Let D be the domain of the function f(x)= sin^-1( log_3x( 6+2 log_3x -5x)). If the range of the function g:D→ mathbbR defined by g(x)=x-[x], ([x] is…

We have been given

$$f(x)={\mathrm{sin}}^{-1}({\mathrm{log}}_{3x}(\frac{6+2{\mathrm{log}}_{3}x}{-5x}))$$f(x)=sin−1(log3x​(−5x6+2log3​x​))

We know that the domain of ${\mathrm{sin}}^{-1}x$sin−1x is $-1\leq x\leq 1$−1≤x≤1.

$$\Rightarrow {\mathrm{log}}_{3x}(\frac{6+2{\mathrm{log}}_{3}x}{-5x})\in \underset{-(1)}{[-11]},\underset{-(2)}{3x>0},\underset{-(3)}{3x\neq 1},\underset{-(4)}{\frac{6+2{\mathrm{log}}_{3}x}{-5x}}>0$$⇒log3x​(−5x6+2log3​x​)∈−(1)[−11]​,−(2)3x>0​,−(3)3x=1​,−(4)−5x6+2log3​x​​>0

$\Rightarrow$⇒ From $(2),x>0,$(2),x>0, from $(3),x\neq \frac{1}{3}$(3),x=31​

From $(4),6+2{\mathrm{log}}_{3}x<0(\because x>0)$(4),6+2log3​x<0(∵x>0)

$$\Rightarrow {\mathrm{log}}_{3}x<-3$$⇒log3​x<−3

$$\Rightarrow x<{3}^{-3}$$⇒x<3−3

$$\Rightarrow x<\frac{1}{27}$$⇒x<271​

From $(2),(3),(4),0<x<\frac{1}{27}...(5)$(2),(3),(4),0<x<271​...(5)

From $(1),-1\leq {\mathrm{log}}_{3x}(\frac{6+2{\mathrm{log}}_{3}x}{-5x})\leq 1$(1),−1≤log3x​(−5x6+2log3​x​)≤1

$$\because 3x\in (0,\frac{1}{9})$$∵3x∈(0,91​)

$$\Rightarrow \frac{1}{3x}\geq \frac{6+2{\mathrm{log}}_{3}x}{-5x}\geq 3x$$⇒3x1​≥−5x6+2log3​x​≥3x

$$\Rightarrow -\frac{5}{3}\leq 6+2{\mathrm{log}}_{3}x\leq -15{x}^{2}$$⇒−35​≤6+2log3​x≤−15x2

$$\Rightarrow -\frac{23}{6}\leq {\mathrm{log}}_{3}x\leq \frac{-15{x}^{2}-6}{2}$$⇒−623​≤log3​x≤2−15x2−6​

$$\Rightarrow {3}^{-\frac{23}{6}}\leq x\leq {3}^{\frac{-15{x}^{2}-6}{2}}$$⇒3−623​≤x≤32−15x2−6​

$$\Rightarrow \frac{1}{{3}^{\frac{23}{6}}}\leq x\leq \frac{1}{{3}^{\frac{15{x}^{2}+6}{2}}}$$⇒3623​1​≤x≤3215x2+6​1​

$\Rightarrow {3}^{\frac{-23}{6}}\leq x<\frac{1}{{3}^{\frac{15{x}^{2}+6}{2}}}......(6)$⇒36−23​≤x<3215x2+6​1​......(6), here ${3}^{\frac{-23}{6}}$36−23​ is very small positive quantity,

From $(5)$(5) and $(6),{3}^{\frac{-23}{6}}<x<\frac{1}{27},[x]=0$(6),36−23​<x<271​,[x]=0

$$\Rightarrow g(x)=x$$⇒g(x)=x

$\Rightarrow$⇒ Range $g(x)$g(x) is domain of $f(x)$f(x)

$\Rightarrow g(x)\in ({3}^{\frac{-23}{6}},\frac{1}{27})\equiv (\alpha ,\beta )$⇒g(x)∈(36−23​,271​)≡(α,β).

$$\Rightarrow {\alpha }^{2}+\frac{5}{\beta }={({3}^{\frac{-23}{6}})}^{2}+\frac{5}{\frac{1}{27}}={({3}^{\frac{-23}{6}})}^{2}+135\approx 135$$⇒α2+β5​=(36−23​)2+271​5​=(36−23​)2+135≈135

Note: This question was bonus in Jee Main April 2023 session.