Q38 / 60 · JEE Advanced 2012 (Paper 2) · Mathematics
Let $f:(-1,1) \rightarrow I R$ 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)
- Correct answerA$1-\sqrt{\frac{3}{2}}$
- Correct answerB$1+\sqrt{\frac{3}{2}}$
- C$1-\sqrt{\frac{2}{3}}$
- D$1+\sqrt{\frac{2}{3}}$
Solution
Given : $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}=\frac{2 \cos ^{2} \theta}{2 \cos ^{2} \theta-1}$
$=\frac{1+\cos 2 \theta}{\cos 2 \theta}=1+\frac{1}{\cos 2 \theta}$
Let $\cos 4 \theta=\frac{1}{3} \Rightarrow 2 \cos ^{2} 2 \theta-1=\frac{1}{3} \Rightarrow \cos 2 \theta=\pm \sqrt{\frac{2}{3}}$
$\therefore f(\cos 4 \theta)=1+\frac{1}{\cos 2 \theta}=1 \pm \sqrt{\frac{3}{2}}$ or $f\left(\frac{1}{3}\right)=1 \pm \sqrt{\frac{3}{2}}$
Hence, the correct answer is options: 'A', 'B'.