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CBSE Class 11 Maths Revision Notes Chapter 5 Linear Inequalities

Linear Inequalities explains how algebraic expressions are compared using symbols such as <, >, ≤ and ≥. For CBSE Class 11 Maths 2026–27, this chapter covers solution sets, number line graphs, one-variable inequalities and two-variable inequalities.

Linear Inequalities is an important chapter in Class 11 Mathematics. In earlier classes, you solved equations where two expressions were equal. In this chapter, you study situations where one expression is greater than, less than, greater than or equal to, or less than or equal to another expression.

For example, if a student needs at least 60 marks on average, or a buyer has a limited budget, the condition is not always written as an equation. It is often written as an inequality.

Use these CBSE Class 11 Maths Revision Notes Chapter 5 to revise types of inequalities, rules for solving inequalities, solution sets, graphical representation on the number line, linear inequalities in one variable and linear inequalities in two variables.

These Class 11 Maths Chapter 5 Notes are useful when you want quick revision of definitions, rules, examples and graph-based questions before practice.

Key Takeaways

Inequality: An inequality compares two numbers or expressions using <, >, ≤ or ≥.
Solution set: The set of all values that make an inequality true is called its solution set.
Sign reversal rule: The inequality sign reverses when both sides are multiplied or divided by a negative number.
Graphical representation: Solutions can be shown on a number line or as a region in the Cartesian plane.

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Class 11 Maths Chapter 5 Notes for Linear Inequalities Revision

These Class 11 mathematics revision notes chapter 5 are arranged for 30-minute revision, so you can revise the chapter in the same order you study it in class.

Start with the meaning of an inequality. Then revise strict inequalities, slack inequalities and double inequalities. After that, understand how to solve linear inequalities in one variable and represent the answer on a number line.

Once that is clear, move to linear inequalities in two variables and learn how to identify the solution region in the Cartesian plane.

Topic	What You Revise
Inequality	Meaning of <, >, ≤ and ≥
Strict inequality	Inequalities using < or >
Slack inequality	Inequalities using ≤ or ≥
Double inequality	Inequality with two comparison signs
Solution set	All values satisfying the inequality
One-variable inequality	Inequality involving one variable
Number line graph	Graphical representation of solution set
Two-variable inequality	Inequality involving x and y
Half-plane	Region represented by a linear inequality
System of inequalities	Common solution of two or more inequalities

What is an Inequality?

An inequality is formed when two real numbers or two algebraic expressions are related by one of these symbols:

<, >, ≤ or ≥

Examples:

3 < 5
7 > 4
x < 6
2x + 3 ≥ 9
x + y ≤ 10

An equation uses the equality sign “=”. An inequality uses comparison signs.

Statement	Type
2x + 3 = 9	Equation
2x + 3 < 9	Inequality
x + y = 10	Equation
x + y ≤ 10	Inequality

Types of Inequalities

Class 11 Maths Linear Inequalities Notes mainly cover strict inequalities, slack inequalities and double inequalities.

Type	Symbol	Example
Strict inequality	< or >	x < 5, y > 2
Slack inequality	≤ or ≥	x ≤ 5, y ≥ 2
Double inequality	Two comparison signs	2 < x < 7

A strict inequality does not include the boundary value.

A slack inequality includes the boundary value.

Example:

x < 5 means x can be any value less than 5, but not 5.

x ≤ 5 means x can be 5 or any value less than 5.

Numerical and Literal Inequalities

Inequalities can be numerical or literal.

Type	Meaning	Example
Numerical inequality	Compares numbers	3 < 5
Literal inequality	Contains variables	x > 4

Examples of literal inequalities:

x < 5
y ≥ 2
3x + 4 ≤ 10
2x - 5 > 7

Linear Inequalities in One Variable

A linear inequality in one variable contains only one variable and its highest power is 1.

General forms:

ax + b < 0
ax + b > 0
ax + b ≤ 0
ax + b ≥ 0

Here, a and b are real numbers and a ≠ 0.

Examples:

2x + 3 < 7
5x - 1 ≥ 9
3 - x ≤ 6

These are linear inequalities in one variable because each has only x and the power of x is 1.

Linear Inequalities in Two Variables

A linear inequality in two variables contains two variables, usually x and y.

General forms:

ax + by < c
ax + by > c
ax + by ≤ c
ax + by ≥ c

Here, a, b and c are real numbers, and a and b are not both zero.

Examples:

2x + 3y < 5
x - y ≥ 4
3x + 2y ≤ 12

These are linear inequalities in two variables because they contain x and y with power 1.

Quadratic Inequalities Are Not Linear

An inequality is not linear if the variable has power 2 or more.

Examples:

x² + 3x + 2 > 0
2x² - 5x ≤ 8

These are quadratic inequalities, not linear inequalities.

In CBSE Class 11 Maths Notes Chapter 5, the main focus is on linear inequalities in one and two variables.

Solution and Solution Set of an Inequality

A solution of an inequality is a value of the variable that makes the inequality true.

The solution set is the set of all such values.

Example:

x < 5

Here, x = 1, x = 2 and x = 4 are solutions.

But x = 5 is not a solution.

The solution set is:

{x : x < 5}

In interval form:

(-∞, 5)

Rules for Solving Linear Inequalities

Solving an inequality is similar to solving an equation, but one rule is different.

