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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    The value of\[f\]at\[x=0\]so that function\[f(x)=\frac{{{2}^{x}}-{{2}^{-x}}}{x},x\ne 0,\]is continuous at\[x=0,\]is:

    A)  0                                            

    B)  log 2

    C)  4                            

    D)        \[{{e}^{4}}\]

    E)  log 4

    Correct Answer: E

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{x}}-{{2}^{-x}}}{x}=\underset{x\to 0}{\mathop{\lim }}\,{{2}^{x}}\log 2+{{2}^{-x}}\log 2\] (by L Hospitals rule) \[=log\text{ }2+log\text{ }2=log\text{ }4\] Since, Function is continuous at\[x=0\]. \[\therefore \]  \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{x}}-{{2}^{-x}}}{x}\] \[\Rightarrow \]               \[f(0)=\log 4\]


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