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Domain of function √a² – x² (a > 0) is
Choice (b) is correct Let y = √a² - x² The function y is defined if a² = x² ≥ 0 ⇒ a² ≥ x² ⇒ x² ≤ a² ⇒ -a ≤ x ≤ a Domain = [-a, a] This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding. For more please visiRead more
Choice (b) is correct
Let y = √a² – x²
The function y is defined if a² = x² ≥ 0 ⇒ a² ≥ x² ⇒ x² ≤ a² ⇒ -a ≤ x ≤ a
Domain = [-a, a]
This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding.
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Domain of real valued function f(x) = 1/2x +1 is
Choice (c). is correct. f(x) = 1/2x + 1 For domain, 2x + 1= 0 ⇒ 2x ≠ - 1 ⇒ x ≠ -(1/2) x can take any real value except x = -(1/2). Domain = R - {-(1/2)} This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understandingRead more
Choice (c). is correct. f(x) = 1/2x + 1
For domain, 2x + 1= 0 ⇒ 2x ≠ – 1 ⇒ x ≠ -(1/2)
x can take any real value except x = -(1/2).
Domain = R – {-(1/2)}
This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding.
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The domain function of the function f given by f(x) = x³ + 2x² – 4x + 5/ x² -2x – 3 is
Choice (c) is correct. The function f is defined for all values of x except when x² - 2x - 3 = 0 x² - 2x - 3 = 0 ⇒ x² + x -3x - 3 = 0 ⇒ x(x +1) -3(x +1) = 0 (x + 1) (x - 3) = 0 ⇒ x =-1 or x = 3 Hence, the domain of the function is the set of all real numbers except x = 3 or x = -1. Domain of funcRead more
Choice (c) is correct. The function f is defined for all values of x except when x² – 2x – 3 = 0
x² – 2x – 3 = 0 ⇒ x² + x -3x – 3 = 0 ⇒ x(x +1) -3(x +1) = 0
(x + 1) (x – 3) = 0 ⇒ x =-1 or x = 3
Hence, the domain of the function is the set of all real numbers except x = 3 or x = -1. Domain of function f = R – {-1, 3}
This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding.
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Relation R is defined as R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3, 4, 5) : x ∈ (0, 1, 2, 3, 4, 5, )}. The domain of the relation R is
Choice (d) is correct. Since, R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3, 4, 5)} R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)} Domain (R) = {1, 2, 3, 4, 5, 6} This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your undRead more
Choice (d) is correct.
Since, R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3, 4, 5)}
R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}
Domain (R) = {1, 2, 3, 4, 5, 6}
This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding.
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Given set A = {2, 3, 4, 5, 6} and set B = {0, 1, 2, 3}. A relation R from set A to set B defined as R = {(x, y): x+y =8} is
Choice (a) is correct. A relation R from set A to set B is given by R = {(x, y) : x + y = 8}. x ∈ A and y ∈ B such that x + y = 8 6 ∈ A and 2 ∈ B such that 6 + 2 = 8 ⇒ (6, 2) ∈ R and 5 ∈ A and 3 ∈ B such that 5 + 3 = 8 ⇒ (5, 3) ∈ R and there is no other pair. This question related to Chapter 2 mRead more
Choice (a) is correct.
A relation R from set A to set B is given by R = {(x, y) : x + y = 8}.
x ∈ A and y ∈ B such that x + y = 8
6 ∈ A and 2 ∈ B such that 6 + 2 = 8 ⇒ (6, 2) ∈ R
and 5 ∈ A and 3 ∈ B such that 5 + 3 = 8 ⇒ (5, 3) ∈ R
and there is no other pair.
This question related to Chapter 2 maths Class 11th NCERT. From the Chapter 2: Relations and Functions. Give answer according to your understanding.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-2