For any invertible square matrix A, the determinant of its inverse is given by: |A⁻¹| = 1/|A| Since |A| = 4, we have: |A⁻¹| = 1/4 Thus, the correct answer is 1/4. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
For any invertible square matrix A, the determinant of its inverse is given by:
Starting with the differential equation: y (dy/dx) + x = C Rearrange to get a derivative alone on one side y (dy/dx) = C − x Multiply by dx y dy = (C − x) dx Integrate both sides ∫ y dy = ∫ (C − x) dx (1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration Multiply the whole equation by 2Read more
Starting with the differential equation:
y (dy/dx) + x = C
Rearrange to get a derivative alone on one side
y (dy/dx) = C − x
Multiply by dx
y dy = (C − x) dx
Integrate both sides
∫ y dy = ∫ (C − x) dx
(1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration
Multiply the whole equation by 2:
y² = 2Cx − x² + K, where K = 2k
Rewrite to bring together like terms:
x² + y² − 2Cx = −K
Complete the square for the x-terms:
(x − C)² + y² = C² − K
This is the equation of a circle with center at (C, 0) and radius √(C² − K).
Thus, the differential equation represents a family of circles.
For any square matrix A of order n, the determinant of its adjugate is given by: |adj A| = |A|^(n − 1) For a 2×2 matrix (n = 2): |adj A| = |A|^(2 − 1) = |A| Given |A| = −7, we have: |adj A| = −7 Thus, the correct answer is −7. Click here for more solution: https://www.tiwariacademy.com/ncert-solutioRead more
For any square matrix A of order n, the determinant of its adjugate is given by:
|adj A| = |A|^(n − 1)
For a 2×2 matrix (n = 2):
|adj A| = |A|^(2 − 1) = |A|
Among the given options, the differential equation y′ y″ + y = sin x is the only one that involves the second derivative (y″), making it a second order differential equation. Thus, the correct answer is: y′ y″ + y = sin x Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maRead more
Among the given options, the differential equation
y′ y″ + y = sin x
is the only one that involves the second derivative (y″), making it a second order differential equation.
Starting with the differential equation: 2x dx − 5y dy = 0 Separate variables: 2x dx = 5y dy Integrate both sides: ∫ 2x dx = ∫ 5y dy x² = (5/2)y² + C Rearrange the equation: x² − (5/2)y² = C This represents a family of conic sections. Since the equation is of the form: (x²) − (constant)·(y²) = C itRead more
Starting with the differential equation:
2x dx − 5y dy = 0
Separate variables:
2x dx = 5y dy
Integrate both sides:
∫ 2x dx = ∫ 5y dy
x² = (5/2)y² + C
Rearrange the equation:
x² − (5/2)y² = C
This represents a family of conic sections. Since the equation is of the form:
(x²) − (constant)·(y²) = C
it represents a family of hyperbolas (for nonzero C).
Given the differential equation: dy/dx + P y = Q The integrating factor μ(x) is defined by: μ(x) = exp(∫ P dx) We are told that: μ(x) = cos²x Taking the natural logarithm of both sides: ln μ(x) = ln(cos²x) = 2 ln|cos x| Differentiate with respect to x: d/dx [ln μ(x)] = d/dx [2 ln|cos x|] = 2 · (−tanRead more
Given the differential equation:
dy/dx + P y = Q
The integrating factor μ(x) is defined by:
μ(x) = exp(∫ P dx)
We are told that:
μ(x) = cos²x
Taking the natural logarithm of both sides:
ln μ(x) = ln(cos²x) = 2 ln|cos x|
Differentiate with respect to x:
d/dx [ln μ(x)] = d/dx [2 ln|cos x|] = 2 · (−tan x) = −2 tan x
The family of parabolas with axes parallel to the y-axis can be expressed as: y = a x² + b x + c which contains three arbitrary constants, all of them, namely a, b, and c. A differential equation must therefore have a general solution containing three arbitrary constants. Three successive differentiRead more
The family of parabolas with axes parallel to the y-axis can be expressed as:
y = a x² + b x + c
which contains three arbitrary constants, all of them, namely a, b, and c.
A differential equation must therefore have a general solution containing three arbitrary constants. Three successive differentiations eliminate all the arbitrary constants. Indeed three successive differentiations of:
y = a x² + b x + c
After this gives:
y′ = 2a x + b
which twice gives:
y″ = 2a
and thrice gives:
y‴ = 0
So the differential equation that represents the family is:
y‴ = 0
This is a third order differential equation.
Thus, the order of the differential equation is 3.
The area bounded by the curve y = cos x, the x-axis, and the given ordinates x = π/2 and x = π is given by the definite integral: A = ∫[π/2 to π] cos x dx Evaluating the integral: ∫ cos x dx = sin x Applying the limits: A = sin(π) - sin(π/2) = 0 - 1 = -1 Since area cannot be negative, we take the abRead more
The area bounded by the curve y = cos x, the x-axis, and the given ordinates x = π/2 and x = π is given by the definite integral:
A = ∫[π/2 to π] cos x dx
Evaluating the integral:
∫ cos x dx = sin x
Applying the limits:
A = sin(π) – sin(π/2)
= 0 – 1
= -1
Since area cannot be negative, we take the absolute value:
A = 1 square unit
We are given that A is a square matrix of order 3 and |A| = -4. We are asked to find the value of |adj(A)|. The formula for the determinant of the adjugate matrix adj(A) for a square matrix A of order n is |adj(A)| = |A|^(n-1) For a matrix of order 3, n = 3. Hence, the above formula becomes |adj(A)|Read more
We are given that A is a square matrix of order 3 and |A| = -4. We are asked to find the value of |adj(A)|.
