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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[lo{{g}_{e}}4=1.3868,\]then the approximate value of\[lo{{g}_{e}}4.01\]is:

    A)  0.13893               

    B)         1.3843 

    C)         1.3893                 

    D)         0.13843

    E)  1.3869

    Correct Answer: C

    Solution :

    \[f(x)={{\log }_{e}}x\] where\[x=4\]and\[\Delta x=4.01-4=0.01\] Now,     \[f(x)=\frac{1}{x}\] \[\therefore \]  \[f(x+\Delta x)=f(x)+f(x)\Delta x\]                 \[={{\log }_{e}}4+\frac{1}{4}\times 0.01\] \[=1.3868+0.0025\] \[=1.3893\]


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