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  1. We find the minimum value of the function f(x) = 2 cos x + x over the closed interval [0, π/2] by determining its critical numbers and then we check the value of the function at the interval's endpoints. Compute the derivative of the function: f'(x) = d/dx(2 cos x + x) = -2 sin x + 1 Set the derivatRead more

    We find the minimum value of the function f(x) = 2 cos x + x over the closed interval [0, π/2] by determining its critical numbers and then we check the value of the function at the interval’s endpoints.

    Compute the derivative of the function:
    f'(x) = d/dx(2 cos x + x) = -2 sin x + 1

    Set the derivative equal to zero to find critical points:
    -2 sin x + 1 = 0

    sin x = 1/2

    The solution to sin x = 1/2 in the interval [0, π/2] is x = π/6.

    Evaluate f(x) at the critical point and the endpoints of the interval:
    – At x = 0:
    f(0) = 2 cos(0) + 0 = 2 × 1 + 0 = 2

    – At x = π/6:
    f(π/6) = 2 cos(π/6) + π/6 = 2 × (√3/2) + π/6 = √3 + π/6

    – At x = π/2:
    f(π/2) = 2 cos(π/2) + π/2 = 2 × 0 + π/2 = π/2

    Compare the values:
    – f(0) = 2
    – f(π/6) = √3 + π/6 ≈ 1.732 + 0.523 = 2.255
    – f(π/2) = π/2 ≈ 1.570

    The minimum value of the function in the interval [0, π/2] is attained at f(0) = 2.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-6

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  2. To find the intervals where the function f(x) = x² - 4x + 6 is strictly increasing, we have to look at its derivative. Find the derivative of the function: f'(x) = d/dx(x² - 4x + 6) = 2x - 4 Determine where the derivative is positive (for strictly increasing behavior): f'(x) > 0 2x - 4 > 0 xRead more

    To find the intervals where the function f(x) = x² – 4x + 6 is strictly increasing, we have to look at its derivative.

    Find the derivative of the function:
    f'(x) = d/dx(x² – 4x + 6) = 2x – 4

    Determine where the derivative is positive (for strictly increasing behavior):
    f'(x) > 0
    2x – 4 > 0
    x > 2
    So, the function is strictly increasing when x > 2.

    Conclusion:
    The function is strictly increasing on the interval (2, ∞).

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-6

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  3. 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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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-6

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    • 25
  4. We are given a 3×3 matrix A that satisfies the equation: A² = 4A - 3I Step 1: Expressing A⁻¹ To find A⁻¹, we rearrange the given equation: A² - 4A + 3I = 0 Factoring, (A - I)(A - 3I) = 0 This implies that A satisfies the equation: (A - I)(A - 3I) = 0 Multiplying both sides by (A - I)⁻¹ (if it existsRead more

    We are given a 3×3 matrix A that satisfies the equation:

    A² = 4A – 3I

    Step 1: Expressing A⁻¹
    To find A⁻¹, we rearrange the given equation:

    A² – 4A + 3I = 0

    Factoring,

    (A – I)(A – 3I) = 0

    This implies that A satisfies the equation:

    (A – I)(A – 3I) = 0

    Multiplying both sides by (A – I)⁻¹ (if it exists),

    A – 3I = (A – I)⁻¹ 0

    Since A – I is invertible, we take its inverse on both sides,

    A⁻¹ = 1/3 (4I – A)

    Step 2: Selecting the Correct Option
    Comparing with the given choices, the correct answer is: 1/3 (4I – A)

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

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  5. A matrix with 5 elements can have different possible orders, provided that the total number of elements is the product of its rows and columns. Let the number of rows be m and the number of columns be n, then: m × n = 5 The possible values of (m, n) that satisfy this equation are: - 1 × 5 (1 row andRead more

    A matrix with 5 elements can have different possible orders, provided that the total number of elements is the product of its rows and columns.

    Let the number of rows be m and the number of columns be n, then:

    m × n = 5

    The possible values of (m, n) that satisfy this equation are:

    – 1 × 5 (1 row and 5 columns)
    – 5 × 1 (5 rows and 1 column)

    Since both 1 × 5 and 5 × 1 are possible, the correct answer is:

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

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