Question

Let $f:(-1,1)\to R$ be such that $f\left(\cos 4\theta \right)=\frac{2}{2-se{c}^2\theta }$ for $\theta \in (0,\frac{\pi }{4})\cup \left(\frac{\pi }{4}\frac{\pi }{2}\right)$.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 1

Answer: (A), (B)

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

Given $f\left(\cos 4\theta \right)=\frac{2}{2-se{c}^2\theta }=\frac{2{\cos }^2\theta }{2{\cos }^2\theta -1}$
$=\frac{1+\cos 2\theta }{\cos 2\theta }=1+\frac{1}{cos2\theta }$
Let $cos4\theta =\frac{1}{3}\Rightarrow 2co{s}^{2}2\theta -1=\frac{1}{3}\Rightarrow \cos 2\theta =\pm \sqrt{\frac{2}{3}}$
$\therefore f\left(\cos 4\theta \right)=1\pm \sqrt{\frac{3}{2}}orf\left(\frac{1}{3}\right)=1\pm \sqrt{\frac{3}{2}}$