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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    If \[y=\frac{1}{{{t}^{2}}+t-2}\] where \[t=\frac{1}{x-1}\], then find the number of points of discontinuities of \[y=f(x),\],

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: C

    Solution :

    [c] \[t=\frac{1}{x-1}\] is discontinuous at \[x=1.\] Also \[y=\frac{1}{{{t}^{2}}+t-2}\] discontinuous at \[t=-2\] and \[t=1\] When \[t=-2,\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}\] When \[t=1,\,\,\frac{1}{x-1}\,\,\,\,\Rightarrow x=2\] So, \[y=f(x)\] is discontinuous at three points \[x=1,\,\,\frac{1}{2},\,\,2\]


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