﻿ Class 11 NCERT Math Solution
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TOPICS
Miscellaneous

Question-1 :- Solution :-
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Question-2 :-  For any two complex numbers z₁ and z₂, prove that Re (z₁.z₂) = Re z₁.Re z₂ – Imz₁.Imz₂.

Solution :-
```  Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
Therefore, z₁.z₂ = (x₁ + iy₁)(x₂ + iy₂)
= x₁.x₂ + x₁.iy₂ + x₂.iy₁ + i²y₁.y₂
= x₁.x₂ + x₁.iy₂ + x₂.iy₁ - y₁.y₂
= (x₁.x₂ + y₁.y₂) + i(x₁.y₂ + x₂.y₁)
Re(z₁.z₂) = x₁.x₂ + y₁.y₂
Re (z₁.z₂) = Re z₁.Re z₂ – Imz₁.Imz₂
```

Question-3 :- Solution :-
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Question-4 :- Solution :-
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Question-5 :-  Convert the following in the polar form: Solution :-
``` Here, z = - 1 + i
Now, -1 = r cos θ, 1 = r sin θ
By squaring and adding, we get
r2 cos2 θ + r2 sin2 θ = (-1)2 + 12
r2 (cos2 θ + sin2 θ) = 1 + 1
r2 (cos2 θ + sin2 θ) = 2
r x 1 = √2
r = √2

Therefore, -1 = r cos θ and 1 = r sin θ
cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4   [lies on II Quadrant]
Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
```
``` Here, z = - 1 + i
Now, -1 = r cos θ, 1 = r sin θ
By squaring and adding, we get
r2 cos2 θ + r2 sin2 θ = (-1)2 + 12
r2 (cos2 θ + sin2 θ) = 1 + 1
r2 (cos2 θ + sin2 θ) = 2
r x 1 = √2
r = √2

Therefore, -1 = r cos θ and 1 = r sin θ
cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4   [lies on II Quadrant]
Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
```

Question-6 :-  Solve 3x2 - 4x + 20/3 = 0.

Solution :-
```  We have, 3x2 - 4x + 20/3 = 0
Hence, 9x2 - 12x + 20 = 0
Now, a = 9; b = -12; c = 20
b2 - 4ac = (-12)2 - 4 x 9 x 20 = 144 - 720 = -576
Therefore, the solutions are given by x = (12 ± √-576)/18 = (12 ± 24i)/18 = 2/3 ± 4i/3
```

Question-7 :-  Solve x2 - 2x + 3/2 = 0.

Solution :-
```  We have, x2 - 2x + 3/2 = 0
Hence, 2x2 - 4x + 3 = 0
Now, a = 2; b = -4; c = 3
b2 - 4ac = (-4)2 - 4 x 2 x 3 = 16 - 24 = -8
Therefore, the solutions are given by x = (4 ± √-8)/4 = (4 ± 2√2i)/4 = 1 ± √2i/2
```

Question-8 :-  Solve 27x2 - 10x + 1 = 0.

Solution :-
```  We have, 27x2 - 10x + 1 = 0.
Now, a = 27; b = -10; c = 1
b2 - 4ac = (-10)2 - 4 x 27 x 1 = 100 - 108 = -8
Therefore, the solutions are given by x = (10 ± √-8)/54 = (10 ± 2√2i)/54 = 5/27 ± √2i/27
```

Question-9 :-  Solve 21x2 - 28x + 10 = 0

Solution :-
```  We have, 21x2 - 28x + 10 = 0
Now, a = 21; b = -28; c = 10
b2 - 4ac = (-28)2 - 4 x 21 x 10 = 784 - 840 = -56
Therefore, the solutions are given by x = (28 ± √-56)/42 = (28 ± √56i)/42 = 2/3 ± √14i/21
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Question-10 :- Solution :-
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Question-11 :- Solution :-
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Question-12 :- Solution :-
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Question-13 :-  Find the modulus and argument of the complex number Solution :-
``` Now, -1/2 = r cos θ, 1/2 = r sin θ
By squaring and adding, we get
r2 cos2 θ + r2 sin2 θ = (-1/2)2 + (1/2)2
r2 (cos2 θ + sin2 θ) = 1/4 + 1/4
r2 (cos2 θ + sin2 θ) = 1/2
r x 1 = 1/√2
r = 1/√2
Modulus = 1/√2

Therefore, -1/2 = r cos θ and 1/2 = r sin θ
cos θ = -√2/2 = -1/√2 and sin θ = √2/2 = 1/√2,
Since cos θ are negative and cosθ are negative in II quadrant,
So, which gives θ = (π - π/4) = 3π/4
Arguments = 3π/4
```

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

Solution :-
```  We have, z = (x – iy) (3 + 5i) = 3x + 5ix - 3iy - 5iy² = (3x + 5y) + i(5x - 3y)
Conjugate, z = (3x + 5y) - i(5x - 3y)
Given that : –6 – 24i
Therefore, (3x + 5y) - i(5x - 3y) = –6 – 24i
Equating real and imaginary parts, we obtain
3x + 5y = -6 ......(i)
5x - 3y = 24 ......(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
9x + 15y + 25x - 15y = -18 + 120
34x = 102
x = 3
Putting the value of x in equation (i), we obtain
3 x 3 + 5y = -6
9 + 5y = -6
5y = -6 - 9
5y = -15
y = -3

Thus, the values of x and y are 3 and –3 respectively.
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Question-15 :- Solution :-
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Question-16 :- Solution :-
```  (x + iy)³ = u + iv
x³ + i³y³ + 3.x².iy + 3.x.(iy)² = u + iv
x³ - iy³ + 3.x².iy - 3.x.y² = u + iv
(x³ - 3xy²) + i(3x²y - y³) = u + iv
Therefore, u = x³ - 3xy², v = 3x²y - y³ ```

Question-17 :-  If α and β are different complex numbers with |β| = 1, then find Solution :-
```  Let α = a + ib and β = x + iy
It is given that, |β| = 1
Therefore, (√x² + y²)x = 1
x² + y² = 1 ```

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

Solution :-
```  |1 - i|x = 2x
(√1² + (-1)²)x = 2x
(√2)x = 2x
2x/2 = 2x
Now, x/2 = x
x = 2x
2x - x = 0
x = 0
Thus, 0 is the only integral solution of the given equation.
Therefore, the number of non-zero integral solutions of the given equation is 0.
```

Question-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 :-
```  (a + ib) (c + id) (e + if) (g + ih) = A + iB
|(a + ib) (c + id) (e + if) (g + ih)| = |A + iB|
|(a + ib)| x |(c + id)| x |(e + if)| x |(g + ih)| = |A + iB|      [|z₁.z₂| = |z₁|.|z₂|]
√a² + b² x √c² + d² x √e² + f² x √g² + h² = √A² + B²
On squaring both sides, we obtain
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
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Question-20 :- Solution :-
``` Therefore, m = 4k, where k is some integer.
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4 × 1).

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