- INTRODUCTION
- PROPERTIES OF INEQUALITIES
- SOME USEFUL INEQUATIONS
- SOLVING INEQUALITIES
- INEQUALITIES AND ABSOLUTE VALUES
- PROPERTIES OF MODULUS
- QUADRATIC INEQUALITIES
Inequalities
An inequality as the name suggests, is a relationship that states that two quantities are not equal. It means that one expression is greater than the other in value.
Inequality can be expressed by using the following symbols:
- x ≠ y means x is not equal to y
- x > y means x is greater than y
- x < y means x is less than y
- x
y means x is greater than or equal to y, or x is
not less than y
- x
y means x is less than or equal to y, or x is not
greater than y
Inequalities of the last two forms are called inequations.
For example, consider the statement “the
total number of people (P) at a given time inside the “City Plaza” mall
should not exceed 1500”. This statement can be expressed mathematically as
P
1500
Equalities involving variables often give us exact values of the variables but inequalities involving variables often give a range of values for the variables.
For example, If 3x > 9, then x > 3, which means all real numbers greater than 3 satisfy the condition.
The properties of inequalities are better understood if numbers are substituted in place of variables. It is important that inequalities be tested for positive as well as negative numbers.
Transitivity
For real numbers a, b and c
- If a > b and b > c, then a > b > c and a > c
- If a < b and b < c, then a < b < c and a < c
Sum and Difference
- If a > b, then a + c > b + c and a – c > b – c
For example,
If a = 12, b = 10 and c = 5
a + c = 17 and b + c = 15
a + c > b + c
a – c = 7 and b – c = 5
a – c > b – c
- If a > b and c > d, then a + c > b + d
For example,
If a = 12, b = 10, c = 2 and d = 1
a + c = 14, b + d = 11
a + c > b + d
If a = 12, b = 10, c = –5, d = –6
a + c = 7, b + d = 4
a + c > b + d
This rule can be generalised as follows.
If a > b, c > d, e > f, and so on, then
a + c + e + … > b + d + f + …
- If a > b and c > d, then
a
c > b
d is not necessarily true.
However, a – d > b – c
For example,
If a = 12, b = 10, c = 7 and d = 6
a – d = 6, b – c = 3
a – d > b – c
Multiplication and Division
- If a > b and c > 0 then a
c > b
c and a/c > b/c
For example,
If a = 12, b = 10, c = 5 and d = 4
ac = 60 and bc = 50
ac > bc
a/d = 3 and b/d = 2.5
a/d > b/d
- If a > b and c < 0 then a
c < b
c and a/c < b/c
For example,
If a = 12, b = 10, c = –5 and d = –4
ac = –60, bc = –50
ac < bc
a/d = –3 and b/d = –2.5
a/d < b/d
It follows that if a > b, then –a < –b
- If a > b and c > d, and a, b, c, d > 0, then ac > bd
Let a = 12, b = 10, c = 5 and d = 4
ac = 60 and bd = 40
ac > bd
This rule can be generalised as follows.
If a, b, c, d, e, f, … are all positive and if a > b, c > d, e > f, and so on,
then ace… > bdf…
- If a > b and a, b, n > 0, then an > bn
- If a > b and a, b, n > 0, then a1/n >; b1/n
- If a > b and a, b, n > 0, then 1/an < 1/bn or a–n < b–n
It follows that if a > b, then 1/a < 1/b
REMEMBER
- Working with inequalities is similar to working with equalities, except for the fact that when both the sides of the inequality are multiplied or divided by a negative number, the inequality gets reversed.
We know that the square of any positive number is always positive. Thus,
(x – y)2
0
x2 + y2 – 2xy
0
x2 + y2
2xy (The two sides of the expression are equal only if x =
y)
We can state a few important results based on this inequation.
- x2 + y2
2xy
Let x2 = a and y2 = b
The arithmetic mean of two numbers is greater than the geometric
mean.
