# CBSE Class 11-Mathematics: Complex Numbers: Inequalities Exercise 1 (For CBSE, ICSE, IAS, NET, NRA 2022)

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**Question 1**:

Solve , when is a natural number is an integer

**Question 2**:

Solve , when

(i) is a natural number

(ii) is an integer

**Question 3**:

Solve , when

(i) is an integer

(ii) is a real number

**Question 4**:

Solve , when

(i) is an integer

(ii) is a real number

**Question 5**:

Solve the given inequality for real

**Question 6**:

Solve the given inequality for real

**Question 7**:

Solve the given inequality for real

**Question 8**

Solve the given inequality for real

**Question 9**

Solve the given inequality for real :

**Question 10**

Solve the given inequality for real :

**Question 11**

Solve the given inequality for real :

**Question 12**:

Solve the given inequality for real :

**Question 13**:

Solve the given inequality for real :

**Question 14**:

Solve the given inequality for real :

**Question 15**:

Solve the given inequality for real :

**Question 16**:

Solve the given inequality for real :

**Question 17**:

Solve the given inequality and show the graph of the solution on number line:

**Question 18**:

Solve the given inequality and show the graph of the solution on number line:

**Question 19**:

Solve the given inequality and show the graph of the solution on number line:

**Question 20**

Solve the given inequality and show the graph of the solution on number line:

**Question 21**:

Ravi obtained and marks in first two-unit test. Find the minimum marks he should get in the third test to have an average of at least marks.

**Question 22**:

To receive Grade βAβ in a course, one must obtain an average of 90 marks or more in five examinations (each of marks) . If SunitaΥs marks in first four examinations are

and , find minimum marks that Sunita must obtain in fifth examination to get grade βAβ in the course.

**Question 23**:

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

**Question 24**

Find all pairs of consecutive even positive integers, both of which are larger than such that their sum is less than

**Question 25**:

The longest side of a triangle is times the shortest side and the third side is shorter than the longest side. If the perimeter of the triangle is at least , find the minimum length of the shortest side

**Question 26**

A man wants to cut three lengths from a single piece of board of length . The second length is to be longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least longer than the second? [Hint: If x is the length of the shortest board, then , and are the lengths of the second and third piece, respectively. Thus, and

**Solution 1**:

The given inequality is .

Dividing both the side by the positive number

(i) It is evident that and are the only natural numbers less than

Hence, in this case, the solution set is .

(ii) The integers less than are .

Solution .

**Solution 2**:

The given inequality is

Dividing both the sides by

Also, as we are dividing by negative integer, the inequality sign will reverse

(i) There is no natural number less than

Thus, there is no solution of the given inequality.

(ii) The integers less than are β¦ , .

Solution

**Solution 3**:

The given inequality is

(adding on both the sides)

(Dividing both the sides by )

(i) The integers less than 2 are β¦ ,

Hence, in this case, the solution set is .

(ii) When x is a real number, the solutions of the given inequality are given by x < 2,

Solution .

**Solution 4**:

The given inequality is

(i) The integers greater than are

Hence, in this case, the solution set is .

(ii) When x is a real number, the solutions of the given inequality are all the real

Numbers, which are greater than .

Solution .

**Solution 5**:

Solution .

**Solution 6**:

(Subtracting on both sides)

(Dividing on both the sides and reversing the inequality sign)

Solution

**Solution 7**:

Solution

**Solution 8**:

(Adding on both sides)

(Subtracting on both sides)

(Dividing on both the sides and reversing the inequality sign)

Solution

**Solution 9**:

Solution .

**Solution 10**:

Solution .

**Solution 11**:

Solution .

**Solution**:

Solution .

**Solution 13**:

Solution .

**Solution 14**

Solution .

**Solution 15**:

Solution .

**Solution 16**:

Solution .

**Solution 17**:

The graphical representation of the solutions of the given inequality is as follows.

**Solution 18**:

The graphical representation of the solutions of the given inequality is as follows.

**Solution 19**

The graphical representation of the solutions of the given inequality is as follows.

**Solution 20**:

**Solution 21**:

Let be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least marks,

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

**Solution 22**:

Let be the marks obtained by Sunita in the fifth examination.

In order to receive grade βAβ in the course, she must obtain an average of marks or more in five examinations.

Therefore,

Thus, Sunita must obtain greater than or equal to marks in the fifth examination.

**Solution 23**:

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

Then, the other integer is .

Since both, the integers are smaller than ,

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

From (i) and (ii) , we obtain.

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

Thus, the required possible pairs are and .

**Solution 24**:

Let y be the smaller of the two consecutive even positive integers. Then, the other integer

is .

Since both, the integers are larger than ,

Also, the sum of the two integers is less than .

But we know that

So

Since is an even number, can take the values, , and .

Thus, the required possible pairs are , and

**Solution 25**:

Let the length of the shortest side of the triangle be .

Then, length of the longest side

Length of the third side

Since the perimeter of the triangle is at least ,

Thus, the minimum length of the shortest side is .

**Solution 26**:

Let the length of the shortest piece be . Then, length of the second piece and the third piece are and respectively.

Since the three lengths are to be cut from a single piece of board of length ,

Also, the third piece is at least longer than the second piece.

**Solution 26**:

Let the length of the shortest piece be z cm. Then, length of the second piece and the third piece are and respectively.

Since the three lengths are to be cut from a single piece of board of length ,

Also, the third piece is at least longer than the second piece.

From both the results

Thus, the possible length of the shortest board is greater than or equal to but less than or equal to .