Crash Course, Mathematics : Inequalities

Writing domain of a variable x in Inequalities

For x a real number, domain is a set of values that x can take.

Example: If x < a  this means   x ∈ (-∞, a)    : here x can take all the values between -∞ and a . the extreme values are not included. So (-∞, a) is the domain of x.

if x < = a  this means  x ∈ (-∞, a]  : here x can take all the values between -∞ and a. -∞ is not included and a is included. Note that I have used square braces for a.

What else you can do inside qs leap ?

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Similar examples:

b < x < a  => a x ∈ (b , a)

b ≤ x ≤ a  => a x ∈ [b, a]

Note: infinity is never included in a domain.*

Changing sign of an inequality:

Rule 1 : Multiplication with a negative number on either side of an inequality reverses the sign of the inequality.

Example :  if  x < a   then

-2 x < -2 a …. this is true for multiplication with any negative number.

example 2:  if x < 2   =>  -x > -2     multiplication by -1

Rule 2 : Taking reciprocals of both side reverses the inequality

Example : if x < a   then 1/x  >  1/a

Generalized rule (Not in the scope of syllabus but good to know):  

  • Sign of an inequality changes if we apply a decreasing function on both sides of an inequality. Sign of inequality remains same if it’s an increasing function or a constant function.

Example: we know that e-x is a decreasing function i.e. it decreases with increase in x.  {ex is called exponential function. e is Euler’s number with value approx. 2.718}

so if  a < b    =>  e-a > e-b

Example 2: cos x  decreases between 0 and π radians (or 0 and 180 degrees)

so for a and b both between 0 and π radians. if a < b   =>  cos a > cos b

Solving Polynomial Inequalities

An inequality of the form a1xn + a2xn-1 + a3xn-2 . . . .  . . + anx0 > or < = 0 is a polynomial inequality.

Example:  x2 – x < 0; x3 – 2x2 -x + 5> 0

How to solve:

Step 1: Bring every term on one side of the inequality

Step 2: Factorise

Step 3: Use number line to find suitable intervals.

Very important Note: Do not cancel variables while solving inequalities

If x2 > x

if we cancel x on both sides and write x > 1 we will be losing a part of the solution

Example 1:

x2 > x

  • x2 – x > 0 {Notice I didn’t cancel x on either side} {Step 1}
  • x (x – 1) > 0 {Step 2 factorization}

there are two points where x (x – 1) = 0; at x = 0 and x = 1. Marking these points on the number line and finding the

<- -∞———- x(x-1)>0—————-|0———–x(x-1) <0————–|1———-x(x-1)>0—–∞ ->

Real number line

Explanation:  we take 3 intervals. -∞< x < 0; 0 < x < 1 & 1 < x < ∞

we look at the signs of x, x-1 and x(x-1) in these three ranges

For x є (-∞,0)

x is negative and x-1 is negative hence x(x-1) is positive

For x є (0, 1)

x is positive and x-1 is negative hence x(x-1) is negative

For x є (1, ∞)

x is positive and x-1 is positive hence x(x-1) is positive

Since we need x(x-1) > 0 i.e. positive

Required range is x є (-∞,0) U (1, ∞)

Example 2:

x2 + 2 ≤ 3x

{note that this is a ≤ sign not < sign. So [] brackets will come for the domain}

Shift x to LHS

x2 – 3x + 2 ≤ 0

x2 – 2x – x + 2 ≤ 0

x(x-2) – 1(x-2) ≤ 0

(x-2) (x-1) ≤ 0             {factorized form}

Zeroes of the equation (x-2) (x-1) are x = 2 and x = 1

Marking 2 and 1 on the number line and determining whether (x-2) (x-1) is positive or negative for the different ranges of x. we have to use the method given in the previous example. Checking signs of x-2 and x-1 in each of the ranges and then multiplying.

<– -∞ ———-Positive————|1————-Negative——————|2———–Positive——–∞–>

But since we need (x-2) (x-1) to be ≤ 0 or negative

Required answer is x є  [1,2] {square braces because of equality sign}

Important note:

If there is an in-equation like x3 < x2 < x, it is recommended to solve the inequalities taking one at a time. Though it is possible to solve both the inequality signs at the same time, chance of making an error is very high. So whether someone is very proficient in Math or not it is highly recommended to solve such equations one by one.

Part 1: x3 < x2 Part 2: x2<x

x3 – x2 < 0

x2(x-1) < 0        using number line we can show that

this is true for all x є (-∞, 1)

x2 – x <0

x(x-1) < 0

true for x є (0, 1)

We take intersection of both the solution to get the answer x є (0, 1)


. * I have never seen infinity included in the domain. It is beyond my knowledge if that is a possibility. But I have seen weirder things happen with infinity in Mathematics so I cannot vouch for it. But nothing to worry about, if such a thing is possible its definitely not in the syllabus.

There are no inequalities in complex numbers.

If you wish to discuss further on “Inequalities” or want to clarify any specific query, connect with me here.

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