Problem #3: Opposite extremes
Consider the following statement, where is a constant.
The function has two local extrema, and their product is negative.
Let be the minimum of positive so that the above statement is false.
Let be the maximum of negative so that the above statement is false.
Let be the minimum of integer so that the above statement is true.
The sum of all the ones that exist among is for rational numbers Find
This problem is a part of the series <Advanced Problems> .
The answer is 490.0.
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1 solution
The given statement is equivalent to the following statement:
Since x = 0 is not the root of the original equation, we may ignore it.
Let us solve the equation for k , obtaining k = − 2 + x 3 1 2 x 2 − 6 x − 2 .
Substitute x 1 = t , so that k = − 2 t 3 − 6 t 2 + 1 2 t − 2 .
Let f ( t ) = − 2 t 3 − 6 t 2 + 1 2 t − 2 . We know that y = f ( t ) ( t = 0 ) and y = k must meet at three distinct points if and only if the statement is true.
Notice, that f ′ ( t ) = − 6 t 2 − 1 2 t + 1 2 = − 6 ( t 2 + 2 t − 2 ) so that the extrema occur at t = − 1 ± 3 .
Since f ( t ) = − 2 t 3 − 6 t 2 + 1 2 t − 2 = ( t 2 + 2 t − 2 ) ( − 2 t − 2 ) + 1 2 t − 6 , we have that the two extrema are − 1 8 ± 4 3 2 .
Hence − 1 8 − 4 3 2 < k < − 1 8 + 4 3 2 , but since t = 0 , we need k = − 2 , for it to be equivalent with the statement.
Therefore, a = − 1 8 + 4 3 2 , b = − 2 , c = − 3 8 , which yields a + b + c = − 5 8 + 4 3 2 , so that p − q = 4 9 0 .
Fyi, using the cubic discriminant turns the entire process into a quadratic inequality solving.