Question

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

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

Answer 1

Answer: (A), (C)

$f\left(\frac{2}{3}\right)=\frac{\pi }{3}$
$f\left(\frac{1}{3}\right)=\frac{\pi }{3}$

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

Let $\cos \theta =x$

and $\theta ={\cos }^{-1}x$

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

$f\left(x\right)={\cos }^{-1}\left(\cos \theta \right)+{\cos }^{-1}[\frac{\cos \theta }{2}+\frac{\sqrt{3}}{2}\sqrt{(1-{\cos }^2\theta )}]$

$f\left(x\right)=\theta +{\cos }^{-1}[\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sqrt{{\sin }^2\theta }]$

$f\left(x\right)=\theta +{\cos }^{-1}[\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta ]$

$f\left(x\right)=\theta +{\cos }^{-1}[\cos \frac{\pi }{3}\cos \theta +\sin \frac{\pi }{3}\sin \theta ]$

$f\left(x\right)=\theta +{\cos }^{-1}\left[\cos \right(\frac{\pi }{3}-\theta \left)\right]$

$f\left(x\right)=\theta +\frac{\pi }{3}-\theta$

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

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

and $f\left(\frac{2}{3}\right)=\frac{\pi }{3}$