Method of Intervals

Consider the sign of the expression
f(x) = (x -a₁)(x -a₂) ... (x -a_n) where a₁ < a₂ < ... < a_n.
           
On the real number line, mark points a₁, a₂, ..., a _n. Start with a positive sign for x > a_n; -ve for a_{n -1} < x < a_n; and then alternating +ve and -ve signs. The logic is that when x> a_n; all factors x -a₁, x -a₂, ..., x -a _n are +ve and hence f(x) is +ve; when a_n-1 < x < a_n, only one factor x -a_n is -ve, and so f(x) is -ve, and so on.

Quadratic fractions and their range

Expressions of the kind (a x² +b x +c)/(p x² +q x +r), where x is any real number, are called quadratic fractions.
Their range can easily be found by putting
       (a x² +b x +c)/(p x² +q x +r) = y      
=> a x² +b x +c = p y x² +q x y +r y
=> x² (a -p y) +x (b -q y) +(c -r y) = 0
Since the value of x is real, discriminant of this equation in x must be non-negative
=> (b -q y)² -4 (a -p y)(c -r y) = 0.
This inequality yields range of y.

Illustrative Examples

Example

Solve: x (x -1)(x -2)(x -3) > 0.
 

Solution

Mark points 0, 1, 2, 3 on real line. By method of intervals, we see that given expression is +ve when x > 3 or 1 < x < 2 or x < 0.
Thus required solutions set is {x < 0; 1 < x < 2; x > 3}.

Example

Solve 1/x < 1.

Solution

Do not fall into this trap: (1/x) < 1    =>   x > 1,
because we cant multiply by negative quantity without  changing the sign of inequality.
Correct solution is:
Since x² > 0 for all real x, x 0.
  1/x < 1    =>  (1/x) x² < x²   =>  x < x²   =>  x² -x > 0
=>   x (x -1) > 0  =>  x < 0 or x > 1, which is the required solution.

Example

For what values of a is the inequality (x² + a x -2)/(x² -x +1) < 2 satisfied for all real values of x?

Solution

(x² +a x -2)/(x² -x + 1) < 2
=> (x² +a x -2)/(x² -x +1) -2 < 0
=>  (x² +a x -2 -2 x² +2 x -2)/(x² -x + 1) < 0
=>    [-x² +x (a +2) -4]/(x² -x +1) < 0
=> -x² -x (a +2) -4 > 0   (as for all real x)
=>  x² -x (a +2) +4 > 0
Now the expression x² -x (a +2) +4 is positive for all real x if
   < 0 (as coeff of x² = 1 > 0)
=> [-(a +2)]² -4. 1. 4 < 0   => a² +4 a +4 -16 < 0
=>  a² +4 a -12 < 0   =>  (a +6)(a -2) < 0
=> -6 < a < 2, which is the required range for a.

Example

Show that for all real values of x, the expression (x² -2x +4)/(x² +2x +4) has greatest value 3 and least value 1/3.

Solution

Let (x² -2 x +4)/(x² +2 x +4) = y
=>  x² -2 x +4 = x² y +2 x y +4 y
=>  x² (1 -y) -2 x (1 +y) +4 (1 -y) = 0.
Since x is real, discriminant of this quadratic equation in x must be non-negative
=>  = b² -4 a c = 0
=>   [-2 (1 +y)]² -4. (1 -y). 4 (1 -y) 0
=>   4 [1 +y² +2 y -4 (1 +y² -2 y)] 0
=> -3 y² +10 y -3 = 0    =>  3 y² -10 y +3 = 0
=> (3 y -1)(y -3) ; 0    =>  1  y 3
=> 1/3 (x² -2 x +4)/(x² +2 x +4) 3 for all real x
Thus the greatest value of given quadratic fraction is 3 and least value is 1/3.

Exercise

1. Solve: (i) x³ -3 x² -x +3 < 0      (ii) x⁴ -5 x² +4 0
2. Solve :
3. Find all real values of x which satisfy
   (i) x³ (x -1)(x -2)² > 0
   (ii) x² (x -1)(x -2) 0
4. Solve : (x² +6 x -11)/(x + 3) < -1
5. Solve : (x² -3 x +240/(x² -3 x +3) < 4
6. Find the range of values of x for which (x² +x + 1)/(x +1) < 1/3, x being real.
7. Find the set of all x for which 2 x/(2 x² +5 x +2) > 1/(x+1)
8. Let y = Find all the real values of x for which y takes real values.
9. Find all integral values of x for which (5 x -1) < (x + 1)² < 7 x -3
10. Solve (i)      (ii)
11. If x is real, show that (x² -x +1)/(x² +x +1) takes values from 1/3 to 3.
12. If x be real, show that the value of (x -1)(x +3)]/[(x -2)(x +4)] cannot lie between 4/9 and 1.
13. If for real values of x, the greatest value of (x² -x +a)/(x² +x +a) is 3, then prove that value of a cannot be less than 1.
14. Show the expression (2x² +4x +1)/(x² +4x +2) can have any real value if x is real.
15. Show that if x is real, the expression (x² -bc)/(2x -b -c) has no real values between b and c.
16. Find the minimum and maximum values of (x² -1)/(x² +1). Is maximum value obtained?
17. Let f(x) =( x² +6 x -8)/( +6 x -8x²). Find the interval of values of a for which f(x) takes all real values for real values of x.

Answers

1. (i) x < -1 or 1 < x < 3         (ii) x -2 or -1 x 1 or x 2
2. x < -1 or x > 1
3. (i) x < 0 or x > 1, x 2        (ii) x = 0 or 1  x   2
4. x < -8 or -3 < x < 1           5. x < -1 or x > 4
6. -1 < x < -1/2                      7. -2 < x < -1 or -2/3 < x < 1/2
8. -1 x < 2 or x > 3            9. x = 3
10. (i) All real numbers            (ii) x -2 or x -1
16. -1, 1; No                         17. 2    14