JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer Let\[\sum\limits_{k=1}^{10}{f(a+k)=16}\left( {{2}^{10}}-1 \right),\] where the function ? satisfies \[\left( x+y \right)=\left( x \right)\left( y \right)\]for all natural numbers x, y and ?(1) = 2. then the natural number 'a' is [JEE Main 9-4-2019 Morning]

    A) 4                                 

    B) 3

    C) 16                               

    D) 2

    Correct Answer: B

    Solution :

    From the given functional equation : \[\left( x \right)={{2}^{x}}\,\,\,\,\,\,\forall x\in N\] \[{{2}^{a+1}}+{{2}^{a+2}}+....+{{2}^{a+10}}=16({{2}^{10}}-1)\] \[{{2}^{a}}(2+{{2}^{2}}+....+{{2}^{10}})=16({{2}^{10}}-1)\] \[{{2}^{a}}.\frac{2\left( {{2}^{10}}-1 \right)}{1}=16\left( {{2}^{10}}-1 \right)\] \[{{2}^{a+1}}=16={{2}^{4}}\] \[a=3\]


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