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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2006

  • question_answer
        The minimum value of \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x,0\le x\le \frac{\pi }{2}\]is

    A)  \[\frac{1}{2\sqrt{2}}\]                  

    B)  \[\frac{1}{4}\]

    C)  \[\frac{-1}{2}\]                                

    D)  \[\frac{1}{2}\]

    Correct Answer: D

    Solution :

                     Given \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[\Rightarrow \]               \[f(x)=1-\frac{1}{2}{{\sin }^{2}}2x\] Also, \[0<si{{n}^{2}}2x\le 1\] \[\therefore \]Minimum value of\[f(x)\]is\[1-\frac{1}{2}=\frac{1}{2}\]


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