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Let  ' ƒ'  be continuous function defined on [-2009 , 2009 ]  such that ƒ(x) is irrational  for each   x belongs to [-2009  ,  2009]


and   ƒ(0)=2+√3+√5  .  the equation         ƒ(2009)x2  + 2ƒ(0)x + ƒ(2009) = 0 has  ?


 


 




  • 1 )  only rational roots 

  • 2 )  only rational roots

  • 3 )  one rational and one irrational root

  • 4 )  imaginary roots     


  •           which is the appropriate answer  explain ?

    6 years ago

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    Answers : (1)

    										

    Dear Chitikala ,


    Well , the catch is that the function is irrational and continuous on the domain . so you think how can a function be both irrational and continuous. if f(x) will have any other value than f(0) , it will leave all the rational points in between so won't be continuous, which implies it won't take any other value than f(0). so f(x) is a constant function over [-2009 , 2009] having the value f(0).


     i.e f(-2009)=f(-2008)........f(0)= f(1)=f(2)=f(3)= ..............f(2009) , so


    equation becomes   x2  + 2x + 1 = 0   . so , roots are 1 and 1 . therefore option 1) is correct.



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    6 years ago

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