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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    Function\[f(x)\]is defined as follows \[f(x)=\left\{ \begin{matrix}    ax-b, & x\le 1  \\    3x, & 1<x<2  \\    b{{x}^{2}}-a, & x\ge 2  \\ \end{matrix} \right.\] lf\[f(x)\] is continuous at\[x=1\], but discontinuous at\[x=2\]then the locus of the point\[(a,\,\,b)\] is a straight line excluding the point where it cuts the line

    A) \[y=3\]                                

    B) \[y=2\]

    C) \[y=0\]                

    D)        \[y=1\]

    Correct Answer: A

    Solution :

    Given;\[f(x)\]is continuous at\[x=1\] \[\therefore \]\[f(1)=RHL\] \[\Rightarrow \]\[f(1)=\underset{x\to 1}{\mathop{\lim }}\,+f(x)\Rightarrow f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1+h)\] \[\Rightarrow \]\[a-b=\underset{h\to 0}{\mathop{\lim }}\,3(1+h)\Rightarrow a-b=3\]   ? (i) Again, given\[f(x)\]is discontinuous at\[x=2\]. \[\therefore \]  \[LHL\ne f(x)\]                                  \[\Rightarrow \]\[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)\ne f(2)\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f(2-h)\ne f(2)\] \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,3(2-h)\ne 4b-a\Rightarrow 6\ne 4b-a\]    ... (ii) Assume,\[6=4b-a\] then from (i) and (ii), we get\[b=3\]. \[\therefore \]locus                       \[y=3\] Which is impossible                       \[(\because \,\,6\ne 4b-a)\] Hence, locus of\[(a,\,\,b)\]is\[x-y=3\]excluding the point when it cuts the line\[y=3\].


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