1 Crore+ students have signed up on EduRev. Have you? Download the App |
If f(x) is differentiable on interval l and such that |f'(x)| ≤ a on l, then f(x) is
For xy ∈ l, by Lagrange’s mean value theorem
where x < c < y
implies f(x)- f(y) = (x - y) f '(c )
implies |f(x) - f(y) | = x - y || f'(c)|
for a given such that |f(x) - f(y)| < ε, x, y ∈ l. Hence f(x) is uniformly continuous on l. We know that every uniformly continuous function is also continuous.
The value of dydx by changing the order of integration is
If f(x) is differentiable on (a, b) and f(a) = 0 and there exists a real number k such that |f'(x) ≤ k |f (x)| on [a, b], then f(x) is
Choose x1 > a such that k(x1 - a) < 1,
let α = sup | f(t) | on a < t < x1 then by Lagrange’s mean value theorem
where a < t< x ≤ x1
implies f(x) - f(a) = (x - a) f '(t)
[since f(a) = 0]
If f(x+y) = f (x) . f(y) for all x and y. Suppose that f(3) = 3 and f'(0) = 11 then , f'(3) is equal to
Given that f'(0) = 11
implies
implies
implies
[Since f(3 + 0) = f(3) - f(0)
=> f(3) = f(3) • (0)]
implies f(0) = 1 Now we have
The area bounded by the curve y = ψ(x), x-axis and the lines x = l , x = m(l <m ) is given by
If f(x) and g(x) are real number function defined on and
, , then which of following is correct?
Here,
cos f(x) - sin (g(x))
using eq. (ii) in eq. (i), then we obtain cos (f (x)) - sin (g (x)) > 0
or
The volume of an object expressed in spherical coordinates is given by
The value of the integral is
Using the transformation x + y = u, y = v. The value of Jacobian (J) for the integral is
Which of the following function is not called the Euler’s integral of the first kind?
Euler’s integral of the first kind is nothing but Beta function. So, here only is not the definition of Beta function.
Consider the following conditions
(i) f(x) is well defined at x = a
(ii) must exist.
(iii) f(x) is continuous.
(iv) f(x) ≠ 0 at x = a
Q. which of these conditions are necessary for a function f(x) to be derivable at a point x = a of its domain?
The necessary condition for a function f(x) to be derivable at a point x = a of its domain.
(i) f(x) is well defined at x = a
(ii) must exist.
(iv) f (x) is continuous.
Let us approaches (0, 0) along the line y = mx which passes through origin. Put y = mx, we get
which depends on m. Hence, does not exists, Therefore f(x, y) is discontinuous at (0,0).
Now,
and
Hence, f(x,y) is discontinuous at (0, 0). But both the partial derivative fx and fy exists at origin.
Let us suppose (x, y) approaches (0, 0) along the line y = mx. Which is a line through the origin. Put y = mx and allows x —> 0, we get
which depends on m, therefore the limit of f(x, y) at (0, 0) does not exists. Hence, f(x, y) is discontinuous at origin.
Now,
since fy exists at origin.
Let f : R2 —> R be defined by
Then, the directional derivative direction of f at (0 , 0) in the direction o f the vector is
we are given that u = and a = (0,0). we have directional derivative at a in the direction u as
L et f : R2 —> R be defined by
Then,
We have,
But f(0 , 0) = 0
Hence,
implies f(x, y) is continuous at (0, 0) Now,
where A = 0, B = 0 which does not depends on h and k and
Hence, y(x, y) is continuous as well as differentiable at (0, 0).
The inverse of function f(x) = x3 + 2 is ____________
To find the inverse of the function equate f(x) then find the value of x in terms of y such that f -1 (y) = x.
Let f : R2 —> R be such that and
exist at all points. Then,
Let f: R2 → R be defined by f(x, y)
Then the value of at the point (0, 0) is
Here, f(x, y)
27 docs|150 tests
|
27 docs|150 tests
|