JEE Advance - Mathematics (2012 - Paper 2 Offline - No. 20)
Let $$f:( - 1,1) \to R$$ be such that $$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$ for $$\theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 4},{\pi \over 2}} \right)$$. Then the value(s) of $$f\left( {{1 \over 3}} \right)$$ is(are)
$$1 - \sqrt {{3 \over 2}} $$
$$1 + \sqrt {{3 \over 2}} $$
$$1 - \sqrt {{2 \over 3}} $$
$$1 + \sqrt {{2 \over 3}} $$
Explanation
$$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$
Let $$\cos 4\theta = t$$
$$ \Rightarrow 2{\cos ^2}2\theta - 1 = t \Rightarrow {\cos ^2}2\theta = {2 \over 3}$$
For $$t = {1 \over 3}$$ we have $${\cos ^2}2\theta = {2 \over 3}$$
$$\cos 2\theta = \sqrt {{2 \over 3}} $$ or $$\cos 2\theta = - \sqrt {{2 \over 3}} $$
$$f(\cos 4\theta ) = {2 \over {2 - {1 \over {{{\cos }^2}\theta }}}} = {{2{{\cos }^2}\theta } \over {2{{\cos }^2}\theta - 1}} = {{1 + \cos 2\theta } \over {\cos 2\theta }} = 1 + {1 \over {\cos 2\theta }}$$
Hence, $$f\left( {{1 \over 3}} \right) = 1 + \sqrt {{3 \over 2}} $$ or $$1 - \sqrt {{3 \over 2}} $$
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