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KVPY Sample Paper KVPY Stream-SX Model Paper-26

  • question_answer
    Let [n] denotes the greatest integer less than or equal to \[\lambda \]. How many real numbers x satisfy the equation \[{{x}^{2}}+10000\left[ x \right]=10000x\]

    A) 198

    B) 199     

    C) 200

    D) 201

    Correct Answer: B

    Solution :

    We have, \[{{x}^{2}}+10000\left[ x \right]=10000x\]\[\Rightarrow {{x}^{2}}+10000\left( x-\{x\} \right)=10000x\]\[\Rightarrow {{x}^{2}}+10000x-10000\{x\}=10000x\]
    \[\Rightarrow \,\{x\}=\frac{{{x}^{2}}}{10000}\Rightarrow 0\le \{x\}<1\]
    \[\therefore \,\,\,\,\,\,\,\,\,0\le \frac{{{x}^{2}}}{10000}<1\]
    \[-100<x<100\]Total number is 199


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