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JEE Advanced · Mathematics · 25. AOD

Let f :0, R be given by fx= 1xxe-t+1tdtt, then

  1. A fx is monotonically increasing on 1,
  2. B fx is monotonically decreasing on (0, 1)
  3. C fx+f1x=0, for all x 0,
  4. D f2x is an odd function of x on R
Verified Solution

Answer & Solution

Correct Answer

(A) fx is monotonically increasing on 1,

Step-by-step Solution

Detailed explanation

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
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