How to Prepare for Quantitative Aptitude for CAT (2014)

Block V: Algebra

Chapter 14. INEQUALITIES

This chapter will seem to be highly mathematical to you when you read the theory contained in the chapter and look at the solved examples. For students weak in Math, there is no need to be disheartened about the seemingly high mathematical content. I would advise you to go through this chapter and internalise the concepts. However, keep in mind the fact that in an aptitude test, the questions will have options, and with options all you will need to do will be check the validity of the inequality for the different options.

In fact, the questions in this chapter have options on all three levels and with option-based solutions, all these questions will seem easy to you.

However, I would advise students aiming to score high marks in Quantitative Ability to try to mathematically solve all the questions on all three levels in this chapter (even though option-based solution will be much easier.)

Two real numbers or two algebraic expressions related by the symbol > (“Greater Than”) or < (“Less Than”) (and also by the signs ≥ or £ ) form an inequality.

A < B, A > B (are plain inequalities)

A ≥ B, A £ B (are called as inequations)

The inequality consists of two sides—the left hand side, A and the right hand side, B. A and B can be algebraic expressions or they can be numbers.

An inequality with the < or > sign is called a strict inequality while an inequality having ≥ or £ sign is called a slack inequality. The expressions A and B have to be considered on the set where A and B have sense simultaneously. This set is called the set of permissible values of the inequality. If the terms on the LHS and the RHS are algebraic equations/identities, then the inequality may or may not hold true for a particular value of the variable/set of variables assumed.

The direction in which the inequality sign points is called the sense of the inequality. If two or several inequalities contain the same sign (< or >) then they are called inequalities of the same sense. Otherwise they are called inequalities of the opposite sense.

Now let us consider some basic definitions about inequalities.

For 2 real numbers a and b

The inequality a > b means that the difference a – b is positive.

The inequality a < b means that the difference a – b is negative.

PROPERTIES OF INEQUALITIES

For any two real numbers a and b, only one of the following restrictions can hold true:

a = b, a > b or a < b

Definitions of Slack Inequalities

The inequality a ≥ b means that a > b or a = b, that is, a is not less than b.

The inequality a £ b means that a < b or a = b, that is, a is not greater than b.

We can also have the following double inequalities for simultaneous situations:

a < b < c, a < b £ c , a £ b < c , a £ b £ c

To Find the Greatest Value of a Product, the Sum of Whose Factors is Constant

Let there be n factors a, b, c,… n, of a composite number and suppose that their sum is constant and equal to S.

Consider the product abc….n, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors (a + b)/2, and (a + b)/2, the product is increased while the sum remains unaltered. Hence, so long as the product contains two unequal factors it can be increased altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case the value of each of the n factors is S/n, and the greatest value of the product is (S/n)ⁿ , or {(a + b + c +….. + n/n}ⁿ

This will be clearer through an example.

Let us define a number as a × b = c such that we restrict a + b = 100 (maximum).

Then, the maximum value of the product will be achieved if we take the value of a and b as 50 each.

Thus 50 × 50 = 2500 will be the highest number achieved for the restriction a + b £ 100.

Further, you can also say that 50 × 50 > 51 × 49 > 52 × 48 > 53 × 47 > 54 × 46 > … > 98 × 2 > 99 × 1

Thus if we have a larger multiplication as

4 × 6 × 7 × 8 this will always be less than 5 × 5 × 7 × 8. [Holds true only for positive numbers.]

Corollary If a, b, c…k, are unequal, {(a + b + c +…+ k)/n}ⁿ > abc … k;

that is , (a + b + c +,…+ k)/n > (abc… k)^1/n .

By an extension of the meaning of the arithmetic and geometric means this result is usually quoted as follows: The arithmetic mean of any number of positive quantities is greater than the geometric mean.

Definition of Solution of an Inequality

The solution of an inequality is the value of an unknown for which this inequality reduces to a true numerical identity. That is, to solve an inequality means to find all the values of the variable for which the given inequality is true.

An inequality has no solution if there is no such value for which the given inequality is true.

Equivalent Inequalities: Two inequalities are said to be equivalent if any solution of one is also a solution of the other and vice versa.

If both inequalities have no solution, then they are also regarded to be equivalent.

To solve an inequality we use the basic properties of an inequality which have been illustrated above.

Notation of Ranges

1. Ranges where the ends are excluded If the value of xis denoted as (1, 2) it means 1 < x< 2 i.e. x is greater than 1 but smaller than 2.

