Find the number of real roots of f ( x ) = 9 9 if
f ( x ) = k = 1 ∑ 2 0 1 8 x − ( 3 k 3 + 3 k 2 + 5 k + 1 1 ) k 2 + k + 1
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Right on the spot!
Yeah same solution. Try out this one too https://brilliant.org/problems/how-many-real-roots/.
Another way:
Let's suppose that x = a + i b is a root, then we can rationalise and by equating the imaginary part on both sides we can get b = 0 .
I was trying something big but the solution is very much basic and defines beauty!
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We have f ( x ) = x − 2 2 3 + x − 5 7 7 + ⋯ + x − b 2 0 1 8 a 2 0 1 8 where a 2 0 1 8 = 2 0 1 8 2 + 2 0 1 8 + 1 and b 2 0 1 8 = 3 ( 2 0 1 8 ) 3 + 3 ( 2 0 1 8 ) 2 + 5 ( 2 0 1 8 ) + 1 1 . Note that 3 k 3 + 3 k 2 + 5 k + 1 1 is an increasing function for k > 0 , so each of the integers in the denominator is increasing.
Note also that f ( x ) is strictly decreasing on its domain (take a derivative; each of the terms in the sum is negative).
So there are 2 0 1 8 vertical asymptotes to y = f ( x ) . For x < 2 2 , f ( x ) is negative. For 2 2 < x < 5 7 , f ( x ) decreases from + ∞ to − ∞ . So it crosses the line y = 9 9 exactly once. Ditto for 5 7 < x < 1 3 4 . And so on. Finally, for x > b 2 0 1 8 , y = f ( x ) decreases from + ∞ to approaching 0 as x → ∞ , so it crosses the line y = 9 9 exactly once there as well.
So there are 2 0 1 7 roots, one between each consecutive pair of vertical asymptotes; and one more root to the right of the righmost vertical asymptote. This is a total of 2 0 1 8 .