To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative. Compute the derivative of the function: y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x Determine where the derivative is positive, so the function is increasing: y'(x) > 0 3x² + 12x > 0 Factor tRead more
To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative.
Compute the derivative of the function:
y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x
Determine where the derivative is positive, so the function is increasing:
y'(x) > 0
3x² + 12x > 0
Factor the expression:
3x(x + 4) > 0
This inequality holds when x 0. Thus, the function is increasing in the intervals (-∞, -4) and (0, ∞).
Conclusion:
The correct intervals where the function is increasing are (-∞, 0) U (4, ∞).
We are given: cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π Step 1: Using the property of inverse cosine The principal value range of cos⁻¹ x is [0, π]. For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means: cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π Step 2: Find the values oRead more
We are given:
cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π
Step 1: Using the property of inverse cosine
The principal value range of cos⁻¹ x is [0, π].
For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means:
cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π
Step 2: Find the values of α, β, and γ
When cos⁻¹ α = π, then cos π = -1, so:
α = -1, β = -1, γ = -1
Step 3: Substitute into the expression
We are given to calculate:
α(β + γ) + β(γ + α) + γ(α + β)
We are given: tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) Step 1: Simplify tan⁻¹(tan 5π/6) The range of tan⁻¹ x is (-π/2, π/2). For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range. The angle 5π/6 lies outside this range. To bring it into the principal range, we use theRead more
We are given:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6)
Step 1: Simplify tan⁻¹(tan 5π/6)
The range of tan⁻¹ x is (-π/2, π/2).
For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range.
The angle 5π/6 lies outside this range. To bring it into the principal range, we use the periodicity of tan and adjust it:
5π/6 – π = -π/6
Thus:
tan⁻¹(tan 5π/6) = -π/6
Step 2: Simplify cos⁻¹(cos 13π/6)
The range of cos⁻¹ x is [0, π].
For any angle θ, cos⁻¹(cos θ) gives the principal value of θ, which must lie in this range.
The angle 13π/6 is outside this range. To bring it into the range, subtract 2π:
13π/6 – 2π = π/6
Thus:
cos⁻¹(cos 13π/6) = π/6
Step 3: Add the two results
Now, add the simplified terms:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) = -π/6 + π/6 = 0
Let us now solve this function step by step to find out the domain of sin⁻¹(3x - 1). Step 1: Domain of sin⁻¹(y) We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1. In the case of sin⁻¹(3x - 1), we must have: -1 ≤ 3x - 1 ≤ 1 Step 2: Solve the inequality 1. Add 1 to all sides: 0 ≤ 3x ≤ 2 2. DivideRead more
Let us now solve this function step by step to find out the domain of sin⁻¹(3x – 1).
Step 1: Domain of sin⁻¹(y)
We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1.
In the case of sin⁻¹(3x – 1), we must have:
-1 ≤ 3x – 1 ≤ 1
Step 2: Solve the inequality
1. Add 1 to all sides:
0 ≤ 3x ≤ 2
2. Divide through by 3:
0 ≤ x ≤ 2/3
Step 3: Domain
The domain of sin⁻¹(3x – 1) is:
[0, 2/3]
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps: Step 1: Simplify sin⁻¹(1/2) The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is: sin⁻¹(1/2) = π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(1/2) = π/6 into the expression: sin[π/3 + sin⁻¹(1/2)] =Read more
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps:
Step 1: Simplify sin⁻¹(1/2)
The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is:
sin⁻¹(1/2) = π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(1/2) = π/6 into the expression:
sin[π/3 + sin⁻¹(1/2)] = sin(π/3 + π/6)
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties: Step 1: Definition of the principal value branch The principal value branch of cosec⁻¹x is defined such that: 1. It includes all possible values of the inverse cosecant function. 2. It avoids discontinuities or undefiRead more
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties:
Step 1: Definition of the principal value branch
The principal value branch of cosec⁻¹x is defined such that:
1. It includes all possible values of the inverse cosecant function.
2. It avoids discontinuities or undefined values (like when cosec x = 0).
Step 2: Range of cosec⁻¹x
For cosec⁻¹x, the principal value is taken from the range:
[-π/2, π/2] – {0}
This excludes 0 because cosec x is undefined at sin x = 0.
