RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    \[{{(3\sqrt{3}+5)}^{7}}=P+F,\]where P is an integer number and F is a proper fraction, then the value of\[F.(P+F)\]is

    A)  \[{{2}^{6}}\]              

    B)  \[{{2}^{7}}\]

    C)  \[{{3}^{6}}\]              

    D)  \[{{3}^{7}}\]

    Correct Answer: B

    Solution :

     Given, \[{{(3\sqrt{3}+5)}^{7}}=P+F\]                ... (i) Let\[{{(3\sqrt{3}-5)}^{7}}=S\] \[\Rightarrow \] \[0<3\sqrt{3}-5\] \[\Rightarrow \]         \[0<\frac{(3\sqrt{3}-5)(3\sqrt{3}+5)}{3\sqrt{3}+5}\] \[\Rightarrow \] \[0<\frac{2}{3\sqrt{3}+5}<1\] \[\Rightarrow \] \[0<S<1\]                   ...(ii) \[\therefore \] \[(P+F)-S{{(3\sqrt{3}+5)}^{7}}-{{(3\sqrt{3}-5)}^{7}}\] \[=2{{[}^{7}}{{C}_{1}}{{(3\sqrt{3})}^{6}}{{.5}^{7}}{{C}_{3}}{{(3\sqrt{3})}^{4}}{{5}^{3}}+....]\] Even integer. Since, P is an integer. So,\[F-S\]is also an integer. \[\therefore \] \[-1<F-S<1\] \[\Rightarrow \] \[F-S=0\] \[\Rightarrow \] \[F=S\] \[\therefore \] \[F.(P+F)=S.(P+F)\] \[={{(3\sqrt{3}-5)}^{7}}{{(3\sqrt{3}+5)}^{7}}\] \[={{[{{(3\sqrt{3})}^{2}}-{{(5)}^{2}}]}^{7}}\] \[={{[27-25]}^{7}}\] \[={{2}^{7}}\]


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