JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+x)}{{{3}^{x}}-1}\]equals to

    A)  \[{{\log }_{e}}3\]                            

    B)  0

    C)  1                                            

    D)  \[{{\log }_{3}}e\]

    Correct Answer: D

    Solution :

                    We have the differential equation, \[{{e}^{\frac{dy}{dx}}}=x+1\] Taking log on both sides, we get \[\frac{dy}{dx}=\log (x+1)\] \[\Rightarrow \]               \[dy=\log (x+1)dx\] On integrating both sides, we get \[\int{dy=\int{\log (x+1)}}dx\] \[\Rightarrow \]\[y=\log (x+1)x-\int{\frac{1}{x+1}}.x\,dx+c\] \[\Rightarrow \]\[y=x\log (x+1)x-\int{\left( 1-\frac{1}{x+1} \right)}dx+c\] \[\Rightarrow \]\[y=x\log (x+1)-x+\log (x+1)+c\]...(i) But given that\[y(0)=5,\]so from Eq. (i), \[5=0-0+0+c\] \[\Rightarrow \]                               \[c=5\] Putting this value of c in Eq. (i), we get\[y=x\text{ }log(x+1)-x+log(x+1)+5\]which is the required solution.


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