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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        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

    A)  \[a=-11\]           

    B)  \[a=-6\]

    C)  \[a=6\]                               

    D)  \[a=11\]

    Correct Answer: D

    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.


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