Engineering Mathematics Miscellaneous


Engineering Mathematics Miscellaneous

Engineering Mathematics

  1. The root of the function f(x) = x3 + x – 1 obtained after first iteration on application of Newton Raphson scheme using an initial guess of x0 = 1 is









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    f (x) = x3 + x–1
    f (1) = 1+1 – 1 = 1
    f’(x) = 3x2 + 1
    f’ (1) = 3+1=4

    x1 = x0 -
    f(x0)
    = 1 -
    1
    f'(x0)4

    = 1 – 0.25
    x1 = 0.75

    Correct Option: C

    f (x) = x3 + x–1
    f (1) = 1+1 – 1 = 1
    f’(x) = 3x2 + 1
    f’ (1) = 3+1=4

    x1 = x0 -
    f(x0)
    = 1 -
    1
    f'(x0)4

    = 1 – 0.25
    x1 = 0.75


  1. Solve the equation x = 10 cos(x) using the Newton-Raphson method. The initial guess is x = π / 4. The value of the predicted root after the first iteration, up to second decimal, is _____.









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    Correct Option: A




  1. Newton-Raphson method is used to find the roots of the equation, x3 + 2x2 + 3x – 1 = 0. If the initial guess is x0 = 1, then the value of x after 2nd iteration is ________.









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    By Newton-Raphson Method,

    1st iteration, x1 = x0
    f(0)
    f'(0)

    = 1 -
    f(1)
    = 1 -
    5
    =
    1

    f'(1)102

    Where f(x) = x3 + 2x2 + 3x – 1 ⇒ f(1) = 5
    f'(x) = 3x2 + 4x + 3 ⇒ f'(1) = 10
    2nd iteration, x2 = x1
    f(x1)
    = 0.5 -
    f(0.5)
    = 0.3043
    f'(x1)f'(0.5)

    Correct Option: A

    By Newton-Raphson Method,

    1st iteration, x1 = x0
    f(0)
    f'(0)

    = 1 -
    f(1)
    = 1 -
    5
    =
    1

    f'(1)102

    Where f(x) = x3 + 2x2 + 3x – 1 ⇒ f(1) = 5
    f'(x) = 3x2 + 4x + 3 ⇒ f'(1) = 10
    2nd iteration, x2 = x1
    f(x1)
    = 0.5 -
    f(0.5)
    = 0.3043
    f'(x1)f'(0.5)


  1. The real root of the equation 5x – 2 cosx – 1 = 0 (up to two decimal accuracy) is _______.









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    Let f(x) =5x – 2 cos x – 1
    ⇒ f'(x) = 5 + 2 sin x
    f(0) = – 3; f(1) = 2.9
    By intermediate value theorem roots lie between 0 and 1.
    Let x0 = 1 rad = 57.32º
    By Newton Raphson method,

    Xn+1 = Xn -
    f(xn)
    f'(xn)

    ⇒ xn+1
    2xnsin xn + 2 cos xn + 1
    5 + 2Sin xn

    ⇒ x1 = 0.5632
    ⇒ x2 = 0.5425
    ⇒ x3 = 0.5424

    Correct Option: A

    Let f(x) =5x – 2 cos x – 1
    ⇒ f'(x) = 5 + 2 sin x
    f(0) = – 3; f(1) = 2.9
    By intermediate value theorem roots lie between 0 and 1.
    Let x0 = 1 rad = 57.32º
    By Newton Raphson method,

    Xn+1 = Xn -
    f(xn)
    f'(xn)

    ⇒ xn+1
    2xnsin xn + 2 cos xn + 1
    5 + 2Sin xn

    ⇒ x1 = 0.5632
    ⇒ x2 = 0.5425
    ⇒ x3 = 0.5424



  1. Let X and Y be two independent random variables. Which one of the relations between expectation (n), variance (Var) and covariance (Cov) given below is FALSE?









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    X and Y are independent
    ∴ (a), (b), (c) are true
    Only (d) is odd one

    Correct Option: D

    X and Y are independent
    ∴ (a), (b), (c) are true
    Only (d) is odd one