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Question

Let f(x)={1x<01+sinx0xπ2 then what is the value of f(x) at x=0?

A
1
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B
1
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C
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D
Does not exist
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Solution

The correct option is D Does not exist
For x<0 , f(x)=1
f1(0)=0
For x0 , f(x)=1+sinx
f1(0+)=cos(0)=1
Therefore f1(0)f1(0+)
So f1(x) at x=0 does not exist
Therefore the correct option is D

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