Question

Let $f:(-1,1)\to R$ be such that $f(\cos 4\theta )=\frac{2}{2-{\sec }^2\theta }$ 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

Answer: (A), (B)

For $\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{4},\frac{\pi }{2}\right)$
Let $\cos 4\theta =\frac{1}{3}.$
$\Rightarrow \cos 2\theta =\pm \sqrt{\frac{1+\cos 4\theta }{2}}=\pm \sqrt{\frac{2}{3}}$
$f\left(\frac{1}{3}\right)=\frac{2}{2-{\sec }^2\theta }=\frac{2{\cos }^2\theta }{2{\cos }^2\theta -1}=1+\frac{1}{\cos 2\theta }$
$f\left(\frac{1}{3}\right)=1-\sqrt{\frac{3}{2}}$ or $1+\sqrt{\frac{3}{2}}$