Question

Let $f:(-1,1) \rightarrow R$ 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 $1+\sqrt{\frac{3}{2}}$