JEE Main & Advanced Mathematics Relations Question Bank Mock Test - Relations and Functions

  • question_answer
    The function \[f:R\to R\] is defined by \[f\left( x \right)={{\cos }^{2}}x+{{\sin }^{4}}x\] for \[x\in R\]. Then the range of \[f(x)\]is

    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 :

    [c] \[y=f(x)=co{{s}^{2}}x+{{\sin }^{4}}x\] \[={{\cos }^{2}}x+{{\sin }^{2}}x(1-co{{s}^{2}}x)\] \[={{\cos }^{2}}x+{{\sin }^{2}}x-{{\sin }^{2}}xco{{s}^{2}}x)\] \[=1-{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{1}{4}{{\sin }^{2}}2x\] \[\therefore \,\,\,\,\frac{3}{4}\le f(x)\le 1\]  \[(\therefore 0\le si{{n}^{2}}2x\le 1)\] \[\therefore \,\,\,f(x)\in [3/4,1]\]     


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