A) \[-\frac{1}{3}\]and 0
B) \[\frac{1}{3}\]and 0
C) \[-\frac{1}{3}\]and \[\frac{1}{3}\]
D) - 3 and 3
Correct Answer: B
Solution :
Let\[y=\frac{\tan 3x}{\tan x}.\]Then \[y=\frac{3\tan x-{{\tan }^{3}}x}{(1-3{{\tan }^{2}}x)}\times \frac{1}{\tan x}\] \[\Rightarrow \]\[y=\frac{3-{{\tan }^{2}}x}{1-3{{\tan }^{2}}x}\]\[\Rightarrow \]\[{{\tan }^{2}}x(3y-1)=y-3\] \[\Rightarrow \]\[{{\tan }^{2}}x=\frac{y-3}{3y-1}\] Since, \[{{\tan }^{2}}x\ge 0\] for all x. \[\therefore \]\[\frac{y-3}{3y-1}\ge 0\] \[\Rightarrow \]\[(y-3)(3y-1)\ge 0\]\[\Rightarrow \]\[y<\frac{1}{3}\]or\[y\ge 3\] \[\Rightarrow \]does not lie between\[\frac{1}{3}\] and 3.
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