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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If \[f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{4}}x}{{{\sin }^{2}}x+{{\cos }^{4}}x}\] for \[x\in R\], then \[f(2002)=\] [EAMCET 2002]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: A

    Solution :

               \[f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{4}}x}{{{\sin }^{2}}x+{{\cos }^{4}}x}\]Þ \[f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{2}}x(1-{{\cos }^{2}}x)}{{{\sin }^{2}}x+{{\cos }^{2}}x(1-{{\sin }^{2}}x)}\]            Þ \[f(x)=\frac{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}\]            Þ \[f(x)=1\] Þ \[f(2002)=1\].


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