From mean value theorem f (b) - f (a) = (b - a) f' (x1) ; a < b if f (x) = \(\frac{1}{x},\) then x1 =

1
\(\sqrt {ab} \)
2
\(\frac{{a + b}}{2}\)
3
\(\frac{{2ab}}{{a + b}}\)
4
\(\frac{{b - a}}{{b + a}}\)

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