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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    The minimum value of\[{{e}^{(2{{x}^{2}}-2x+1){{\sin }^{2}}x}}\]is

    A)  0           

    B)  1

    C)  2            

    D)  3

    Correct Answer: B

    Solution :

    Given, \[y={{e}^{(2{{x}^{2}}-2x+1){{\sin }^{2}}x}}\] \[2{{x}^{2}}-2x+1>0\]for all\[x,\] \[y\]is minimum for\[sin\text{ }x=0\]. \[\Rightarrow \] \[x=0\] \[\therefore \] \[{{y}_{\min }}={{e}^{0}}=1\]


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