- a2 + b2
2ab
If a and b both are positive or negative, ab > 0
The sum of a positive number and its reciprocal is always greater than
2.
- a2 + b2
2ab
Let a + b > 0
(a2 + b2)(a + b)
2ab(a + b)
a3 + ab2 + a2b +
b3
2a2b + 2ab2
a3 + b3
ab(a + b) (The two sides of the expression are equal only
if x = y)
- x3 + y3
xy(x + y)
Let x = 
and y = 
- Let a ≠ b ≠ c Then,
a2 + b2 > 2ab
b2 + c2 > 2bc
c2 + a2 > 2ac
Adding,
2(a2 + b2 + c2) > 2(ab + bc + ac)
a2 + b2 + c2 >
(ab + bc + ac)
- (a + b)(b + c)(a + c) = 2abc + b(a2 + c2) + a(b2 + c2) + c(b2 + a2)
Also, a2 +
c2
2ac,
a2 + b2
2ab and
b2 + c2
2bc
If a, b and c are positive,
b(a2 +
c2)
2abc,
c(a2 +
b2)
2abc and
a(b2 +
c2)
2abc
2abc
+ b(a2 + c2) +
a(b2 + c2) +
c(b2 + a2)
2abc + 2abc + 2abc + 2abc
(a + b)(b + c)(a + c)
8abc, if a, b and c are positive.
- The arithmetic mean of 1, 2, 3, …, n is
Since Arithmetic Mean
Geometric Mean,
(n + 1)n
2n (n!)
- The arithmetic mean of 2, 4, 6, … , 2n is
Since Arithmetic Mean
Geometric Mean,
2
4
6
…
2n
(n + 1)n
- Using the binomial theorem,
- If x and y are positive and x > y, then log x > log y
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A = x2(2x + 1) + y2(2y + 1) B = 2xy(x + y + 1) where x and y are distinct positive numbers.
Is A greater than B? Solution: A = 2x3 + x2 + 2y3 + y2
x3 + y3 > xy(x + y) and (x2 + y2)/2 > xy
Hence, A is greater than B. |
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The sum of the squares of two numbers m and n is 5 and the sum of the squares of other two numbers p and q is 1. What can be said about the possible range of values of the sum (mp + nq) if all the four numbers are unequal?
Solution: m2 + n2 = 5 and p2 + q2 = 1
As arithmetic mean > geometric mean for unequal numbers, m2 + p2 > 2mp and n2 + q2 > 2nq
Thus, m2 + p2 + n2 + q2 > 2mp + 2nq
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Solution:
Since Arithmetic Mean
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Maximizing the product of n numbers when the sum is known.
Consider any two numbers a and b.
Let S = a + b and P = ab
(a + b)2 – (a – b)2 = 4ab
S2 – (a – b)2 = 4P
P will be maximum when (a – b)2 = 0
The product of two numbers, when their sum is known, is maximum when they are
equal.
Also, S2 = 4P + (a – b)2
S is minimum when (a – b)2 = 0
The sum of two numbers, when their product is known, is minimum when they are
equal.
Now consider n numbers a, b, c, d, …, k, such that a + b + c + d + … + k = S
As stated earlier, the product of two numbers with a fixed sum is maximum when the two numbers are made equal.
Consider a and b. The product ab will be maximised, while keeping their sum constant, when a and b are both made equal to (a + b)/2.
Thus, the product of n numbers a, b, c, d, …, k is maximum when each of the n numbers is made equal to S/n.
This result also states that the arithmetic mean of n quantities is greater than their geometric mean when the n quantities are unequal. The arithmetic mean of n quantities is equal to their geometric mean when the n quantities are equal.
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The sum of three numbers is 42. Find the maximum value that the product of these three numbers can have.
