(1)
It is given that k is a constant. If 2x - (2k+1) = 5x + 3k -2 is an equation in x and its solution is positive, find the range of values of k.
(2)
If 3kx - (x^2) - 1 < 0 for any real number x and k is a constant, find the range of values of k.
F.4 math---inequalities1
2009-12-26 8:55 pm
回答 (2)
2009-12-26 9:09 pm
✔ 最佳答案
1)2x - (2k+1) = 5x + 3k -22 - 3k - 2k - 1 = 3x
1 - 5k = 3x
x = (1 - 5k)/3
So x = (1 - 5k)/3 > 0
1 - 5k > 0
5k < 1
k < 1/5
2)3kx - (x^2) - 1 < 0
x^2 - 3kx + 1 > 0
The equation x^2 - 3kx + 1 = 0 has no real roots,
△ = (-3k)^2 - 4 < 0
(3k - 2)(3k + 2) < 0
-2/3 < k < 2/3
2009-12-26 9:24 pm
(1) 3x+5k-1=0
x>0
5k=-3x+1
So,k<1/5
(2) It is a quadratic equation.
(x^2)-3kx+1>0
(x^2)-3kx+1=y can plot a parabolic curve
y>0 means the curve is well above x-axis.
So, (x^2)-3kx+1=0 has no solution.
That means b^2-4ac<0
x>0
5k=-3x+1
So,k<1/5
(2) It is a quadratic equation.
(x^2)-3kx+1>0
(x^2)-3kx+1=y can plot a parabolic curve
y>0 means the curve is well above x-axis.
So, (x^2)-3kx+1=0 has no solution.
That means b^2-4ac<0
收錄日期: 2021-04-13 17:00:08
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