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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    If\[f(x)\left\{ \begin{matrix}    a{{x}^{2}}+b, & 0\le x\le 1  \\    x+3, & 1<x\le 2  \\    4, & x=1  \\ \end{matrix} \right.\], then the value of \[(a,\,\,b)\]for which\[f(x)\]cannot be continuous at\[x=1\]is

    A)  (2, 2)                                    

    B)  (3, 1)

    C)  (4, 0)                    

    D)         (5, 2)

    Correct Answer: D

    Solution :

    We have, \[\underset{h\to 0}{\mathop{\lim }}\,f(1-h)=\underset{h\to 0}{\mathop{\lim }}\,{{(1-h)}^{2}}+b=a+b\] \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,f(1+h)=\underset{h\to 0}{\mathop{\lim }}\,(1+h)+3=4\] and\[f(1)=4\] \[\therefore \]\[f(x)\]will not be continuous at\[x=1\], if\[a+b\ne 4\]


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