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sin (tan⁻¹x), where |x| < 1, is equal to
We are given to determine sin(tan⁻¹x), where |x| < 1 . Step 1: Let us take θ = tan⁻¹x . This gives us the following equations: tan(θ) = x Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below: -Opposite side= x Adjacent siRead more
We are given to determine sin(tan⁻¹x), where |x| < 1 .
Step 1: Let us take θ = tan⁻¹x .
This gives us the following equations:
tan(θ) = x
Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below:
-Opposite side= x
Adjacent side = 1
Step2: Applying Pythagorean theorem
To find the hypotenuse, use the Pythagorean theorem:
Hypotenuse = √(1² + x²) = √(1 + x²)
Step 3: Calculate sin(θ)
We know that:
sin(θ) = Opposite / Hypotenuse = x / √(1 + x²)
Final Answer:
Thus, the value of sin(tan⁻¹x) is:
x / √(1 + x²)
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If sin⁻¹ x > cos⁻¹ x, then x, should lie in the interval
We are given to find the interval where sin⁻¹x > cos⁻¹x. Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π]. For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy: sin⁻¹x > cos⁻¹x Step 2:Read more
We are given to find the interval where sin⁻¹x > cos⁻¹x.
Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π].
For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy:
sin⁻¹x > cos⁻¹x
Step 2: Use the identity sin⁻¹x + cos⁻¹x = π/2 From the identity:
sin⁻¹x + cos⁻¹x = π/2 We can subtract cos⁻¹x from both sides to get:
sin⁻¹x = π/2 – cos⁻¹x Thus, for sin⁻¹x > cos⁻¹x, we need:
π/2 – cos⁻¹x > cos⁻¹x That is:
π/2 > 2cos⁻¹x which gives:
cos⁻¹x 1/√2
Hence, the required condition sin⁻¹x > cos⁻¹x holds when x belongs to:
(1/√2, 1)
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If y = cot⁻¹ x, x < 0, then
We are given that y = cot⁻¹x and x < 0. We need to find the range of y. Step 1: Recall the range of cot⁻¹x The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x. Step 2: Analyze the condition x < 0 When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), becaRead more
We are given that y = cot⁻¹x and x < 0. We need to find the range of y.
Step 1: Recall the range of cot⁻¹x
The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x.
Step 2: Analyze the condition x < 0
When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), because the cotangent function is negative in this interval.
Final Answer:
Thus, the range of y is π/2 < y ≤ π.
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A relation R is defined on N. Which of the following is the reflexive relation?
In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself. Let's examine each of the relations listed: (a) R = {(x, y) : x > y, x, y ∈Read more
In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself.
Let’s examine each of the relations listed:
(a) R = {(x, y) : x > y, x, y ∈ ℕ}
For this relation to be reflexive, we must have x > x for all x ∈ ℕ. However, this is never the case since x is never greater than itself. So this relation is not reflexive.
(b) R = {(x, y) : x + y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require that x + x = 10 for all x ∈ ℕ. This yields 2x = 10, or x = 5. So, the only element that satisfies this is x = 5. Hence, the relation is **not reflexive** for all elements of ℕ, but only for x = 5.
(c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}
For this relation to be reflexive, we must have x * x to be a square number for all x ∈ ℕ. Since x * x = x² is always a square number for all natural numbers x, this relation is reflexive.
(d) R = {(x, y) : x + 4y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require x + 4x = 10 for all x ∈ ℕ. This simplifies to 5x = 10, or x = 2. Hence, the only element satisfying this condition is x = 2, so this relation is not reflexive for all elements of ℕ, only for x = 2.
Conclusion:
The only reflexive relation is:
– (c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}.
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Let the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4}. Then [1], the equivalence class containing 1, is:
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as: R = {(a, b) : |a - b| is a multiple of 4} This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to beRead more
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as:
R = {(a, b) : |a – b| is a multiple of 4}
This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to be all elements b ∈ A such that |1 – b| is a multiple of 4.
Step 1: Determine what values of b make |1 – b| a multiple of 4.
The possible values of b such that |1 – b| is a multiple of 4 are those for which:
|1 – b| = 4k for some integer k.
This gives us the following conditions:
1 – b = 4k or b – 1 = 4k
Thus, b = 1 + 4k for some integer k. Now let’s check the values of b in the set A = {0, 1, 2, 3,., 12}.
For k = 0, b = 1.
For k = 1, b = 1 + 4 = 5.
For k = 2, b = 1 + 8 = 9.
For k = -1, b = 1 – 4 = -3 (which is outside of A).
For k = -2, b = 1 – 8 = -7 (which is outside of A).
Thus, the equivalence class of 1, [1], includes the elements 1, 5, and 9.
Conclusion:
The equivalence class that contains 1 is {1, 5, 9}.
Hence, the correct answer is:
– (a) {1, 5, 9}.
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