Let D be the domain of the function f(x)=sin−1log3x6+2log3x−5x. If the range of the function g:D→R defined by g(x)=x−[x],([x] is the greatest integer function), is (α,β), then α2+5β is equal to

Correct Question is f(x)=log3x6+2log3x−5xFor log3x6+2log3x−5xto exist x>0 and 6+2log3x<0⇒x<127∴0<x<127Now, (α,β)=(0,127)∴α2+5β=5(27)=135

Let D be the domain of the function f(x)=sin−1log3x6+2log3x−5x. If the range of the function g:D→R defined by g(x)=x−[x],([x] is the greatest integer function), is (α,β), then α2+5β is equal to

Correct Question is f(x)=log3x6+2log3x−5xFor log3x6+2log3x−5xto exist x>0 and 6+2log3x<0⇒x<127∴0<x<127Now, (α,β)=(0,127)∴α2+5β=5(27)=135

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

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Detailed Solution

Correct Question is $f (x) = (\log_{3 x} (\frac{6 + 2 \log_{3} x}{- 5 x}))$
For $(\log_{3 x} (\frac{6 + 2 \log_{3} x}{- 5 x}))$ to exist x>0 and $6 + 2 \log_{3} x < 0 \Rightarrow x < \frac{1}{27}$
$∴$ $0 < x < \frac{1}{27}$
Now, $(α, β) = (0, \frac{1}{27})$
$∴$ $α^{2} + \frac{5}{β} = 5 (27) = 135$