Functions of Several Variables MCQ Quiz in বাংলা - Objective Question with Answer for Functions of Several Variables - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

α = β = 1/4 and x = y then \(\frac{x^α y^β}{\sqrt{x^2+y^2}}\) = \(\frac{\sqrt{x.x}}{\sqrt{x^2+x^2}}\)= \(\frac{\sqrt x}{{x}\sqrt 2}\) → ∞ as x → 0

So (1) is false

Let x = rcosθ, y = rsinθ then

\(\lim_{(x, y)\to(0. 0)}\)\(\frac{x^{α} y^{β}}{\sqrt{x^2+y^2}}\) = \(\lim_{r\to0}\frac{r^{α}\cos^{α}\theta r^{β}\sin^{β}\theta}{r}\) = \(\lim_{r\to0}r^{α+β-1}\cos^{α}\theta \sin^{β}\theta\) = 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): \( \{(\alpha, \beta): \alpha>2, \beta>2\} \)

In this case α + β > 1 satisfies.

(2) is true.

(4): Let α = 1/2, β = 1/2

Then α + 4β > 1

But \(\lim_{(x, y)\to(0. 0)}\)\(\frac{x^{α} y^{β}}{\sqrt{x^2+y^2}}\) = \(\lim_{r\to0}r^{α+β-1}\cos^{α}\theta \sin^{β}\theta\) = \(\lim_{r\to0}r^{\frac12+\frac12-1}\cos^{\frac12}\theta \sin^{\frac12}\theta\) = \(\cos^{\frac12}\theta \sin^{\frac12}\theta\) ≠ 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:

\(f_{xy} = \lim_{h \to 0} \lim_{k \to 0} \frac{F(h,k)}{hk} \)

and

\(f_{yx} = \lim_{k \to 0} \lim_{h \to 0} \frac{F(h,k)}{hk} \)

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,  \(F(h, k) = \frac{hk^2}{h+ k} - 0 - 0 + 0 = \frac{hk^2}{h+k} \)

\(f_{xy} = \lim_{h \to 0} \lim_{k \to 0} \frac{hk^2}{hk(h+k)} \)

\(= \lim_{h \to 0} \lim_{k \to 0} \frac{k}{h+k} \)

= 0

\(f_{yx} = \lim_{k \to 0} \lim_{h \to 0} \frac{hk^2}{hk(h+k)} \)

\(= \lim_{k \to 0} \frac{k}{k} \)

= 1

Therefore, \(-5 (5 \frac{\partial^2 f}{\partial x \partial y} -5\frac{\partial^2 f}{\partial y \partial x})\) = \(-5 (5f_{xy} - 5 f_{yx}) \)  at (0,0) = -5(0 -5 ×1) = 25

Hence Option(3) is the correct answer.

Concept:

 Arithmetic Mean \(\geq \) Geometric Mean

A.M \(\geq \) G.M

Explanation:

Option(1): \(\frac{(x+y)^k}{2^k} \leq max{(x^k , y^k) } \) for all x, y>0 and all k \(\geq \)1.

For k=1

\(\frac{(x+y)}{2} < max{(x , y) } \)

Let for k=t

\(\frac{(x+y)^t}{2^t} \leq max{(x^t , y^t) } \) holds.

For k = t+1

LHS: \(\frac{(x+y) ^{t+1}}{2 ^{t+1}} \) = \(\frac{(x+y)^t}{2^t} \) \(\cdot \)\(\frac{(x+y)}{2} \) \(\leq max{(x^t , y^t) } \)\(\cdot \)\(max(x,y) \) \(\leq \) \(max{(x^{t +1}, y^{t+1} )} \) =RHS

Hence : \(\frac{(x+y)^k}{2^k} \leq max{(x^k , y^k) } \) for all x, y>0 is true by Principal of mathematical induction.

so, Option(1) is correct.

Option(2): 

\(\sin{ \frac{x+y}{2}} \leq \frac{(\sin x+\sin y)}{2} \) for all x, y>0

this does not hold for x= \(\frac{\pi}{6} \) and  y= \(\frac{\pi}{3} \).

LHS: sin \(\pi \) = 0 & RHS:  \(\sqrt{3} +1 \)

So, Option (2) is not correct.

Option(3):

Arithmetic mean for \(e^{x} \) and  \(e^y \)=  \(\frac{e^x + e^y }{2} \)

Geometric mean for \(e^{x} \) and  \(e^y \)=  \(\sqrt {e^x \cdot e^y } \)

since A.M \(\geq \) G.M

so, \(\frac{e^x + e^y }{2} \)\(\geq \)\(\sqrt {e^x \cdot e^y } \)

Option(3) is correct.

Option(4):

Arithmetic mean for \(\log x \) and  \(\log y\) = \(\frac{​​​​\log x+\log y}{2} \)

Geometric mean for \(\log x \) and  \(\log y\) = \(​​​​\sqrt{\log x \cdot \log y} \) = \(​​​​\frac{1}{2} {\log( x + y)} \)

since A.M \(\geq \) G.M

\(\frac{​​​​\log x+\log y}{2} \)\(\geq \)\(​​​​\frac{1}{2} {\log( x + y)} \)

Option(4) is not correct.

