1. Tardigrade
  2. Question
  3. Mathematics
  4. If the function, f(x)=x3-6x2+ax+b satisfies Rolles theorem in the interval [1, 3] and f( (2√3+1/√3) )=0, then

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Q. If the function, $ f(x)={{x}^{3}}-6{{x}^{2}}+ax+b $ satisfies Rolles theorem in the interval [1, 3] and $ f\left( \frac{2\sqrt{3}+1}{\sqrt{3}} \right)=0, $ then

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Solution:

We have $ f(x)=\left\{ \begin{matrix} \frac{|x-4|}{x-4}, & x\ne 4 \\ 0, & x=4 \\ \end{matrix} \right. $ Now, RHL $ \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(4+h) $ $ =\underset{h\to 0}{\mathop{\lim }}\,\frac{|4+h-4|}{4+h-4}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|}{h} $ $ =\underset{h\to 0}{\mathop{\lim }}\,1=1 $ and LHL $ \underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(4-h) $ $ =\underset{h\to 0}{\mathop{\lim }}\,\frac{|4-h+4|}{4-h+4} $ $ =\underset{h\to 0}{\mathop{\lim }}\,\frac{|-h|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,(-1)=-1 $ Since, $ LHL\ne RHL $ ie, $ \underset{x\to {{4}^{-1}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x) $ Thus, limit does not exist.