The value of x for which $C o s^{- 1} (Cos4) > 3 x^{2} - 4 x$ is

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$(0, \frac{2 + \sqrt{6 π - 8}}{3})$

$(- 2, 2)$

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

$(\frac{2 + \sqrt{6 π - 8}}{3}, 0)$

answer is D.

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Detailed Solution

$C o s^{- 1} C o s (2 π - 4) > 3 x^{2} - 4 x 2 π - 4 > 3 x^{2} - 4 x \Rightarrow 3 x^{2} - 4 x + 4 - 2 π < 0 \Rightarrow x \in (\frac{2 - \sqrt{6 π - 8}}{3}, \frac{2 + \sqrt{6 π - 8}}{3})$

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