A) \[a<{{x}_{1}}\le b\]
B) \[a\le {{x}_{1}}<b\]
C) \[a<{{x}_{1}}<b\]
D) \[a\le {{x}_{1}}\le b\]
Correct Answer: C
Solution :
According to mean value theorem, In interval [a, b] for f (x) \[\frac{f(b)-f(a)}{b-a}=f'(c)\], where \[a<c<b\] \[\therefore a<{{x}_{1}}<b\].
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