Solution: x + y + z = 42 The product of the three numbers is maximum when x = y = z = 42/3 = 14
The maximum value of the product = 14 |
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If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is: [IIFT 2008]
Solution: The arithmetic mean of n numbers is greater than or equal to their geometric mean. As a, b and c are in Arithmetic Progression, their arithmetic mean = b
Hence, option 2. |
Solution of an inequality is the set of values of the variable(s) that satisfy the inequality. The solution involves reducing one side of the inequality to one single variable and algebraically solving to find the required range of values.Solutions of inequalities are represented using two kinds of notations.
The first one uses <, >,
and
signs to denote the range of values that a variable can take. For example,
a
x < y.
The second one uses brackets to denote the
range. A round bracket on either or both the sides denotes that the number on
that side is not included in the range. A square bracket on either or both the
sides indicates that the number on that side is included in the range. Thus,
a
x < y is same as [a, y).
For example,
If x lies between 5 and 8, the solution can be written as 5 < x < 8 or (5, 8).
5
x
8 can also be written as [5, 8].
5
x < 8 can be written as [5, 8).
5 < x
8 can be written as (5, 8].
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Find the range of solutions for x for the inequation 7x + 6 < 18.
Solution: 7x + 6 < 18
All values of x less than 12/7 will satisfy the given inequality. |
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For which values of x is 7x + 9 < 8x – 19?
Solution: 7x + 9 < 8x
(Since we are dividing both sides of the inequation by a negative quantity, the sense of the inequality will change.) Here all values of x greater than 28 will satisfy the given inequality. |
|
If
Solution: We can add two inequalities of the same type. Hence, to find the range of x + y, we add the extreme values of both the inequalities.
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If x > 2 and y > – 1, Then which of the following statements is necessarily true? [CAT 2000]
(1) xy > –2 (2) –x < 2y (3) xy < –2 (4) –x > 2y
Solution: y > –1
Hence, option 2. |
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If 32 < x < 54 and
Solution: Since we cannot subtract one inequality from the
other, we find the range of x + ( To find the range of
So to find the range of x
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Solving a System of Inequalities
A number of inequalities in one unknown, which can be solved to find a common set of values that satisfy all inequalities, are called a System of Inequalities.
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Find the range of values of x that satisfy the following in equations. –17 –22 –19
Solution: –17
–22
–19
The common range of values that satisfies the three equations is [–5, –4]
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x and y are real numbers satisfying the conditions 2 < x < 3 and –8 < y < –7. Which of the following expressions will have the least value? [CAT 2001] (1) x2y (2) xy2 (3) 5xy (4) None of these
Solution: Option 2 can easily be eliminated as it will give positive value, while both option 1 and 3 will be negative.
Hence, option 3. |
x2 will fall in the range 4 < x2
< 9.
|
Given that [CAT 2003 Leaked Test]
(1) (3)
Solution: w = vz/u From the given range of values for u, v and z, we have, Maximum possible value of w is 4 when
v is 1, z is Also minimum value of w is Hence, option 2. |
The absolute value or modulus of a number x denoted by |x| is defined as its distance from zero on the number line. Since distance cannot be negative, the absolute value of any number is always positive.
Mathematically, the absolute value of x is defined as,
|x| = x, if x
0
|x| = x,
if x
0
- If |x| < p then this means x
< p and x > p,
which can be written as p
< x < p, and can be stated as ‘x lies between
negative p and positive p’.
- If |x| > p, then this can be interpreted as
x > p or x < p
which can also be stated as ‘x is either greater than positive p
or x is less than negative p’.
