Let f:R→R and g:R→R be defined as:
f(x)={logex,x>0e-x,x≤0 and g(x)={x,x≥0ex,x<0.
Then, gof:R→R is [2024]
(1)
g(f(x)) ={f(x),f(x)≥0ef(x),f(x) < 0
={logex,x≥1e-x,x≤0elogex=x, 0<x<1
gof(x)={e-x ,x≤0x ,0<x<1logex,x≥1
Now, range of gof=[0,∞)≠R
Hence, not one-one and not onto.