Similarly, if we denote the range of values of x as –(7, –2)U(3, 21), this means that the value of x can be denoted as –7 < x < –2 and 3 < x < 21. This would mean that the inequality is satisfied between the two ranges and is not satisfied outside these two ranges.

Based on this notation write the ranges of x for the following representations:

(1, +•) U (–•, –7)

(–•, 0) U (4, +•), (–•, 50) U (–50, +•)

2. Ranges where the Ends are Included

[2, 5] means 2 £ x £ 5

3. Mixed ranges

(3, 21] means 3 < x £ 21

Solving Linear Inequalities in one Unknown

A linear inequality is defined as an inequality of the form

ax + b I 0

where the symbol ‘I’ represents any of the inequalities < , >, ≥ , £.

For instance if ax + b £ 0, then ax £ –b

Æ x £ –b/a if a > 0 and x ≥ –b/a if a < 0

Example: Solve the inequality 2(x – 3) –1 > 3(x – 2) – 4(x + 1)

Æ 2x – 7 > 3x – 6 – 4x – 4 Æ 3x > –3. Hence, x > –1

This can be represented in mathematical terms as (–1, +•)

Example: Solve the inequality 2(x – 1) + 1 > 3 – (1 – 2x) Æ 2x – 1 > 2 + 2x Æ 0.x > 3 Æ This can never happen. Hence, no solution.

Example: Solve the inequality 2(x – 1) + 1 < 3 – (1 – 2x)

Gives: 0.x < 3.

This is true for all values of x

Example: Solve the inequality ax > a.

This inequality has the parametre a that needs to be investigated further.

If a > 0, then x > 1

If a < 0, then x < 1

Solving Quadratic Inequalities

A quadratic inequality is defined as an inequality of the form:

ax² + bx + c I 0 (a π 0)

where the symbol I represents any of the inequalities <, >, ≥, £.

For a quadratic expression of the form ax² + bx + c, (b² – 4ac) is defined as the discriminant of the expression and is often denoted as D. i.e. D = b² – 4ac

The following cases are possible for the value of the quadratic expression:

Case 1: If D < 0

1.If a < 0 then ax² + bx + c < 0 for all x

2.If a > 0 then ax² + bx + c > 0 for all x.

In other words, we can say that if D is negative then the values of the quadratic expression takes the same sign as the coefficient of x².

This can also be said as

If D < 0 then all real values of x are solutions of the inequalities ax² + bx + c > 0 and ax² + bx + c ≥ 0 for a > 0 and have no solution in case a < 0.

Also, for D < 0, all real values of x are solutions of the inequalities ax² + bx + c < 0 and ax² + bx + c £ 0 if a < 0 and these inequalities will not give any solution for a > 0.

Case 2: D = 0

If the discriminant of a quadratic expression is equal to zero, then the value of the quadratic expression takes the same sign as that of the coefficient of x² (except when x = –b/2a at which point the value of the quadratic expression becomes 0).

We can also say the following for D = 0:

1.The inequality ax² + bx + c > 0 has as a solution any x π –(b/2a) if a > 0 and has no solution if a < 0.

2.The inequality ax² + bx + c < 0 has as a solution any x π –(b/2a) if a < 0 and has no solution if a > 0.

3.The inequality ax² + bx + c ≥ 0 has as a solution any x if a > 0 and has a unique solution x = –b/2a if a < 0.

4.The inequality ax² + bx + c £ 0 has as a solution any x if a < 0 and has a unique solution x = –b/2a for a > 0.

Case 3: D > 0

If x₁ and x₂ are the roots of the quadratic expression then it can be said that:

1.For a > 0, ax² + bx + c is positive for all values of x outside the interval [x₁, x₂] and is negative for all values of x within the interval (x₁, x₂). Besides for values of x = x₁ or x = x₂, the value of the quadratic expression becomes zero (By definition of the root).

2.For a < 0, ax² + bx + c is negative for all values of x outside the interval [x₁, x₂] and is positive for all values of x within the interval (x₁, x₂). Besides for values of x = x₁ or x = x₂, the value of the quadratic expression becomes zero (By definition of the root).

Here are a few examples illustrating how quadratic inequalities are solved.

Solve the following inequalities.

Example 1: x² – 5x + 6 > 0

Solution: (a) The discriminant D = 25 – 4 × 6 > 0 and a is positive (+1); the roots of the quadratic expression are real and distinct: x₁ = 2 and x₂ = 3.

