+2 x-3}{4 x^2-4 x-3}}$
$x+0.5>0, x+0.5 \neq 1 \quad \Rightarrow \quad x \in(-0.5, \infty) \ x \neq 0.5$
$ $\frac{x^2+2 x-3}{4 x^2-4 x-3}>0$ or $\frac{(x+3)(x-1)}{(2 x-3)(2 x+1)}>0$
$\text { or } \quad x \in(-\infty,-3) \cup\left(-\frac{1}{2}, 1\right) \cup\left(\frac{3}{2}, \infty\right)$
$(A) \cap(B) \therefore$ Domain of $f(x): x \in\left(-\frac{1}{2}, 1\right) \cup\left(\frac{3}{2}, \infty\right)-\left\{\frac{1}{2}\right\}$
(iv) $f(x)=\frac{5}{\left[\frac{x-1}{2}\right]}-3^{\sin ^{-1} x^2}+\frac{(7 x+1) !}{\sqrt{x+1}}$
${\left[\frac{x-1}{2}\right] \neq 0 \quad \Rightarrow \quad x \notin[1,3)} \\$
$\text { \& } \quad x \in[-1,1] \\$
$\text { \& } \quad x+1>0 \quad \Rightarrow \quad x \in(-1, \infty) \\$
$\text { \& } 7 x+1 \in W \\$
$\therefore \quad \text { Domain }\left\{-\frac{1}{7}, 0, \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right\} \\$
(v) $3^y=2^{4 x^2-1}-2^{x^4}>0 \\$
$4 x^2-1>x^4 \quad \Rightarrow \quad\left(x^2\right)^2-4 x^2+1<0 \\$
$\left(x^2-2\right)^2+1-4<0 \\$
$\sqrt{2-\sqrt{3}}<|x|<\sqrt{2+\sqrt{3}} \\$
$x \in(-\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}) \cup(\sqrt{2-\sqrt{3}}, \sqrt{2+\sqrt{3}}) \\$
$x \in\left(\frac{-\sqrt{3}-1}{\sqrt{2}}, \frac{-\sqrt{3}+1}{\sqrt{2}}\right) \cup\left(\frac{\sqrt{3}-1}{\sqrt{2}}, \frac{\sqrt{3}+1}{\sqrt{2}}\right)$
