Functions of Several Variables MCQ Quiz in தமிழ் - Objective Question with Answer for Functions of Several Variables - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

α = β = 1/4 and x = y then  = =  → ∞ as x → 0

So (1) is false

Let x = rcosθ, y = rsinθ then

 =  =  = 0 if α + β - 1 > 0

i.e., if α + β > 1

(3) is false. Also by taking similar example as (1), we can show that (3) is false.

(2): 2, \beta>2\} \)

In this case α + β > 1 satisfies.

(2) is true.

(4): Let α = 1/2, β = 1/2

Then α + 4β > 1

But  =  =  =  ≠ 0

(4) is false.

Explanation:

it is common for many functions that are radially unbounded to also satisfy v(0) = 0,

especially in the field of machine learning and optimization, given that many loss/objective functions coincide with this property. 

Concept:

and

where F(h, k) = f(h, k) - f(h,0) - f(0,k) + f(0,0).

Explanation: 

Since F(h, k) = f(h, k) - f(h,0) - f(0,k) + f(0,0)

Now,  

= 0

= 1

Therefore,  =   at (0,0) = -5(0 -5 ×1) = 25

Hence Option(3) is the correct answer.

Concept:

 Arithmetic Mean  Geometric Mean

A.M  G.M

Explanation:

Option(1):  for all x, y>0 and all k 1.

For k=1

Let for k=t

 holds.

For k = t+1

LHS:  =      =RHS

Hence :  for all x, y>0 is true by Principal of mathematical induction.

so, Option(1) is correct.

Option(2): 

 for all x, y>0

this does not hold for x=  and  y= .

LHS: sin  = 0 & RHS:  

So, Option (2) is not correct.

Option(3):

Arithmetic mean for  and  =  

Geometric mean for  and  =  

since A.M  G.M

so, 

Option(3) is correct.

Option(4):

Arithmetic mean for  and   = 

Geometric mean for  and   =  = 

since A.M  G.M

Option(4) is not correct.

Hence Option(2) and Option(4) are Answers.

Explanation:

(1): fx(0, 0) =  =  =  = 0

So, the partial derivative fx exist at (0, 0).

(1) is true

(3): fy(0, 0) =  =  =  = 0

So, the partial derivative fy exist at (0, 0).

(3) is true.

(3): fx(x, y) = 

                 = 

Now,  =  does not exist as  does not exist

Hence the partial derivative fx is not continuous at (0, 0).

(2) is false

(4): fy(x, y) = 

                 = 

Now,  =  does not exist as  does not exist

Hence the partial derivative fy is not continuous at (0, 0).

(4) is false.

Explanation:

The constraint g(x, y) = 1 represents the line x + y = 1 .

Substituting y = 1 x into  , we get:

 
To find the extrema,

we compute the derivative of f(x, 1 x) :

 
Setting the derivative to zero:

 
Substituting    into y = 1 x , then   .

The minimum value of f(x, y) occurs at (x, y) = ,

Determine if This is a Minimum or Maximum: 

the second derivative of f(x, 1 x) :

Since the second derivative is positive. 

The function f(x, 1 x) has a local minimum at x =, and hence a global minimum on the constraint line x + y = 1 .

Substitute x =   into f(x, 1 x) to find the minimum value:

Thus, the minimum value of f(x, y) subject to the constraint g(x, y) = 1 is   ,

and it occurs at.

Check the Behavior at Boundary Points

We now check the behavior of f(x, y) at some boundary points where g(x, y) = 1 .

When x = 1 , we get y = 0 , so:

f(1, 0) = = 1

When x = 0 , we get y = 1 , so:

f(0, 1) =  = 1

These values are larger than the minimum value of at ,

confirming that the minimum occurs at  .

⇒ The function f has global extreme values at the points where    or   .

Hence Option(2) is the correct Answer. 

Explanation:

(1) & (2): The function f(x, y) is continuous if  = f(0, 0)

 =   

                             =  = 0 = f(0, 0)

  (putting x = r cosθ, y = r sinθ )

Hence f(x, y) is continuous at (0, 0)

(1) is correct and (2) is incorrect.

(3) & (4): fx(0, 0) =  =  =  = 0

So, the partial derivative fx exist at (0, 0).

(3) & (4) both are false.

Concept:

and 

Explanation:

We are given the function  , defined as

We are tasked with determining the truth of various statements about the

partial derivatives, continuity, and differentiability of this function at (0, 0) .

Option 1: To check whether the partial derivative with respect to x exists at (0, 0) ,

we calculate it as follows

When y = 0 , the function reduces to

Thus, 

Hence,  exists and is equal to 0.

Option 2: We calculate the partial derivative with respect to y at (0, 0)

using the limit definition:

When x = 0 , the function is defined as f(0, y) = 0 . Thus,

Therefore,  also exists and is equal to 0.

Option 3: Along x = my² 

 = 

                                =  = 1/m which is different for different value of m

Thus, f is not continuous at (0, 0) .

Option 4: A function is differentiable at (0, 0) if it is continuous and its partial derivatives exist and

are continuous in a neighborhood of (0, 0) . Since the function is not continuous at (0, 0) , it cannot

be differentiable there. Thus, f is not differentiable at (0, 0) .

Hence correct options are 1), 2), 3) and 4).

Concept:

A function of two variable f(x, y) is differentiable at (0, 0) if  = 0 where r(h, k) = f((0, 0)+(h, k))- f(0, 0) -   

 Explanation:

(1): The function f(x, y) is continuous if  = f(0, 0)

 =   

                             =  = 0 = f(0, 0)

  (putting x = r cosθ, y = r sinθ )

Hence f(x, y) is continuous  at (0, 0)

(1) is correct

(2): f_x(0, 0) =  =  =  = 0

So, the partial derivative fx exist at (0, 0).

(2) is false

(3): f_x(x, y) = 

                 = 

Now,  =  does not exist as  does not exist

Hence the partial derivative fx is not continuous at (0, 0).

(3) is false

(4): fy(0, 0) =  =  =  = 0

 =  = 0

So, r(h, k) = f((0, 0)+(h, k))- f(0, 0) -   

               = f(h, k) - 0 - 0 =  

Now,  =  

Putting h = r cosθ, k = r sinθ we get 

=  = 0

Hence f is differentiable at (0, 0).

(4) is correct

Concept:

The range of a function f(x, y) refers to all the possible values f(x, y) could be

Solution: 

Here given  f : ℝ2 → ℝ2 and f(x,y) = (ex cos(y), ex sin(y)) 

We know that the range of cos y and sin y lie between -1 to 1 

and range of ex is (0, ∞) 

Hence the range of given function f(x,y) is ℝ2 \ (0,0) 

The point in ℝ2 that does NOT lie in the range of f is (0,0)

number of such points is 1

(2) is correct