NCERT Solutions for Class 11 Maths Chapter 5- Complex Numbers and Quadratic Equations Miscellaneous Exercise

The textbook of Class 11 contains an exercise at the end of each chapter. This exercise contains questions that cover all the topics in the chapter. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations is based on the following topics:

  1. Complex Numbers
  2. Algebra of Complex Numbers
  3. The Modulus and the Conjugate of a Complex Number
  4. Argand Plane and Polar Representation
  5. Quadratic Equations

Mathematics is a subject that needs more and more practice to understand the problem solving method. Practising with the help of NCERT Solutions for Class 11 Maths will help the students in understanding the concepts and problems better, in turn helping them to score high marks in the exams.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 5- Complex Numbers and Quadratic Equations Miscellaneous Exercise

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Access other exercise solutions of Class 11 Maths Chapter 5- Complex Numbers and Quadratic Equations

Exercise 5.1 Solutions 14 Questions

Exercise 5.2 Solutions 8 Questions

Exercise 5.3 Solutions 10 Questions

Access Solutions for Class 11 Maths Chapter 5 Miscellaneous Exercise

1.

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 1

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 2

2. For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re zRe z2 – Im z1 Im z2

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 3

3. Reduce to the standard form

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 4

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 5

4.

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 6

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 7

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 8

5. Convert the following in the polar form:

(i) NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 9, (ii) NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 10

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 11

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 12

Solve each of the equation in Exercises 6 to 9.

6. 3x2 – 4x + 20/3 = 0

Solution:

Given quadratic equation, 3x2 – 4x + 20/3 = 0

It can be re-written as: 9x2 – 12x + 20 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576

Hence, the required solutions are

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 13

7. x2 – 2x + 3/2 = 0

Solution:

Given quadratic equation, x2 – 2x + 3/2 = 0

It can be re-written as 2x2 – 4x + 3 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 2, b = –4, and c = 3

So, the discriminant of the given equation will be

D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8

Hence, the required solutions are

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 14

8. 27x2 – 10x + 1 = 0

Solution:

Given quadratic equation, 27x2 – 10x + 1 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 27, b = –10, and c = 1

So, the discriminant of the given equation will be

D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8

Hence, the required solutions are

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 15

9. 21x2 – 28x + 10 = 0

Solution:

Given quadratic equation, 21x2 – 28x + 10 = 0

On comparing it with ax2 + bx = 0, we have

a = 21, b = –28, and c = 10

So, the discriminant of the given equation will be

D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56

Hence, the required solutions are

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 16

10. If z1 = 2 – i, z2 = 1 + i, find

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 17

Solution:

Given, z1 = 2 – i, z2 = 1 + i

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 18

11.

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 19

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 20

12. Let z1 = 2 – i, z2 = -2 + i. Find

(i) NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 22, (ii) NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 23

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 24

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 25

13. Find the modulus and argument of the complex number

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 26

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 27

14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let’s assume z = (x – iy) (3 + 5i)

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 28

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of and y are 3 and –3 respectively.

15. Find the modulus of

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 29

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 30

16. If (x + iy)3 = u + iv, then show that

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 31

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 32

17. If α and β are different complex numbers with |β| = 1, then find

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 34

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 35

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 36

18. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 37

Therefore, 0 is the only integral solution of the given equation.

Hence, the number of non-zero integral solutions of the given equation is 0.

19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 38

20. If, then find the least positive integral value of m.

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 39

Solution:

NCERT Solutions Class 11 Mathematics Chapter 5 misc.ex - 40

Thus, the least positive integer is 1.

Therefore, the least positive integral value of m is 4 (= 4 × 1).

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