If the domain of the real valued function $f(x) = \frac{1}{\sqrt{\log_{\frac{1}{3}}\left(\frac{x-1}{2-x}\right)}}$ is $(a, b)$,then $2b =$

If $f:[2, \infty) \rightarrow B$ defined by $f(x)=x^2-4x+5$ is a bijection,then $B$ is equal to

Let $f(x) = \frac{x^2-6x+5}{x^2-5x+6}$. Match the conditions / expressions in Column $I$ with statements in Column $II$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

Let the domain of the function $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2\left(2-6\log_{27}(2x+5)\right)$ be $(\gamma, \delta)$. Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to . . . . . . .

Let $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$ and $f: A \rightarrow R$ be defined by $f(x) = \frac{|x|}{x}$ for $x \in A$. Then,the range of $f$ is

The set of all real values of $x$ for which the real-valued function $f(x) = \left(1 + \frac{1}{x}\right)^x$ is defined,is