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If z = 1 – cos 2 θ + i sin 2 θ, then ∣ z ∣
Choice (c) is correct. Here, z = 1 - cos 2 θ + i sin 2 θ ∣ z ∣ = ∣1 - cos 2 θ + i sin 2 θ∣ = √(1 - cos 2θ)² + (sin 2θ)² = √ 1 -2 cos 2 θ + sin² 2θ = √1 - 2 cos2θ + 1 = √2 -2 cos 2 θ = √2(1- cos 2θ ) = √2(2 sin² θ) = 2 ∣ sin θ ∣ This question related to Chapter 4 maths Class 11th NCERT. From the ChRead more
Choice (c) is correct.
Here, z = 1 – cos 2 θ + i sin 2 θ
∣ z ∣ = ∣1 – cos 2 θ + i sin 2 θ∣ = √(1 – cos 2θ)² + (sin 2θ)²
= √ 1 -2 cos 2 θ + sin² 2θ = √1 – 2 cos2θ + 1
= √2 -2 cos 2 θ = √2(1- cos 2θ ) = √2(2 sin² θ) = 2 ∣ sin θ ∣
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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For the following LPP Maximise Z = 3x + 4y subject to constraints x – y ≥ – 1, x ≤ 3 x ≥ 0, y ≥ 0 The maximum value is
To solve the LPP, we first plot the constraints: 1. x - y ≥ -1 ⟹ y ≤ x + 1 2. x ≤ 3 3. x ≥ 0 4. y ≥ 0 Now, we find the corner points of the feasible region: - From x = 0 and y = 0, the point is (0, 0). - For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4). - From the intersecRead more
To solve the LPP, we first plot the constraints:
1. x – y ≥ -1 ⟹ y ≤ x + 1
2. x ≤ 3
3. x ≥ 0
4. y ≥ 0
Now, we find the corner points of the feasible region:
– From x = 0 and y = 0, the point is (0, 0).
– For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4).
– From the intersection of x = 3 and y = x + 1, we get (3, 4).
Now, evaluate Z = 3x + 4y at the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (3, 4), Z = 3(3) + 4(4) = 9 + 16 = 25
The maximum value of Z occurs at (3, 4) and is 25.
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The value of i¹⁰⁰² is
Choice (b) is correct. i¹⁰⁰² = i¹⁰⁰⁰ × i² = (i⁴ )²⁵⁰ × (-1) = 1²⁵⁰ × (-1) = -1 This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.cRead more
Choice (b) is correct.
i¹⁰⁰² = i¹⁰⁰⁰ × i²
= (i⁴ )²⁵⁰ × (-1) = 1²⁵⁰ × (-1) = -1
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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Which of the following options define imaginary number?
Choice (c) is correct. We know that the square root of negative number is imaginary number. This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding. For more please visit here: https://www.tiwRead more
Choice (c) is correct. We know that the square root of negative number is imaginary number.
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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An atom with 3 protons and 4 neutrons will have a valency of
(c) An atom with 3 protons and 4 neutrons (lithium) will have a valency of 1. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
(c) An atom with 3 protons and 4 neutrons (lithium) will have a valency of 1.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-4/
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