Question

The domain of sin^-1[log³(x/3)]is?

Answer 1

We're asked to find domain of the function,

$\longrightarrow f(x)=\sin^{-1}\left(\log^3\left(\dfrac{x}{3}\right)\right)$

Taking $y=f(x),$

$\longrightarrow y=\sin^{-1}\left(\log^3\left(\dfrac{x}{3}\right)\right)$

Taking sine,

$\longrightarrow \sin y=\log^3\left(\dfrac{x}{3}\right)$

Taking cube root,

$\longrightarrow \sqrt[3]{\sin y}=\log\left(\dfrac{x}{3}\right)$

Taking antilog, we get,

$\longrightarrow \dfrac{x}{3}=e^{\sqrt[3]{\sin y}}$

Multiplying by 3, we get,

$\longrightarrow x=3e^{\sqrt[3]{\sin y}}\quad\quad\dots(1)$

We know range of sine function is $[\,-1,\ 1\,].$

$\Longrightarrow \sin y\in[\,-1,\ 1\,]$

Taking cube root,

$\longrightarrow \sqrt[3]{\sin y}\in\left[\,\sqrt[3]{-1},\ \sqrt[3]{1}\,\right]$

$\longrightarrow \sqrt[3]{\sin y}\in[\,-1,\ 1\,]$

Taking antilog,

$\longrightarrow e^{\sqrt[3]{\sin y}}\in[\,e^{-1},\ e^1\,]$

$\longrightarrow e^{\sqrt[3]{\sin y}}\in\left[\,\dfrac{1}{e},\ e\,\right]$

Multiplying by 3 we get,

$\longrightarrow 3e^{\sqrt[3]{\sin y}}\in\left[\,\dfrac{3}{e},\ 3e\,\right]$

From (1),

$\longrightarrow \underline{\underline{x\in\left[\,\dfrac{3}{e},\ 3e\,\right]}}$

This is the domain of the function.

Answer 2

Answer:

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