Linear Inequalities Class 11 Notes CBSE Maths Chapter 5 (Free PDF Download)

Section–A (1 Mark Questions)

1. Solve for real $x: \frac{\left | x-2 \right |}{x-2}\geq 0$   

Ans. Since $\frac{|x-2|}{x-2} \geq 0$, for $x-2 \geq 0$, and

$$ \begin{aligned} & x-2 \neq 0 . \\ & \Rightarrow x \in(2, \infty) \end{aligned} $$

2. Solve for real x: -3x+17<-13.

Ans. $$ \text {} \begin{aligned} & -3 x+17<-13 \\ & \Rightarrow-3 x<-30 \\ & \Rightarrow 3 x>30 \\ & \Rightarrow x>10 \\ & \Rightarrow x \in(10, \infty) \end{aligned} $$

3. Fill in the blanks with appropriate sign of inequality. If $x\geq -3$, then x+5……2. 

Ans. $x \geq-3$

$$ \begin{aligned} & \Rightarrow x+5 \geq-3+5 \\ & \Rightarrow x+5 \geq 2 . \end{aligned} $$

4. Fill in the blanks with appropriate sign of inequality. If p>0 and q<0, the (p+q).....p. 

Ans. Since, $p$ is positive and $q$ is negative, Therefore, $(p+q)$ is always smaller than $p$.

5. Fill in the blanks with appropriate sign of inequality. If $\frac{2}{x+2}>0$ , then x…..-2.

Ans. If $\frac{2}{x+2}>0$

$$\Rightarrow x>-2$$

Section–B (2 Marks Questions)

6. The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then find the possible breadth of rectangle.

Ans. Let $x$ be the breadth of rectangle Then, length of rectangle will be $3 x$ Therefore,

$$ \begin{aligned} \text { perimeter } & =2(\text { Length }+ \text { Breadth }) \\ & =2(3 x+x) \end{aligned} $$

Given, minimum perimeter $=160 \mathrm{~cm}$

$$ \begin{aligned} & \Rightarrow 2(3 x+x) \geq 160 \\ & \Rightarrow x \geq 20 \end{aligned} $$

Hence, breadth $\geq 20 \mathrm{~cm}$.

7. $C(x)=600+\frac{5}{2}x$ and $R(x)=4x$ are respectively, the cost and revenue function of a cassette company where x is the number of cassettes produced. Find the number of cassettes to be produced and sold so as to realize a profit. 

Ans. Profit $=P(x)=R(x)-C(x)$

Given: $P(x)>0$

$$ \begin{aligned} & \Rightarrow 4 x-600-\frac{5}{2} x>0 \\ & \Rightarrow \frac{3}{2} x-600>0 \\ & \Rightarrow \frac{3}{2} x-600>0 \\ & \Rightarrow \frac{3}{2} x>600 \Rightarrow x>400 \end{aligned} $$

The number of cassettes to be sold to earn some profit is greater than 400 .

8. Solve for x : $\frac{\left [ 2(x-1) \right ]}{5}\leq \frac{\left [ 3(2+x) \right ]}{7}$ .

Ans. Given: $\frac{[2(x-1)]}{5} \leq \frac{[3(2+x)]}{7}$

$$ \begin{aligned} & \Rightarrow \frac{(2 x-2)}{5} \leq \frac{(6+3 x)}{7} \\ & \Rightarrow 7(2 x-2) \leq 5(6+3 x) \\ & \Rightarrow 14 x-14 \leq 30+15 x \\ & \Rightarrow 14 x-15 x \leq 30+14 \\ & \Rightarrow-x \leq 44 \Rightarrow x \geq-44 \end{aligned} $$

$\therefore$ The solution of the given inequality is $[-44, \infty)$.

