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If P is a 3 x 3 matrix such that P’ = 2P + I, where P’ is the transpose of P, then
We are given that P is a 3 × 3 matrix such that P' = 2P + I, where P' is the transpose of P. Step 1: Take the transpose of both sides We take the transpose of both sides of the equation P' = 2P + I: (P')' = (2P + I)' Since the transpose of the transpose of a matrix is the matrix itself, we get: P =Read more
We are given that P is a 3 × 3 matrix such that P’ = 2P + I, where P’ is the transpose of P.
Step 1: Take the transpose of both sides
We take the transpose of both sides of the equation P’ = 2P + I:
(P’)’ = (2P + I)’
Since the transpose of the transpose of a matrix is the matrix itself, we get:
P = 2P’ + I
Step 2: Plug P’ = 2P + I into this equation
Next, plug the expression P’ = 2P + I into the equation:
P = 2(2P + I) + I
P = 4P + 2I + I
P = 4P + 3I
Step 3: Move terms around
Now we move the terms around in the equation.
P – 4P = 3I
-3P = 3I
P = -I
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If y = tan⁻¹ (e²ˣ), then dy/dx is equal to
We are given y = tan⁻¹(e²ˣ) and need to find dy/dx. Step 1: Differentiate both sides with respect to x We differentiate the equation y = tan⁻¹(e²ˣ) using the chain rule. The derivative of tan⁻¹(u) with respect to u is 1/(1 + u²), so: dy/dx = 1 / (1 + (e²ˣ)²) * d/dx(e²ˣ) Step 2: Differentiate e²ˣ TheRead more
We are given y = tan⁻¹(e²ˣ) and need to find dy/dx.
Step 1: Differentiate both sides with respect to x
We differentiate the equation y = tan⁻¹(e²ˣ) using the chain rule. The derivative of tan⁻¹(u) with respect to u is 1/(1 + u²), so:
dy/dx = 1 / (1 + (e²ˣ)²) * d/dx(e²ˣ)
Step 2: Differentiate e²ˣ
The derivative of e²ˣ with respect to x is:
d/dx(e²ˣ) = 2e²ˣ
Step 3: Substitute into the derivative
Substitute this back into the expression for dy/dx:
dy/dx = 1 / (1 + e⁴ˣ) * 2e²ˣ
dy/dx = 2e²ˣ / (1 + e⁴ˣ)
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The function f(x) = [x], where [x] is the greatest integer function that is less than or equal to x, is continuous at
The function f(x) = ⌊x⌋, where ⌊x⌋ is the greatest integer function (also known as the floor function), gives the greatest integer less than or equal to x. Continuity of the floor function: The floor function is discontinuous at integer values of x. This is because, at any integer n, the function juRead more
The function f(x) = ⌊x⌋, where ⌊x⌋ is the greatest integer function (also known as the floor function), gives the greatest integer less than or equal to x.
Continuity of the floor function:
The floor function is discontinuous at integer values of x. This is because, at any integer n, the function jumps from n-1 to n. Hence, the function is not continuous at integer points.
Continuity at non-integer points:
At non-integer points, the function is continuous since it is a constant between integers.
Checking the given points:
– At x = 4, f(x) is not continuous since it jumps at an integer value.
– At x = -2, f(x) is also not continuous since it jumps at an integer value.
– At x = 1.5, f(x) = 1, which is continuous since it is not at an integer point.
– At x = 1, the function f(x) is discontinuous because it hops at the integer value.
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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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