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# Engineering

Let \sum_{k=1}^{10}f\left ( a+k \right )=16\left ( 2^{10-1} \right )  , where the function f satisfies f\left ( x+y \right )=f\left ( x \right )f\left ( y \right ) for all natural numbers x,y and f\left ( 1 \right )=2 \; . Then the natural numbers 'a' is :

  • Option 1)

    2

  • Option 2)

    16

  • Option 3)

    4

  • Option 4)

     3

 

Answers (1)
S solutionqc

f\left ( 1 \right )=2

and f\left ( x+y \right )=f\left ( x \right )f\left ( y \right ) 

put x=1,y=1

f\left ( 2 \right )=f\left ( 1 \right )f\left ( 1 \right )=2\times 2=4=2^{2}

f\left ( 3 \right )=f\left ( 1 \right )f\left ( 2 \right )=2^{3}                                      \left ( x=1,y=2 \right )

f\left ( 4 \right )=f\left ( 1 \right )f\left ( 3 \right )=2^{4}                                            \left ( x=1,y=3 \right )

 

Similarly   f\left ( 10 \right )=2^{10}

Now,           \sum_{k=1}^{10}f\left ( a+k \right )=16\left ( 2^{10}-1 \right )

                  \Rightarrow \sum_{k=1}^{10}f\left ( a \right )f\left ( k \right )=16\left ( 2^{10}-1 \right )

               \Rightarrow f\left ( a \right )\sum_{k=1}^{10}f\left ( k \right )=16\left ( 2^{10}-1 \right )

                 \Rightarrow f\left ( a \right )\left [ f\left ( 1 \right )+f\left ( 2 \right )+.........+f\left ( 10 \right ) \right ]=16\left ( 2^{10}-1 \right )

                \Rightarrow f\left ( a \right )\left [ 2^{2}+2^{2}+....+2^{10} \right ]=16\left [ 2^{10}-1 \right ]

                  \Rightarrow f\left ( a \right )\left [ \frac{2\left ( 2^{10}-1 \right )}{2-1} \right ]=16\left [ 2^{10}-1 \right ]

                    f\left ( a \right )=8=2^{3}

                            a=3

 

 


Option 1)

2

Option 2)

16

Option 3)

4

Option 4)

 3

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