9. Solve the inequality: $\frac{x}{5}<\frac{(3 x-2)}{4}-\frac{(5 x-3)}{5}$.

Ans. $\frac{x}{5}<\frac{(3 x-2)}{4}-\frac{(5 x-3)}{5}$

$\Rightarrow \frac{x}{5}<\frac{(5(3 x-2)-4(5 x-3))}{4 \times 5}$

$\Rightarrow \frac{x}{5}<\frac{(15 x-10-20 x+12)}{20}$

$\Rightarrow \frac{x}{5}<\frac{(2-5 x)}{20}$

$\Rightarrow 4 x<2-5 x$

$\Rightarrow 9 x<2 \Rightarrow x<\frac{2}{9}$

$\therefore$ The solution of the given inequality is

$$\left(-\infty, \frac{2}{9}\right) \text {. }$$

10. The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.

Ans. Let $x$ be the marks obtained by student in the annual examination. Then,

$$\frac{62+48+x}{3} \geq 60 \text { or } 110+x \geq 180$$ or $x \geq 70$

Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

11. A company manufactures cassettes, its cost and revenue functions are C(x)=26000+30x and R(x)=43x, respectively, where x is the number of cassettes produced and sold in week. How many cassettes must sold by the company to realize some profit.

Ans. Given that: Cost function, $C(x)=26000+30 x$

And revenue function $R(x)=43 x$

Now for profit

$$ \begin{aligned} & P(x)=R(x)-C(x)>0 \\ & \Rightarrow R(x)>C(x) \\ & \Rightarrow 43 x>26000+30 x \\ & \Rightarrow 43 x-30 x>26000 \\ & \Rightarrow 13 x>26000 \Rightarrow x>2000 \end{aligned} $$

Hence, the number of cassettes to be manufactured for some profit must be more than 2000 .

12. Solve the following system of equations in R 2x-7>5-x, 11-5x$\leq$1.

Ans. Given, 

$\Rightarrow 2x+x>5+7$

$\Rightarrow 3x>12\Rightarrow x>4$

$\therefore x\epsilon (4,\infty )$.....(1)

And 11-5x$\leq$ 1

$\Rightarrow -5x\leq 1-11$

$\Rightarrow -5x\leq - 10\Rightarrow x\geq 2$

$\therefore x\epsilon [2,\infty )$

From (1) and (2) we get $x\epsilon (4,\infty )$ 

PDF Summary - Class 11 Maths Linear Inequalities Notes (Chapter 5)

An inequality is a relationship that exists between two values that aren't equal.

For example, \[\text{x  >  9}\]. Here there is a relation between \[\text{x}\] & \[\text{9}\].

Any two algebraic expressions or real numbers related by symbol ‘\[\text{ < }\]’, ‘\[\text{ > }\]’, ‘\[\le \]’ or ‘\[\ge \]’ form inequality.

Inequalities can be used to solve problems in science, mathematics, statistics, economics, optimization problems, psychology, and other fields.

Example of Inequality in Daily Life: 

Rina and Samira have \[\text{Rs}\text{. 5,000}\] & wants to buy t-shirts and shoes for trekking. The cost price of t-shirt and shoes is \[\text{Rs}\text{. 250}\] and \[\text{Rs}\text{. 550}\] respectively. We can write the above statement mathematically using inequalities, as follows; 

Let the number of t-shirt they can buy be \[\text{x}\] & number of shoes be \[\text{y}\]. 

Then, the total amount spent by them is

\[\text{250x + 550y}\le 50\text{00}\]

Here, the total amount is upto  \[\text{Rs}\text{. 5,000}\].

The above given statement consists of two statements as,

\[\text{250x + 550y} < 50\text{00}\] which is an inequality and

\[\text{250x + 550y}=50\text{00}\] is an equation

Notations:

• The notation \[\text{a  <  b}\] means, \[\text{a}\] is less than \[\text{b}\].

• The notation \[\text{a  >  b}\] means, \[\text{a}\] is greater than \[\text{b}\].

• The notation \[\text{a }\ne \text{ b}\] means, \[\text{a}\] is not equal to \[\text{b}\].

• The notation \[\text{a }\le \text{ b}\] means, \[\text{a}\] is less than or equal to \[\text{b}\].

• The notation \[\text{a }\ge \text{ b}\] means, \[\text{a}\] is greater than or equal to \[\text{b}\].

Types of Inequalities: 

• Numerical Inequalities: 

Relationship between numbers. 

