A) \[\left( 0,\frac{2+\sqrt{6\pi -8}}{3} \right)\]
B) \[\left( \frac{2-\sqrt{6\pi -8}}{3},0 \right)\]
C) \[(-2,2)\]
D) \[\left( \frac{2-\sqrt{6\pi -8}}{3},\frac{2+\sqrt{6\pi -8}}{3} \right)\]
Correct Answer: D
Solution :
Now, \[{{\cos }^{-1}}(\cos 4)={{\cos }^{-1}}\{\cos (2\pi -4)\}.\] \[=2\pi -4\] \[\therefore \]\[(2\pi -4)>(3{{x}^{2}}-4x)\] \[\Rightarrow \]\[3{{x}^{2}}-4x-(2\pi -4)<0\] \[\Rightarrow \]\[\frac{2-\sqrt{6\pi -8}}{3}<x<\frac{2+\sqrt{6\pi -8}}{3}\]
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