2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}\]for\[x\ne 0\]and \[f(0)=k\]and\[f(x)\] is continuous at\[x=0\]then k is equal to:

    A)  \[a+b\]    

    B)  \[a-b\]

    C)         \[a\]                    

    D)         \[b\]

    E)  \[b-a\]

    Correct Answer: A

    Solution :

          \[f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}\] \[f(x)\]is continuous at\[x=k\]and\[f(0)=k\]. \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+ax)-\log (1-bx)}{x}\]                                                 \[\left( \frac{0}{0}form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{1+ax}.a+\frac{b}{1-bx} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{a}{1+ax}+\frac{b}{1-bx} \right)=a+b\] \[\therefore \]  \[a+b=f(0)=k\]                 


You need to login to perform this action.
You will be redirected in 3 sec spinner