For  f(x)=ln⁡x2+α7x  Roll’s theorem is applicable on  [3, 4] with  c∈(3,4)  such that  f′(c)=0.  The value of  f′′(c) is

For  f(x)=lnx2+α7x  Roll's theorem is applicable on  [3, 4] with  c(3,4)  such that  f(c)=0 The value of  f′′(c) is

  1. A

    112

  2. B

    112

  3. C

    16

  4. D

    16

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    Solution:

    Roll's theorem is applicable to  f(x)=lnx2+α7x on [3,4]

         f(3)=f(4)     ln9+α21=ln16+α28

     9+α21=16+α2836+4α=48+3αα=12

    Now,  f(x)=lnx2+α7x

     f(x)=lnx2+12ln7x f(x)=2xx2+121x=x212xx2+12 f(x)=0x212=0x=±12 c=12

    Now,  f(x)=x212xx2+12

    xx2+12f′′(x)=2x2x2+12x212x2+12+2x2x2x2+122f′′(x)=x4+48x2+144x2x2+122f′′(12)=112

     

     

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