Question

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Find the possible values of
$\sf \: sin {}^{ - 1} (1 - x) + cos {}^{ - 1} \sqrt{x - 2}$


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Answer 1

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

Given expression of inverse Trigonometric function is

$\rm :\longmapsto\: {sin}^{ - 1}(1 - x) + {cos}^{ - 1} \sqrt{x - 2}$

We know,

$\boxed{ \tt{ \: - 1 \leqslant {sin}^{ - 1}x \leqslant 1 \: }}$

So,

$\red{\rm :\longmapsto\: {sin}^{ - 1}(1 - x) \: is \: defined \: if \: }$

$\rm \implies\: - 1 \leqslant (1 - x) \leqslant 1$

$\rm :\longmapsto\: - 1 - 1 \leqslant - x \leqslant 1 - 1$

$\rm :\longmapsto\: - 2 \leqslant - x \leqslant 0$

$\bf\implies \:0 \leqslant x \leqslant 2$

Now, We know

$\boxed{ \tt{ \: - 1 \leqslant {cos}^{ - 1}x \leqslant 1 \: }}$

So,

$\red{\rm :\longmapsto\: {cos}^{ - 1} \: \sqrt{x - 2} \: is \: defined \: if \: }$

$\rm :\longmapsto\: - 1 \leqslant \sqrt{x - 2} \leqslant 1$

On squaring both sides, we get

$\rm :\longmapsto\:0 \leqslant x - 2 \leqslant 1$

$\rm :\longmapsto\:0 + 2 \leqslant x - 2 + 2 \leqslant 1 + 2$

$\bf\implies \:2 \leqslant x \leqslant 3$

So, It means

$\rm :\longmapsto\: {sin}^{ - 1}(1 - x) + {cos}^{ - 1} \sqrt{x - 2} \: is \: defined \: when$

$\rm :\longmapsto\:x \in \: [0,2] \: \cap \: [2,3]$

$\bf\implies \:x = 2$

So, The possible value of

$\rm :\longmapsto\: {sin}^{ - 1}(1 - x) + {cos}^{ - 1} \sqrt{x - 2}$

$\rm \:  =  \: {sin}^{ - 1}(1 - 2) + {cos}^{ - 1} \sqrt{2 - 2}$

$\rm \:  =  \: {sin}^{ - 1}( - 1) + {cos}^{ - 1} \sqrt{0}$

$\rm \:  =  \: - \: {sin}^{ - 1}1 + {cos}^{ - 1} 0$

$\rm \:  =  \: - \: \dfrac{\pi}{2} + \dfrac{\pi}{2}$

$\rm \:  =  \: 0$

Hence,

• The possible value is 0.

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More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Domain \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}x & \sf [ - 1, \: 1] \\ \\ \sf {cos}^{ - 1}x & \sf [ - 1, \: 1] \\ \\ \sf {tan}^{ - 1}x & \sf R\\ \\ \sf {cosec}^{ - 1}x & \sf ( - \infty , - 1] \cup \: [1, \: \infty )\\ \\ \sf {sec}^{ - 1}x & \sf ( - \infty , - 1] \cup \: [1, \: \infty )\\ \\ \sf {cot}^{ - 1}x & \sf R \end{array}} \\ \end{gathered}$

Answer 2

Question:-

Find the possible values of

$\sf \: sin {}^{ - 1} (1 - x) + cos {}^{ - 1} \sqrt{x - 2}$

Solution:-

We have,

$\sf \: f(x) = sin {}^{ - 1} (1 - x) + cos {}^{ - 1} \sqrt{x - 2}$

We must have

$\sf \: - 1 \leqslant (1 - x) \leqslant 1$

$\sf \: = > - 2 \leqslant - x \leqslant 0$

$\sf \: = > 0 \leqslant x \leqslant 2 \: and \: x - 2 \geqslant 0$

$\sf = > 0 \leqslant x \leqslant 2 \: and \: x \geqslant 2$

$\sf \therefore \: x = 2$

Thus,expression is defined only for x = 2.

$\sf \therefore f(2) = sin {}^{ - 1} ( - 1) + cos {}^{ - 1} (0)$

$\sf \: = ( - \frac{\pi}{2} ) + ( \frac{\pi}{2} )$

$\sf \: = 0$

Answer:-

The possible value of

$\sf \: sin {}^{ - 1} (1 - x) + cos {}^{ - 1} \sqrt{x - 2} \: \: \: \: is \: 0.$