Q. The value of $x$ for which $\cos^{- 1}{(\cos{4) > 3x^{2} - 4x}}\$is
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) $\left( \frac{2 - \sqrt{6\pi - 8}}{3},\frac{2 + \sqrt{6\pi - 8}}{3} \right)$
Explanation
(d)
Now, $\cos^{- 1}{(\cos{4) = \cos^{- 1}{\left\{ \cos(2\pi - 4) \right\} = 2\pi - 4}}}$
$\Rightarrow 2\pi - 4 > 3x^{2} - 4x$
$\Rightarrow 3x^{2} - 4x - (2\pi - 4) < 0$
$\Rightarrow \frac{2 - \sqrt{6\pi - 8}}{3} < x < \frac{2 + \sqrt{6\pi - 8}}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.