We are asked to find sin[π/3 – sin⁻¹(-1/2)]. Step 1: Simplify sin⁻¹(-1/2) The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2]. The angle that satisfies this condition is: sin⁻¹(-1/2) = -π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(-1/Read more
We are asked to find sin[π/3 – sin⁻¹(-1/2)].
Step 1: Simplify sin⁻¹(-1/2)
The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2].
The angle that satisfies this condition is:
sin⁻¹(-1/2) = -π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(-1/2) = -π/6 into the expression:
sin[π/3 – (-π/6)] = sin[π/3 + π/6]
To solve the problem given, let's go through the properties of the inverse tangent function: Step 1: Define tan⁻¹x = y The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x Step 2: Range of tan⁻¹x The principalRead more
To solve the problem given, let’s go through the properties of the inverse tangent function:
Step 1: Define tan⁻¹x = y
The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x
Step 2: Range of tan⁻¹x
The principal range of the inverse tangent function tan⁻¹x (or arctan) is:
-π/2 ≤ y ≤ π/2
This signifies that y is in the interval (-π/2, π/2).
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)
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)
The interval, in which function y = x³ + 6x² + 6 is increasing is
To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative. Compute the derivative of the function: y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x Determine where the derivative is positive, so the function is increasing: y'(x) > 0 3x² + 12x > 0 Factor tRead more
To find the interval for which the function y = x³ + 6x² + 6 is increasing, we will look at its derivative.
Compute the derivative of the function:
y'(x) = d/dx(x³ + 6x² + 6) = 3x² + 12x
Determine where the derivative is positive, so the function is increasing:
y'(x) > 0
3x² + 12x > 0
Factor the expression:
3x(x + 4) > 0
This inequality holds when x 0. Thus, the function is increasing in the intervals (-∞, -4) and (0, ∞).
Conclusion:
The correct intervals where the function is increasing are (-∞, 0) U (4, ∞).
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If cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) is equal to
We are given: cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π Step 1: Using the property of inverse cosine The principal value range of cos⁻¹ x is [0, π]. For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means: cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π Step 2: Find the values oRead more
We are given:
cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π
Step 1: Using the property of inverse cosine
The principal value range of cos⁻¹ x is [0, π].
For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means:
cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π
Step 2: Find the values of α, β, and γ
When cos⁻¹ α = π, then cos π = -1, so:
α = -1, β = -1, γ = -1
Step 3: Substitute into the expression
We are given to calculate:
α(β + γ) + β(γ + α) + γ(α + β)
Replace α = -1, β = -1, and γ = -1:
(-1)((-1) + (-1)) + (-1)((-1) + (-1)) + (-1)((-1) + (-1))
Simplify each term:
(-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6
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The value of tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) is
We are given: tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) Step 1: Simplify tan⁻¹(tan 5π/6) The range of tan⁻¹ x is (-π/2, π/2). For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range. The angle 5π/6 lies outside this range. To bring it into the principal range, we use theRead more
We are given:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6)
Step 1: Simplify tan⁻¹(tan 5π/6)
The range of tan⁻¹ x is (-π/2, π/2).
For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range.
The angle 5π/6 lies outside this range. To bring it into the principal range, we use the periodicity of tan and adjust it:
5π/6 – π = -π/6
Thus:
tan⁻¹(tan 5π/6) = -π/6
Step 2: Simplify cos⁻¹(cos 13π/6)
The range of cos⁻¹ x is [0, π].
For any angle θ, cos⁻¹(cos θ) gives the principal value of θ, which must lie in this range.