- For a real number x, |x| = max
{x, x}
- For all real numbers x,
|x|2 = x2 = |x|2
Explanation
By definition,
|x| = x, if x
0
= x,
if x < 0
In either case,
|x|2 = x2
|x|2
= (x)2
= x2
Hence, |x|2 =
x2 = |x|2
- For any real number x, |x| =|x|
Explanation:
|x|
= max{x,
(x)}
= max {x,
x}
= max {x, x}
= |x|
- For real numbers, x and y, |xy| = |x| |y|
Explanation:
|xy|2 = (xy)2
= x2 y2
= (|x|)2 (|y|)2
= (|x| |y|)2
As, |xy| and |x||y| are both non negative, therefore, taking the positive square root on both sides we have, |xy| = |x| |y|
- Triangle Inequality
For all real numbers x and y,
|x + y|
|x| + |y|
- For all real numbers x and y,
|x
y|
||x|
|y||
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Find the value of x, if |x
Solution: |x This can be written as
This means all values of x between |
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Find the value of x, if |6x
Solution: |6x This can be written as (6x
This means that all values of x, except values between and including 13/3 and 1/3, will satisfy the inequality. |
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If |b| [CAT 2003 Re-Test]
(1) a – xb < 0(2) a – xb (3) a – xb > 0(4) a – xb
Solution: |b| Consider the case when b is negative: i.e. say b = –k, where k Then, x = –|a|b = –|a|
Now, If a > 0, then a – xb = a + |a|k2 > 0 since all the terms will be positive If a < 0 (say a = –2), then
a – xb = –2 + 2k2 However, if a = 0, then a – xb = 0 + 0 = 0 Hence, when b is negative, a –
xb Now, consider the case when b is positive: i.e. say b = +k, where k Then, x = –|a|b = –|a|
This is the same value of xb as we got in the previous case. Hence, the same conclusions will hold.
Hence, option 2. |
A quadratic inequality is an
inequality of the form ax2 + bx + c < 0 or
ax2 + bx + c > 0 or ax2 +
bx + c
0 or ax2 + bx + c
0.
If y = ax2 +
bx + c where a, b and c are real and a
≠ 0, then y represents a parabola whose axis is parallel to the
y–axis. Let 
and 
be the two roots of the equation where
< .
y may be positive, negative or zero depending on the value of x.
If a > 0, then the parabola opens upwards and if a < 0, the
parabola opens downwards. The parabola intersects the x-axis at
and ,
if
and
are real. If
and
are imaginary, the parabola does not intersect the x-axis.
For a quadratic expression of the form
ax2 + bx + c = 0, the discriminant is given by
∆ = b2
4ac.
The nature of the roots of the quadratic equation depends on the value of the discriminant.
Consider the following cases.
Case 1: ∆ < 0
Roots of ax2 + bx + c = 0 are imaginary, when ∆ < 0.
Thus the value of y = ax2 + bx + c has the same sign as a when ∆ < 0.
Case 2: ∆ = 0
Roots of ax2 + bx + c = 0 are real and equal when ∆ = 0.
Thus when ∆ = 0, the value of y =
ax2 + bx + c has the same sign as a,
except when x = b/2a.
Case 3: ∆ > 0
Roots of ax2 + bx + c = 0 are real and distinct when ∆ > 0.
Thus, if
and
are the two roots of the equation ax2 + bx + c =
0 then,
if a > 0, ax2 +
bx + c is negative in the interval (,
),
zero at x =
and x = ,
and positive otherwise;
if a < 0, ax2 +
bx + c is positive in the interval (,
),
zero at x =
and x = ,
and negative otherwise.
REMEMBER
- If a quadratic equation ax2 + bx +
c = 0; a > 0 has real roots
and
such that
< .
- If ax2 + bx + c < 0, then
< x < ,
i.e. x will lie between the roots.
- If ax2 + bx + c > 0, then
x <
or x > ,
i.e. x will not lie between the roots.
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Solve x2 + 8x + 15 < 0.
Solution: x2 + 8x +15 < 0
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Solve x2 + 10x
Solution: x2 + 10x
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Solve x2 – 7x + 15 > 0.
Solution: x2 – 7x + 15 > 0 Δ = 49
Also, the coefficient of x2 is positive.
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Solve
Solution:
Δ = 25 – 4(
Also, the coefficient of x2 is negative.
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x2 – 12x + 72 < 0 For what values of x does the inequality hold?
Solution: x2 – 12x + 72 < 0 Δ = 144 – 4(72) < 0 Coefficient of x2 is positive.