Rule	Effect on Inequality Sign
Add the same number on both sides	Sign does not change
Subtract the same number from both sides	Sign does not change
Multiply both sides by a positive number	Sign does not change
Divide both sides by a positive number	Sign does not change
Multiply both sides by a negative number	Sign reverses
Divide both sides by a negative number	Sign reverses

The last rule is the most important.

Example:

-2x < 8

Divide both sides by -2.

Since we divide by a negative number, the sign reverses.

x > -4

Strict Inequality and Slack Inequality on Number Line

The number line helps students see the solution set clearly.

Inequality	Point on Number Line	Meaning
x < a	Open circle at a	a is not included
x > a	Open circle at a	a is not included
x ≤ a	Filled circle at a	a is included
x ≥ a	Filled circle at a	a is included

Example:

x < 3 means all real numbers less than 3.

On the number line, place an open circle at 3 and shade to the left.

x ≥ 2 means all real numbers greater than or equal to 2.

On the number line, place a filled circle at 2 and shade to the right.

Algebraic Solution of Linear Inequalities in One Variable

An algebraic solution means solving the inequality step by step using rules.

Example 1: Solve 5x - 3 < 7

Solution:

5x - 3 < 7

Add 3 on both sides:

5x < 10

Divide by 5:

x < 2

So, the solution set is:

(-∞, 2)

Example 2: Solve 4x + 3 < 6x + 7

Solution:

4x + 3 < 6x + 7

Subtract 6x from both sides:

-2x + 3 < 7

Subtract 3 from both sides:

-2x < 4

Divide by -2 and reverse the sign:

x > -2

So, the solution set is:

(-2, ∞)

Solving Double Inequalities

A double inequality has two comparison signs.

Example:

-8 ≤ 5x - 3 < 7

Solve it together.

-8 ≤ 5x - 3 < 7

Add 3 to all parts:

-5 ≤ 5x < 10

Divide all parts by 5:

-1 ≤ x < 2

So, the solution set is:

[-1, 2)

System of Linear Inequalities in One Variable

A system of inequalities has two or more inequalities that must be true at the same time.

Example:

3x - 7 < 5 + x

11 - 5x ≤ 1

Solve the first inequality:

3x - 7 < 5 + x

2x < 12

x < 6

Solve the second inequality:

11 - 5x ≤ 1

-5x ≤ -10

Divide by -5 and reverse the sign:

x ≥ 2

Common solution:

2 ≤ x < 6

So, the solution set is:

[2, 6)

Word Problems on Linear Inequalities

Many Linear Inequalities Class 11 Notes questions are based on real-life conditions such as minimum marks, maximum budget, temperature range and length restrictions.

Example: Marks Problem

A student scored 70 and 75 marks in two tests. Find the minimum marks needed in the third test to get an average of at least 60.

Let x be the marks in the third test.

Average ≥ 60

(70 + 75 + x)/3 ≥ 60

145 + x ≥ 180

x ≥ 35

So, the student needs at least 35 marks in the third test.

Linear Inequalities in Two Variables and Cartesian Plane

A linear inequality in two variables represents a region in the Cartesian plane.

Example:

x + y ≤ 5

First replace the inequality sign with “=”.

x + y = 5

This gives the boundary line.

The inequality x + y ≤ 5 represents one side of that line.

The solution set contains all points (x, y) that satisfy the inequality.

Boundary Line in Linear Inequalities

A boundary line separates the Cartesian plane into two half-planes.

Inequality	Boundary Line Type
ax + by < c	Dashed line
ax + by > c	Dashed line
ax + by ≤ c	Solid line
ax + by ≥ c	Solid line

Use a solid line when the boundary points are included.

Use a dashed line when the boundary points are not included.

Half-Plane and Solution Region

A straight line divides the Cartesian plane into two half-planes.

For a linear inequality in two variables, one half-plane contains the solution region.

The solution region is the set of all points that satisfy the inequality.

Example:

x + y ≤ 5

The boundary line is:

x + y = 5

Now test a point, usually (0, 0).

0 + 0 ≤ 5

This is true.

So, the half-plane containing (0, 0) is the solution region.

Steps for Graphical Representation of Linear Inequalities in Two Variables

Follow these steps:

Replace the inequality sign with “=” and draw the boundary line.
Use a solid line for ≤ or ≥.
Use a dashed line for < or >.
Choose a test point not on the line.
Substitute the test point in the inequality.
If the point satisfies the inequality, shade that side.
If it does not satisfy the inequality, shade the other side.

Graphical Solution of a System of Linear Inequalities

A system of linear inequalities has more than one inequality.

The solution is the common region that satisfies all inequalities.

Example:

x + y ≤ 5
x ≥ 0
y ≥ 0

The first inequality gives a half-plane below or on the line x + y = 5.

The inequalities x ≥ 0 and y ≥ 0 restrict the region to the first quadrant.

The common shaded region is the solution set.

