A) \[\frac{f(2)-f(0)}{2-0}\]
B) \[\frac{f(2)-f(-1)}{2-(-1)}\]
C) \[\frac{f(1)-f(-1)}{1-(-1)}\]
D) None of these
Correct Answer: A
Solution :
\[f(x)=2{{x}^{2}}-|x|+4\] Again define the function, \[f(x)=\left\{ \begin{matrix} 2{{x}^{2}}+x+4,\,-1\le x\le 0 \\ 2{{x}^{2}}-x+4,0\le x\le 2 \\ \end{matrix} \right.\] Case I For interval \[[-1,0]\] The function \[f(x)=2{{x}^{2}}+x+4,\] are continuous and differentiable in its domain. Because it is a polynomial function. Now, we check, the log range?s condition, \[f'(c)=\frac{f(0)-f(-1)}{0-(-1)}\] \[=\frac{4-(2-1+4)}{1}\] \[(4c+1)=-1\] \[\Rightarrow \] \[4c=-2\] \[\Rightarrow \] \[c=-1/2\] Which lies in the interval \[(-1,\,0)\]. Case II. For interval \[[0,2]\] The function \[f(x)=2{{x}^{2}}-x+4\] are continuous and differentiable in its domain. Because it is a polynomial function. Now, by log range?s mean value theorem. \[f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{(8-2+4)-4}{2}\] \[4c-1=\frac{6}{2}=3\,\,\Rightarrow \,\,c=1\] Which lies in the interval \[(0,\,\,2).\]
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