VI. MARCO TEÓRICO
6.4 Sentido de pertenencia e identidad
In this chapter the quadratic inequalities and fractional inequalities of single vari- able are discussed.
Any quadratic inequality can be arranged in one of the following forms (i)f(x)>0; (ii)f(x)≥0; (iii)f(x)<0; (iv)f(x)≤0, wheref(x) =ax2+bx+cwitha6= 0. Below for the convenience of discussion, we assumea >0, i.e. the curve of the quadratic functiony =ax2+bx+cis a parabola opening upwards.
Basic Methods for Solving Quadratic Inequalities
(I) When the equationf(x) = 0has two real rootsx1≤x2, the solution set of the inequalities (i) to (iv) are
(i)x < x1orx > x2; (ii)x≤x1orx≥x2; (iii)x1< x < x2; (iv)x1≤x≤x2; respectively.
Geometrically,x1, x2are thex-coordinates of the points of intersection of the curvey = f(x)with thex-axis, and the solution set is the range of thex-coordinates of the points on the curve with positive y-coordinates (for (i)), or with non-negative y-coordinates (for (ii)), with negative y- coordinates (for (iii)), or with non-positivey-coordinates (for (iv)).
84 Lecture 28 Quadratic Inequalities and Fractional Inequalities
(II) When the equationf(x) = 0has no real solution, thenf(x)>0for any realx, so the solution set of inequalities (i) and (ii) are both the whole real axis, and no solution for (iii) and (iv).
Geometrically, the equationf(x) = 0has no real solution means the whole curve ofy=f(x)is above thex-axis, so it does not intersect with thex- axis, hence any point on the curve has a positivey-coordinate.
(III) For fractional inequalities of the forms
(i)g(x)>0; (ii)g(x)≥0; (iii)g(x)<0; (iv)g(x)≤0, where g(x) = x−a
x−c witha 6= c, the problems will become those dis- cussed in (I) when multiplying both sides of the inequality by(x−c)2. (IV) If there are more than one linear factors in the denominator or numerator of
the fractional expressiong(x), then it is needed to discuss the sign ofg(x)
by partitioning the range of xinto several intervals, where the partition points are given by letting each linear factor be zero.
Notice that, if the factor(x−a)2n+1withn≥1occurs in the numerator of denominator ofg(x), it can be replaced by(x−c)without changing the solution set, and if(x−a)2n withn ≥1occurs in the the numerator of g(x), it can be removed at first, and then determine ifais in the resultant solution set. If(x−a)2noccurs in the denominator ofg(x), then remove it first, and remove the pointafrom the resultant solution set if any. Thus, the construction ofg(x)to be considered can be simplified much. (V) In mathematical olympiad competitions, inequalities containing parame-
ters often occur. One kind of the common problems is to determine the values or ranges of the parameters based on information contained in the given inequality and other given conditions.
Examples
Example 1. Solve the inequality(x−2)4(x−5)5(x+ 3)3<0.
Solution Since(x−2)4(x−5)5(x+ 3)3 <0 ⇔(x−5)(x+ 3)<0, so the solution set is
{−3< x <5} − {2}, or equivalently, {−2< x <2} ∪ {2< x <5}.
Example 2. Solve the inequality(x2−x−1)2≥(x2+x−3)2.
Solution By factorization the given inequality can be simplified.
(x2−x−1)2≥(x2+x−3)2⇔(x2−x−1)2−(x2+x−3)2≥0
Lecture Notes on Mathematical Olympiad 85
Thus, the solution set is{x≤1}. Example 3. Solve the inequality x
2−x−2 x2−3x+ 1 ≥0.
Solution The inequality is equivalent to the systems
x2−x−2≥0, x2−3x+ 1>0 or x2−x−2≤0, x2−3x+ 1<0. x2−x−2≥0, x2−3x+ 1>0 ⇔(x−2)(x+ 1)≥0,(x−3− √ 5 2 )(x− 3 +√5 2 )>0 ⇔ {{x≤ −1} ∪ {x≥2}} ∩ ( {x < 3− √ 5 2 } ∪ {x > 3 +√5 2 } ) ⇔ {x≤ −1} ∪ {x > 3 + √ 5 2 }. x2−x−2≤0, x2−3x+ 1<0 ⇔(x−2)(x+ 1)≤0,(x−3− √ 5 2 )(x− 3 +√5 2 )<0 ⇔ {{−1≤x≤2}} ∩ ( {3− √ 5 2 < x < 3 +√5 2 } ) ⇔ {3− √ 5 2 < x≤2}.
