A) 2 and \[\frac{1}{2}\]
B) 3 and \[\frac{1}{3}\]
C) 4 and \[\frac{1}{4}\]
D) 5 and \[\frac{1}{5}\]
Correct Answer: B
Solution :
\[y=\frac{3\tan x-{{\tan }^{3}}x}{(1-3{{\tan }^{2}}x)\,\,\tan x}\] |
\[y=\frac{3-{{\tan }^{2}}x}{1-3{{\tan }^{2}}x}\] |
\[{{\tan }^{2}}x=\frac{3-y}{1-3y}\] |
\[if\,\,{{\tan }^{2}}x\ge 0\] |
\[\frac{3-y}{1-3y}\ge 0\] |
Never lies between \[\frac{1}{3}\]and 3. |
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