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Given that u = x^{m }y^{n}
Taking logarithm of both sides, we get log u = m log x + n log y Differentiating with respect to x,we get
f is a homogeneous function of degree one
It is a homogeneous function of degree n
It is a homogeneous function of
degree 2.
If an error of 1% is made in measuring the major and minor axes of an ellipse, then the percentage error in the area is approximately equal to
Let 2 a and 2 b be the major and minor axes of the ellipse
Consider the Assertion (A) and Reason (R) given below:
Reason (R): Given function u is homogeneous of degree 2 in x and y.
Of these statements
Given that u = xyf(y/x) Since it is a homogeneous function of degree 2.
If u = x log xy, where x^{3} + y^{3} + 3xy = 1, then du/dx is equal to
Given that u = x log xy ... (i)
The given function is homogeneous of degree 2.
If a < 0, then f(x) = e^{ax} + e^{ax} is decreasing for
The least value of a for which f(x) = x^{2} + ax + 1 is increasing on ] 1, 2, [ is
f'(x) = (2x + a)
The minimum distance from the point (4, 2) to the parabola y^{2} = 8x is
Let the point closest to (4, 2) be (2t^{2},4)
The coordinates of the point on the parabola y = x^{2} + 7x + 2 which is closest to the straight line y = 3x  3, are
Let the required point be P(x, y). Then, perpendicular distance of P(x, y) from y  3x  3 = 0 is
The shortest distance of the point (0, c), where 0 ≤ c ≤ 5, from the parabola y = x^{2} is
Let A (0,c) be the given point and P (x, y) be any point on y = x^{2}
f (x) = (1 / x)^{x}
f’ (x) = (1 / x)^{x} (log (1 / x) – 1))
f’ (x) = 0
log (1 / x) – 1 = log e
1 / x = e
x = 1 / e
The maximum value of function is e^{1/e}.
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