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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    If\[[x]\]denotes the greatest integer less than or equal to x, then the value of\[\int_{0}^{2}{(|x-2|+[x])}dx\]is equal to

    A)  2                                            

    B)  3

    C)  1                            

    D)         4

    E)  \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    \[\int_{0}^{1}{(|x-2|+[x])}dx\] \[=\int_{0}^{2}{(|x-2|dx+\int_{0}^{2}{[x]}}dx\] \[=-\int_{0}^{2}{(x-2)dx+\int_{0}^{1}{0}\,}dx+\int_{1}^{2}{1\,dx}\] \[=-\left[ \frac{{{x}^{2}}}{2}-2x \right]_{0}^{2}+0+[x]_{1}^{2}\] \[=-[2-4]+[2-1]\] \[=2+1=3\]         


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