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

done JEE Main Online Paper (Held On 23 April 2013) Total Questions - 1

• question_answer1)                 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                                           
  done clear
  B)
                                           3
  done clear
  C)
                                           2                                           
  done clear
  D)
                                           1
  done clear
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