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Corner points of the feasible region determined by the system of linear constraints are (0,3), (1,1) and (3,0). Let Z = 4x + 5y be the objective function. The minimum value of Z occurs at
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function. 1. For the point (0, 3): Z = 4(0) + 5(3) Z = 15 2. For the point (1, 1): Z = 4(1) + 5(1) Z = 9 3. For the point (3, 0): Z = 4(3) + 5(0) Z = 12 The point (1, 1) is where the miniRead more
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function.
1. For the point (0, 3):
Z = 4(0) + 5(3)
Z = 15
2. For the point (1, 1):
Z = 4(1) + 5(1)
Z = 9
3. For the point (3, 0):
Z = 4(3) + 5(0)
Z = 12
The point (1, 1) is where the minimum value of Z = 9 occurs.
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Which of the following is not correct?
Choice (c) is correct. As sec θ ≤ - 1 or sec θ ≥ 1, so sec θ cannot be equal to 1/2. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncert-sRead more
Choice (c) is correct.
As sec θ ≤ – 1 or sec θ ≥ 1, so sec θ cannot be equal to 1/2.
This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
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Tangent function is negative in
Choice (b) is correct. As tangent function is negative in 2nd and 4th quadrants. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncert-solutRead more
Choice (b) is correct. As tangent function is negative in 2nd and 4th quadrants.
This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
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If tan θ = 3 and θ lies in third quadrant, then the value of sin θ is
Choice (d) is correct. Given that , tan θ = 3 we know that, sec² θ = 1 + tan² θ ⇒ sec² θ = 1 + 3² = 1 + 9 = 10 ⇒ cos² θ = 1/10 We know that sin² θ = 1 - cos² θ sin² θ = 1 - 1/10 = 10 - 1/10 = 9/10 ⇒ sin θ = ± √9/10 = ± 3/√10 ⇒ sin θ = -3√10 This question related to Chapter 3 maths Class 11th NCERT.Read more
Choice (d) is correct. Given that , tan θ = 3
we know that, sec² θ = 1 + tan² θ
⇒ sec² θ = 1 + 3² = 1 + 9 = 10 ⇒ cos² θ = 1/10
We know that sin² θ = 1 – cos² θ
sin² θ = 1 – 1/10 = 10 – 1/10 = 9/10
⇒ sin θ = ± √9/10 = ± 3/√10 ⇒ sin θ = -3√10
This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
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Which of the following is correct?
Choice (b) is correct. We know that sine function is an increasing function in first quadrant and 1 radian = 57.296 degrees. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding. For more please visit hereRead more
Choice (b) is correct. We know that sine function is an increasing function in first quadrant and 1 radian = 57.296 degrees. This question related to Chapter 3 maths Class 11th NCERT. From the Chapter 3: Trigonometric Functions. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-3