A) \[R-\{\,1,\,\,\,11\}\]
B) \[R-\{\,2,\,\,\,11\}\]
C) \[R-\{\,11\}\]
D) \[R-\{\,1,\,\,2\}\]
Correct Answer: A
Solution :
Given function, \[f(x)=|x-2|\,\,{{\log }_{10}}\,(x-1)\] The function is defined as, \[f(x)=\left\{ \begin{matrix} -(x-2)\,{{\log }_{10}}(x-1), & 1<x\le 2 \\ 0, & x=2 \\ (x-2)\,{{\log }_{10}}\,(x-1), & 2>x \\ \end{matrix} \right.\] At \[(x=2),\] \[Lf'=(2)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{f(2-h)-f(2)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-(2-h-2)\,{{\log }_{10}}\,(1-h)-0}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\,{{\log }_{10}}(1-h)}{h}={{\log }_{10}}1=0\] \[Rf'(2)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{f(2+h)-f(2)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{(2+h-2){{\log }_{10}}(1+h)-0}{h}\] \[[\because \,\,Lf'\,(2)=Rf'(2)]\] So, function is differentiable at \[(x=2).\] At \[(x=1),\] the function is not defined. So, at \[(x=1),\] the function is not differentiable. i.e, function is differentiable at \[R-\{1,11\}.\]
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