The angle 13π/6 is outside this range. To bring it into the range, subtract 2π:
13π/6 – 2π = π/6
Thus:
cos⁻¹(cos 13π/6) = π/6
Step 3: Add the two results
Now, add the simplified terms:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) = -π/6 + π/6 = 0
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The domain of the function sin⁻¹ (3x -1) is
Let us now solve this function step by step to find out the domain of sin⁻¹(3x - 1). Step 1: Domain of sin⁻¹(y) We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1. In the case of sin⁻¹(3x - 1), we must have: -1 ≤ 3x - 1 ≤ 1 Step 2: Solve the inequality 1. Add 1 to all sides: 0 ≤ 3x ≤ 2 2. DivideRead more
Let us now solve this function step by step to find out the domain of sin⁻¹(3x – 1).
Step 1: Domain of sin⁻¹(y)
We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1.
In the case of sin⁻¹(3x – 1), we must have:
-1 ≤ 3x – 1 ≤ 1
Step 2: Solve the inequality
1. Add 1 to all sides:
0 ≤ 3x ≤ 2
2. Divide through by 3:
0 ≤ x ≤ 2/3
Step 3: Domain
The domain of sin⁻¹(3x – 1) is:
[0, 2/3]
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sin[π/3 + sin⁻¹(1/2)] is equal to
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps: Step 1: Simplify sin⁻¹(1/2) The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is: sin⁻¹(1/2) = π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(1/2) = π/6 into the expression: sin[π/3 + sin⁻¹(1/2)] =Read more
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps:
Step 1: Simplify sin⁻¹(1/2)
The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is:
sin⁻¹(1/2) = π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(1/2) = π/6 into the expression:
sin[π/3 + sin⁻¹(1/2)] = sin(π/3 + π/6)
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Simplify sin(π/2)
From the unit circle, sin(π/2) = 1.
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Which of the following is the principal value branch of cosec⁻¹x?
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties: Step 1: Definition of the principal value branch The principal value branch of cosec⁻¹x is defined such that: 1. It includes all possible values of the inverse cosecant function. 2. It avoids discontinuities or undefiRead more
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties:
Step 1: Definition of the principal value branch
The principal value branch of cosec⁻¹x is defined such that:
1. It includes all possible values of the inverse cosecant function.
2. It avoids discontinuities or undefined values (like when cosec x = 0).
Step 2: Range of cosec⁻¹x
For cosec⁻¹x, the principal value is taken from the range:
[-π/2, π/2] – {0}
This excludes 0 because cosec x is undefined at sin x = 0.
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sin [π/3 – sin⁻¹ (-1/2)] is equal to
We are asked to find sin[π/3 – sin⁻¹(-1/2)]. Step 1: Simplify sin⁻¹(-1/2) The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2]. The angle that satisfies this condition is: sin⁻¹(-1/2) = -π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(-1/Read more
We are asked to find sin[π/3 – sin⁻¹(-1/2)].
Step 1: Simplify sin⁻¹(-1/2)
The value sin⁻¹(-1/2) is the angle whose sine is -1/2, and the range of sin⁻¹ is [-π/2, π/2].
The angle that satisfies this condition is:
sin⁻¹(-1/2) = -π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(-1/2) = -π/6 into the expression:
sin[π/3 – (-π/6)] = sin[π/3 + π/6]
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Calculate sin(π/2)
From the unit circle, sin(π/2) = 1.
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If tan⁻¹ x = y, then:
To solve the problem given, let's go through the properties of the inverse tangent function: Step 1: Define tan⁻¹x = y The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x Step 2: Range of tan⁻¹x The principalRead more
To solve the problem given, let’s go through the properties of the inverse tangent function:
Step 1: Define tan⁻¹x = y
The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x
Step 2: Range of tan⁻¹x
The principal range of the inverse tangent function tan⁻¹x (or arctan) is:
-π/2 ≤ y ≤ π/2
This signifies that y is in the interval (-π/2, π/2).
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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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