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

  • question_answer
    The range of following function is \[f(x)=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{...\infty }}}}\]

    A) \[\left[ 0,\text{ }1 \right]\]

    B) \[\left[ 0,\text{ }1/2 \right]\]

    C) \[\left[ 0,\text{ }2 \right]\]

    D) none of these

    Correct Answer: C

    Solution :

    [c] Given 
    \[f(x)=\sqrt{(1-\cos x)(1-\cos x)\sqrt{(1-\cos x)\sqrt{...\infty }}}\]
    \[={{(1-\cos x)}^{\frac{1}{2}}}{{(1-\cos x)}^{\frac{1}{4}}}{{(1-\cos x)}^{\frac{1}{8}}}...\infty \]
    \[={{(1-\cos x)}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\infty }}\]
    \[={{(1-\cos x)}^{\frac{1/2}{1-(1/2)}}}\]
    \[=1-\cos x\]
    Thus, the range of f(x) is [0, 2).


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