Let  ddx(F(x))=esin⁡xx,x>0 if ∫14 2esin⁡x2xdx=F(k)−F(1) ), then one of the possible values of k is, 

Let  ddx(F(x))=esinxx,x>0 if 142esinx2xdx=F(k)F(1) ), then one of the possible values of k is, 

  1. A

    4

  2. B

    8

  3. C

    16

  4. D

    32

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    Solution:

    We have,

    ddx(F(x))=esinxxesinxxdx=F(x)               …(i)

    Now, 

    14esinx2xdx=14esinx2x2dx2=116esinttdt, where t=x2

    142esinx2xdx=[F(t)]16=F(16)F(1)      [using (i)]

    hence , k=16

     

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