1.3 Easy Ways Through Absolute Values
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Solution Observe that for k = 1, 2, . . . , 50, we have |x − k| + x − (101 − k) ≥ 101 − 2k, with equality if x ∈ [k, 101 − k]. Adding these inequalities yields E(x) ≥ 2500, and the equality holds for all x such that [k, 101 − k] = [50, 51]. x∈ 1≤k≤50
Problem 1.38 Find the values of a for which the equation (x − 1)2 = |x − a| has exactly three solutions. Solution Observe that (x − 1)2 = |x − a| if and only if x − a = ±(x − 1)2 , that is, if and only if a = x ± (x − 1)2 . The number of solutions of the equation is equal to the number of intersection points between the line y = a and the graphs of the functions f (x) = x + (x − 1)2 = x 2 − x + 1 and g(x) = x − (x − 1)2 = −x 2 + 3x − 1. The graph of f is a parabola with vertex B( 12 , 34 ) and the graph of g a parabola with vertex C( 32 , 54 ). Now, since the equation f (x) = g(x) is a quadratic with one real root, it follows that the graphs are tangent to each other at point A(1, 1). We deduce that the line y = a intersects the two graphs at three points if and only is it passes through one of the points A, B, C; that is, if a ∈ { 34 , 1, 54 }. Try to use some of the ideas above in solving the following problems: Problem 1.39 Solve the equation |x − 3| + |x + 1| = 4. Problem 1.40 Show that the equation |2x − 3| + |x + 1| + |5 − x| = 0.99 has no solutions. Problem 1.41 Let a, b > 0. Find the values of m for which the equation |x − a| + |x − b| + |x + a| + |x + b| = m(a + b) has at least one real solution.