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 ?

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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