A) \[\left( 0,\frac{\pi }{3} \right)\]
B) \[\left( 0,\frac{\pi }{6} \right)\]
C) \[\left( 0,\frac{\pi }{4} \right)\]
D) \[\left( \frac{\pi }{4},\frac{\pi }{2} \right)\]
Correct Answer: D
Solution :
\[g'(x)=f'(ta{{n}^{2}}x-2tanx+4)2se{{c}^{2}}x(tanx-1)\] \[f''(x)>0\Rightarrow f'(x)\]is increasing \[{{\tan }^{2}}x-2\tan x+4={{(tanx-1)}^{2}}+3\] \[\Rightarrow \]\[{{\tan }^{2}}-2\tan x+4\ge 3\] \[f'(ta{{n}^{2}}x-2tanx+4)\ge f'(3)=0\] \[g'(x)>0iff\,tanx-1>0\Rightarrow x\in (\pi /4,\pi /2)\]You need to login to perform this action.
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