Let D be the domain of the function f(x)= ^-1( _3x(6+2 _3x-5x)).

Question

Let $D$ be the domain of the function $f(x)={\sin }^{-1}({\log }_{3x}(\frac{6+2{\log }_{3}x}{-5x}))$. If the range of the function $g:D	o R$ defined by $g(x)=x-[x],([x]$ is the greatest integer function), is $(\alpha ,\beta )$, then ${\alpha }^{2}+\frac{5}{\beta }$ is equal to

Accepted Answer

First, the function $f(x)={\sin }^{-1}({\log }_{3x}(\frac{6+2{\log }_{3}x}{-5x}))$ has several restrictions :

Since the arcsine function ${\sin }^{-1}(x)$ is only defined for $-1\le x\le 1$, this means that ${\log }_{3x}(\frac{6+2{\log }_{3}x}{-5x})$ must be between -1 and 1.

For the logarithm to be defined, $3x>0\Rightarrow x>0$ .......(1)

because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x\ne \frac{1}{3}$ .......(2)

as the base cannot be 1.

Moreover, the inner function of the logarithm $\frac{6+2{\log }_{3}x}{-5x}$ must be greater than 0.

$\Rightarrow$ $6+2{\log }_{3}x0)$

$ $$Rightarrow \& {\log }_{3}x$$

The next step is to solve the inequality for $-1\le {\log }\_{3x}(\frac{6+2{\log }\_{3}x}{-5x})\le 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x\le \frac{6+2{\log }\_{3}x}{-5x}\le \frac{1}{3x}$.

Solving the inequality $15{x}\^{2}+6+2{\log }\_{3}x\ge 0$, we find that $x\in (0,\frac{1}{27})$ ........(4)

Likewise, solving the inequality $6+2{\log }\_{3}x+\frac{5}{3}\ge 0$, we find that $x\ge {3}\^{-\frac{23}{6}}$ ........(5)

Combining Equations (3), (4), and (5), the intersection of all these intervals is $x\in [{3}\^{-\frac{23}{6}},\frac{1}{27})$.

Now, consider the function $g(x)=x-[x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\alpha ,\beta )$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\alpha ={3}\^{-\frac{23}{6}}$ and $\beta =\frac{1}{27}$.

Finally, substitute these values into the equation ${\alpha }\^{2}+\frac{5}{\beta }$:

${\alpha }\^{2}+\frac{5}{\beta }=({3}\^{-\frac{23}{6}}{)}\^{2}+\frac{5}{\frac{1}{27}}={3}\^{-\frac{23}{3}}+135$ = 135.0002198.

Since ${3}\^{-\frac{23}{3}}$ = 0.0002198 is an extremely small number, it's approximately 0. So, $

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