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CBSE Online Class 11 Maths NCERT Solutions-Linear Inequality

Online Class 11 Maths NCERT Solutions-Linear Inequality

Online Class 11 Maths NCERT Solutions-Linear Inequality

Class 11 Maths NCERT Solutions: Takshila Learning offers NCERT solutions for Class 11 Maths. NCERT Solutions being an important part of learning for those who want to score well in their Class 11 Maths exam. The series of chapters includes solutions to every question and help students to prepare well and score good marks. Now get easy access to CBSE Class 11 Online classes with NCERT solutions of all the chapters in a chapter wise manner.

For those students who have trouble finding solutions to their exercise can refer Class 11 Maths online classes as the content and solutions are prepared by expert teachers. For a better understanding of concepts, 2D-3D animated videos, as well as recorded lectures, are available. For practice, we provide question bank with solutions, objective as well as subjective, study material will help the students to achieve maximum marks in their exam. Regularly we share notes in the form of articles and blogs, Here, we discuss ‘Linear inequalities’.

 

LINEAR INEQUALITIES

A statement involving the symbols ‘>’, ‘<’, ‘ ≥’, ‘≤’ is called an inequality. For example 5 > 4,  x ≤ 4,  x + y ≥ 7.

 

A solution of inequality:

The value(s) of the variable(s) which makes the inequality a true statement is called its solutions. The set of all solutions of an inequality is called the solution set of the inequality.

 

Rules to solve an inequality:

  • Add(or subtract) the same quantity to (from) both sides without changing the sign of inequality.

 

  • Multiply (or divide) both sides by the same positive quantity without changing the sign of inequality. However, if both sides of the inequality are multiplied (or divided)by the same negative quantity the sign of inequality is reversed, i.e., ‘>’ change into ‘<’ and vice versa.

Result

If a is any positive real number, i.e., a > 0, then

(i) | x| < a ⇔– a <x <a

| x| ≤ a ⇔– a ≤ x ≤ a

 

(ii)  | x| > a ⇔x < – a or x >a

| x| ≥ a ⇔x ≤ – a or x ≥ a

 

 

Example 1: Solve the inequality  3x– 5 <x + 7, when

(i)  x is a natural number (ii) x is a whole number

(iii)  x is an integer (iv) x is a real number

 

Solution: We have 3x – 5 <x + 7

⇒3x <x + 12                                           (Add 5 to both sides)

⇒2x < 12                                                 (Subtractx from both sides)

x< 6 (Dividing by 2 on both sides)

 

(i) Solution set is {1, 2, 3, 4, 5}

(ii) Solution set is {0, 1, 2, 3, 4, 5}

(iii) Solution set is {….– 3, – 2, –1, 0, 1, 2, 3, 4, 5}

(iv) Solution set is {x :x R and x < 6}, i.e., any real number less than 6.

Example 2: The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to get some profit?

 

Solution: We have profit = Revenue – Cost

= (60x + 2000) – (20x + 4000)

= 40x – 2000

To earn some profit, 40x – 2000 > 0

x> 50

Hence, the manufacturer must sell more than 50 items to get some profit.

Example 3: Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 38.

 

Solution: Let x be the smaller of the two consecutive odd natural number so that the other one is x +2. Then, according to given condition we have

X > 10 and

x + ( x+ 2) < 38

Solving  we get

2x + 2 < 38

2x < 36

i.e., x < 18

So , we get

10 < x < 18

Since x is an odd number,  x can take the values 11, 13, 15, and 17. So, the required possible pairs will be

(11, 13), (13, 15), (15, 17), (17, 19)

Example 4: The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then

(A) breadth > 20 cm (B) length < 20 cm

(C) breadth  ≥ 20 cm (D) length ≤ 20 cm

 

Solution: (C) is the correct choice.

Let x is breadth then length is 3x and the perimeter is 2(3x+x).

Acc to given condition

2 (3x + x) ≥ 160 ⇒x ≥ 20.

Example 5: Solve │x+2│≤ 9

Solution:

-9≤ x+2 ≤ 9 (Since | x| ≤ a ⇔– a ≤ x ≤ a)

-11≤  x ≤ 7

Example 6: Solve │3-4x│≥9

Solution:

3 – 4x ≤ – 9 or3 – 4x ≥ 9 (Since | x| ≥ a ⇔  x≤ – a or x ≥ a)

– 4x ≤ – 12or – 4x ≥ 6

X ≥ 3     or      x ≤ -3/2

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September 5, 2017

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