A) \[1-\frac{\sqrt{15}}{6}\]
B) \[1+\sqrt{15}\]
C) \[1-\frac{\sqrt{21}}{6}\]
D) \[1+\sqrt{21}\]
Correct Answer: C
Solution :
From mean value theorem \[{f}'(c)=\frac{f(b)-f(a)}{b-a}\] \[a=0,\,f(a)=0\] Þ \[b=\frac{1}{2},\,f(b)=\frac{3}{8}\] \[{f}'(x)=(x-1)(x-2)+x(x-2)+x(x-1)\] \[{f}'(c)=(c-1)(c-2)+c(c-2)+c(c-1)\] = \[{{c}^{2}}-3c+2+{{c}^{2}}-2c+{{c}^{2}}-c\] \[{f}'(c)=3{{c}^{2}}-6c+2\] According to mean value theorem, \[{f}'(c)=\frac{f(b)-f(a)}{b-a}\] Þ \[3{{c}^{2}}-6c+2=\frac{(3/8)-0}{(1/2)-0}\,=\frac{3}{4}\] Þ \[3{{c}^{2}}-6c+\frac{5}{4}=0\] \[c=\frac{6\pm \sqrt{36-15}}{2\times 3}=\frac{6\pm \sqrt{21}}{6}\]\[=1\pm \frac{\sqrt{21}}{6}\].
You need to login to perform this action.
You will be redirected in
3 sec
