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If u = xm yn then
Given that u = xm yn
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
Match the List–I with List–II.
It is a homogeneous function of
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 x3 + y3 + 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) = eax + e-ax is decreasing for
f(x) = x2e-x is increasing in the interval
The least value of a for which f(x) = x2 + ax + 1 is increasing on ] 1, 2, [ is
f'(x) = (2x + a)
The minimum distance from the point (4, 2) to the parabola y2 = 8x is
Let the point closest to (4, 2) be (2t2,4)
The co-ordinates of the point on the parabola y = x2 + 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 = x2 is
Let A (0,c) be the given point and P (x, y) be any point on y = x2
The maximum value of ( 1/x)x is
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 e1/e.
The maximum value of f ( x) = (1 + cos x) sin x is
The greatest value of
on the interval [0, π/2] is
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