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`

Let `f(x)="max"(1+s in x ,1,1-cosx),x in [0,2pi],a n dg(x)=max{1,|x-1|

1.	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`
Answer» Correct Answer - 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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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`

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