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|A| = |kA|, where A is a square matrix of order 2, then sum of all possible values of k is
To solve this, let's break it down step by step. We are given the equation |A| = |kA|, where A is a square matrix of order 2. That means it is a 2x2 matrix. We are trying to find the sum of all possible values of k. Key Concepts: 1. Determinant of a matrix: For a square matrix A, the determinant isRead more
To solve this, let’s break it down step by step.
We are given the equation |A| = |kA|, where A is a square matrix of order 2. That means it is a 2×2 matrix. We are trying to find the sum of all possible values of k.
Key Concepts:
1. Determinant of a matrix: For a square matrix A, the determinant is denoted as |A|.
2. Scalar multiplication and determinant:** Let A be any square matrix of order n and k be any scalar, then |kA| = k^n |A|.
Here:
– A is of the order of 2. Thus, n = 2.
– The determinant |kA| is defined as:
|kA| = k² |A|
Now, from |A|=|kA| we have:
|A| = k² |A|
If |A| ≠ 0, then divide both sides by |A|. Then
1 = k²
Hence, k = ±1
If |A| = 0, the equation holds true for all value of k.
Therefore, sum of all the possible values of k
= 1 + (-1)
= 0
Hence, sum of all possible values of k is zero.
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The area of the region bounded by the lines y = mx, x = 1, x = 2 and x-axis is 6 sq. units, then m is equal to
The given region is bounded by the lines y = mx, x = 1, x = 2, and the x-axis. The area enclosed by these boundaries is 6 square units. The definite integral of mx from x = 1 to x = 2 gives us the area under the line y = mx, which is then used to calculate the value of m. We solve the integral. We gRead more
The given region is bounded by the lines y = mx, x = 1, x = 2, and the x-axis. The area enclosed by these boundaries is 6 square units.
The definite integral of mx from x = 1 to x = 2 gives us the area under the line y = mx, which is then used to calculate the value of m.
We solve the integral. We get the area as (3m/2). Therefore, we equate it to 6 and solve for m.
Solving
We multiply both sides by 2 to get 3m=12.
Divide it by 3, and thus we get m = 4.
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If d/dx [f(x)] = ax + b and f(0) = 0, then f(x) is equal to
We are given that d/dx [f(x)] = ax + b Step 1: Integrate both sides To find f(x), integrate the given derivative: f(x) = ∫ (ax + b) dx Using standard integration rules: ∫ ax dx = (a x²)/2 and ∫ b dx = bx Thus, f(x) = (a x²)/2 + bx + C where C is the constant of integration. Step 2: Use the given conRead more
We are given that
d/dx [f(x)] = ax + b
Step 1: Integrate both sides
To find f(x), integrate the given derivative:
f(x) = ∫ (ax + b) dx
Using standard integration rules:
∫ ax dx = (a x²)/2 and ∫ b dx = bx
Thus,
f(x) = (a x²)/2 + bx + C
where C is the constant of integration.
Step 2: Use the given condition f(0) = 0
Substituting x = 0 in the equation:
0 = (a(0)²)/2 + b(0) + C
0 = C
Thus, C = 0, so the final function is:
f(x) = (a x²)/2 + bx
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Black holes in orbit around a normal star are detected gas into the black hole, which can reach temperatures greater than 10⁶K. Assuming that the infalling gas can be modelled as a black body radiator, then the wavelength of maximum power lies
We use Wien's Displacement Law to find the wavelength of maximum power emission for a black body: ln λₘₐₓ = b / T where b is Wien's constant, approximately 2.9 × 10⁻³ m·K, and T is the temperature in Kelvin. We know that the gas temperature is T > 10⁶ K, so we compute: λₘₐₓ = (2.9 × 10⁻⁹ m) / (10Read more
We use Wien’s Displacement Law to find the wavelength of maximum power emission for a black body: ln
λₘₐₓ = b / T
where b is Wien’s constant, approximately 2.9 × 10⁻³ m·K, and T is the temperature in Kelvin. We know that the gas temperature is T > 10⁶ K, so we compute:
λₘₐₓ = (2.9 × 10⁻⁹ m) / (10⁶ K)
λₘₐₓ = 2.9 nm
Since X-rays have wavelengths in the range of 0.01 nm to 10 nm, the maximum emission falls in the X-ray region.
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The area of the region bounded by the curve y = √x -1 and the lines x = 1 and x = 5 is
To determine the area of the region bounded by the curve y = √x - 1 and the lines x = 1 and x = 5, we must set up the definite integral. The area is given by: A = ∫₁⁵ (√x - 1) dx Step 1: Solve the integral We can split the integral into two parts: A = ∫₁⁵ √x dx - ∫₁⁵ 1 dx First integral: ∫ √x dx = ∫Read more
To determine the area of the region bounded by the curve y = √x – 1 and the lines x = 1 and x = 5, we must set up the definite integral.
The area is given by:
A = ∫₁⁵ (√x – 1) dx
Step 1: Solve the integral
We can split the integral into two parts:
A = ∫₁⁵ √x dx – ∫₁⁵ 1 dx
First integral:
∫ √x dx = ∫ x^(1/2) dx = (2/3) x^(3/2)
Evaluating this from 1 to 5:
[(2/3) x^(3/2)]₁⁵ = (2/3) (5^(3/2) – 1^(3/2)) = (2/3) (5√5 – 1)
Second integral:
∫ 1 dx = x
Evaluating this from 1 to 5:
[x]₁⁵ = 5 – 1 = 4
Step 2: Combine the results
The total area is:
A = (2/3) (5√5 – 1) – 4
Simplifying this expression gives the final result:
A = 13/3 square units
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