Question

Find the domain of the following function

$f(x) = \sqrt{ {sin}^{ - 1} log_{2}(x) }$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{ {sin}^{ - 1} (log_{2}x) }$

We know,

$\boxed{\tt{ Domain \: of \: {sin}^{ - 1}x \: is \: x \: \in \: [ - 1,1] \: }}$

So,

$\rm \implies\: - 1 \leqslant log_{2}(x) \leqslant 1$

We know

$\boxed{\tt{ log_{b}(a) = c \: \rm \implies\:a = {b}^{c} \: }}$

So, using this, we get

$\rm \implies\: {2}^{ - 1} \leqslant x \leqslant {2}^{1}$

$\rm \implies\: \dfrac{1}{2} \leqslant x \leqslant 2 - - - (1)$

Again,

$\rm :\longmapsto\: log_{2}(x) \geqslant 0$

$\rm \implies\:x \geqslant {2}^{0}$

$\rm \implies\:x \geqslant 1 - - - (2)$

From equation (1) and (2), we concluded that

$\rm \implies\:1 \leqslant x \leqslant 2$

$\rm \implies\:x \: \in \: [1, \: 2]$

Hence,

$\boxed{\tt{Domain \: of \: f(x) = \sqrt{ {sin}^{ - 1}( log_{2}x) } \: is \: x \in \: [1,2] \: }}$

Answer 2

$\large\underline{\sf{Solution-}}$

function Are

$\rm :\longmapsto\:f(x) = \sqrt{ {sin}^{ - 1} (log_{2}x) }$

Given

$\boxed{\tt{ Domain \: of \: {sin}^{ - 1}x \: is \: x \: \in \: [ - 1,1] \: }}$

Then

$\rm \implies\: - 1 \leqslant log_{2}(x) \leqslant 1$

Given

$\boxed{\tt{ log_{b}(a) = c \: \rm \implies\:a = {b}^{c} \: }}$

So, using this, we get

$\rm \implies\: {2}^{ - 1} \leqslant x \leqslant {2}^{1}$

$\rm \implies\: \dfrac{1}{2} \leqslant x \leqslant 2 - - - (1)$

Again,

$\rm :\longmapsto\: log_{2}(x) \geqslant 0$

$\rm \implies\:x \geqslant {2}^{0}$

$\rm \implies\:x \geqslant 1 - - - (2)$

From equation (1) and (2),

$\rm \implies\:1 \leqslant x \leqslant 2$

$\rm \implies\:x \: \in \: [1, \: 2]$

Answer

$\boxed{\tt{Domain \: of \: f(x) = \sqrt{ {sin}^{ - 1}( log_{2}x) } \: is \: x \in \: [1,2] \: }}$