Question

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

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

Answer 1

Answer: (A), (D)

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

We have, $f\left(x\right)=co{s}^{-1}\left(x\right)+co{s}^{-1}(\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2})$

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

$=co{s}^{-1}\left(x\right)+co{s}^{-1}\left(cos\right(\frac{\pi }{3}-\theta \left)\right)$

$=co{s}^{-1}\left(x\right)+\frac{\pi }{3}-\theta$

Now $\theta =co{s}^{-1}\left(x\right)$

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