JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    The function \[f:R\to R\] is defined by \[f(x)={{\cos }^{2}}x+{{\sin }^{4}}x\] for \[x\in R\], then \[f(R)=\]            [EAMCET 2002]

    A)                    \[\left( \frac{3}{4},\ 1 \right]\]

    B)             \[\left[ \frac{3}{4},\ 1 \right)\]

    C)            \[\left[ \frac{3}{4},\ 1 \right]\]

    D)            \[\left( \frac{3}{4},\ 1 \right)\]

    Correct Answer: C

    Solution :

               \[y=f(x)={{\cos }^{2}}x+{{\sin }^{4}}x\]            Þ \[y=f(x)={{\cos }^{2}}x+{{\sin }^{2}}x(1-{{\cos }^{2}}x)\]            Þ \[y={{\cos }^{2}}x+{{\sin }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x\]            Þ  \[y=1-{{\sin }^{2}}x{{\cos }^{2}}x\] Þ  \[y=1-\frac{1}{4}.{{\sin }^{2}}2x\]            \ \[\frac{3}{4}\le f(x)\le 1\], \[(\because 0\le {{\sin }^{2}}2x\le 1)\]            Þ  \[f(R)\in [3/4,\,\,1]\].


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