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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    Let \[[t]\] denote the greatest integer \[\le t\] and \[\underset{x\to 0}{\mathop{\lim }}\,\,x\left[ \frac{4}{x} \right]=A.\] Then the function, \[f(x)=[{{x}^{2}}]\,\,sin(\pi x)\] is discontinuous, when x is equal to [JEE MAIN Held on 09-01-2020 Evening]

    A) \[\sqrt{A+21}\]

    B) \[\sqrt{A}\]

    C) \[\sqrt{A+1}\]

    D) \[\sqrt{A+5}\]

    Correct Answer: C

    Solution :

    \[\underset{x\to 0}{\mathop{Lim}}\,\,\,x\left[ \frac{4}{x} \right]=A\] \[\Rightarrow \,\,\,\,\,\underset{x\to 0}{\mathop{Lim}}\,\,\,\,x\left( \frac{4}{x}-\left\{ \frac{4}{x} \right\} \right)=A\] \[\Rightarrow \,\,\,\,\,4-\underset{x\to 0}{\mathop{Lim}}\,\,\,\,x\,\left\{ \frac{4}{x} \right\}=A\,\,\,\,\,\,\,\Rightarrow A=4\] Now,  at \[x=\sqrt{A+1}\] i.e. \[x=\sqrt{5},\] \[f(x)=[{{x}^{2}}]\,\sin \pi x\] is discontinuous Whereas at \[x=5,\] 2 and 3, \[f(x)\] is continuous.


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