Functions of Several Variables MCQ Quiz in मल्याळम - Objective Question with Answer for Functions of Several Variables - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

 f(x, y) = $\{\begin{array}{ll}\frac{2xy}{x^2+y^2},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0).\end{array}$

f(x, y) is not continuous at (0,0) as

 $\underset{(x,y)\to (0,0)}{lim}\frac{2xy}{x^2+y^2}$

= $\underset{x\to 0}{lim}\frac{2x.mx}{x^2+m^2{x}^2}$ (along y = mx)

= $\frac{2m}{1+m^2}$, which is not unique

g(x, y) = $\sum _{n=1}^{\mathrm{\infty }}\frac{f((x-n),(y-n))}{2^n}$.

Now $f(r\cos \theta ,r\sin \theta )=\frac{2r^2\cos \theta \sin \theta }{r^2}$

= 2sinθcosθ = sin2θ  

so $|f(x-n,y-n)|\le 1\mathrm{\forall }n$

$\frac{|f(x-n,y-n)|}{2^n}\le \frac{1}{2^n}\mathrm{\forall }n$

 Now we will check the continuity of g(x,y)

 g(x, y) = $\sum _{n=1}^{\mathrm{\infty }}\frac{f((x-n),(y-n))}{2^n}$

$=\frac{f((x-1),(y-1))}{2^1}+\frac{f((x-2),(y-2))}{2^2}$

$+\frac{f((x-3),(y-3))}{2^3}+...$

Therefore g(x,y) is continuous on ${\mathbb{R}}^2$\$\{(k,k){\}}_{k\in \mathbb{N}}$

Option (4) is correct. 

Options (2) and (3) are incorrect.

h(y) = g(c,y)$=\frac{f((c-1),(y-1))}{2^1}+\frac{f((c-2),(y-2))}{2^2}$

$+\frac{f((c-3),(y-3))}{2^3}+...$

If c ∉ $\mathbb{N}$ then also h(y) is continuous.

 The function h(y) = g(c, y) is continuous on $\mathbb{R}$ for all c.

∴ Option (1) and Option (4) are correct.

Given:

 f(x, y) = $\sqrt{|xy|}$ 

Concept:

Apply definition of partial derivative .

Calculation:

 f(x, y) = $\sqrt{|xy|}$ 

Now,

${\mathrm{f}}_{\mathrm{x}}(0,0)=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{h}\to 0}\frac{\mathrm{f}(0+\mathrm{h},0)-\mathrm{f}(0,0)}{\mathrm{h}}=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{h}\to 0}\frac{0-0}{\mathrm{h}}=0$

And

${\mathrm{f}}_{\mathrm{y}}(0,0)=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{h}\to 0}\frac{\mathrm{f}(0,0+\mathrm{h})-\mathrm{f}(0,0)}{\mathrm{h}}=\mathrm{l}\mathrm{i}{\mathrm{m}}_{\mathrm{h}\to 0}\frac{0-0}{\mathrm{h}}=0$

Hence both partial derivative exists and equal to 0(zero) .

Concept:

A function of two variable f(x, y) is differentiable at (0, 0) if $\underset{h\to 0}{lim}\frac{||r(h,k)||}{||(h,k)||}$ = 0 where r(h, k) = f((0, 0)+(h, k))- f(0, 0) - $(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$  

Explanation:

$f(x,y)=x^{\frac{1}{3}}{y}^{\frac{1}{3}}$

(1): Directional derivative at (0, 0) in the direction of U = (u, v) is

D_U(0, 0) = $\underset{h\to 0}{lim}\frac{f((0,0)+h(u,v))-f(0,0)}{h}$ = $\underset{h\to 0}{lim}\frac{(hu{)}^{\frac{1}{3}}(hv{)}^{\frac{1}{3}}}{h}$ = $\underset{h\to 0}{lim}\frac{u^{\frac{1}{3}}{v}^{\frac{1}{3}}}{h^{\frac{1}{3}}}$ exist only if u = 0 or v = 0

(1) is correct

(2):  fx(0, 0) = $\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}$ = $\underset{h\to 0}{lim}\frac{0}{h}$ = 0

The partial derivative fx exists at (0, 0)

(2) is false

(3): $\underset{(x,y)\to (0,0)}{lim}f(x,y)$ 

= $\underset{(x,y)\to (0,0)}{lim}{x}^{\frac{1}{3}}{y}^{\frac{1}{3}}$ 

= $\underset{r\to 0}{lim}{r}^{\frac{2}{3}}{\cos }^{\frac{1}{3}}\theta {\sin }^{\frac{1}{3}}\theta$ (putting x = rcosθ, y = rsinθ)