Common Mistakes in Linear Inequalities

Mistake	Correct Approach
Not reversing the sign when dividing by a negative number	Reverse the inequality sign
Treating < and ≤ the same way	Use open circle for < and filled circle for ≤
Using a solid line for strict inequalities	Use dashed line for < and >
Forgetting real-life restrictions	Check if values must be natural numbers or positive
Shading the wrong side of a line	Test a point before shading

Important Formulas and Rules for Linear Inequalities

Concept	Formula or Rule
One-variable linear inequality	ax + b < 0, ax + b > 0, ax + b ≤ 0, ax + b ≥ 0
Two-variable linear inequality	ax + by < c, ax + by > c, ax + by ≤ c, ax + by ≥ c
Sign reversal rule	Multiply or divide by negative number, reverse sign
Strict inequality	Boundary not included
Slack inequality	Boundary included
Solution set	All values satisfying inequality
Boundary line	Replace inequality sign by =
Test point	Usually (0, 0), if not on boundary line

Solved Examples on Linear Inequalities

Example 1: Solve 7x + 3 < 5x + 9

Solution:

7x + 3 < 5x + 9

Subtract 5x from both sides:

2x + 3 < 9

Subtract 3 from both sides:

2x < 6

Divide by 2:

x < 3

So, the solution set is:

(-∞, 3)

Example 2: Solve 3x + 8 > 2

Solution:

3x + 8 > 2

Subtract 8 from both sides:

3x > -6

Divide by 3:

x > -2

So, the solution set is:

(-2, ∞)

Example 3: Solve -5 ≤ (5 - 3x)/2 ≤ 8

Solution:

-5 ≤ (5 - 3x)/2 ≤ 8

Multiply all parts by 2:

-10 ≤ 5 - 3x ≤ 16

Subtract 5 from all parts:

-15 ≤ -3x ≤ 11

Divide all parts by -3 and reverse signs:

5 ≥ x ≥ -11/3

Write in increasing order:

-11/3 ≤ x ≤ 5

So, the solution set is:

[-11/3, 5]

Example 4: Find the Solution Region

Find the solution region for:

x + 2y ≤ 6

Solution:

First draw the boundary line:

x + 2y = 6

Use a solid line because the inequality is ≤.

Test the point (0, 0):

0 + 2(0) ≤ 6

0 ≤ 6, which is true.

So, shade the region containing (0, 0).

This shaded region is the solution set.

Example 5: Minimum Marks Problem

A student scored 62 and 48 marks in two exams. Find the minimum marks needed in the annual exam to get an average of at least 60.

Let x be the marks in the annual exam.

(62 + 48 + x)/3 ≥ 60

110 + x ≥ 180

x ≥ 70

So, the student must score at least 70 marks.

Quick Highlights of CBSE Class 11 Maths Notes Chapter 5

Topic	Quick Revision Point
Inequality	Comparison using <, >, ≤ or ≥
Strict inequality	Uses < or >
Slack inequality	Uses ≤ or ≥
Double inequality	Contains two comparison signs
Solution	Value that satisfies the inequality
Solution set	Set of all possible solutions
One-variable inequality	Contains one variable
Two-variable inequality	Contains x and y
Number line	Used for one-variable solutions
Boundary line	Line formed by replacing inequality with =
Half-plane	One side of a boundary line
Solution region	Region satisfying the inequality
System of inequalities	Two or more inequalities solved together

Important Terms from CBSE Class 11 Maths Revision Notes Chapter 5

The terms below cover the main definitions students need while revising this chapter.

Term	Meaning
Inequality	Statement comparing two expressions
Numerical inequality	Inequality involving numbers
Literal inequality	Inequality involving variables
Strict inequality	Inequality using < or >
Slack inequality	Inequality using ≤ or ≥
Double inequality	Inequality with two comparison signs
Solution	Value that makes an inequality true
Solution set	Set of all solutions
Linear inequality	Inequality with variables of degree 1
Algebraic solution	Solving using algebraic rules
Graphical representation	Showing solution on number line or plane
Number line	Line used to show one-variable solutions
Cartesian plane	Plane used for two-variable graphs
Boundary line	Line separating two half-planes
Half-plane	One of the two regions formed by a line
Solution region	Region containing all solutions

Useful Links for Class 11 Maths

Section	Useful Links
Syllabus	CBSE Class 11 Maths Syllabus
Revision Notes	CBSE Class 11 Maths Revision Notes
Maths Notes	CBSE Class 11 Maths Revision Notes Chapter 1
Maths Notes	CBSE Class 11 Maths Revision Notes Chapter 2
NCERT Solutions	NCERT Solutions Class 11 Maths
Sample Papers	CBSE Sample Papers for Class 11 Maths
Important Questions	Important Questions Class 11 Maths
NCERT Books	NCERT Books for Class 11 Maths

Q.1

$\begin{array}{l} If α and β are different complex numbers with | β | = 1, \\ then find | \frac{β - α}{1 - \bar{α} β} | . \end{array}$

