Solved papers for JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

done JEE Main Online Paper (Held on 9 April 2013) Total Questions - 1

• question_answer1)                                 Let\[f(x)=\frac{{{x}^{2}}-x}{{{x}^{2}}+2x},x\ne 0,-2.\]Then\[\frac{d}{dx}\left[ {{f}^{-1}}(x) \right]\] (wherever it is defined) is equal to :        JEE Main Online Paper (Held On 09 April 2013)                         
  A)
                  frac{-1}{{{(1-x)}^{2}}}\]                
  done clear
  B)
                  \[\frac{3}{{{(1-x)}^{2}}}\]                
  done clear
  C)
                  \[\frac{1}{{{(1-x)}^{2}}}\]                
  done clear
  D)
                  \[\frac{-3}{{{(1-x)}^{2}}}\]                                           
  done clear
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