Q.

Let f,g:R→R be defined as:                                                                [2024]

f(x)=|x-1| and g(x)={ex,x≥0x+1,x≤0.

Then the function f(g(x)) is

1 onto but not one-one.  
2 neither one-one nor onto.  
3 both one-one and onto.  
4 one-one but not onto.  

Ans.

(2)

f(g(x))=|g(x)-1|

={|ex-1|,x≥0|x+1-1|,x≤0

={ex-1,x≥0-x,x≤0

f(g(x)) is neither one-one nor onto as negative numbers have no pre-image.