Test: Functions Of One,Two Or Three Real Variables -1 - Mathematics MCQ

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.

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]

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

Here,
cos f(x) - sin (g(x))


using eq. (ii) in eq. (i), then we obtain cos (f (x)) - sin (g (x)) > 0
or 

Euler’s integral of the first kind is nothing but Beta function. So, here only is not the definition of Beta function.

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 f_x 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 f_y exists at origin.

we are given that u = and a = (0,0). we have directional derivative at a in the direction u as 

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).

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.

Here, f(x, y)