Ans

$\begin{array}{l} Let α = a + ib and β = c + id \\ \bar{α} = a - ib and | β | = \sqrt{c^{2} + d^{2}} = 1 \\ \Rightarrow c^{2} + d^{2} = 1 . .. (1) \\ β - α = c + id - a - ib \\ = c - a + i (d - b) \\ | β - α | = \sqrt{{(c - a)}^{2} + {(d - b)}^{2}} \\ = \sqrt{c^{2} + a^{2} - 2 ac + d^{2} + b^{2} - 2 db} \\ = \sqrt{c^{2} + d^{2} + a^{2} + b^{2} - 2 ac - 2 db} \\ = \sqrt{c^{2} + d^{2} + a^{2} + b^{2} - 2 ac - 2 db} \\ = \sqrt{1 + a^{2} + b^{2} - 2 ac - 2 db} . .. (2) [by (1)] \\ 1 - \bar{α} β = 1 - (\bar{a + id}) (c + id) \\ =1 - (a - id)(c + id) \\ =1 - (ac + iad - ibc + db) \\ =(1 - ac - bd) + i(ad - bc) \\ | 1 - \bar{α} β | = \sqrt{{(1 - ac - ad)}^{2} + {(ad - bc)}^{2}} \\ = \sqrt{1 + a^{2} c^{2} + b^{2} d^{2} - 2 ac + 2 acbd - 2 bd + a^{2} d^{2} + b^{2} c^{2} - 2 adbc} \\ = \sqrt{1 + a^{2} (c^{2} + d^{2}) + b^{2} (c^{2} + d^{2}) - 2 ac - 2 bd} \\ = \sqrt{1 + a^{2} + b^{2} - 2 ac - 2 bd} . .. (3) \\ Now |\frac{β - α}{1 - αβ}| = \frac{\sqrt{1 + a^{2} + b^{2} - 2 ac - 2 bd}}{\sqrt{1 + a^{2} + b^{2} - 2 ac - 2 bd}} = 1 [by (2) and (3)] \\ ∴ |\frac{β - α}{1 - \bar{α} β}| = 1 \end{array}$

Q.2

$If {(\frac{1 + i}{1 - i})}^{m} = - 1, then find the least positive integral value of m .$

Ans

$\begin{array}{l} We have, \\ (\frac{1 + i}{1 - i}) = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} \\ = \frac{{(1 + i)}^{2}}{1^{2} - i^{2}} \\ = \frac{1 + 2 i + i^{2}}{1 - (- 1)} = \frac{1 + 2 i + (- 1)}{1 + 1} \\ = \frac{2 i}{2} = i \\ ∴ {(\frac{1 + i}{1 - i})}^{m} = - 1 \\ \Rightarrow {(i)}^{m} = i^{2} \\ \Rightarrow m = 2 \\ Thus the least integral value of m is 2 . \end{array}$

Q.3

$Convert the complex number z = \frac{i - 1}{\cos \frac{π}{3} + i \sin \frac{π}{3}} in the polar form.$

Ans

$\begin{array}{l} z = \frac{i - 1}{\frac{1}{2} + \frac{\sqrt{3}}{2} i} \\ = \frac{2 (i - 1)}{1 + \sqrt{3} i} \times \frac{1 - \sqrt{3 i}}{1 - \sqrt{3 i}} \\ = \frac{2 (i + \sqrt{3} - 1 + \sqrt{3 i})}{1 + 3} = \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} i \\ Now r cosθ = \frac{\sqrt{3} - 1}{2}, r sinθ = \frac{\sqrt{3} + 1}{2} \\ On squaring and adding, we obtain \\ r^{2} = {(\frac{\sqrt{3} - 1}{2})}^{2} + {(\frac{\sqrt{3} + 1}{2})}^{2} = \frac{2 [{(\sqrt{3})}^{2} + 1]}{4} \\ = \frac{2 \times 4}{4} = 2 \\ \Rightarrow r = \sqrt{2}, which gives \cos θ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ \sin θ = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ tanθ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{1 + 1 / \sqrt{3}}{1 - 1 / \sqrt{3}} = \frac{\tan 45 ° + \tan 30 °}{1 - \tan 45 ° \tan 30 °} = \tan (45 ° + 30 °) \\ ∴ θ = \frac{π}{4} + \frac{π}{6} = \frac{5 π}{12} \\ Hence the polar form is \\ \sqrt{2} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12}) \end{array}$

Q.4

$Convert the complex number - \sqrt{3} + i in the polar form and represent it in Argand Plane .$

Ans

$\begin{array}{l} Let z = - \sqrt{3} + i = r (\cos θ + i \sin θ) \\ Comparing real part and imaginary part, we get \\ r cosθ = - \sqrt{3} and r sinθ = 1 \\ On squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- \sqrt{3})}^{2} + 1^{2} \\ \Rightarrow r^{2} = 4 \\ \Rightarrow r = 2 \\ ∴ cosθ = \frac{- \sqrt{3}}{2} and sinθ = \frac{1}{2} \\ The value of θ satistying both the equation is by, θ = \frac{5 π}{6} \\ Hence Z = 2 (\cos \frac{5 π}{6} + isin \frac{5 π}{6}) \end{array}$

Q.5

$Convert the complex number \frac{- 16}{1 + i \sqrt{3}} into polar form .$

Ans

$\begin{array}{l} The given complex number \frac{- 16}{1 + i \sqrt{3}} = \frac{- 16}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} \\ = \frac{- 16 (1 - i \sqrt{3})}{1 - {(i \sqrt{3})}^{3}} = \frac{- 16 (1 - i \sqrt{3})}{1 + 3} \\ = - 4 (1 - i \sqrt{3}) \\ = - 4 + i 4 \sqrt{3} \end{array}$

