JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relations and Functions-I

  • question_answer
    The domain of the function\[f(x)=\sqrt{x-\sqrt{1-{{x}^{2}}}}\] is

    A) \[\left[ -1,-\frac{1}{\sqrt{2}} \right]\cup \left[ \frac{1}{\sqrt{2}},1 \right]\]

    B) \[[-1,1]\]

    C) \[\left( -\infty ,-\frac{1}{2} \right]\cup \left[ \frac{1}{\sqrt{2}},+\infty  \right)\]

    D) \[\left[ \frac{1}{\sqrt{2}},1 \right]\]

    Correct Answer: D

    Solution :

    [d] For f(x) to be defined, we must have
    \[x-\sqrt{1-{{x}^{2}}}\ge 0\] or \[x\ge \sqrt{1-{{x}^{2}}}>0\] or \[{{x}^{2}}\ge 1-{{x}^{2}}\] or \[{{x}^{2}}\ge \frac{1}{2}.\]
    Also, \[1-{{x}^{2}}\ge 0\,\,or\,\,{{x}^{2}}\le 1.\]
    Now. \[{{x}^{2}}\ge \frac{1}{2}\Rightarrow \left( x-\frac{1}{\sqrt{2}} \right)\left( x+\frac{1}{\sqrt{2}} \right)\ge 0\]
    \[\Rightarrow x\le -\frac{1}{\sqrt{2}}\] or \[x\ge \frac{1}{\sqrt{2}}\]
    Also, \[{{x}^{2}}\le 1\Rightarrow (x-1)(x+1)\le 0\Rightarrow -1\le x\le 1\]
    Thus, \[x>0,{{x}^{2}}\ge \frac{1}{2}\] and \[{{x}^{2}}\le 1\Rightarrow x\in \left[ \frac{1}{\sqrt{2}},1 \right]\]


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