If limx→∞ 1+ax+bx22x=e2, then

If limx1+ax+bx22x=e2, then

  1. A

    a=1,b=2

  2. B

    a=2,b=1

  3. C

    a=1,bR

  4. D

    a=b=1

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    Solution:

    We have,

    limx1+ax+bx22x=elimxax+bx2×2x =elimx2a+2bx=e2a limx1+ax+bx22x=e2e2a=e22a=2a=1.

    Hence, a=1 and bR

    limxaf(x)g(x)=limxaf′′(x)g′′(x)

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