Engineering Mathematics Miscellaneous

Engineering Mathematics Miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Engineering Mathematics

Engineering Mathematics Miscellaneous
  1. At x = 0, the function f(x) = x3 + 1 has








  1. View Hint View Answer Discuss in Forum

    Function f(x) = x 3 + 1 has a point of inflection at x = 0, since in the graph sign of the curvature (i.e., concavity) is changed.

    Correct Option: D

    Function f(x) = x 3 + 1 has a point of inflection at x = 0, since in the graph sign of the curvature (i.e., concavity) is changed.

  1. The area enclosed between the parabola y = x2 and the straight line y = x is








  1. View Hint View Answer Discuss in Forum

    Given equations are
    y = x2 ...(i)
    y = x ...(ii)
    From equations (i) and (ii)
    x 2 – x = 0
    ⇒ x(x – 1) = 0
    ⇒ x = 0, 1

    Area enclosed = dydx = ∫1x = 0 dx ∫y = x²y = x dy

    = [y]y = x²y = xdx = (x² - x) dx

    =
    x3
    		 -
    x2
    		1
    320

    =
    1
    		 -
    1
    		 =
    2 - 3
    		 = -
    1
    3266


    Correct Option: B

    Given equations are
    y = x2 ...(i)
    y = x ...(ii)
    From equations (i) and (ii)
    x 2 – x = 0
    ⇒ x(x – 1) = 0
    ⇒ x = 0, 1

    Area enclosed = dydx = ∫1x = 0 dx ∫y = x²y = x dy

    = [y]y = x²y = xdx = (x² - x) dx

    =
    x3
    		 -
    x2
    		1
    320

    =
    1
    		 -
    1
    		 =
    2 - 3
    		 = -
    1
    3266


  1. Consider a spatial curve in three-dimensional space given in parametric form by x(t) = cos t, y(t) = sin t,
    z(t) =
    2
    		t , 0 ≤ t ≤
    π
    		 The length of the curve is ___________.
    π2








  1. View Hint View Answer Discuss in Forum

    The length of the curve


    = 1.8622

    Correct Option: B

    The length of the curve


    = 1.8622

  1. The best approximation of the minimum value attained by e–x sin (100 x) for x > 0 is ______.








  1. View Hint View Answer Discuss in Forum

    f(x) = e– x sin(100x)
    f '(x) = – e– xsin(100x) + e– xcos(100x) × 100
    for minima f '(x) = 0
    tan(100 x) = 100

    x =
    1
    		tan-1(100) = 0.0156
    100

    f(x) = e– 0.0156 sin(100 × 0.0156) = 0.9844

    Correct Option: A

    f(x) = e– x sin(100x)
    f '(x) = – e– xsin(100x) + e– xcos(100x) × 100
    for minima f '(x) = 0
    tan(100 x) = 100

    x =
    1
    		tan-1(100) = 0.0156
    100

    f(x) = e– 0.0156 sin(100 × 0.0156) = 0.9844

  1. In the Taylor series expansion of ex about x = 2 the coefficient of (x – 2)4 is








  1. View Hint View Answer Discuss in Forum

    Taylor series expansion of f(x) about a is given by

    f(x) = f(a) +
    (x - a)
    		f '(a) +
    (x - a)2
    		f ''(a) + .......
    1!2!

    coefficient of (x - a)4 is
    f ''''(a)
    4!

    Now f(x) = ex
    ⇒ f ''''(x) = ex
    ⇒ f ''''(a) = ea
    Hence for a = 2,
    f ''''(a)
    		 =
    e2
    4!4!

    Correct Option: C

    Taylor series expansion of f(x) about a is given by

    f(x) = f(a) +
    (x - a)
    		f '(a) +
    (x - a)2
    		f ''(a) + .......
    1!2!

    coefficient of (x - a)4 is
    f ''''(a)
    4!

    Now f(x) = ex
    ⇒ f ''''(x) = ex
    ⇒ f ''''(a) = ea
    Hence for a = 2,
    f ''''(a)
    		 =
    e2
    4!4!