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Let f(x)="max"(1+s in x ,1,1-cosx),x in ...

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`

A

`g(f(0))=1`

B

`g(f(1))=1`

C

`f(g(1))=1+sin1`

D

`f(g(0))=1+sin1`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`f(x)={{:(1+sinx,0lexle3pi//4),(1-cosx,3pi//4lexle3pi//2),(1,3pi//2lexle2pi):}`
`g(x)={{:(1-x,xle0),(1,0lexle2),(1-1,xle2):}`
`f(0)=1impliesg(f(0))=1`
`f(1)=1+sin1`
`because0lt1lt(3pi)/(4)`
`impliesf(f(1))=1` `because1lt1+sin1lt2`
Again `g(1)=1impliesf(g(1))=1+sin1`
`g(0)=1impliesf(g(0))=1+sin1`
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