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BCECE Engineering BCECE Engineering Solved Paper-2009

  • question_answer
    \[f(x)=\left\{ \begin{matrix}    a{{x}^{2}}+b, & b\ne 0,x\le 1  \\    b{{x}^{2}}+ax+c, & x>1  \\ \end{matrix} \right.\] then\[f(x)\] is continuous and differentiable at \[x=1\]if

    A) \[c=0,\,a=2b\]

    B)         \[a=b,\,c\in R\]

    C)  \[a=b,c=0\]

    D)  \[a=b,c\ne 0\]

    Correct Answer: A

    Solution :

    \[f(x)\]is continuous at \[x=1\] \[\therefore \]  \[f(1)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\] \[\Rightarrow \]               \[a+b=b+a+c\] \[\Rightarrow \]               \[c=0\] Also, \[f(x)\]is difference at \[x=1\] \[Lf(x)=Rf(x)\] \[\therefore \]\[\frac{d}{dx}(a{{x}^{2}}+b)=\frac{d}{dx}(b{{x}^{2}}+ax+c)\] \[\Rightarrow \]\[2ax=2bx+a\] At           \[x=1\] \[\Rightarrow \]\[2a=2b+a\] \[\Rightarrow \]\[a=2b\]


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