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Solutions ⇒ Class 11th ⇒ Mathematics ⇒ 6. Linear Inequalities

Solutions 6. Linear Inequalities - Exercise 6.1 | Class 11 Mathematics - Toppers Study

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6. Linear Inequalities

| Exercise 6.1 |

Solutions 6. Linear Inequalities - Exercise 6.1 | Class 11 Mathematics - Toppers Study


Exercise 6.1


Q1. Solve 24x < 100, when

(i) x is a natural number

(ii) x is an integer 

Solution: 

The given inequality is 24x < 100. 

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than

∴ when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than

Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.  

Q2. Solve –12x > 30, when

(i) x is a natural number

(ii) x is an integer 

Solution: 

The given inequality is –12x > 30. 

(i) There is no natural number less than

Thus, when x is an integer, the solutions of the given inequality are …, –5, –4, –3.

Hence, in this case, the solution set is {…, –5, –4, –3}. 

Q3. Solve 5x– 3 < 7, when

(i) x is an integer

(ii) x is a real number 

Soluution: 

The given inequality is 5x– 3 < 7. 

Q5. Solve the given inequality for real x: 4x + 3 < 5x + 7

Solution :

4x + 3 < 5x + 7

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–4, ∞). 

Q23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11. 

Solution: 

Let x be the smaller of the two consecutive odd positive integers.

Then, the other integer is x + 2.

Since both the integers are smaller than 10, x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8 … (1)

Also, the sum of the two integers is more than 11.

∴x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

⇒ x > 9/2

⇒ x > 4.5   ....... (2)

From (1) and (2), we get .

Since x is an odd number, x can take the values, 5 and 7.

Therefore, the required possible pairs are (5, 7) and (7, 9). 

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Mathematics Class - 11th

NCERT Maths book for CBSE Students.

books

Study Materials List:

Solutions ⇒ Class 11th ⇒ Mathematics
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability

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