The formula for the determinant of the adjugate matrix adj(A) for a square matrix A of order n is
|adj(A)| = |A|^(n-1)
For a matrix of order 3, n = 3. Hence, the above formula becomes
|adj(A)| = |A|^(3-1) = |A|²
We know that |A| = -4. So,
The differential equation given is: sin(x) + cos(dy/dx) = y² To know the degree of the differential equation, we have the following steps: 1. Rearrange the formula: The formula contains dy/dx that is not raised to any power directly. However, with the term cos(dy/dx) the degree calculation is compliRead more
The differential equation given is:
sin(x) + cos(dy/dx) = y²
To know the degree of the differential equation, we have the following steps:
1. Rearrange the formula: The formula contains dy/dx that is not raised to any power directly. However, with the term cos(dy/dx) the degree calculation is complicated since it deals with a trigonometric function of the derivative.
2. Degree of a differential equation: The degree of a differential equation is defined to be the highest power of the highest-order derivative after the equation has been made free from any irrational terms, fractional powers, or trigonometric functions involving derivatives.
3. This equation contains a trigonometric function, cos(dy/dx), whose argument is a first derivative, not an algebraic expression, not a polynomial, not a simple rational term. Since it’s placed within the argument of a trigonometric function, the degree is undefined.
If A is any square matrix of order 3 x 3 such that |A| = 4, then the value of |A⁻¹| is
For any invertible square matrix A, the determinant of its inverse is given by: |A⁻¹| = 1/|A| Since |A| = 4, we have: |A⁻¹| = 1/4 Thus, the correct answer is 1/4. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
For any invertible square matrix A, the determinant of its inverse is given by:
|A⁻¹| = 1/|A|
Since |A| = 4, we have:
|A⁻¹| = 1/4
Thus, the correct answer is 1/4.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
The differential equation y dy/dx + x = C represents
Starting with the differential equation: y (dy/dx) + x = C Rearrange to get a derivative alone on one side y (dy/dx) = C − x Multiply by dx y dy = (C − x) dx Integrate both sides ∫ y dy = ∫ (C − x) dx (1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration Multiply the whole equation by 2Read more
Starting with the differential equation:
y (dy/dx) + x = C
Rearrange to get a derivative alone on one side
y (dy/dx) = C − x
Multiply by dx
y dy = (C − x) dx
Integrate both sides
∫ y dy = ∫ (C − x) dx
(1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration
Multiply the whole equation by 2:
y² = 2Cx − x² + K, where K = 2k
Rewrite to bring together like terms:
x² + y² − 2Cx = −K
Complete the square for the x-terms:
(x − C)² + y² = C² − K
This is the equation of a circle with center at (C, 0) and radius √(C² − K).
Thus, the differential equation represents a family of circles.
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If A is any square matrix of order 2 x 2 such that |A| = -7, then the value of |adj. A| is
For any square matrix A of order n, the determinant of its adjugate is given by: |adj A| = |A|^(n − 1) For a 2×2 matrix (n = 2): |adj A| = |A|^(2 − 1) = |A| Given |A| = −7, we have: |adj A| = −7 Thus, the correct answer is −7. Click here for more solution: https://www.tiwariacademy.com/ncert-solutioRead more
For any square matrix A of order n, the determinant of its adjugate is given by:
|adj A| = |A|^(n − 1)
For a 2×2 matrix (n = 2):
|adj A| = |A|^(2 − 1) = |A|
Given |A| = −7, we have:
|adj A| = −7
Thus, the correct answer is −7.
Click here for more solution:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
Which of the following is a second order differential equation?
Among the given options, the differential equation y′ y″ + y = sin x is the only one that involves the second derivative (y″), making it a second order differential equation. Thus, the correct answer is: y′ y″ + y = sin x Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maRead more
Among the given options, the differential equation
y′ y″ + y = sin x
is the only one that involves the second derivative (y″), making it a second order differential equation.
Thus, the correct answer is: y′ y″ + y = sin x
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9
Solution of the differential equation 2x dx – 5y dy = 0 represents
Starting with the differential equation: 2x dx − 5y dy = 0 Separate variables: 2x dx = 5y dy Integrate both sides: ∫ 2x dx = ∫ 5y dy x² = (5/2)y² + C Rearrange the equation: x² − (5/2)y² = C This represents a family of conic sections. Since the equation is of the form: (x²) − (constant)·(y²) = C itRead more
Starting with the differential equation:
2x dx − 5y dy = 0
Separate variables:
2x dx = 5y dy
Integrate both sides:
∫ 2x dx = ∫ 5y dy
x² = (5/2)y² + C
Rearrange the equation:
x² − (5/2)y² = C
This represents a family of conic sections. Since the equation is of the form:
(x²) − (constant)·(y²) = C
it represents a family of hyperbolas (for nonzero C).
