Let f : ( - 1 , 1 ) be such that f ( cos 4 ) = 2 ? 2 - sec 2? for ( 0 , 4 ) ( 4 , 2 ) . Then the value(s) of f ( 1 3 ) is (are) [2012]

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) [2012]

Smart Achievers · Accepted Answer

(1, 2)

$Given: f (\cos 4 θ) = \frac{2}{2 - \sec^{2} θ} = \frac{2 \cos^{2} θ}{2 \cos^{2} θ - 1}$

$= \frac{1 + \cos 2 θ}{\cos 2 θ} = 1 + \frac{1}{\cos 2 θ}$

$Let$ $\cos 4 θ = \frac{1}{3}$ $\Rightarrow 2 \cos^{2} 2 θ - 1 = \frac{1}{3}$ $\Rightarrow \cos 2 θ = \pm \sqrt{\frac{2}{3}}$

$∴$ $f (\cos 4 θ) = 1 + \frac{1}{\cos 2 θ} = 1 \pm \sqrt{\frac{3}{2}}$ or $f (\frac{1}{3}) = 1 \pm \sqrt{\frac{3}{2}}$

Solution

Q.
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) [2012]

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

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

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

4 $1 + \sqrt{\frac{2}{3}}$

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Solution

Q. Let f:(-1,1)→ℝ be such that f(cos4θ)=2 2-sec2θ for θ∈(0,π4)∪(π4,π2). Then the value(s) of f(13) is (are) [2012]

1 1-32

2 1+32

3 1-23

4 1+23