= 0 = f(0, 0)

Hence f is continuous at (0, 0)

(3) is correct

(4): fy(0, 0) = $\underset{k\to 0}{lim}\frac{f(0,k)-f(0,0)}{k}$ = $\underset{k\to 0}{lim}\frac{0}{k}$ = 0

$(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$ = $\left(00\right)\left(\begin{array}{c}h\\ k\end{array}\right)$ = 0

So, r(h, k) = f((0, 0)+(h, k))- f(0, 0) - $(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$  

               = f(h, k) - 0 - 0 = $h^{\frac{1}{3}}{k}^{\frac{1}{3}}$ 

Now, $\underset{h\to 0}{lim}\frac{||r(h,k)||}{||(h,k)||}$ = $\underset{(h,k)\to (0,0)}{lim}\frac{h^{\frac{1}{3}}{k}^{\frac{1}{3}}}{\sqrt{h^2+k^2}}$ 

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

= $\underset{r\to 0}{lim}\frac{r^{\frac{2}{3}}{\cos }^{\frac{1}{3}}\theta {\sin }^{\frac{1}{3}}\theta }{r}$ = $\underset{r\to 0}{lim}\frac{{\cos }^{\frac{1}{3}}\theta {\sin }^{\frac{1}{3}}\theta }{r^{\frac{1}{3}}}$ does not exist 

Hence f is not differentiable at (0, 0).

(4) is correct

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:

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

and

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

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

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

$=\underset{h\to 0}{lim}\underset{k\to 0}{lim}\frac{k}{h+k}$

= 0

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

$=\underset{k\to 0}{lim}\frac{k}{k}$

= 1

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

Hence Option(3) is the correct answer.

Concept:

 Arithmetic Mean $\ge$ Geometric Mean

A.M $\ge$ G.M

Explanation:

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

For k=1

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

Let for k=t

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

For k = t+1

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

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

so, Option(1) is correct.

Option(2): 

$\sin \frac{x+y}{2}\le \frac{(\sin x+\sin y)}{2}$ 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 $\ge$ G.M

so, $\frac{e^x+e^y}{2}$$\ge$$\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 $\ge$ G.M

$\frac{\log x+\log y}{2}$$\ge$$\frac{1}{2}\log (x+y)$

Option(4) is not correct.

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

Explanation:

$f(x,y)=\{\begin{array}{ll}(x+y{)}^2\cos \frac{1}{x+y}& \text{ if }x\ne y\\ 0& \text{ if }x=y.\end{array}$

(1): fx(0, 0) = $\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}$ = $\underset{h\to 0}{lim}\frac{h^2\cos \frac{1}{h}-0}{h}$ = $\underset{h\to 0}{lim}h\cos \frac{1}{h}$ = 0

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

(1) is true

(3): fy(0, 0) = $\underset{k\to 0}{lim}\frac{f(0,k)-f(0,0)}{k}$ = $\underset{k\to 0}{lim}\frac{k^2\cos \frac{1}{k}-0}{k}$ = $\underset{k\to 0}{lim}k\cos \frac{1}{k}$ = 0

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

(3) is true.

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

                 = $\{\begin{array}{ll}2(x+y)\cos \frac{1}{x+y}+\sin \frac{1}{x+y}& \text{ if }x\ne y\\ 0& \text{ if }x=y.\end{array}$

Now, $\underset{(x,y)\to (0,0)}{lim}{f}_x(x,y)$ = $\underset{(x,y)\to (0,0)}{lim}(2(x+y)\cos \frac{1}{x+y}+\sin \frac{1}{x+y})$ does not exist as $\underset{(x,y)\to (0,0)}{lim}\sin \frac{1}{x+y}$ does not exist

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

(2) is false

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

                 = $\{\begin{array}{ll}2(x+y)\cos \frac{1}{x+y}+\sin \frac{1}{x+y}& \text{ if }x\ne y\\ 0& \text{ if }x=y.\end{array}$

Now, $\underset{(x,y)\to (0,0)}{lim}{f}_y(x,y)$ = $\underset{(x,y)\to (0,0)}{lim}(2(x+y)\cos \frac{1}{x+y}+\sin \frac{1}{x+y})$ does not exist as $\underset{(x,y)\to (0,0)}{lim}\sin \frac{1}{x+y}$ 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) :

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

$4x-2=0\Rightarrow x=\frac{1}{2}.$
 
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) = $(\frac{1}{2},\frac{1}{2})$ ,

Determine if This is a Minimum or Maximum: 

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

$\frac{d^2}{d{x}^2}(2x^2-2x+1)=4$

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:

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

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

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

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 $(\frac{1}{2},\frac{1}{2})$ ,

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

⇒ 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. 

