The real valued function f ( x ) = x e x - 1 + x 2 + 1 defined on R \ { 0 } is

The real valued function f(x)=xex-1+x2+1 defined on R\{0} is
  1. An odd function
  2. An even function
  3. Both even & odd function
  4. Neither even nor odd function

Solution

Given f(x)=xex-1+x2+1

=2x+xex-x+2ex-22(ex-1)

=ex(x+2)+x-22(ex-1)

Then, f(-x)=e-x(-x+2)-x-22(e-x-1)

=-x+2ex-x-221ex-1

=-x+2-xex-2ex21-ex

=-x+2-exx+221-ex

=x-2+exx+22ex-1

 ⇒f(x)=f(-x)

Therefore, f(x) is an even function. 

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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