${sin}^{- 1} (sin 5) > x^{2} - 4 x$ holds if

x = 2 - \sqrt{9 - 2 π}

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x = 2 + \sqrt{9 - 2 π}

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x > 2 + \sqrt{9 - 2 π}

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x ϵ (2 - \sqrt{9 - 2 π}, 2 + \sqrt{9 - 2 π})

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Solution

The correct option is D $x ϵ (2 - \sqrt{9 - 2 π}, 2 + \sqrt{9 - 2 π})$
$\frac{3 π}{2} < 5 < 2 π$ . Since the range of ${sin}^{- 1} x$ is $[- \frac{π}{2}, \frac{π}{2}]$ and since $sin x$ is negative in $[\frac{3 π}{2}, 2 π]$ ,
${sin}^{- 1} (sin 5) = 5 - 2 π$
Hence,
$5 - 2 π > x^{2} - 4 x$
$5 - 2 π > (x - 2)^{2} - 4$
$9 - 2 π > (x - 2)^{2}$
$x - 2 < | \sqrt{9 - 2 π} |$
$x \in (2 - \sqrt{9 - 2 π}, 2 + \sqrt{9 - 2 π})$

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