2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    If\[f''(x)>0\forall x\in R\]and\[f'(3)=0\]and\[g(x)=f(ta{{n}^{2}}x-2tanx+4),\]\[0<x<\pi /2\]then g(x) is increasing in

    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.
You will be redirected in 3 sec spinner