CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    \[{{\log }_{e}}\frac{1+3x}{1-2x}\]is equal to

    A)  \[-5x-\frac{5{{x}^{2}}}{2}-\frac{35{{x}^{3}}}{3}-...\]

    B)  \[-5x+\frac{5{{x}^{2}}}{2}-\frac{35{{x}^{3}}}{3}+...\]

    C)  \[5x-\frac{5{{x}^{2}}}{2}\,+\frac{35{{x}^{3}}}{3}+...\]

    D)  \[5x+\frac{5{{x}^{2}}}{2}+\frac{35{{x}^{3}}}{3}-...\]

    E)  \[x+\frac{3{{x}^{2}}}{2}+\frac{5{{x}^{3}}}{4}+...\]

    Correct Answer: C

    Solution :

    \[{{\log }_{e}}\left( \frac{1+3x}{1-2x} \right)={{\log }_{e}}(1+3x)-{{\log }_{e}}(1-2x)\] \[=\left[ 3x-\frac{(3{{x}^{2}})}{2}+\frac{{{(3x)}^{3}}}{3}-..... \right]\]                                 \[+\left[ 2x+\frac{{{(2x)}^{2}}}{2}+\frac{{{(2x)}^{3}}}{3}+.... \right]\] \[=5x-\frac{5{{x}^{2}}}{2}+\frac{35{{x}^{3}}}{3}-.....\]


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