A) \[\frac{4-\sqrt{7}}{2}\]
B) \[\frac{4-\sqrt{5}}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{\sqrt{7}-2}{3}\]
Correct Answer: A
Solution :
[a] \[\because \,\,\,\,f'(c)=\frac{f(b)-f(a)}{b-a}\] \[\therefore \,\,\,\,\,\,\,\,\,\,3{{c}^{2}}-8c+8=\frac{f(1)-f(0)}{1-0}\] \[3{{c}^{2}}-8c+8=16-11\] \[3{{c}^{2}}-8c+3=0\] \[\therefore \,\,\,\,c=\frac{8\pm 2\sqrt{7}}{6}\] \[\therefore \,\,\,\,c=\frac{4-\sqrt{7}}{3}\] as \[c\in (0,1)\]
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