Trigonometrical Ratios And Identities Ques 20
Let $f:(-1,1) \rightarrow R$ be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}$ for $\theta \in ( 0, \frac{\pi}{4} )\cup (\frac{\pi}{4}, \frac{\pi}{2})$. Then, the value(s) of $f (\frac{1}{3})$ is/are
(a) $1-\sqrt{\frac{3}{2}}$
(b) $1+\sqrt{\frac{3}{2}}$
(c) $1-\sqrt{\frac{2}{3}}$
(d) $1+\sqrt{\frac{2}{3}}$
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Answer:
Correct Answer: 20.(a,b)
Solution:
Formula:
$f(\cos 4 \theta) =\frac{2}{2-\sec ^{2} \theta} $ $\quad$ …….(i)
$\text { At } \quad \cos 4 \theta =\frac{1}{3} $
$\Rightarrow \quad 2 \cos ^{2} 2 \theta-1 =\frac{1}{3} $
$\Rightarrow \quad \cos ^{2} 2 \theta =\frac{2}{3} $
$\Rightarrow \quad \cos 2 \theta = \pm \sqrt{\frac{2}{3}} $ $\quad$ …….(ii)
$\therefore f(\cos 4 \theta) =\frac{2 \cdot \cos ^{2} \theta}{2 \cos ^{2} \theta-1} $
$ =\frac{1+\cos 2 \theta}{\cos 2 \theta} $
[from Eq. (ii)]
$\Rightarrow \quad f(\frac{1}{3}) =1 \pm \sqrt{\frac{3}{2}}$
BETA
