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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    Range of the function \[\frac{1}{2-\sin 3x}\] is [AMU 1999]

    A)                    [1, 3]

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

    C)            (1, 3)

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

    Correct Answer: B

    Solution :

               \[f(x)=\frac{1}{2-\sin 3x},\,\,\sin 3x\in [-1,\,\,1]\]            Hence \[f(x)\] lies in \[\left[ \frac{1}{3},\,\,1 \right]\].


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