INEQUALITIES

The following are some very useful points to remember:

  •     a < b => either a < b or a = b
  •     a < b and b < c => a < c
  •     a < b => a + c < b + c ∀ c ε R
  •     a < b => -a > -b i.e. inequality sign reverses if both sides are multiplied by a  negative number
  •     a < b and c < d => a + c < b + d and a - d < b - c
  •     a < b => ma < mb if m > 0 and ma > mb if m < 0
  •     0 < a < b => ar < br if r > 0 and ar > br if r < 0
  •     (a+(1/2)) > 2 ∀ a > 0 and equality holds for a = 1
  •     (a+(1/2)) < -2 ∀ a > 0 and equality holds for a = -1

SINOSODIAL CURVE METHOD

In order to solve inequalities of the form (P(x)/Q(x))  >  0, (P(x)/Q(x))  <  0, where P(x) and Q(x) are polynomials, we use the following method:

If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation, then within this interval the polynomial itself takes on values having the same sign. Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write

      (P(x)/Q(x)) = f(x) (((x-α1)(x-α2)(x-α3)....(x-αn))/((x-β1)(x-β2)(x-β3)....(x-βm)))

where α1, α2, ......, αn, β1, β2, ........, βm are distinct real numbers. Then f(x) = 0 for x = α1, α2, ......, αn and f(x) is not defined for x = β1, β2, ........, βm. Apart from these (m + n) real numbers f(x) is either positive or negative. Now arrange α1, α2, ......, αn, β1, β2, ........, βm in an increasing order say c1, c2, c3, c4, c5, ........, cm+n. Plot them on the real line. An draw a curve starting from right of cm+n along the real line which alternately changes its position at these points. This curve is known as the sinosodial curve.

                          sinosodial-curve

 

Illustration:      Let f(x) = ((x-3)(x+2)(x+5))/((x+1)(x-7)). Find the intervals where f(x) is positive or negative.

Solution:           The relevant sinosodial curve of the given function is

                                     relevant-sinosodial-curve

                        f(x) > 0 ∀ x ε (-5, -2)υ (-1, 3) υ (7, ∞) and

                        f(x) < 0 ∀ x ε (-∞, -5) υ (-2, -1) υ (3, ).

Exercise

        (i)     Let f(x) = (x2-3x+2)/(x2-1). Find the intervals where f(x) is negative.

        (ii)    If ((x-1)(x-2)2(x-3)3)/((x-4)2(x-5)6) > 0, then find the values of x.

 

 

LOGARITHM

 *    The expression logb a is meaningful for a > 0 and for either 0 < b < 1 or b > 1.

 *    a = blogb a

 *    loga b = logc b/logc a

 *    logb a = 1/loga b provided both a and b are non-unity

 *    logb a1 > logb a2              inequalities

ABSOLUTE VALUE

Let x ε R. Then the magnitude of x is called it's absolute value and is, in general, denoted by |x|. Thus |x| can be defined as,

                                           absolute-value

Note that x = 0 can be included either with positive values of x or with negative values of x. As we know all real numbers can be plotted on the real number line, |x| in fact represents the distance of number 'x' from the origin, measured along the number-line. Thus, |x| > 0. Secondary, any point 'x' lying on the real number line will have its coordinate as (x, 0). Thus its distance from the origin is √x2.

Hence |x| = √x2. Thus we can defined |x| as |x| = √x2 or |x| = max (x, -x)

e.g. if x = 2.5, then |x| = 2.5, if x = 3.8 then |x| = 3.8.

Basic Properties of |x|

 *  | |x| | = |x|

 *  |x| > a => x < a or x < -a if a ε R+ and x ε R if a ε R-

 *  |x| < a => -a < x < a if a ε R+ and no solution if a ε R- υ {0}

 *  |x + y| < |x| + |y|

 *  |x - y| > |x| ~ |y|

 *  The last two properties can be put in one compact form namely,

     |x| ~ |y| < |x + y| < |x| + |y|

 *  |xy| = |x|  |y|

 *  |x/y| = |x/y| y ≠ 0

Illustration:      Solve the inequality for real values of x: |x - 3| > 5.

Solution:           |x -3| > 5 => x - 3 < -5 or x - 3 > 5

                        => x < -2 or x > 8 => x ε (-∞, -2) υ (8, ∞).

 GREATEST INTEGER AND FRACTIONAL PART

Let x ε R then [x] denotes the greatest integer less than or equal to x and {x} denotes the fractional part of x and is given by {x} = x - [x]. Note that 0 < {x} < 1.

e.g. x = 2.69 => 2 < x < 3 => [x] = 2, x = -3.63 => -4 < x < -3 => [x] = -4

It is obvious that if x is integer, then [x] = x.

 Basic Properties of greatest Integer and Fractional Part

 * [[x]] = [x]x, [{x}] = 0, {[x]} = 0

 * x - 1 < [x] < x, 0 < {x} < 1

 * [n + x] = n + [x] where n ε I

   * [x] + [-x] eqn1  

 * {x} + {-x}  eqn2

 * [x+y] eqn3    Hence [x] + [y] < [x + y] < [x] + [y] + 1 

 * [[x]/n] = [x/n] , n ε N,  x ε R

 

Illustration:     

             If y = 3[x] + 1 = 2[x - 3] + 5, then find the value of [x + y].

Solution:      

          We are given that 3[x] + 1 = 2([x] -3) + 5

           => [x] = -2  =>  y = 3(-2) + 1 = -5.

           Hence [x + y] = [x] + y = -2 - 5 = -7.

Illustration:     

         Solve the equation |2x-1| = 3[x] + 2{x} for x.

Solution:            

Case I: For x < 1/2, |2x -1| = 1 - 2x  =>  1 - 2x = 3[x] + 2{x}

           => 1 -2x = 3(x - {x}) + 2{x}  =>  {x} = 5x - 1.

           Now 0 < {x} < 1 => = 0 < 5x - 1 < 1

           =>  1/5  <  x  < 2/5 => [x] = 0  =>  x = {x}

           =>  x = 5x - 1  =>  x = ¼, which is a solution.

           Case II: For > ½, |2x - 1|= 2x - 1

           => 2x - 1 = 3[x] + 2{x}  =>  2x - 1 = 3(x - {x}) + 2{x}.

           {x} = x + 1

           Now 0 < {x} < 1  =>  0 < x + 1 < 1  =>  -1 < x < 0

           Which is not possible since x > 1/2.

           Hence x = ¼ is the only solution

Even though inequalities do not fetch an independent question but the concept is invariably used in 2-3 questions every year. This can also help in calculations in Physics and Physical Chemistry, so this cannot be ignored. For more free online study material and live classroom programmes visit askIITians.com.

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