Concept:

$\frac{\partial f}{\partial x}(0,0)=\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}$

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

Explanation:

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

$f(x,y)=\{\begin{array}{ll}\frac{y\sqrt{x^2+y^2}}{x},& \text{if }x\ne 0\\ 0,& \text{if }x=0\end{array}$

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

$\frac{\partial f}{\partial x}(0,0)=\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}$

When y = 0 , the function reduces to

$f(x,0)=0\text{for all}x\ne 0$

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

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:

$\frac{\partial f}{\partial y}(0,0)=\underset{h\to 0}{lim}\frac{f(0,h)-f(0,0)}{h}$

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

$\frac{\partial f}{\partial y}(0,0)=\underset{h\to 0}{lim}\frac{0-0}{h}=0$

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

Option 3: Along x = my² 

$\underset{(x,y)\to (0,0)}{lim}\frac{y\sqrt{x^2+y^2}}{x}$ = $\underset{y\to 0}{lim}\frac{y\sqrt{m^2{y}^4+y^2}}{m{y}^2}$

                                = $\underset{y\to 0}{lim}\frac{y^2\sqrt{m^2{y}^2+1}}{m{y}^2}$ = 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 $\underset{h\to 0}{lim}\frac{||r(h,k)||}{||(h,k)||}$ = 0 where r(h, k) = f((0, 0)+(h, k))- f(0, 0) - $(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$  

 Explanation:

$f(x,y)=\{\begin{array}{ll}(x-y{)}^2\sin \frac{1}{x-y}& \text{ if }x\ne y\\ 0& \text{ if }x=y.\end{array}$

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

$\underset{(x,y)\to (0,0)}{lim}f(x,y)$ = $\underset{(x,y)\to (0,0)}{lim}(x-y{)}^2\sin \frac{1}{x-y}$  

                             = $\underset{r\to 0}{lim}{r}^2(\cos \theta -\sin \theta {)}^2\sin \frac{1}{r(\cos \theta -\sin \theta )}$ = 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) = $\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}$ = $\underset{h\to 0}{lim}\frac{h^2\sin \frac{1}{h}-0}{h}$ = $\underset{h\to 0}{lim}h\sin \frac{1}{h}$ = 0

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

(2) is false

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

                 = $\{\begin{array}{ll}2(x-y)\sin \frac{1}{x-y}-\cos \frac{1}{x-y}& \text{ if }x\ne y\\ 0& \text{ if }x=y.\end{array}$

Now, $\underset{(x,y)\to (0,0)}{lim}{f}_x(x,y)$ = $\underset{(x,y)\to (0,0)}{lim}(2(x-y)\sin \frac{1}{x-y}-\cos \frac{1}{x-y})$ does not exist as $\underset{(x,y)\to (0,0)}{lim}\cos \frac{1}{x-y}$ does not exist

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

(3) is false

(4): fy(0, 0) = $\underset{k\to 0}{lim}\frac{f(0,k)-f(0,0)}{k}$ = $\underset{k\to 0}{lim}\frac{k^2\sin \frac{1}{k}-0}{k}$ = $\underset{k\to 0}{lim}k\sin \frac{1}{k}$ = 0

$(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$ = $\left(00\right)\left(\begin{array}{c}h\\ k\end{array}\right)$ = 0

So, r(h, k) = f((0, 0)+(h, k))- f(0, 0) - $(\frac{\partial f}{\partial x}(0,0)\frac{\partial f}{\partial y}(0,0))\left(\begin{array}{c}h\\ k\end{array}\right)$  

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

Now, $\underset{h\to 0}{lim}\frac{||r(h,k)||}{||(h,k)||}$ = $\underset{(h,k)\to (0,0)}{lim}\frac{(h-k{)}^2\sin \frac{1}{h-k}}{\sqrt{h^2+k^2}}$ 

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

= $\underset{r\to 0}{lim}\frac{r^2(\cos \theta -\sin \theta {)}^2\sin \frac{1}{r(\cos \theta -\sin \theta )}}{r}$ = 0

Hence f is differentiable at (0, 0).

(4) is correct