Question

Let $F:\left( { - 1,1} \right) \to R$ be such that $f\left( {\cos4\theta } \right) = \dfrac{2}{{2 - {{\sec }^2}\theta }}for \theta \in \left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right).$Then the value(s) of $f\left( {\dfrac{1}{3}} \right)$ is (are)

Answer 1

Here we need to have the knowledge of functions and we also need to use some of the trigonometric formulas in this problem mainly we need to know the formula of cos2x. Doing this will solve your problem and will give you the right answer.

Complete step-by-step answer :
For $\theta \in \left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)$
We know the general formula $\cos 2x = 2{\cos ^2}x - 1$.
Therefore we can say that
$\cos 4\theta = 2{\cos ^2}2\theta - 1 \\\ {\text{or}} \\\ \cos 2\theta = \pm \sqrt {\dfrac{{\cos 4\theta + 1}}{2}} \\\$
Suppose $\cos 4\theta = \dfrac{1}{3}$
Then by applying the formula of $\cos 2\theta$ above will be
$\\\ \Rightarrow \cos 2\theta = \pm \sqrt {\dfrac{{1 + \cos 4\theta }}{2}} = \pm \sqrt {\dfrac{2}{3}} \\\$
And then using the function we get the equation as,
$f\left( {\dfrac{1}{3}} \right) = \dfrac{2}{{2 - {{\sec }^2}\theta }} = \dfrac{{2{{\cos }^2}\theta }}{{2{{\cos }^2}\theta - 1}}$ as ${\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}$
On solving the above equation we get,
$\dfrac{{\cos 2\theta + 1}}{{\cos 2\theta }} = 1 + \dfrac{1}{{\cos 2\theta }}$
Then, on putting the value of $\cos 2\theta$ we get the value as:
$f\left( {\dfrac{1}{3}} \right) = 1 - \sqrt {\dfrac{3}{2}} or 1 + \sqrt {\dfrac{3}{2}}$

Note : In maths, the relation is defined as the collection of ordered pairs, which contains an object from one set to the other set. For instance, X and Y are the two sets, and ‘a’ is the object from set X and b is the object from set Y, then we can say that the objects are related to each other if the ordered pairs (a, b) is to be in relation. Functions - The relation that defines the set of inputs to the set of outputs is called the functions. In function, each input in the set X has exactly one output in the set Y.