JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Let \[\alpha \] and \[\beta \] be the roots of the equation \[{{x}^{2}}-x-1=0\].  If \[{{p}_{k}}={{(\alpha )}^{k}}+{{(\beta )}^{k}}\], \[k\ge 1,\]then which one of the following statements is not true? [JEE MAIN Held on 07-01-2020 Evening]

    A) \[{{p}_{3}}={{p}_{5}}{{p}_{4}}\]

    B) \[({{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}+{{p}_{5}})=26\]

    C) \[{{p}_{5}}=11\]

    D) \[{{p}_{5}}={{p}_{2}}\cdot {{p}_{3}}\]

    Correct Answer: D

    Solution :

    [d] \[\because \,\,\,\,\,\alpha ,\beta \]are roots of \[{{x}^{2}}x1=0\] ...(i)
    \[\therefore \,\,\,\,{{\alpha }^{2}}-\alpha -1=0\]
    \[\Rightarrow \,\,\,\,{{\alpha }^{n+2}}-{{\alpha }^{n+1}}-{{\alpha }^{n}}=0\]                                ...(ii)
    Similarly, \[{{\beta }^{n+2}}-{{\beta }^{n+1}}-{{\beta }^{n}}=0\]              ...(iii)
    From eq. (ii) + (iii), we get
    \[{{\alpha }^{n+2}}+{{\beta }^{n+2}}=({{\alpha }^{n+1}}+{{\beta }^{n+1}})+({{\alpha }^{n}}+{{\beta }^{n}})\]
    \[\therefore \,\,\,\,{{p}_{n+2}}={{p}_{n+1}}+{{p}_{n}}\]
    For \[n=0,\,\,{{p}_{0}}={{\alpha }^{0}}+{{\beta }^{0}}=2\]
    For \[n=1,\text{ }{{p}_{1}}=\alpha +\beta =1\]
    and      \[{{p}_{2}}={{p}_{0}}+{{p}_{1}}=2+1=3\]

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