If  a limx→1 x11−x+b=e−1(a≥1,b≥0) then

If  a limx1x11x+b=e1(a1,b0) then

  1. A

    a=1,b=e1

  2. B

    a=2,b=e1

  3. C

    a=1,b=e1

  4. D

    a=1,b=0

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    Solution:

    We have,

    alimx1x1/1x+b=e1alimx1[1+(x1)]1/1x+b=e1alimx1ex-11x+b=e1a e1+b=e1a=1 and b=0

     

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