By the property of quadratic inequalities, we get that the expression is positive outside the interval [2, 3]. Hence, the solution is x < 2 and x > 3.

We can also see it as x² – 5x + 6 = (x – 2) (x – 3) and the given inequality takes the form (x – 2) (x – 3) > 0.

The solutions of the inequality are the numbers x < 2 (when both factors are negative and their product is positive) and also the numbers x > 3 (when both factors are positive and, hence, their product is also positive).

Answer: x < 2 and x > 3.

Example 2: 2x² + x + 1 ≥ 0

Solution: The discriminant D = 1 – 4 ◊ (– 2) = 9 > 0; the roots of the quadratic expression are real and distinct:

x₁, ₂ = =

hence, x₁ = – 1/2 and x₂ = 1, and consequently, –2x² + x + 1 = –2(x + 1/2) × (x – 1). We have

–2(x + 1/2) (x – 1) ≥ 0 or (x + 1/2) (x – 1) £ 0

(When dividing both sides of an inequality by a negative number, the sense of the inequality is reversed). The inequality is satisfied by all numbers from the interval

[–1/2, 1]

Please note that this can also be concluded from the property of quadratic expressions when D > 0 and a is negative.

Answer: –1/2 £ x £ 1.

Example 3: 2x² + x – 1 < 0

Solution: D = 1 – 4 ◊ (–2) ◊ (–1) < 0, the coefficient of x² is negative. By the property of the quadratic expression when D < 0 and a is negative –2x² + x – 1 attains only negative values.

Answer: x can take any value.

Example 4: 3x² – 4x + 5 < 0

Solution: D = 16 – 4 × 3 × 5 < 0, the coefficient of x² is positive. The quadratic expression 3x² – 4x + 5 takes on only positive values.

Answer: There is no solution.

Example 5: 4x² + 4x + 1 > 0.

Solution: D = 16 – 4 × 4 = 0. The quadratic expression 4x² + 4x + 1 is the square (2x + 1)², and the given inequality takes the form (2x + 1)² > 0. It follows that all real numbers x, except for x = –1/2, are solutions of the inequality.

Answer: x π –1/2.

Example 6: Solve the inequality (a – 2)x² – x – 1 ≥ 0

Here, the value of the determinant D = 1 – 4(–1)(a – 2) = 1 + 4(a – 2) = 4a – 7

There can then be three cases:

Case 1: D < 0 Æ a < 7/4

Then the coefficient of x² Æ a – 2 is negative.

Hence, the inequality has no solution.

Case 2: D = 0

a = 7/4. Put a = 7/4 in the expression and then the inequality becomes

–(x – 2)² ≥ 0. This can only happen when x = 2.

Case 3: D > 0

Then a > 7/4 and a π 2, then we find the roots x₁ and x₂ of the quadratic expression:

x₁ = and x₂ =

Using the property of quadratic expression’s values for D > 0 we get

If a – 2 < 0, the quadratic expression takes negative values outside the interval [x₁, x₂]. Hence, it will take positive values inside the interval (x₁, x₂).

If a – 2 > 0, the quadratic expression takes positive values outside the interval [x₁, x₂] and becomes zero for x₁ and x₂.

If a – 2 = 0, then we get a straight linear equation. –x – 1 ≥ 0 Æ x £ – 1.

System of Inequalities in One Unknown

Let there be given several inequalities in one unknown. If it is required to find the number that will be the solution of all the given equalities, then the set of these inequalities is called a system of inequalities.

The solution of a system of inequalities in one unknown is defined as the value of the unknown for which all the in-equalities of the system reduce to true numerical inequalities.

To solve a system of inequalities means to find all the solutions of the system or to establish that there is none.

Two systems of inequalities are said to be equivalent if any solution of one of them is a solution of the other, and vice versa. If both the systems of inequalities have no solution, then they are also regarded to be equivalent.

Example 1: Solve the system of inequalities:

3x – 4 < 8x + 6

2x – 1 > 5x – 4

11x – 9 £ 15x + 3

Solution: We solve the first inequality:

3x – 4 < 8x + 6

–5x < 10

x > –2

It is fulfilled for x > –2.

Then we solve the second inequality

2x – 1 > 5x – 4

–3x > – 3

x < 1

It is fulfilled for x < 1.

