Let f   : 0 ,   ∞ → R be given by f x =   ∫ 1 x x e - t + 1 t d t t , then

Let f :0, R be given by fx= 1xxe-t+1tdtt, then
  1. fx is monotonically increasing on 1, 
  2. fx is monotonically decreasing on (0, 1)
  3. fx+f1x=0, for all x 0, 
  4. f2x is an odd function of x on R

Solution

fx=2e-x+1xx
Which is increasing in 1, 
Also, fx+f1x=0
gx=f2x= 2-x2xe-t+1tt dt
g-x= 2x2-xe-t+1tt dt= -gx
Hence, an odd function

Asked in: JEE Advanced 2014 (Paper 1)

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