Test: Functions Of One,Two Or Three Real Variables -2 - IIT JAM MCQ

We know Rank A = Rank AT= 2
and also we know Rank (AB) £ min {Rank A, Rank B} as here we are taking product of matrix with its transpose.
Thus Rank (ATA) = 2

for x =0 and y =0 the value of f(x,y) = 0 hence the partial differentiation of the f(x,y) will be the zero so the value of f_x(0, 0) and, f_y(0, 0) are 0 respectively.

Limit along x → 0 and y → 0:

Limit along y → 0 and x → 0

Limit along the y = mx, then

Since the limit along any path is the same, the limit exists.

The graph of the greatest integer function is

Clearly the function y = f ( x ) = [x] is discontinuous at the points 0, ±1, ± 2, ± 3,.. the points of discontinuity are countable.

The product of two uniformly continuous functions may not be uniformly continuous we know that f (x) = x + 1 and g (x) = x - 1 are uniformly continuous on R but f(x) • g(x) = x² - 1 is not uniformly continuous on R.

∵ Equation of normal at (x, y) is
Y − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y² = 2x X
⇒ y² = 2x ( x + y dy/dx)
⇒ dy/dx = (y² − 2x²)/(2xy) = ((y/x)² − 2)/ ( 2y/x)
Put y = vx, we get dx/dy = v + x dv/dx
Then, v + x dv/dx = v²−2/2v
On integrating both sides, we get
ln (2 + v²) + ln|x| = ln c
⇒ ln (|x|(2 + v²)) = ln c
⇒ |x| ( 2 + y²/x²) = c
∵ It passes through (2, 1), then 2 (2 + 1 4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y²/x²) = 9/2
⇒ 2x² + y² = 9/2 |x|
⇒ 4x² + 2y² = 9|x|

Let f be the function defined on R by setting f (x) =   when x is not an integer , f(x ) = 0, then f is continuous, when x is an integer.

f_x = f_z z_x + f_w w_x      ...(1) 
g_x = g_z z_x + g_w w_x            ...(2) 
Multiply (1) and (2) by g_w and f_w respectively, then subtract, we obtain 
f_x g_w – g_x f_w = (f_z g_w – g_z f_w)z_x 
but f_z g_w = f_w g_z = 1, then 
z_x = f_x g_w – g_x f_w

Since, α is the repeated root of the polynomial equation.
f(x) = 0, then f(a) = 0 and f'(α) = 0
Clearly , f(x) = sin x is continuous in the interval [0 ,π] and differentiable in the open interval [0 , π]
Also , f(0) = f(π) = 0, then by Rolle’s theorem
f '(c) = 0 => cos c = 0
c = π/2 

Sin x is differentiable and xⁿ is also differentiable.
Hence, sin xⁿ is also differentiable.

 Given that

 
It is clear that f(x) is not continuous at the points x = nπ, n ∈ z, therefore it is not differentiable at the points.
Which f(x) > 1/2, and f (x) ≥ - 1, also there points are not countable. 

Solution : c)

Hence Monotonicaly increasing

Clearly, f(x) = cos x is continuous as well as differentiable in interval 
Then by Rolle’s theorem we have

Given that f(x) 
Since, f(x) is differentiable at x = 0 therefore, we have