From mean value theorem \[f(b)-f(a)=\] \[(b-a)f'({{x}_{1}});\] \[a

1.	From mean value theorem \[f(b)-f(a)=\] \[(b-a)f'({{x}_{1}});\] \[a
A.	\[\sqrt{ab}\]
B.	\[\frac{a+b}{2}\]
C.	\[\frac{2ab}{a+b}\]
D.	\[\frac{b-a}{b+a}\]
Answer» B. \[\frac{a+b}{2}\]

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From mean value theorem \[f(b)-f(a)=\] \[(b-a)f'({{x}_{1}});\] \[a

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