$\begin{array}{l} Let - 4 = rcosθ, 4 \sqrt{3} = rsinθ \\ By squaring and adding, we get \\ 16 + 48 = r^{2} (\cos^{2} θ + \sin^{2} θ) \\ \Rightarrow r^{2} = 64, i.e., r = 8 \\ Hence cosθ = \frac{- 1}{2}, sinθ = \frac{\sqrt{3}}{2} \\ θ = π - \frac{π}{3} = \frac{2 π}{3} \\ Thus, the required polar form is: \\ 8 (\cos \frac{2 π}{3} + isin \frac{2 π}{3}) \end{array}$

Q.6

$If x + iy = \frac{a + ib}{a - ib}, prove that x^{2} + y^{2} = 1$

Ans

$\begin{array}{l} x + iy = \frac{a + ib}{a - ib} \\ or x + iy = \frac{(a + ib) (a + ib)}{(a - ib) (a + ib)} \\ = \frac{a^{2} - b^{2} + 2 abi}{a^{2} + b^{2}} \\ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} + \frac{2 ab}{a^{2} + b^{2}} \\ ∴ x - iy = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} - \frac{2 ab}{a^{2} + b^{2}} j \\ Now, x^{2} + y^{2} = (x + iy) (x - iy) \\ = \frac{{(a^{2} - b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} + \frac{4 a^{2} b^{2}}{(a^{2} + b^{2})} \\ = \frac{{(a^{2} + b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} = 1 \end{array}$

Q.7

$\begin{array}{l} Find the modulus and the argument of the following : \\ (i) {(2 - i)}^{2} \\ (ii) \frac{1 + i}{1 - i} \end{array}$

Ans

$\begin{array}{l} (i) {(2 - i)}^{2} = 4 - 1 - 4 i = - 3 - 4 i \\ \Rightarrow | - 3 - 4 i | = \sqrt{9 + 16} = 5 \\ ∴ modulus = 5 \\ Since (- 3, - 4 i) lies in third quadrant, therefore its \\ argument θ is given by \\ θ = π + \tan^{- 1} (\frac{4}{3}) \\ (ii) \frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{1 - 1 + 2 i}{1 + 1} = i = 0 + i \\ | 0 + i | = \sqrt{1} = 1 \\ ∴ modulus = 1 \\ and argument θ is given by \\ θ = \tan^{- 1} (\frac{1}{0}) \\ \Rightarrow θ = \frac{π}{2} \end{array}$

Q.8

$\begin{array}{l} If z_{1} = 2 + i, z_{2} = 2 - 3 i and z_{3} = 4 + 5 i, evaluate \\ (i) Re (\frac{z_{1} \cdot \bar{z_{2}}}{z_{3}}) \\ (ii) Im (\frac{z_{1} \cdot \bar{z_{2}}}{z_{3}}) \end{array}$

Ans

$\begin{array}{l} (i) \frac{z_{1} \cdot \bar{z_{2}}}{z_{3}} = \frac{(2 + i) (2 - \bar{3 i})}{4 + 5 i} \\ = \frac{(2 + i) (2 + 3 i) (4 - 5 i)}{16 + 25} \\ = \frac{44}{41} + \frac{27}{41} i \\ ∴ Re (\frac{z_{1} \cdot \bar{z_{2}}}{z_{3}}) = \frac{44}{41} \\ (ii) Im (\frac{z_{1} \cdot \bar{z_{2}}}{z_{3}}) = \frac{27}{41} \end{array}$

Q.9

$Solve equation 2 z = | z | + 2 i$

Ans

$\begin{array}{l} Let z = x + iy where x, y \in R \\ | z | = \sqrt{x^{2} + y^{2}} \\ given 2 z = | z | + 2 i \\ \Rightarrow 2 (x + iy) = \sqrt{x^{2} + y^{2}} + 2 i \\ \Rightarrow 2 x = \sqrt{x^{2} + y^{2}} and 2 y = 2 \\ \Rightarrow 4 x^{2} = x^{2} + y^{2} and y = 1 \\ \Rightarrow 3 x^{2} = 1 and y = 1 \\ \Rightarrow x = \pm \frac{1}{\sqrt{3}} and y = 1 \\ But z = - \frac{1}{\sqrt{3}} + i does not satisty the given equation 2 z = | z | + 2 i \\ ∴ z = \frac{1}{\sqrt{3}} + i \end{array}$

Q.10

$Solve: \sqrt{5} x^{2} + x + \sqrt{5} = 0$

Ans

$Here the discriminant b^{2} - 4 ac of the equation is 1^{2} - 4 \times \sqrt{5} \times \sqrt{5} = 1 - 20 = - 19 ∴ the solutions are \frac{- 1 \pm \sqrt{- 19}}{2 \sqrt{5}} = \frac{- 1 \pm i \sqrt{19}}{2 \sqrt{5}}$

Q.11

$Express in the standard form a + ib : {[i^{18} + {(\frac{1}{i})}^{25}]}^{3}$

Ans

${[i^{18} + {(\frac{1}{i})}^{25}]}^{3} = {[i^{4 \times 4 + 2} + (- i)^{25}]}^{3} = {[i^{2} - {(i)}^{6 \times 4 + 1}]}^{3} = {[- 1 - i]}^{3} = - {[1 + i]}^{3} = - [1 + 3 i + 3 i^{2} + i^{3}] = - [1 + 3 i - 3 - i] = - [- 2 + 2 i] = 2 - 2 i$

Q.12 Find the multiplicative inverse of 2 + 3i.

