The range of the function $f (x) = {sin}^{- 1} (x^{2} - 2 x + 2)$

ϕ

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[- \frac{π}{2}, \frac{π}{2}]

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\frac{π}{2}

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none of these

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Solution

The correct option is D $\frac{π}{2}$
$f (x) = {sin}^{- 1} (x^{2} - 2 x + 2)$

Domain of inverse sin function is $[- 1, 1]$

$\Rightarrow - 1 \leq x^{2} - 2 x + 2 \leq 1$

On solving these inequalities separately, we get

$\Rightarrow x^{2} - 2 x + 2 \leq 1$

$\Rightarrow x^{2} - 2 x + 1 \leq 0$

$\Rightarrow (x - 1)^{2} \leq 0$

This inequality only holds good for $x = 1$

Substitute $x = 1$ in the above expression , we get $f (x) = s i n^{- 1} (1) = \frac{π}{2}$

Therefore, Range of $f (x) = \frac{π}{2}$

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\begin{matrix} f (x) = \frac{1 - sin x}{{(\frac{π}{2} - x)}^{2}}, & f o r & x \neq \frac{π}{2} = 3, & f o r & x = \frac{π}{2} \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

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