Question

If $f\left(x\right)={\cos }^{-1}x+{\cos }^{-1}\left\{\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right\}$, then

• A. $f\left(\frac{2}{3}\right)=\frac{\mathrm{\pi }}{3}$
• B. $f\left(\frac{2}{3}\right)=2{\cos }^{-1}\left(\frac{2}{3}\right)-\frac{\mathrm{\pi }}{3}$
• C. $f\left(\frac{1}{3}\right)=\frac{\mathrm{\pi }}{3}$
• D. $f\left(\frac{1}{3}\right)=\frac{\mathrm{\pi }}{3}+2{\cos }^{-1}\left(\frac{1}{3}\right)$

Answer 1

Answer: (A), (C)

$f\left(\frac{1}{3}\right)=\frac{\mathrm{\pi }}{3}$

Explanation for the correct option.

Find the values of $f\left(\frac{1}{3}\right)$ and $f\left(\frac{2}{3}\right)$.

In the function $f\left(x\right)={\cos }^{-1}x+{\cos }^{-1}\left\{\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\right\}$ substitute $x=\cos \theta$.

$\begin{array}{rcl}f\left(x\right)& =& {\cos }^{-1}\cos \theta +{\cos }^{-1}\left\{\frac{\cos \theta }{2}+\frac{1}{2}\sqrt{3-3{\left(\cos \theta \right)}^2}\right\}\\ & =& \theta +{\cos }^{-1}\left\{\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sqrt{1-{\cos }^2\theta }\right\}\\ & =& \theta +{\cos }^{-1}\left\{\cos \frac{\mathrm{\pi }}{3}\cos \theta +\sin \frac{\mathrm{\pi }}{3}\sin \theta \right\}\left[\because {\sin }^2\theta =1-{\cos }^2\theta ,\cos \frac{\pi }{3}=\frac{1}{2},\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\right]\\ & =& \theta +{\cos }^{-1}\left(\cos \left(\frac{\mathrm{\pi }}{3}-\theta \right)\right)\left[\because \cos \left(A+B\right)=\mathrm{cosAcosB}+\mathrm{sinAsinB}\right]\\ & =& \theta +\frac{\mathrm{\pi }}{3}-\theta \\ & =& \frac{\mathrm{\pi }}{3}\end{array}$

So for all $x$, $f\left(x\right)=\begin{array}{rcl}& & \frac{\mathrm{\pi }}{3}\end{array}$.

Thus, $f\left(\frac{1}{3}\right)=\frac{\mathrm{\pi }}{3}$ and $f\left(\frac{2}{3}\right)=\frac{\mathrm{\pi }}{3}$.

Hence, the correct options are A and C.