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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    Let\[f\] be a composite function of \[x\] defined by  \[f(u)=\frac{1}{{{\operatorname{u}}^{2}}+\operatorname{u}-2},\operatorname{u}(x)=\frac{1}{x-1}.\]                 Then the number of points \[x\] where f of is discontinuous is:     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                   4                                           

    B)                                          3

    C)                                          2                                           

    D)                                          1

    Correct Answer: B

    Solution :

                     \[\mu (x)=\frac{1}{x-1},\]which is discontinous at\[x=1\] \[f(u)=\frac{1}{{{u}^{2}}+u-2}=\frac{1}{(u+2)(u-1)},\] which is discontinous at \[u=-2,1\] when \[u=-2,\]then\[\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}\] when \[u=1,\]then\[\frac{1}{x-1}=1\Rightarrow x=2\] Hence given composite function is discontinous at three points, \[x=1,\frac{1}{2}\]and 2.


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