And, finally, we solve the third inequality:

11x – 9 £ 15x + 3

–4x £ 12

x ≥ – 3

It is fulfilled for x ≥ – 3. All the given inequalities are true for – 2 < x < 1.

Answer: – 2 < x < 1.

Example 2: Solve the inequality < 1

We have –1 < 0 Æ < 0

This means that the fraction above has to be negative. A fraction is negative only when the numerator and the denominator have opposite signs.

Hence, the above inequality is equivalent to the following set of 2 inequalities:

x – 2 > 0 and x – 2 < 0

and  x + 1 < 0   x + 1 > 0

From the first system of inequalities, we get

x > 2 or x < –1. This cannot happen simultaneously since these are inconsistent.

From the second system of inequalities we get

x < 2 or x > –1 i.e –1 < x < 2

Inequalities Containing a Modulus

Result:

| x | £ a, where a > 0 means the same as the double inequality

–a £ x £ a

This result is used in solving inequalities containing a modulus.

Example 1: | 2x – 3 | £ 5

This is equivalent to –5 £ 2x – 3 £ 5

i.e. 2x – 3 ≥ –5 and  2x – 3 £ 5

2x ≥ –2   x £ 4

x ≥ –1

The solution is

–1 £ x £ 4

Example 2: | 1 – x | > 3

| 1 – x | = | x – 1 |

Hence,  | x – 1 | > 3 Æ x – 1 > 3  i.e x > 4

or x – 1 < –3 or x < –2

Answer: x > 4 or x < –2.

WORKED-OUT PROBLEMS

Problem 14.1 Solve the inequality < 1.

Solution < 1 ¤ – 1 < 0 ¤ < 0 ¤ > 0.

This can happen only when both the numerator and denominator take the same sign (Why?)

Case 1: Both are positive: x – 1 > 0 and x > 0 i.e. x > 1.

Case 2: Both are negative: x – 1 < 0 and x < 0 i.e. x < 0.

Answer: (––• , 0) U (1 + •)

Problem 14.2 Solve the inequality .

Solution ¤ £ 0 ¤ £ 0 ¤ £ 0.

The function f(x) = becomes negative when numerator and denominator are of opposite sign.

Case 1: Numerator positive and denominator negative: This occurs only between –2 < x < – 1.

Case 2: Numerator negative and Denominator Positive: Numerator is negative when (x – 2) and (x + 1) take opposite signs. This can be got for:

Case A: x – 2 < 0 and x + 1 > 0 i.e. x < 2 and x > –1

Case B: x – 2 > 0 and x + 1 < 0 i.e. x > 2 and x < –1. Cannot happen.

Hence, the answer is –2 < x £ 2.

Problem 14.3 Solve the inequality .

Solution ¤ ≥ 0 ¤ £ 0. The function f(x) =

The numerator being a quadratic equation with D < 0 and a > 0, we can see that it will always be positive for all values of x. (From the property of quadratic inequalities).

Further, for the expression to be negative, the denominator should be negative.

That is, x² – 3x < 0. This will occur when x < 3 and x is positive.

Answer: (0, 3)

Suppose F(x) = (x – x₁)^k1 (x – x₂)^k2...(x₁ – x_n)^kn, where k₁, k₂ …, k_n are integers. If k_j is an even number, then the func-tion (x – x_j)^kj does not change sign when x passes through the point x_j and, consequently, the function F(x) does not change sign. If k_p is an odd number, then the function (x – x_p)^kp changes sign when x passes through the point x_p and, consequently, the function F(x) also changes sign.

Problem 14.4 Solve the inequality (x – 1)² (x + 1)³ (x – 4) < 0.

Solution The above inequality is valid for

Case 1: x + 1 < 0 and x – 4 > 0

That is, x < –1 and x > 4 simultaneously. This cannot happen together.

Case 2: x + 1 > 0 and x – 4 < 0

That is x > –1 or x < 4, i.e. –1 < x < 4 is the answer

Problem 14.5 Solve the inequality £ 0

Solution The above expression becomes negative when the numerator and the denominator take opposite signs.

That means that the numerator is positive and the denominator is negative or vice versa.

Case 1: Numerator positive and denominator negative: The sign of the numerator is determined by the value of x + 1 and that of the denominator is determined by x – 2.

This condition happens when x < 2 and x > –1 simultaneously.

Case 2: Numerator negative and denominator positive: This happens when x < –1 and x > 2 simultaneously. This will never happen.