Thus, the correct answer is a hyperbola.
Click here for more:
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If cos²x is an integrating factor of the differential equation dy/dx + Py = Q, then P can be
Given the differential equation: dy/dx + P y = Q The integrating factor μ(x) is defined by: μ(x) = exp(∫ P dx) We are told that: μ(x) = cos²x Taking the natural logarithm of both sides: ln μ(x) = ln(cos²x) = 2 ln|cos x| Differentiate with respect to x: d/dx [ln μ(x)] = d/dx [2 ln|cos x|] = 2 · (−tanRead more
Given the differential equation:
dy/dx + P y = Q
The integrating factor μ(x) is defined by:
μ(x) = exp(∫ P dx)
We are told that:
μ(x) = cos²x
Taking the natural logarithm of both sides:
ln μ(x) = ln(cos²x) = 2 ln|cos x|
Differentiate with respect to x:
d/dx [ln μ(x)] = d/dx [2 ln|cos x|] = 2 · (−tan x) = −2 tan x
But we also have:
d/dx [ln μ(x)] = P
Thus, we find:
P = −2 tan x
Therefore, the correct answer is:
−2 tan x
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The order of the differential equation of family of parabolas whose axes are parallel to y-axis is
The family of parabolas with axes parallel to the y-axis can be expressed as: y = a x² + b x + c which contains three arbitrary constants, all of them, namely a, b, and c. A differential equation must therefore have a general solution containing three arbitrary constants. Three successive differentiRead more
The family of parabolas with axes parallel to the y-axis can be expressed as:
y = a x² + b x + c
which contains three arbitrary constants, all of them, namely a, b, and c.
A differential equation must therefore have a general solution containing three arbitrary constants. Three successive differentiations eliminate all the arbitrary constants. Indeed three successive differentiations of:
y = a x² + b x + c
After this gives:
y′ = 2a x + b
which twice gives:
y″ = 2a
and thrice gives:
y‴ = 0
So the differential equation that represents the family is:
y‴ = 0
This is a third order differential equation.
Thus, the order of the differential equation is 3.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9
The area bounded by the curve y = cosx, x-axis, ordinates x = π/2 and x = π is
The area bounded by the curve y = cos x, the x-axis, and the given ordinates x = π/2 and x = π is given by the definite integral: A = ∫[π/2 to π] cos x dx Evaluating the integral: ∫ cos x dx = sin x Applying the limits: A = sin(π) - sin(π/2) = 0 - 1 = -1 Since area cannot be negative, we take the abRead more
The area bounded by the curve y = cos x, the x-axis, and the given ordinates x = π/2 and x = π is given by the definite integral:
A = ∫[π/2 to π] cos x dx
Evaluating the integral:
∫ cos x dx = sin x
Applying the limits:
A = sin(π) – sin(π/2)
= 0 – 1
= -1
Since area cannot be negative, we take the absolute value:
A = 1 square unit
Thus, the correct answer is: 1 sq. unit
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Given that A is a square matrix of order 3 and |A| = -4, then |adj. A| is equal to:
We are given that A is a square matrix of order 3 and |A| = -4. We are asked to find the value of |adj(A)|. The formula for the determinant of the adjugate matrix adj(A) for a square matrix A of order n is |adj(A)| = |A|^(n-1) For a matrix of order 3, n = 3. Hence, the above formula becomes |adj(A)|Read more
We are given that A is a square matrix of order 3 and |A| = -4. We are asked to find the value of |adj(A)|.
The formula for the determinant of the adjugate matrix adj(A) for a square matrix A of order n is
|adj(A)| = |A|^(n-1)
For a matrix of order 3, n = 3. Hence, the above formula becomes
|adj(A)| = |A|^(3-1) = |A|²
We know that |A| = -4. So,
|adj(A)| = (-4)² = 16
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Degree of the differential equation sin x + cos (dy/dx) = y² is
The differential equation given is: sin(x) + cos(dy/dx) = y² To know the degree of the differential equation, we have the following steps: 1. Rearrange the formula: The formula contains dy/dx that is not raised to any power directly. However, with the term cos(dy/dx) the degree calculation is compliRead more
The differential equation given is:
sin(x) + cos(dy/dx) = y²
To know the degree of the differential equation, we have the following steps:
1. Rearrange the formula: The formula contains dy/dx that is not raised to any power directly. However, with the term cos(dy/dx) the degree calculation is complicated since it deals with a trigonometric function of the derivative.
2. Degree of a differential equation: The degree of a differential equation is defined to be the highest power of the highest-order derivative after the equation has been made free from any irrational terms, fractional powers, or trigonometric functions involving derivatives.
3. This equation contains a trigonometric function, cos(dy/dx), whose argument is a first derivative, not an algebraic expression, not a polynomial, not a simple rational term. Since it’s placed within the argument of a trigonometric function, the degree is undefined.
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9