JEE Advanced · Mathematics · 7. Trigonometry
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)
- 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 & Solution
Correct Answer
(B) \(1+\sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
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 \(f\left(\frac{1}{3}\right)=1 \pm \sqrt{\frac{3}{2}}\)
\(=\frac{1+\cos 2 \theta}{\cos 2 \theta}=1+\frac{1}{\cos 2 \theta}\)
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 \(f\left(\frac{1}{3}\right)=1 \pm \sqrt{\frac{3}{2}}\)
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