The value of $x$ for which $c o s^{- 1} (cos 4) > 3 x^{2} - 4 x$ is
Note: All angles are measured in radians.

(0, \frac{2 + \sqrt{6 π - 8}}{3})

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(- \frac{2 + \sqrt{6 π - 8}}{3}, 0)

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(- 2, 2)

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(\frac{2 - \sqrt{6 π - 8}}{3}, \frac{2 + \sqrt{6 π - 8}}{3})

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Solution

The correct option is D $(\frac{2 - \sqrt{6 π - 8}}{3}, \frac{2 + \sqrt{6 π - 8}}{3})$
${cos}^{- 1} (cos 4) > 3 x^{2} - 4 x$

$\Rightarrow {cos}^{- 1} (cos (2 π - 4)) > 3 x^{2} - 4 x$ ............. $cos (2 π - θ) = cos θ$

$\Rightarrow 2 π - 4 > 3 x^{2} - 4 x$

$\Rightarrow 3 x^{2} - 4 x - (2 π - 4) < 0$

$\Rightarrow \frac{4 - \sqrt{16 + 24 (π - 2)}}{6} < x < \frac{4 + \sqrt{16 + 24 (π - 2)}}{6}$

$\Rightarrow \frac{4 - 2 \sqrt{4 + 6 (π - 2)}}{6} < x < \frac{4 + 2 \sqrt{4 + 6 (π - 2)}}{6}$

$\Rightarrow \frac{2 - \sqrt{6 π - 8}}{3} < x < \frac{2 + \sqrt{6 π - 8}}{3}$

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