A) \[\frac{3}{4}\]
B) \[\frac{4}{3}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{3}\]
E) \[\frac{5}{3}\]
Correct Answer: B
Solution :
Given curve is \[f(x)+x{{(x-1)}^{2}},x\in [0,2]\] \[\Rightarrow \] \[f(x)=x(3x-4)+1\] \[\therefore \] \[f(c)=c(3c-4)+1\] Also, \[f(0)=0,\]\[f(2)=2\] By mean value theorem \[f(0)-f(2)=f(c)(0-2)\] \[\Rightarrow \] \[0-2=(c(3c-4)+1)(-2)\] \[\Rightarrow \] \[3{{c}^{2}}-4c+1=1\] \[\Rightarrow \] \[c(3c-4)=0\] \[\Rightarrow \] \[c=0,\frac{4}{3}\] \[\Rightarrow \] \[c=\frac{4}{3}\] As \[0\cancel{\in }(0,2)\]
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