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Consider the function f(x) = x^{2} – x – 2. The maximum value of f(x) in the closed interval [–4, 4] is
∴ f (x )has minimum at x= 1 / 2 It Shows that a maximum value that will be at x = 4 or x =  4
At x = 4, f (x )= 10
∴ At x= −4, f (x ) = 18
∴ At x= −4, f (x ) has a maximum.
The function f(x) = x^{3} 6x^{2}+ 9x+25 has
The function f(x,y) = 2x^{2} +2xy – y^{3} has
Equation of the line normal to function f(x) = (x8)^{2/3}+1 at P(0,5) is
The distance between the origin and the point nearest to it on the surface z^{2} = 1 + xy is
or pr – q^{2} = 4 – 1 = 3 > 0 and r = +ve
so f(xy) is minimum at (0,0)
Hence, minimum value of d^{2} at (0,0)
d2 = x^{2} + y^{2} + xy + 1 = (0)^{2} + (0)^{2} + (0)(0) + 1 = 1
Then the nearest point is
z^{2} = 1 + xy = 1+ (0)(0) = 1
or z = 1
Given a function
The optimal value of f(x, y)
For the function f(x) = x^{2}ex, the maximum occurs when x is equal to
So maximum two extrema and three zero crossing
Consider the function y = x^{2} – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is
Consider the function y = x^{2} – 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is
Correct Answer : b
The magnitude of the gradient of the function f = xyz^{3} at (1,0,2) is
The expression curl (grad f), where f is a scalar function, is
If the velocity vector in a two – dimensional flow field is given by the vorticity vector, curl
The angle between two unitmagnitude coplanar vectors P (0.866, 0.500, 0) and Q (0.259, 0.966, 0) will be
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