Let f : S → S where S = 0 , ∞ be a twice differentiable function such that f x + 1 = x f x . If…

Let f:SS where S=0, be a twice differentiable function such that fx+1=xfx. If g:SR be defined as gx=logefx, then the value of g''5-g''1 is equal to :

  1. 205144
  2. 197144
  3. 187144
  4. 1

Solution

lnfx+1=lnxfx

lnfx+1=lnx+lnfx

gx+1=lnx+gx

gx+1-g(x)=lnx

  g''x+1-g''x=-1x2

Put x=1,2,3,4

g''2-g''1=-112 ...1

g''3-g''2=-122 ...2

g''4-g''3=-132 ...3

g''5-g''4=-142 ...4

Add all the equations we get

g''5-g''1=-112-122-132-142

g"5-g"1=205144

Asked in: JEE Main 2021 (16 Mar Shift 2)

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