Let f:(-1,1) rightarrow mathbb{R} be such tha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(AB) Given $f(\cos 4 	heta) = \frac{2}{2-\sec^2 	heta}$.We know that $\sec^2 	heta = \frac{1}{\cos^2 	heta} = \frac{2}{1+\cos 2 	heta}$.Substitut

Vedclass · Accepted Answer

$1+\sqrt{\frac{2}{3}}$

Vedclass · Answer

(AB) Given $f(\cos 4 	heta) = \frac{2}{2-\sec^2 	heta}$.We know that $\sec^2 	heta = \frac{1}{\cos^2 	heta} = \frac{2}{1+\cos 2 	heta}$.Substituting this into the expression for $f(\cos 4 	heta)$:$f(\cos 4 	heta) = \frac{2}{2 - \frac{2}{1+\cos 2 	heta}} = \frac{2(1+\cos 2 	heta)}{2(1+\cos 2 	heta) - 2} = \frac{1+\cos 2 	heta}{\cos 2 	heta} = 1 + \frac{1}{\cos 2 	heta}$.Now,let $\cos 4 	heta = \frac{1}{3}$.Using the identity $\cos 4 	heta = 2 \cos^2 2 	heta - 1$,we have $2 \cos^2 2 	heta - 1 = \frac{1}{3} \Rightarrow 2 \cos^2 2 	heta = \frac{4}{3} \Rightarrow \cos^2 2 	heta = \frac{2}{3}$.Thus,$\cos 2 	heta = \pm \sqrt{\frac{2}{3}}$.Substituting this back into the expression for $f(\cos 4 	heta)$:$f\left(\frac{1}{3}ight) = 1 + \frac{1}{\pm \sqrt{2/3}} = 1 \pm \sqrt{\frac{3}{2}}$.Therefore,the values are $1+\sqrt{\frac{3}{2}}$ and $1-\sqrt{\frac{3}{2}}$.

List-$I$	List-$II$
$A$. $\frac{x}{e^x-1} + \frac{x}{2} + 4; x \neq 0$	$I$. is neither odd nor even function
$B$. $\tan^{-1}(\log\|x+\sqrt{x^2+1}\|), x > 0$	$II$. is an even function
$C$. For $3 < x < 5, \|x-2\|+\|x-3\|+\|x-5\|$	$III$. is an odd function
$D$. $\sin 2x + \sin^2 x + \cos 3x, \forall x \in \mathbb{R}$	$IV$. is the identity function
	$V$. is a constant function

Let $f:(-1,1) \rightarrow \mathbb{R}$ 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):

Similar Questions

If $f(x) = \log_e \left( \frac{1-x}{1+x} \right)$,$|x| < 1$,then $f\left( \frac{2x}{1+x^2} \right)$ is equal to

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. For $x \in \mathbb{R}$,let $f(x) = [x] \sin(\pi x)$. Then,

Let $[x]$ denote the integral part of $x \in R$. Let $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$. Then the function $h(x) = f(g(x))$:

Consider two sets $A = \{ x \in \mathbb{Z} : |(| x - 3| - 3)| \leq 1 \}$ and $B = \{ x \in \mathbb{R} - \{1, 2\} : \frac{(x - 2)(x - 4)}{x - 1} \log_{e}(|x - 2|) = 0 \}$. Then the number of onto functions $f: A \rightarrow B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module