CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Let\[[x]\]denote the greatest integer less than or equal to\[x\]. If\[x={{(\sqrt{3}+1)}^{5}},\]then M is equal to

    A)  75                                         

    B)  50

    C)  76                         

    D)         51

    E)  152

    Correct Answer: E

    Solution :

    Now,\[{{(\sqrt{3}+1)}^{5}}={{(\sqrt{3})}^{2}}{{+}^{5}}{{C}_{1}}{{(\sqrt{3})}^{4}}{{+}^{5}}{{C}_{2}}{{(\sqrt{3})}^{3}}\] \[{{+}^{5}}{{C}_{3}}{{(\sqrt{3})}^{2}}{{+}^{5}}{{C}_{4}}(\sqrt{3}){{+}^{5}}{{C}_{5}}\] \[=9\sqrt{3}+45+30\sqrt{3}+30+5\sqrt{3}+1\] \[=76+44\sqrt{3}\] \[\therefore \] \[[{{(\sqrt{3}+1)}^{2}}]=[76+44\sqrt{3}]\]                 \[=[76]+[44\times 1.732]\]                 \[=76+[76.2]\]                 \[=76+76=152\]


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