NCERT Solutions For Class 11 Math Chapter – 5 Miscellaneous Exercise

NCERT Solutions For Class 11 Math Chapter – 5  Miscellaneous Exercise

1.

Solution:

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

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Solution:

3. Reduce to the standard form

Solution:

4.

Solution:

5. Convert the following in the polar form:

(i) , (ii) 

Solution:

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

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

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

9. 21x² – 28x + 10 = 0

Solution:

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

On comparing it with ax² + bx + c = 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

10. If z₁ = 2 – i, z₂ = 1 + i, find

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

11.

Solution:

12. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) , (ii) 

Solution:

13. Find the modulus and argument of the complex number

Solution:

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)

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 x and y are 3 and –3 respectively.

15. Find the modulus of

Solution:

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

Solution:

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

Solution:

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

Solution:

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:

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

Solution:

Thus, the least positive integer is 1.

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