Question

If ${\cos }^{-1}(4x^3-3x)=2\pi -3{\cos }^{-1}x$, then $x$ lies in interval

• A. $[-1,-\frac{1}{2}]$
• B. $\left|x\right|<\frac{1}{2}$
• C. $\left[\frac{1}{2},1\right]$
• D. None of these

Answer 1

The correct option is B $\left|x\right|<\frac{1}{2}$

Let $f\left(x\right)={\cos }^{-1}(4x^3-3x)\forall [-1,1]$Using theory part we can define
the function $f\left(x\right)$ as follows $f\left(x\right)={\cos }^{-1}(4x^3-3x)=\{\begin{array}{ccc}3{\cos }^{-1}x-2\pi & \forall & -1\le x\le -\frac{1}{2}\\ 2\pi -3{\cos }^{-1}x& \forall & \left|x\right|<\frac{1}{2}\\ 3{\cos }^{-1}x& \forall & \frac{1}{2}\le x\le 1\end{array}$