A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{6}\]
Correct Answer: A
Solution :
We have \[\frac{f(1)\,-f(0)}{g(1)\,-g(0)}\,\,=\frac{'f(c)}{g'(c)}\] \[\Rightarrow \,\,\frac{1-0}{1-0}\,=\frac{2c}{3{{c}^{2}}}\,\Rightarrow \,\frac{3c}{2}=1\,(As\,\,0<c<1)\] \[\Rightarrow c=\frac{2}{3}\]You need to login to perform this action.
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