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Zorluk: Çok zorInequalities and Absolute Value Equations

For how many integer values of kk does the equation ∣x−2∣+∣x+4∣=kx+1|x - 2| + |x + 4| = kx + 1 have no real solutions for xx?

Cevap: 4

Cevap

4
The equation has no real solutions when the line y = kx + 1 lies completely below the graph of f(x) = |x - 2| + |x + 4|. Analyzing the three piecewise regions of f(x) shows that f(x) = 6 on [-4, 2], with linear rays extending upward outside this interval. Setting the line to remain below f(x) forces -1.25 < k <= 2. The four integers in this range are -1, 0, 1, and 2.

Adım Adım Çözüm

1
Decompose the sum of absolute values into a piecewise linear function
f(x) = -2x - 2 for x < -4; f(x) = 6 for -4 <= x < 2; f(x) = 2x + 2 for x >= 2
The critical points x = -4 and x = 2 split the real number line into three intervals where each absolute value expression maintains a constant sign.
2
Set up conditions for zero intersections with the line g(x) = kx + 1
g(x) must remain strictly below f(x) for all x
Any intersection point between y = f(x) and y = g(x) corresponds to a real solution of the equation.
3
Evaluate boundary points and slope constraints for each interval
From the middle interval and right ray: k <= 2. From the left ray: k > -1.25.
For x >= 2, the line slope k cannot exceed the ray slope of 2. For x <= -4, g(-4) < 6 requires -4k + 1 < 6, which yields k > -1.25.
4
Determine the allowable interval for k and count integer values
-1.25 < k <= 2, giving integer values k in {-1, 0, 1, 2}
The integer values strictly inside (-1.25, 2] are -1, 0, 1, and 2, making a total of 4 integer values.

Anahtar Kavram

Piecewise analysis of absolute value functions and linear line intersection conditions
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Konular

Inequalities and Absolute Value Equations

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