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

Explanation:

\(f(x, y)= \begin{cases}(x+y)^2 \cos \frac{1}{x+y} & \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

(1): fx(0, 0) = \(\lim_{h\to0}{f(h,0)-f(0,0)\over h}\) = \(\lim_{h\to0}{h^2\cos\frac 1h-0\over h}\) = \(\lim_{h\to0}h\cos\frac 1h\) = 0

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

(1) is true

(3): fy(0, 0) = \(\lim_{k\to0}{f(0,k)-f(0,0)\over k}\) = \(\lim_{k\to0}{k^2\cos\frac 1k-0\over k}\) = \(\lim_{k\to0}k\cos\frac 1k\) = 0

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

(3) is true.

(3): fx(x, y) = \(\begin{cases}2(x+y) \cos \frac{1}{x+y} -(x+y)^2\sin{1\over x+y}\left(-1\over(x+y)^2\right)& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

                 = \(\begin{cases}2(x+y) \cos \frac{1}{x+y} +\sin{1\over x+y}& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

Now, \(\lim_{(x, y)\to(0, 0)}f_x(x, y)\) = \(\lim_{(x, y)\to(0, 0)}\left(2(x+y) \cos \frac{1}{x+y} +\sin{1\over x+y}\right)\) does not exist as \(\lim_{(x, y)\to(0, 0)}\sin{1\over x+y}\) does not exist

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

(2) is false

(4): fy(x, y) = \(\begin{cases}2(x+y) \cos \frac{1}{x+y} -(x+y)^2\sin{1\over x+y}\left(-1\over(x+y)^2\right)& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

                 = \(\begin{cases}2(x+y) \cos \frac{1}{x+y} +\sin{1\over x+y}& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

Now, \(\lim_{(x, y)\to(0, 0)}f_y(x, y)\) = \(\lim_{(x, y)\to(0, 0)}\left(2(x+y) \cos \frac{1}{x+y} +\sin{1\over x+y}\right)\) does not exist as \(\lim_{(x, y)\to(0, 0)}\sin{1\over x+y}\) 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 \(f(x, y) = x^2 + y^2 \) , we get:

\(f(x, 1 - x) = x^2 + (1 - x)^2 = 2x^2 - 2x + 1. \)
 
To find the extrema,

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

\(\frac{d}{dx} \left( 2x^2 - 2x + 1 \right) = 4x - 2. \)
 
Setting the derivative to zero:

\(4x - 2 = 0 \quad \Rightarrow \quad x = \frac{1}{2}. \)
 
Substituting  \(x = \frac{1}{2} \)  into y = 1 x , then  \( y = \frac{1}{2} \) .

The minimum value of f(x, y) occurs at (x, y) = \(\left( \frac{1}{2}, \frac{1}{2} \right) \) ,

Determine if This is a Minimum or Maximum: 

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

\(\frac{d^2}{dx^2} \left( 2x^2 - 2x + 1 \right) = 4 \)

Since the second derivative is positive. 

The function f(x, 1 x) has a local minimum at x =\( \frac{1}{2} \), and hence a global minimum on the constraint line x + y = 1 .

Substitute x = \(\frac{1}{2} \)  into f(x, 1 x) to find the minimum value:

\(f\left( \frac{1}{2}, \frac{1}{2} \right) = 2\left( \frac{1}{2} \right)^2 - 2\left( \frac{1}{2} \right) + 1 = \frac{1}{2} - 1 + 1 = \frac{1}{2} \)

Thus, the minimum value of f(x, y) subject to the constraint g(x, y) = 1 is \(\frac{1}{2} \)  ,

and it occurs at\( \left( \frac{1}{2}, \frac{1}{2} \right) \).

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) = \(1^2 + 0^2 \) = 1

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

f(0, 1) = \( 0^2 + 1^2 \) = 1

These values are larger than the minimum value of \(\frac{1}{2} \) at \(\left( \frac{1}{2}, \frac{1}{2} \right) \) ,

confirming that the minimum occurs at \( \left( \frac{1}{2}, \frac{1}{2} \right) \) .

⇒ The function f has global extreme values at the points where  \(x = \pm 1, y = 0 \)  or  \(x = 0, y = \pm 1 \) .

Hence Option(2) is the correct Answer. 

Explanation:

\(f(x, y)= \begin{cases}(x-y)^2 \cos \frac{1}{x-y} & \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

(1) & (2): The function f(x, y) is continuous if \(\lim_{(x, y)\to(0, 0)}f(x, y)\) = f(0, 0)

\(\lim_{(x, y)\to(0, 0)}f(x, y)\) = \(\lim_{(x, y)\to(0, 0)}(x-y)^2\cos{1\over x-y}\)  

                             = \(\lim_{r\to0}r^2(\cosθ-\sinθ)^2\cos{1\over r(\cosθ-\sinθ)}\) = 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) = \(\lim_{h\to0}{f(h,0)-f(0,0)\over h}\) = \(\lim_{h\to0}{h^2\cos\frac 1h-0\over h}\) = \(\lim_{h\to0}h\cos\frac 1h\) = 0

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

(3) & (4) both are false.

