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Question

If f(x)={x+1xϵ[1,0]x2+1xϵ(0,1), then the value of f1(0)+f1(1)+f1(2)f(1)+f(0)+f(1) is-

A
0
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B
1
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C
2
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D
13
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Solution

The correct option is B 0
f(x)={x+1x(1,0)x2+1x(0,1)
x={f1(x+1)x[1,0]f1(x2+1)x(0,1)
f1(0):- put x=1
f1(0)=x=1
f1(1) put x=0
f1=0
f1(2):- put x=1
f1(2)=1
f(0)=1
f(1)=1+1=2
f(1)=1+1=0
f1(0)+f1(1)+f1(2)f(1)+f(0)+f(1)=1+0+13
=03
=0.

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