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 => a^r < b^r if r > 0 and a^r > b^r 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 downloads:

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-α_n))/((x-β₁)(x-β₂)(x-β₃)....(x-β_m)))

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

                        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 log_b a is meaningful for a > 0 and for either 0 < b < 1 or b > 1.

 *    a = b^{log_b a}

 *    log_a b = log_c b/log_c a

 *    log_b a = 1/log_a b provided both a and b are non-unity

 *    log_b a₁ > log_b a₂    

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,

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] 

 * {x} + {-x} 

 * [x+y]   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