Ans

$z = 2 + 3 i \bar{z} = 2 - 3 i and | z | = 2^{2} + 3^{2} = 13 ∴ the multiplicative inverse of 2 + 3 i is given by z^{- 1} = \frac{\bar{z}}{| z |^{2}} = \frac{2 - 3 i}{13} = \frac{2}{13} - \frac{3}{13} i$

Q.13 If 2x + i (x – y) = 5, where x and y are real numbers, find the values of x and y.

Ans

We have 2x+ i(x – y) = 5
or 2x+ i(x – y) = 5 + 0.i
Comparing the real and imaginary parts, we get
2x = 5 and x – y = 0
⇒ x = 5/2 and x = y
Thus x = y = 5/2.

Q.14

$Evaluate {(- \sqrt{- 1})}^{4 n + 3}, where n \in N$

Ans

${(- \sqrt{- 1})}^{4 n + 3} = {(- i)}^{4 n + 3} = {(- i)}^{4 n} {(- i)}^{3} = 1 \times i = i$

Q.15

$Show that: z + \bar{z} = 2 Re (z)$

Ans

$Let z = a + ib Then, \bar{z} = a - ib So, z + \bar{z} = (a + ib) + (a - ib) = 2 a = 2 Re (z)$

Q.16

$Express \frac{1}{3 - 4i} in the standard form a+ib.$

Ans

$\frac{1}{3 - 4 i} = \frac{1}{3 - 4 i} \times \frac{3 + 4 i}{3 + 4 i} = \frac{3 + 4 i}{3^{2} + 4^{2}} = \frac{1}{25} (3 + 4 i) = \frac{3}{25} + \frac{4}{25} i$

Q.17 Find the multiplicative inverse of 3 + 2i.

Ans

$Let z = 3 + 2 i Then its multiplicative inverse is given by \frac{1}{z} = \frac{1}{3 + 2 i} = \frac{1}{3 + 2 i} \times \frac{3 - 2 i}{3 - 2 i} = \frac{3 - 2 i}{3^{2} + 2^{2}} = \frac{3 - 2 i}{13} = \frac{3}{13} - \frac{2}{13} i$

Q.18

$Find the modulus of z = \frac{1 + i}{1 - i} .$

Ans

$z = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^{2}}{1 + 1} = \frac{1 + i^{2} + 2 i}{2} = \frac{1 - 1 + 2 i}{2} = i = 0 + i | z | = \sqrt{0^{2} + 1^{2}} = 1$

Q.19

$Find the argument of 1 + \sqrt{3} i .$

Ans

$\arg (z) = α = \tan^{- 1} |\frac{Im (z)}{Re (z)}| = \tan^{- 1} |\frac{\sqrt{3}}{1}| = \frac{π}{3}$

Q.20 Solve the equation 25x² + 9 = 0.

Ans

$Here, \Rightarrow 25 x^{2} + 9 = 0 \Rightarrow 5^{2} x^{2} - 9 i^{2} = 0 \Rightarrow (5 x)^{2} - (3 i)^{2} = 0 \Rightarrow (5 x - 3 i) (5 x + 3 i) = 0 \Rightarrow x = \frac{3}{5} i and x = - \frac{3}{5} i$

Q.21

$\begin{array}{l} Find the smallest positive integral value of n for which {(\frac{1 + i}{1 - i})}^{n} is purely a real number \\ with negative real part. \end{array}$

Ans

$\begin{array}{l} We have, \\ {(\frac{1 + i}{1 - i})}^{n} = {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{n} \\ = {(\frac{1 + 2 i + i^{2}}{1^{2} - i^{2}})}^{n} \\ = {(\frac{1 + 2 i - 1}{1 + 1})}^{n} \\ = {(\frac{2 i}{2})}^{n} \\ = {(i)}^{n} \\ ∵ {(\frac{1 + i}{1 - i})}^{n} is purely real number with -ve real part \\ \Rightarrow {(i)}^{n} = is purely real number with -ve real part \\ \Rightarrow {(i)}^{n} = - 1 = i^{2} \\ \Rightarrow n = 2 \end{array}$

Q.22 Write the additive inverse of –3 + 4i.

Ans

The additive inverse of –3 + 4i is 3 – 4i.

Q.23 Write the conjugate of complex number –5 + 3i.

Ans

The conjugate of complex number –5 + 3i is –5 – 3i.

Q.24 Write the multiplicative inverse of i.

Ans

$The multiplicative inverse of i = \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^{2}} = \frac{i}{- 1} = - i$

Q.25

$Evaluate (i)^{2009} + {(i)}^{2010} + {(i)}^{2011} + {(i)}^{2012} where i = \sqrt{- 1}$

Ans

$\begin{array}{l} ∵ 2009 = 502 \times 4 + 1 \\ 2010 = 502 \times 4 + 2 \\ 2011 = 502 \times 4 + 3 \\ 2012 = 502 \times 4 + 4 \\ (i)^{2009} = {(i^{4})}^{502} \times i^{1} = 1 \times i = i \\ (i)^{2010} = {(i^{4})}^{502} \times i^{2} = 1 \times - 1 = - 1 [\begin{array}{l} ∵ i^{2} = - 1 \\ i^{3} = - i \\ i^{4} = 1 \end{array}] \\ {(i)}^{2011} = {(i^{4})}^{502} \times i^{3} = 1 \times - i = - i \\ (i)^{2012} = {(i^{4})}^{502} \times i^{4} = 1 \times 1 = 1 \\ ∴ {(i)}^{2009} + {(i)}^{2010} + {(i)}^{2011} + {(i)}^{2012} = i - 1 + - i + 1 = 0 \end{array}$