Hence, the answer is –1 £ x < 2.

LEVEL OF DIFFICULTY (I)

Solve the following inequalities:

Directions for Questions 49 to 53: Find the largest integral x that satisfies the following inequalities.

Directions for Questions 54 to 69: Solve the following inequalities:

LEVEL OF DIFFICULTY (II)

Directions for Questions 3 to 16: Solve the following polynomial inequalities

Directions for Questions 11 to 67: Solve the following polynomial and quadratic inequalities

Directions for Questions 68 to 92: Solve inequalities based on modulus

Directions for Questions 93 to 95: Solve the following irrational inequalities.

Level of Difficulty (I)

Solutions and Shortcuts

While practically solving inequalities remember the following:

1.The answer to an inequality question is always in the form of a range and represents the range of values where the inequality is satisfied.

2.In the cases of all continuous functions, the point at which the range of the correct answer will start, will always be a solution of the same function if written as an equation.

This rule is only broken for non-continuous functions.

Hence, if you judge that a function is continuous always check the options for LHS = RHS at the starting point of the option.

3.The correct range has to have two essential properties if it has to be the correct answer:

(a)The inequality should be satisfied for each and every value of the range.

(b)There should be no value satisfying the inequality outside the range of the correct option.

Questions on inequalities are always solved using options and based on (3) (a) and (3) (b) above we would reject an option as the correct answer if:

(i)we find even a single value not satisfying the inequality within the range of a single option.

(ii)we can reject given option, even if we find a single value satisfying the inequality but not lying within the range of the option under check.

I will now show you certain solved questions on this pattern of thinking.

Level of Difficulty (I)

1.At x = 0, inequality is not satisfied. Thus, option (c) is rejected. Also x = 0 is not a solution of the equation. Since, this is a continuous function, the solution cannot start from 0. Thus options (a) and (b) are not right. Further, we see that the given function is quadratic with real roots. Hence, option (d) is also rejected.

2.At x = 0, inequality is satisfied. Hence, options (b) and (c) are rejected. x = 3 gives LHS = RHS. and x = – 0.66 also does the same. Hence. roots of the equation are 3 and – 0.66.

Thus, option (a) is correct.

3.At x = 0, inequality is not satisfied.

Hence, options (b), (c) and (d) are rejected. At x = 2, inequality is not satisfied. Hence, option (a) is rejected.

Thus, option (d) is correct.

4.The given quadratic equation has imaginary roots and is hence always positive.

Thus, option (d) is correct

5.At x = 0 inequality is not satisfied. Thus option (d) is rejected.

x = –1 and x = 15 are the roots of the quadratic equation. Thus, option (c) is correct.

6.At x = 0, inequality is satisfied.

Thus, options, (c) and (d) are rejected.

At x = 1, inequality is satisfied

Hence, we choose option (a).

7.At x = 0 inequality is satisfied, option (b) is rejected.

At x = 2, inequality is satisfied, option (c) is rejected.

At x = 5, LHS = RHS.

At x = –1, LHS = RHS.

Thus, option (d) is correct.

8.At x = 0 inequality is satisfied.

Thus, options (a), (b), and (d) are rejected.

Option (c) is correct.

10.At x = 1 and x = 3 LHS = RHS.

At x = 2 inequality is satisfied.

At x = 0.1 inequality is not satisfied.

At x = 2.9 inequality is satisfied.

At x = 3.1 inequality is not satisfied.

Thus, option (a) is correct.

12.The options need to be converted to approximate values before you judge the answer. At x = 0, inequality is satisfied.

Thus, option (a) is rejected.

Option (c) is correct.

13.At x = 0, inequality is not satisfied, option (a) is rejected.

At x = 5, inequality is not satisfied, option (b) is rejected.

At x = 2 inequality is not satisfied.

Options (d) are rejected.

Option (c) is correct.

14.At x = 0, inequality is satisfied, option (a) rejected.

At x = 10, inequality is not satisfied, option (c) rejected.

At x = –5, LHS = RHS.

Also at x = 5, inequality is satisfied and at x = 6, inequality is not satisfied.

Thus, option (d) is correct.

15.At x = 2, inequality is satisfied.

At x = 0, inequality is not satisfied.

At x = 1, inequality is not satisfied but LHS = RHS.

At x = 3, inequality is not satisfied but LHS = RHS.

Thus, option (b) is correct.

Solve other questions of LOD I and LOD II in the same fashion.