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Q.

Two functions f:RR and g:RR are defined as follows f(x)=0(x rational )1(x irrational ),g(x)=1(x rational )0(x irrational ) then (fg)(π)+(gf)(e)=

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a

-1

b

0

c

1

d

2

answer is A.

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Detailed Solution

(fg)(π)+(gf)(π)=f(0)+g(1)=01=1

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Two functions f:R→R and g:R→R are defined as follows f(x)=0(x rational )1(x irrational ),g(x)=−1(x rational )0(x irrational ) then (f∘g)(π)+(g∘f)(e)=