Branch Cutting

Algebra Level 5

Let $f(x)=x(1+a|x|)$ for $a \in \mathbb R$ and $A$ be the solution set of inequality $f(x+a)<f(x)$ . If $\left[-\frac{1}{2},\frac{1}{2}\right]⊆A$ , what is the range of $a$ ?

\left(-1,\frac{1-\sqrt{5}}{2}\right)

(-1,0)

\left(\frac{1-\sqrt{5}}{2},0\right)

\left(0,\frac{1+\sqrt{5}}{2}\right)

1 solution

Pierre Carrette
Aug 16, 2019

First let us define c = -a for clarity (as c will be the "center" of derived curves). Then f(x) = x(1-c|x|). The inequality becomes x(1-c|x|) > (x-c)(1-c|x-c|). This simplifies into c(x|x|) < c(1 + (x-c)|x-c|). Both sides are x^3-shape curves except that the exponent is 2 (instead of 3) (see x|x| in black in the plot). The inflection point is (0,0) for left hand side and (c,c) for right hand side.

Let us consider positive/negative values of c separately. After dividing by c (thus, inflection point (c,1) for right hand side), we get

1) For c<0 : x|x| > 1 + (x-c)|x-c|.This inequality is never satisfied because the right hand side curve is always higher than the left side one (as seen in cyan in the plot).

2) For c>0 : x|x| < 1 + (x-c)|x-c|. For c>sqrt(2) (purple for c=2), the inequality cannot be satisfied for similar reason as for the case of c<0. At c=sqrt(2) (red), the only solution is x=sqrt(2)/2. For smaller c, two solutions for x (x1 and x2 with x1<x2) are found. As c decreases toward zero, x1 (resp. x2) monotonically de(resp. in)creases. This means that x2>sqrt(2)/2>1/2 for all 0<c<sqrt(2). Question: For what value of c would x1<=-1/2? For c=1, x1=0 (blue). For c<1, x1 is the intersection of the negative curvature part of the left and right hand side curves, i.e. x1=-(1-c^2)/c. Imposing x1=-1/2, we get c=-(1-sqrt(5))/2 (green).

All and all, we get, x in [-1/2, 1/2] for 0<c<-(1-sqrt(5))/2. Thus, (1-sqrt(5))/2 < a < 0.

Branch Cutting

1 solution

0 pending reports