1g ust be a const function
g(x) = 5
Let g be a continuous function and attains only rational values. If g(0) = 5, then the roots of the equation
(g(2009))x2 + (g(2008))x + (g(2010)) = 0 are
1. Real and equal
2. Real and unequal
3. Rational
4. Imaginary
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4 Answers
106g is a constant function ..
as g attains only rational values and in the vicinity of any one rational value there are irrational values (and rational values but not continuous)
So g must be constant
Hence g(2009) = g(2008) = g(2010) = g(0) = 5
So the equation is x2+x+1 = 0
This has two imaginary roots