JEE Main & Advanced Sample Paper JEE Main - Mock Test - 37

  • question_answer
    Least value of the function \[f\left( x \right)={{e}^{\sin x-2{{\sin }^{2}}x}}\] is

    A) \[\frac{1}{\sqrt[3]{e}}\]             

    B)        \[\frac{1}{{{e}^{6}}}\]

    C) \[{{e}^{3}}\]              

    D)        \[\frac{1}{{{e}^{3}}}\]

    Correct Answer: D

    Solution :

    [d] \[{{e}^{f(x)}}\] is increasing or decreasing according as f(x) increases or decreases Let \[f\left( x \right)=sinx-2si{{n}^{2}}x\] \[\Rightarrow f'\left( x \right)=cosx-4sinxcosx=\left( 1-4sinx \right)cosx\] \[f'\left( x \right)=0\Rightarrow sin\text{ }x=\frac{1}{4}or\,sinx=\pm 1\] (corresponding to \[cos\,x=0\]) Also, \[f''\left( x \right)=-sin\,x-4cos2x\] \[=-\sin x-4+8\,{{\sin }^{2}}x\] \[f''\left( x \right)>0\,if\,sinx=\pm 1\,and\text{ }f''\left( x \right)<0\text{ }if\text{ }sinx=\frac{1}{4}\] Minima occurs when \[sinx=\pm 1\] when \[sin\text{ }x=1\] then \[f\left( x \right)=-1\]


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