Inequality in Inequality
Let x be a real number such that − x 2 + 3 x − 5 ( 2 k + 1 ) x 2 − ( 2 k − 1 ) x + ( 2 k + 1 ) > 0 is true for constant k . What is the supremum of k ?
The answer is -1.5.
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1 solution
The inequality b 2 − 4 a c < 0 actually becomes − ( 6 k + 1 ) ( 2 k + 3 ) < 0 . The solution to this inequality is k < − 3 / 2 and k > − 1 / 6 .
Therefore, the set of all k that works is k < − 3 / 2 .
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I obtained the same range for k as well. The problem needs to be adjusted to k = (-infinity, -3/2) for the correct answer. This allows both a negative leading coefficient (2k+1) AND a negative discriminant for the quadratic numerator.
many thanks for correcting
@Lingga Musroji ,you should change your answer according to question.
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i have changed the question to fit the answer
Consider the denominator: D = b 2 − 4 a c = ( 3 ) 2 − 4 ( − 1 ) ( − 5 ) = − 1 1 < 0 and the coeff. of x 2 is equal to − 1 < 0 . So the denominator is definite negative.
Since the whole expression must be positive, then the numerator must be definite negative too.
b 2 − 4 a c < 0 ( 6 k + 1 ) ( 2 k + 3 ) < 0 ( 6 k + 1 ) ( 2 k + 3 ) < 0 − 2 3 < k < − 6 1
2 k + 1 < 0 k < − 2 1
− 2 3 < k < − 2 1
a + b + 1 . 2 5 = − 2 3 − 2 1 + 1 . 2 5 = − 0 . 7 5