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Question

Let f(x)=max{1+sinx,1,1cosx},x[0,2π] and g(x)=max{1,|x1|},xR , then

A
g(f(0))=1
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B
g(f(1))=1
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C
f(g(1))=1
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D
f(g(0))=sin1
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Solution

The correct options are
B g(f(1))=1
C g(f(0))=1
f(x)=(max[1+sinx,1,1cos(x)] for xϵ[0,2π]
f(0)
=(max(1+0,1,0))
=1
f(1)
=(max(1+sin(1)),1,1cos1)
=1+sin(1)
g(x)=max(1,|x1|)
Hence,
g(f(0))
=g(1)
=max(1,0)
=1
And, g(f(1))
=max(1,|1+sin(1)1|)
=max(1,|sin(1)|)
=1
g(0)=1
f(g(0))=1+sin1

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