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Solution:
In this case we cannot cross multiply since the sign of variable m is not known. So we rewrite it as,
Since the numerator is negative and the inequation is positive, the denominator has to be negative. m + 1 < 0
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Solve
Solution:
Since LHS is negative, we have 2 cases.
Case 1: When the numerator is negative and denominator is positive, 9x + 3 < 0 and (x + 1)(x x < So x <
Case 2: When the numerator is positive and denominator is negative, 9x + 3 > 0 and (x + 1)(x
x > So, Hence, |
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Solution:
The above expression can be negative in two cases.
Case 1: The numerator is negative and the denominator is positive.
(x – 5) > 0
(i) and (ii) are not possible simultaneously.
Case 2: The numerator is positive and the denominator is negative.
(x – 5) < 0
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Solve the inequality (p – 4)x2 – 3x + 1 > 0.
Solution: Δ = 9 – 4(1)(p – 4)
As the value of p is not known, we cannot find the exact value of Δ.
Case 1: Δ < 0
Case 2: Δ = 0
i.e. at x = 3/2(p – 4) i.e. at x = 2/3
Case 3: Δ > 0
We find the roots of the equation at p < 25/4 and p ≠ 4.
If (p – 4) > 0, the inequality is true
when x < If (p – 4) < 0, the inequality is true
when |
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Solve the inequality x2 + x – 4 > |3x + 4|.
Solution: |3x + 4| < x2 + x – 4
If 3x + 4 < x2 + x – 4, x2
If x2 + 4x > 0
From (i) and (ii), x < |
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Find the range of values of x for which the following inequality is true. (|x – 2|
Solution: (|x – 2| The LHS is negative in two cases.
Case 1: (|x – 2| If |x – 2| < 5
If |x
(i) and (ii) cannot be true simultaneously.
Case 2: (|x – 2| If |x – 2| > 5
If |x
From (iii) and (iv), 7 < x < 10 and |
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Solution:
If x + 1 But, x + 3
If x + 1 > 0,
But x >
From (i) and (ii),
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x2 – 17|x| + 72 < 0 Find the range of values of x that satisfy the inequality.
Solution: x2 – 17|x| + 72 < 0
(|x| This can be true in two cases.
Case 1: (|x|
Case 2: (|x|
This is not possible.
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Solution:
x2 – x + 1 > 0 for all values of x as the coefficient of x2 is positive and Δ for the equation is negative.
(x – 5)(x + 4) < 0
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Solution:
The numerator of the LHS has only even powers of y.
y2 – y – 72 < 0
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Solution:
This is possible in two cases.
Case 1: Both the numerator and the denominator are negative.
Case 2: Both the numerator and the denominator are positive.
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IMPORTANT
- During the exam, multiple choice questions on inequalities can often be solved by evaluating options. However, having an in-depth understanding of inequalities is important.
REMEMBER
- If a > b and b > c, then a > b > c and a > c
- If a < b and b < c, then a < b < c and a < c
- If a > b, then a + c > b + c and a – c > b – c
- If a > b and c > d, then a + c > b + d
- If a > b and c > d, then a – d > b – c
- If a > b and c > 0 then a
c > b
c and a/c > b/c
- If a > b and c < 0 then a
c < b
c and a/c < b/c
- If a > b and c > d, and a, b, c, d > 0, then ac > bd
- If a > b and a, b, n > 0, then an > bn
- If a > b and a, b, n > 0, then a1/n >b1/n
- If a > b and a, b, n > 0, then 1/an < 1/bn or a–n < b–n
- If a > b, then 1/a < 1/b
- a3 + b3
ab(a + b), if a and b are positive.
- a2 + b2 + c2 > (ab + bc + ac)
- (a + b)(b + c)(c + a)
8abc if a, b and c are all positive.
- (n + 1)n
2n (n!)
- 2
4
6
…
2n
(n + 1)n
- Given the sum of two positive quantities, their product is the maximum when they are equal.
- Given the product of two positive quantities, their sum is the least when they are equal.
- If x and y are positive and x > y, then log x > log y.
