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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    The value of\[f(0)\], so that the function\[f(x)=\frac{2-{{(256-7x)}^{1/8}}}{{{(5x+3x)}^{1/5}}-2},\,\,x\ne 0\]is continuous everywhere is given by

    A)  \[-1\]                                   

    B)  \[1\]

    C) \[{{2}^{6}}\]                      

    D)         None of these

    Correct Answer: D

    Solution :

    For\[f(x)\]to be continuous everywhere, we must have                 \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,f(x)\]                 \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-{{(256-7x)}^{1/8}}}{{{(5x+32)}^{1/5}}-2}\]                 \[=\lim \frac{\frac{7}{8}{{(256-7x)}^{-7/8}}}{{{(5x+32)}^{-4/5}}}\]                 \[=\frac{7}{8}\times \frac{{{2}^{-7}}}{{{2}^{-4}}}=\frac{7}{64}\]


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