RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    The value of b for which the function \[f(x)=\left\{ \begin{matrix}    5x-4, & 0<x\le 1  \\    4{{x}^{2}}+3bx, & 1<x<2  \\ \end{matrix} \right.\]is continuous at every point of its domain is

    A)  \[-1\]              

    B)  0

    C)  1              

    D)  13/3

    Correct Answer: A

    Solution :

     \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(5x-4)=1\] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(4{{x}^{2}}+3bx)=4+3b\] \[\Rightarrow \]\[4+3b=1\]\[\Rightarrow \]\[b=-1\]


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