Let `f(x)="max"(1+s in x ,1,1-cosx),x in [0,2pi],a n dg(x)=max{1,|x-1|}, x in Rdot` Then `g(f(0))=1` (b) `g(f(1))=1` `f(f(1))=1` (d) `f(g(0))+1sin1`

Options

A. `g(f(0))=1`
B. `g(f(1))=1`
C. `f(f(1))=1`
D. `f(g(0))=1+sin 1`

Answer

A, B, D

Explanation

`f(0)=max {1+sin 0,1,1-cos 0}=1`
`g(0)=max{1,|0-1|}=1`
`f(1)=max [1+sin1,1,1-cos1}=1+sin1`
`g(f(0))=g(1)=max{1,|1-1|}=1`
`f(g(0))=f(1)=1+sin1`
`g(f(1))=g(1+sin1)=max{1,|1+sin1-1|}=1`