Q.26

$Evaluate {(\frac{1 + i}{1 - i})}^{2011}, where i = \sqrt{- 1}$

Ans

$\begin{array}{l} We have, \\ {(\frac{1 + i}{1 - i})}^{2011} = {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{2011} \\ = {(\frac{1 + 2 i + i^{2}}{1^{2} - i^{2}})}^{2011} \\ = {(\frac{1 + 2 i - 1}{1 + 1})}^{2011} \\ = {(\frac{2 i}{2})}^{2011} \\ = {(i)}^{2011} \\ = {(i^{4})}^{502} \times i^{3} [∵ i^{4} = 1, i^{3} = - i] \\ = 1 \times - i = - i \end{array}$

Q.27

$\begin{array}{l} Find the smallest positive integral value of n for which {(\frac{1 + i}{1 - i})}^{n} is purely an imaginary \\ number with negative real part. \end{array}$

Ans

$\begin{array}{l} We have, \\ {(\frac{1 + i}{1 - i})}^{n} = {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{n} \\ = {(\frac{1 + 2 i + i^{2}}{1^{2} - i^{2}})}^{n} \\ = {(\frac{1 + 2 i - 1}{1 + 1})}^{n} \\ = {(\frac{2 i}{2})}^{n} \\ = {(i)}^{n} \\ ∵ {(\frac{1 + i}{1 - i})}^{n} is purely an imaginary number with -ve real part \\ \Rightarrow {(i)}^{n} = is purely an imaginary number with -ve real part \\ \Rightarrow {(i)}^{n} = - i \\ \Rightarrow n = 3 \end{array}$

Q.28 Find the square root of complex number 5 + 12i.

Ans

$\begin{array}{l} We have \\ 5 + 12 i = 9 - 4 + 12 i \\ = {(3)}^{2} - {(2)}^{2} + 2 \times 6 i \\ = {(3)}^{2} + i^{2} {(2)}^{2} + 2 \times (3) \times (2 i) \\ = {(3)}^{2} + {(2 i)}^{2} + 2 (3) (2 i) \\ = {(3 + 2 i)}^{2} \\ \sqrt{5 + 12 i} = \sqrt{{(3 + 2 i)}^{2}} \\ = \pm (3 + 2 i) \end{array}$

Q.29

$Find the value of | z | if {iz}^{3} + z^{2} - z + i = 0$

Ans

$\begin{array}{l} We have \\ {iz}^{3} + z^{2} - z + i = 0 \\ On dividing both sides by i \\ z^{3} + \frac{1}{i} z^{2} - \frac{1}{i} z + 1 = 0 \\ \Rightarrow z^{3} - {iz}^{2} + iz - i^{2} = 0 [∵ \frac{1}{i} = - i, - i^{2} = 1] \\ \Rightarrow z^{2} (z - i) + i (z - i) = 0 \\ \Rightarrow (z - i) (z^{2} + i) = 0 \\ \Rightarrow (z - i) = 0 or z^{2} + i = 0 \\ \Rightarrow z = i \\ \Rightarrow | z | = 1 \\ Also z^{2} = - i \\ \Rightarrow |z^{2}| = | - i | \\ \Rightarrow | z |^{2} = 1 \\ \Rightarrow | z | = 1 \\ In both cases, value of | z | = 1 \end{array}$

Q.30

$Put the complex number (\frac{1 + i}{1 - i}) in polar form.$

Ans

$\begin{array}{l} We have, \\ z = (\frac{1 + i}{1 - i}) = (\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}) \\ = (\frac{1 + 2 i + i^{2}}{1^{2} - i^{2}}) \\ = (\frac{1 + 2 i - 1}{1 + 1}) \\ = (\frac{2 i}{2}) \\ = (i) = 0 + i \\ \Rightarrow z = 0 + i \\ r = | z | = 1 \\ ∵ tanθ = \frac{y}{X} \\ \Rightarrow tanθ = \frac{1}{0} = \infty \\ \Rightarrow tanθ = \tan \frac{π}{2} \\ \Rightarrow θ = \frac{π}{2} \\ In polar form, z = r (\cos θ + i \sin θ) \\ \Rightarrow z = 1 (\cos \frac{π}{2} + i \sin \frac{π}{2}) \end{array}$

