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Question

Mathematics Question on Functions

Let f:(−1,1)→f:(-1,1)\rightarrow R be such that f(cos4θ)=22−sec2θf(cos 4 \theta)=\frac{2}{2-sec^2 \theta} for θ∈(0,π4)∪(π4,π2).\theta \in \Bigg(0,\frac{\pi}{4}\Bigg)\cup\Bigg(\frac{\pi}{4},\frac{\pi}{2}\Bigg). Then, the value(s) of f(13)f\Bigg(\frac{1}{3}\Bigg) is are

A

1−321-\sqrt{\frac{3}{2}}

B

1+321+\sqrt{\frac{3}{2}}

C

1−231-\sqrt{\frac{2}{3}}

D

1+231+\sqrt{\frac{2}{3}}

Answer

1+321+\sqrt{\frac{3}{2}}

Explanation

Solution

f(cos 4\theta)=\frac{2}{2-sec^2 \theta}\hspace15mm ...(i) At \hspace15mm cos4\theta=\frac{1}{3} \Rightarrow 2 cos^2 \theta-1=\frac{1}{3} \Rightarrow \hspace15mm cos^22\theta=\frac{2}{3}\Rightarrow cos 2\theta=\pm\sqrt{\frac{2}{3}}\hspace15mm ...(ii) \therefore \hspace15mm f(cos4\theta)=\frac{2.cos^2 \theta}{2 cos^2 \theta-1}= frac{1+cos^\theta}{cos 2\theta} \Rightarrow \hspace15mm f\Big(\frac{1}{3}\Big)=1\pm\sqrt{\frac{3}{2}}\hspace15mm [from E(ii)]