For example, \[\text{8  <  19}\]

• Literal or Variable Inequalities: 

Relationship between variables or between a variable and number. 

Example, \[\text{x  <  19}\]

• Double Inequalities: 

Relationship from two side. 

For example, \[\text{19  <  x  <  25}\]

• Strict Inequalities: 

An inequality that employs symbols  \[\text{ < }\] or \[\text{ > }\]

The symbols \[\le \] and  \[\ge \] are not used.

For example, \[\text{y  <  4}\]; \[\text{1 <  4}\]

• Slack Inequalities. 

An inequality that employs symbols  \[\le \] or \[\ge \].

Example, \[\text{y }\le \text{ 7}\]

• Linear Inequalities in One Variable: 

A one-variable inequality involving a linear function.

Example, \[\text{y  <  4}\]

• Linear Inequalities in Two Variables: 

An inequality involving a two-variable linear function.

Example, \[\text{5x+ 7y  >  4}\]

• Quadratic Inequalities: 

An inequality which employs a quadratic function.

Example, \[\text{7}{{\text{x}}^{2}}+3x\le \text{ 4}\] 

Solution for linear inequality in one variable:

Solution & Solution Set:

• Solution: 

Values of \[x\], which make inequality true statement. 

Example - \[5\] is a solution for \[x < 10\]

• Solution Set: 

The set of values of x is known as its solution set.

Example - \[\left\{ 1,2,3,4 \right\}\] is solution set for \[x < 5\] where \[x\] is natural Number.

Rules of Inequality:

• Both sides of an inequality can have equal numbers added to (or taken from) them without changing the sign of the inequality. 

For example, \[x < 5\] is same as \[x+2 < 5+2\]

• Both sides of an inequality can be multiplied (or divided) by the same positive number without affecting the sign of inequality.

For example, \[x-y < 3\] is same as \[\left( x-y \right)\times 2 < 3\times 2\]

• However, the sign of inequality is flipped or reversed when both sides are multiplied or divided by a negative value.

For example, \[x+y < 6\] is same as \[\left( x+y \right)\times \left( -2 \right) > 6\times \left( -2 \right)\]

Question: 

Solve \[30x < 160\] when

1. \[x\] is a natural number,

2. \[x\] is an integer,

3. \[x\] is real number

Ans:

On dividing the inequality by \[30\]according to rule \[2\], we get

\[\dfrac{30x}{30} < \dfrac{160}{30}\] or

\[x < \dfrac{16}{3}\]

Case \[1\]: If \[x\] is a natural number, then the solution set is  \[\left\{ 1,2,3,4,5 \right\}\].

Case \[2\]: If \[x\] is an integer, then the solution set is \[\left\{ .....-4,-3,-2,-1,0,1,2,3,4,5 \right\}\].

Case \[3\]: If \[x\] is a real number, then the solution set is  \[\left( -\infty ,\dfrac{16}{3} \right)\]. 

By representing the case \[3\] solution on a number line, we get

Question: 

Solve \[\mathbf{7x+2\le 5x+8}\]. Show the graph of the solutions on the number line.

Ans:

By subtracting \[2\] from both side, we get \[7x\le 5x+6\]

By subtracting \[5x\] from both side, we get \[2x\le 6\]

On dividing \[2\] both side, we get \[x\le 3\]

We can represent this in the Number line below.

Graphical Solution of Linear Inequalities in \[2\] variables:

• The Cartesian plane is divided into two equal sections by a line.

• Each component is referred to as a half plane. 

• A non-vertical line divides the plane into lower and upper half planes, while a vertical line divides it into left and right half planes.

• In the Cartesian plane, a point will either lie on a line or in one of the half planes. 

• The solution zone is the area that contains all of the solutions to an inequality.

• To find the half plane represented by an inequality, simply choose any point \[\left( \text{a, b} \right)\] (not online) and see if it meets the inequality. 

• If it does, the inequality represents the half plane and shades the region that contains the point; if it does not, the inequality represents the half plane that does not contain the point. 