Q.31

$Calculate the value of \sqrt{8 + 6 i} + \sqrt{8 - 6 i}$

Ans

$\begin{array}{l} We have, \\ 8 + 6 i = 9 - 1 + 6 i \\ = {(3)}^{2} - {(1)}^{2} + 2 \times 3 i \\ = {(3)}^{2} + i^{2} + 2 \times (3) \times (i) \\ = {(3)}^{2} + {(i)}^{2} + 2 (3) (i) \\ = {(3 + i)}^{2} \\ \sqrt{8 + 6 i} = \sqrt{{(3 + i)}^{2}} \\ \sqrt{8 + 6 i} = (3 + i) \\ Also, \\ 8 - 6i = 9 - 1 - 6i \\ = {(3)}^{2} - {(1)}^{2} - 2 \times 3 i \\ = {(3)}^{2} + i^{2} - 2 \times (3) \times (i) \\ = {(3)}^{2} + {(i)}^{2} - 2 (3) (i) \\ = {(3 - i)}^{2} \\ \sqrt{8 - 6 i} = \sqrt{{(3 - i)}^{2}} \\ \sqrt{8 - 6 i} = (3 - i) \\ \Rightarrow \sqrt{8 + 6 i} + \sqrt{8 - 6 i} = 3 + i + 3 - i = 6 \end{array}$

Q.32

$Find the real value of θ such that \frac{3 + 2 isinθ}{1 - 2 isinθ} is purely imaginary.$

Ans

$\begin{array}{l} First, we shall put \frac{3 + 2 i sinθ}{1 - 2 i sinθ} in A + iB form \\ We have, \\ \frac{3 + 2 i sinθ}{1 - 2 i sinθ} = \frac{(3 + 2 i sinθ)}{(1 - 2 i sinθ)} \times \frac{(1 + 2 i sinθ)}{(1 + 2 i sinθ)} \\ = \frac{3 + 2 i sinθ + 6 i sinθ + 4 i^{2} \sin^{2} θ}{1 - 4 i^{2} \sin^{2} θ} \\ = \frac{3 + 8 i sinθ - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} \\ = \frac{3 - 4 \sin^{2} θ + i 8 sinθ}{1 + 4 \sin^{2} θ} \\ = \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} + i \frac{8 sinθ}{1 + 4 \sin^{2} θ} \\ For complex number to be purely imaginary Real part = 0 \\ \Rightarrow \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} = 0 \\ \Rightarrow 3 - 4 \sin^{2} θ = 0 \\ \Rightarrow 4 \sin^{2} θ = 3 \\ \Rightarrow \sin^{2} θ = \frac{3}{4} \\ \Rightarrow sinθ = \pm \frac{\sqrt{3}}{2} \\ case - I \\ sinθ = \frac{\sqrt{3}}{2} \\ \Rightarrow sinθ = \sin \frac{π}{3} \\ \Rightarrow θ = \frac{π}{3} \\ case - I I \\ sinθ = - \frac{\sqrt{3}}{2} \\ \Rightarrow sinθ = - \sin \frac{π}{3} \\ \Rightarrow sinθ = \sin (π + \frac{π}{3}) \\ \Rightarrow sinθ = \sin \frac{4 π}{3} \\ \Rightarrow θ = \frac{4 π}{3} \end{array}$

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FAQs (Frequently Asked Questions)

A linear inequality is an inequality in which the highest power of the variable is 1. Examples include 2x + 3 < 7 and x + y ≤ 5.

A strict inequality uses < or > and does not include the boundary value. A slack inequality uses ≤ or ≥ and includes the boundary value.

The inequality sign is reversed when both sides are multiplied or divided by a negative number. For example, -2x < 8 becomes x > -4.

A solution set is the set of all values that make an inequality true. For x < 5, the solution set is all real numbers less than 5.

First draw the boundary line by replacing the inequality sign with “=”. Then test a point. If the point satisfies the inequality, shade that side. Otherwise, shade the other side.

CBSE Class 11 Maths Revision Notes Chapter 5 Linear Inequalities

Key Takeaways

Class 11 Maths Chapter 5 Notes for Linear Inequalities Revision

What is an Inequality?

Types of Inequalities

Numerical and Literal Inequalities

Linear Inequalities in One Variable

Linear Inequalities in Two Variables

Quadratic Inequalities Are Not Linear

Solution and Solution Set of an Inequality

Rules for Solving Linear Inequalities

Strict Inequality and Slack Inequality on Number Line

Algebraic Solution of Linear Inequalities in One Variable

Example 1: Solve 5x - 3 < 7

Example 2: Solve 4x + 3 < 6x + 7

Solving Double Inequalities

System of Linear Inequalities in One Variable

Word Problems on Linear Inequalities

Example: Marks Problem

Linear Inequalities in Two Variables and Cartesian Plane

Boundary Line in Linear Inequalities

Half-Plane and Solution Region

Steps for Graphical Representation of Linear Inequalities in Two Variables

Graphical Solution of a System of Linear Inequalities

Common Mistakes in Linear Inequalities

Important Formulas and Rules for Linear Inequalities

Solved Examples on Linear Inequalities

Example 1: Solve 7x + 3 < 5x + 9

Example 2: Solve 3x + 8 > 2

Example 3: Solve -5 ≤ (5 - 3x)/2 ≤ 8

Example 4: Find the Solution Region

Example 5: Minimum Marks Problem

Quick Highlights of CBSE Class 11 Maths Notes Chapter 5

Important Terms from CBSE Class 11 Maths Revision Notes Chapter 5

Useful Links for Class 11 Maths

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FAQs (Frequently Asked Questions)

1. What is a linear inequality in Class 11 Maths?

2. What is the difference between strict and slack inequality?

3. When do we reverse the inequality sign?

4. What is a solution set in linear inequalities?

5. How do you graph a linear inequality in two variables?

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