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Let \mathrm{f:[0,1] \rightarrow[0,1]} be a continuous function. Then
 

Option: 1

\mathrm{f(x)=x} for at least one \mathrm{0 \leq x \leq 1}
 


Option: 2

\mathrm{f(x)} will be differentiable in \mathrm{[0,1]}
 


Option: 3

\mathrm{f(x)+x=0 }for at least one x such that \mathrm{0 \leq x \leq 1}
 


Option: 4

 none of these


Answers (1)

Clearly, \mathrm{0 \leq f(0) \leq 1}  and \mathrm{0 \leq f(1) \leq 1}. As \mathrm{f(x)}  is continuous, \mathrm{f(x)} attains all values between \mathrm{f(0)} to \mathrm{f(1)}, and the graph will have no breaks. So, the graph will cut the line \mathrm{y=x} at one point \mathrm{x} at least where \mathrm{0 \leq x \leq 1}. So, \mathrm{f(x)=x} at that point.

Posted by

Sumit Saini

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