A) \[\frac{1}{2}\]
B) \[\frac{1}{5}\]
C) \[\frac{1}{10}\]
D) None of these
Correct Answer: C
Solution :
As \[f(x)\] is differentiable in-[2, 5], therefore it is also continuous in\[[2,\,\,5]\]. Hence, by mean value theorem, there exist a real number \[c\] in \[(2,\,\,5)\] such that \[f'(c)=\frac{f(5)-f(2)}{5-2}\] \[\Rightarrow \] \[f'(c)=\frac{\frac{1}{2}-\frac{1}{5}}{3}=\frac{1}{10}\] \[\left[ \because \,\,f(2)=\frac{1}{5}\,\,and\,\,f(5)=\frac{1}{2}\,\,are\,\,given \right]\]You need to login to perform this action.
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