Let f:R and g:R be defined as f(x)= cases e x, & x>0\ e-x, & x 0 cases and g(x)= casesx, & x 0\ ex, & x<0 cases. Then, gof :R is :
- one-one but not onto
- neither one-one nor onto
- onto but not one-one
- both one-one and onto
Answer
(2) neither one-one nor onto
Solution
x 0: f(x)=e-x 1 g(f(x))=e-x
0<x<1: f(x)= ex<0 g(f(x))=e ex=x
x 1: f(x)= ex 0 g(f(x))= ex
[figure]
g(f(0))=1=g(f(e)) ⇒ not one-one
Range =[0, ) ⇒ not onto
Answer: (2)