Question

Let $f:(-1,1) \rightarrow \operatorname{IR}$ 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

SOLUTION : $\cos 4 \theta=\frac{1}{3} \Rightarrow 2 \cos ^2 2 \theta-1=\frac{1}{3} \Rightarrow \cos ^2 2 \theta=\frac{2}{3} \Rightarrow \cos 2 \theta= \pm \sqrt{\frac{2}{3}}$

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

Hence Option A&B is Correct.