1. 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.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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  2. 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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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9

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  3. 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.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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  4. 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

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9

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  5. 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.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9

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