Let f-1-1- be such that fcos 4-2-2-2- for -0- -4-4- -2- Then the values of f1-3 is are

The correct options are A 1 √(32) B 1+√(32) For θϵ #10;(0, #160;π4)∪ ( #160;π4, #160;π2) #10;Suppose cos 4θ= #160;13 #10; ...

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Standard X

Mathematics

Trigonometric Ratios

Let f:-1,1→...

Question

Let $f : (- 1, 1) \to R$ be such that $f (cos 4 θ) = \frac{2}{2 - {sec}^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$ . Then the value(s) of $f (\frac{1}{3})$ is (are)

1 - \sqrt{\frac{3}{2}}

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

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1 - \sqrt{\frac{2}{3}}

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

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Solution

The correct options are
A $1 - \sqrt{\frac{3}{2}}$
B $1 + \sqrt{\frac{3}{2}}$
For $θ ϵ (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$
Suppose $c o s 4 θ = \frac{1}{3}$
$\Rightarrow c o s 2 θ = \pm \sqrt{\frac{1 + c o s 4 θ}{2}} = \pm \sqrt{\frac{2}{3}}$

and,
$f (\frac{1}{3}) = \frac{2}{2 - s e c^{2} θ} = \frac{2 c o s^{2} θ}{2 c o s^{2} θ - 1}$
$= 1 + \frac{1}{c o s 2 θ}$

then,
$f (\frac{1}{3}) = 1 - \sqrt{\frac{3}{2}}$ or $1 + \sqrt{\frac{3}{2}}$

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Let f:(−1,1)→R be such that f(cos4θ)=22−sec2θ for θ∈(0,π4)∪(π4, π2). Then the value(s) of f(13) is (are)

The correct options are A 1−√32 B 1+√32For θϵ(0,π4)∪(π4,π2) Suppose cos4θ=13 ⇒cos2θ=±√1+cos4θ2=±√23 and,f(13)=22−sec2θ=2cos2θ2cos2θ−1 =1+1cos2θthen, f(13)=1−√32 or 1+√32

Let $f : (- 1, 1) \to R$ be such that $f (cos 4 θ) = \frac{2}{2 - {sec}^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$ . Then the value(s) of $f (\frac{1}{3})$ is (are)