Thus, the solution set is{x≤ −1} ∪ {3− √
5
2 < x≤2} ∪ {x >
3 +√5 2 }. Example 4. Solve the inequality x−2
x+ 3 >−1. Solution x−2 x+ 3 > −1 ⇔ 2x+ 1 x+ 3 > 0 ⇔ (2x+ 1)(x+ 3) > 0, so the solution set is {x <−3} ∪ {x >−1 2}. Example 5. Solve the inequality (x+ 1)(x−2)
(x−4) >0, wherex6= 4. Solution List the following table
Range ofx x <−1 −1< x <2 2< x <4 4< x Sign of (x+ 1)(x−2)
86 Lecture 28 Quadratic Inequalities and Fractional Inequalities
Therefore the solution set isS= (−1,2)∪(4,+∞). Example 6. Find the solution set of the inequality x+ 1
x−1 > 6 x. (x 6= 1 and x6= 0.) Solution Changex+ 1 x−1 > 6 xto the form x+ 1 x−1− 6 x >0, then x+ 1 x−1 − 6 x>0 ⇔ x(x+ 1)−6(x−1) x(x−1) >0⇔ x2−5x+ 6 x(x−1) >0, ⇔ (x−2)(x−3) x(x−1) >0.
Based on the following table
Range ofx (−∞,0) (0,1) (1,2) (2,3) (3,+∞)
Sign of(x−2)(x−3)
x(x−1) + − + − +
the solution set is
(−∞,0)∪(1,2)∪(3,+∞).
Example 7. Solve the quadratic inequalityax2−(a+ 1)x+ 1<0, whereais a parameter.
Solution Sincea6= 0,ax2−(a+ 1)x+ 1<0⇔a(x−1
a)(x−1)<0. (i) Ifa >1, the solution set is1
a < x <1. (ii) Ifa= 1, no solution.
(iii) If0< a <1, the solution set is1< x < 1 a. (iv) Ifa <0, the inequality becomes(x−1
a)(x−1)>0, the solution set is
{x < 1
a} ∪ {x >1}.
Example 8. Given that the inequalitykx2−kx−1<0holds for any realx, then (A)−4< k≤0, (B)−4≤k≤0, (C)−4< k <0, (D)−4≤k <0. Solution It is obvious that the inequality holds whenk= 0.
Whenk <0, the curve ofy=kx2−kx−1is open downwards, and is below thexaxis, so the equationkx2−kx−1 = 0has no real roots, i.e. its discriminant is negative.
Lecture Notes on Mathematical Olympiad 87
Thus,−4< k≤0, the answer is (A).
Example 9.Given that the solution set of the quadratic inequalityax2+bx+c >0 is1< x <2. Find the solution set of the inequalitycx2+bx+a <0.
Solution The first inequality has the solution set1 < x < 2 implies that a <0, and x2+b ax+ c a<0⇔(x−1)(x−2)<0⇔x 2−3x+ 2<0. Therefore b a =−3, c
a = 2orb =−3a, c = 2aanda <0. Then the second inequality becomes
a(2x2−3x+ 1)<0, i.e.2x2−3x+ 1>0. Thus,(2x−1)(x−1)>0, and its solution set is{x < 1
2} ∪ {x >1}.
Testing Questions
(A)
1. Solve the inequality(2 +x)(x−5)(x+ 1)>0.
2. Solve the inequalityx2(x2−4)<0.
3. Solve the inequalityx3≤6x−x2.
4. Solve the inequalityx−1>(x−1)(x+ 2).
5. Solve the inequality(x3−1)(x3+ 1)>0.
6. Solve the inequality 2x−4 x+ 3 >
x+ 2
2x+ 6. 7. Find the solution set of the inequality x
x+ 2 ≥
1
x.
8. Find the solution set of the inequality x−1
x2 ≤0.
9. Find the solution set of the inequality x(2x−1) 2
(x+ 1)3(x−2) >0.
10. Find the solution set of the inequality 2x2
88 Lecture 28 Quadratic Inequalities and Fractional Inequalities
Testing Questions (B)
1. (CHINA/2002) Find the positive solutions of the inequality x2+ 3 x2+ 1 + x2−5 x2−3 ≥ x2+ 5 x2+ 3 + x2−3 x2−1.
2. (CHINA/2001) Solve the fractional inequality
x+ 2 4x+ 3 − x 4x+ 1 > x 4x−1 − x−2 4x−3. 3. Given thata, bare positive constant witha < b. If the inequality
a3+b3−x3≤(a+b−x)3+m
holds for any realx, find the minimum value of the parameterm.
4. Given that the quadratic functionf(x) =x2−2ax+ 6≥afor−2≤x≤2, find the range of the constanta.
5. Given that the inequality
1 8(2a−a
2)≤x2−3x+ 2≤3−a2