• For convenience, the point \[\left( 0,0 \right)\] is preferred.

Example: \[\text{x + 2y} > 9\]

Ans:

Steps for find solution region for a linear inequality in \[2\] variables 

1. Substitute an equal sign for the inequality sign and plot the graph. Plot a graph for \[\text{x + 2y}=9\]. 

Red line represent \[\text{x + 2y}=9\].

2. Take any point on the graph. Here we took \[\left( 8,1 \right)\] and check if satisfies the linear inequality. 

In this case \[\text{x + 2y} > 9\]. 

If yes, then the region where this assumed point lies is the solution region.

3. When solving a Slack inequality \[\left( \ge \text{or}\le  \right)\] use solid line, since the points on the line are part of the solution set.

4. Use dotted line in the case of Strict inequality \[\left(  > \text{or }  < \right)\], since points on the line are not included in solution set. 

In the case of several linear inequalities, the solution region is the area that is shared by all of the inequalities.

Question: 

Solve the following system of inequalities graphically \[\text{5x + 4y}\le 40\], \[\text{x} > 2\] and \[\text{y}\ge 3\]. 

Ans:

Step \[1\]: Draw lines for \[\text{5x + 4y}=40\], \[\text{x}=2\] and \[\text{y}=3\].

Step \[2\]: For each of these linear inequalities, find the solution zone.

Step \[3\]: Locate a common area. The solution region is a common region. 

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Revision Notes for Class 11 Maths Chapter 5 Linear Inequalities

Mathematically, a linear inequality refers to inequality that includes a linear function. That being said, an inequality is called linear when each variable takes place in 1st degree only and there is no term including the product of the variables. A linear inequality consists of one of the symbols of inequality ie: [> is greater than], [< is less than] or [≤ is less than or equal to]. There are certain types of linear Inequalities which are as follows:-

1. Linear inequalities in One Variable

An inequality which takes into account a linear function in one variable is called a linear inequality in one variable. For E.g. y < 5;

2. Linear inequalities in two variables

An inequality which typically includes a linear function in two variables is called a linear inequality in two variables. For E.g. 2x + 3y < 5

3. Quadratic inequalities

An inequality which takes into account a quadratic function is called a linear inequality in quadratic. E.g. y3 + 3y ≤ 5

Note: You will learn more important elements about linear inequalities in Class 11 revision notes Maths Ch 5 designed by Vedantu experts. You will also learn Solutions and Graphical Solution of Linear Inequalities in your Class 11 Maths revision notes Chapter 5.

Solution of System of Linear Inequalities In Maths Class 11 Linear Inequalities Notes

Both sides of an inequality can be added to (or subtracted from) the equal numbers. The same positive number can be multiplied (or divided) on both sides of an inequality. However, when both sides are divided or multiplied by a negative number, then we see a reverse in inequality.

Solution of an Inequality

The value of x, which prompts an inequality to be a true statement, is known as solutions of the inequality.

In order to represent x > a (or x < a) on a number line, we need to put a circle on the number and outline it dark to the right (or left) of the number a.

To represent x ≥ a or x ≤ a on a number line, we have to put a dark circle on the number and outline it dark the line to the right (or left) of the number x.

If an inequality contains the symbol of ≤ or ≥, then the points on the line are also involved in the solutions of the inequality. Moreover, the graph of the inequality lies right (above) or left (below) the graph of the equality which is depicted by a dark line that satiates an arbitrary point in that part.

If an inequality consists of a < or > symbol, then the points on the line are not involved in the solutions of the inequality. In this case, the graph of the inequality lies to the left (below) or right (above) of the graph of the simultaneous equality denoted by a dotted line that satiates an arbitrary point in that part.

In order to eliminate the denominator when we are not known about the sign of the value of the denominator may be +ve or -ve:, then we have to multiply by the square of the denominator. Remember that the Square of the denominator is always positive. More so, the sign of inequality does not change, when we multiply an equation by a positive number.

To gain a brief information of all the topics covered in Chapter 5 Linear Inequalities, download Class 11 Revision Notes For Linear Inequalities of Chapter 5 now through the link provided.