Given,
f:(0,1)→R be a function defined by f(x)=1−e−x1, and g(x)=(f(−x)−f(x))
Now using g(x)=f(−x)−f(x) we get,
g(x)=1−ex1−ex−1ex
⇒g(x)=1−ex1+ex
Now differentiating the above function with respect to x we get,
⇒g′(x)=(1−ex)22ex>0
Since g′(x) is always positive
So,g(x) is one-one in (0,1)
And g(x) is increasing in (0,1)