Concept:

\(\frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}\)

and \(\frac{\partial f}{\partial y}(0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h}\)

Explanation:

We are given the function \(f: \mathbb{R}^2 \to \mathbb{R}\) , defined as

\(f(x, y) = \begin{cases} \frac{y \sqrt{x^2 + y^2}}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}\)

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

\(\frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h}\)

When y = 0 , the function reduces to

\(f(x, 0) = 0 \quad \text{for all} \, x \neq 0\)

Thus, \(\frac{\partial f}{\partial x}(0, 0) = \lim_{h \to 0} \frac{0 - 0}{h} = 0\)

Hence, \(\frac{\partial f}{\partial x}(0, 0)\) exists and is equal to 0.

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

using the limit definition:

\(\frac{\partial f}{\partial y}(0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h}\)

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

\(\frac{\partial f}{\partial y}(0, 0) = \lim_{h \to 0} \frac{0 - 0}{h} = 0\)

Therefore, \(\frac{\partial f}{\partial y}(0, 0)\) also exists and is equal to 0.

Option 3: Along x = my² 

\(\lim_{(x, y)\to(0, 0)}\frac{y \sqrt{x^2 + y^2}}{x}\) = \(\lim_{y\to0}\frac{y \sqrt{m^2y^4 + y^2}}{my^2}\)

                                = \(\lim_{y\to0}\frac{y^2 \sqrt{m^2y^2 + 1}}{my^2}\) = 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 \(\lim_{h\to0}\frac{||r(h, k)||}{||(h, k)||}\) = 0 where r(h, k) = f((0, 0)+(h, k))- f(0, 0) - \(\left({\partial f\over \partial x}(0, 0)\quad {\partial f\over \partial y}(0, 0)\right)\begin{pmatrix}h\\k\end{pmatrix}\)  

 Explanation:

\(f(x, y)= \begin{cases}(x-y)^2 \sin \frac{1}{x-y} & \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

(1): The function f(x, y) is continuous if \(\lim_{(x, y)\to(0, 0)}f(x, y)\) = f(0, 0)

\(\lim_{(x, y)\to(0, 0)}f(x, y)\) = \(\lim_{(x, y)\to(0, 0)}(x-y)^2\sin{1\over x-y}\)  

                             = \(\lim_{r\to0}r^2(\cosθ-\sinθ)^2\sin{1\over r(\cosθ-\sinθ)}\) = 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) = \(\lim_{h\to0}{f(h,0)-f(0,0)\over h}\) = \(\lim_{h\to0}{h^2\sin\frac 1h-0\over h}\) = \(\lim_{h\to0}h\sin\frac 1h\) = 0

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

(2) is false

(3): f_x(x, y) = \(\begin{cases}2(x-y) \sin \frac{1}{x-y} +(x-y)^2\cos{1\over x-y}\left(-1\over(x-y)^2\right)& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

                 = \(\begin{cases}2(x-y) \sin \frac{1}{x-y} -\cos{1\over x-y}& \text { if } x \neq y \\ 0 & \text { if } x=y.\end{cases}\)

Now, \(\lim_{(x, y)\to(0, 0)}f_x(x, y)\) = \(\lim_{(x, y)\to(0, 0)}\left(2(x-y) \sin \frac{1}{x-y} -\cos{1\over x-y}\right)\) does not exist as \(\lim_{(x, y)\to(0, 0)}\cos{1\over x-y}\) does not exist

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

(3) is false

(4): fy(0, 0) = \(\lim_{k\to0}{f(0,k)-f(0,0)\over k}\) = \(\lim_{k\to0}{k^2\sin\frac 1k-0\over k}\) = \(\lim_{k\to0}k\sin\frac 1k\) = 0

\(\left({\partial f\over \partial x}(0, 0)\quad {\partial f\over \partial y}(0, 0)\right)\begin{pmatrix}h\\k\end{pmatrix}\) = \(\left(0\quad 0\right)\begin{pmatrix}h\\k\end{pmatrix}\) = 0

So, r(h, k) = f((0, 0)+(h, k))- f(0, 0) - \(\left({\partial f\over \partial x}(0, 0)\quad {\partial f\over \partial y}(0, 0)\right)\begin{pmatrix}h\\k\end{pmatrix}\)  

               = f(h, k) - 0 - 0 = \((h-k)^2\sin{1\over h-k}\) 

Now, \(\lim_{h\to0}\frac{||r(h, k)||}{||(h, k)||}\) = \(\lim_{(h, k)\to(0,0)}{(h-k)^2\sin{1\over h-k}\over \sqrt{h^2+k^2}}\) 

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

= \(\lim_{r\to0}{r^2(\cosθ-\sinθ)^2\sin{1\over r(\cosθ-\sinθ)}\over r}\) = 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