Question

Let f : (-1, 1) $\to$ R be such that f $\left(\cos 4\theta \right)$ = $\frac{2}{2-{\sec }^2\theta }$ for $\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{4},\frac{\pi }{2}\right).$ Then, the value(s) of $f\left(\frac{1}{3}\right)$ is are

• A. $1-\sqrt{\frac{3}{2}}$
• B. $1+\sqrt{\frac{3}{2}}$
• C. $1-\sqrt{\frac{2}{3}}$
• D. $1+\sqrt{\frac{2}{3}}$

Answer 1

The correct option is A $1-\sqrt{\frac{3}{2}}$

$f\left(\cos 4\theta \right)=\frac{2}{2-{\sec }^2\theta }=\frac{2}{2-\frac{1}{{\cos }^2\theta }}=\frac{2{\cos }^2\theta }{2{\cos }^2\theta -1}f\left(\cos y\theta \right)=\frac{2{\cos }^2\theta -1+1}{\cos \left(2\theta \right)}f\left(\cos y\theta \right)=\frac{\cos \left(2\theta \right)+1}{\cos \left(2\theta \right)}=1+\frac{1}{\cos \left(2\theta \right)}\cos 4\theta =\frac{1}{3}2{\cos }^{2}2\theta -1=\frac{1}{3}2{\cos }^{2}2\theta =1-\frac{1}{3}{\cos }^{2}2\theta =\frac{2}{3}$

$\cos 2\theta =-\sqrt{\frac{2}{3}}$ as $\theta \in (\frac{\pi }{4},\frac{\pi }{2})$

$\therefore f\left(\frac{1}{3}\right)=1+\frac{1}{(-\sqrt{\frac{2}{3}})}=1-\sqrt{\frac{3}{2}}$