Partial Differentiation | IIT JAM 2017 | Problem 5

Try this problem from IIT JAM 2017 exam (Problem 5) based on Partial Differentiation. It deals with calculating the partial derivative of a multi-variable function.

Partial Differentiation | IIT JAM 2017 | Problem 5

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function. If $g(u, v)=f\left(u^{2}-v^{2}\right),$ then
$\frac{\partial^{2} g}{\partial u^{2}}+\frac{\partial^{2} g}{\partial v^{2}}=$

$4\left(u^{2}-v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)$
$4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)$
$2 f^{\prime}\left(u^{2}-v^{2}\right)+4\left(u^{2}-v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)$
$2(u-v)^{2} f^{\prime \prime}\left(u^{2}-v^{2}\right)$

Key Concepts

Real Analysis

Function of Multi-variable

Partial Differentiation

Check the Answer

Answer: $4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)$

IIT JAM 2017, Problem 5

Real Analysis : Robert G. Bartle

Try with Hints

Here $g$ is a function of $u$ and $v$, to calculate $\frac{\partial g}{\partial u}$ we will differentiate the function $g$ with respect to $u$ keeping $v$ as constant.

and to calculate $\frac{\partial g}{\partial v}$ we will differentiate the function $g$ with respect to $v$ keeping $u$ as constant.

and $\frac{\partial^2 g}{\partial u^2}= \frac{\partial }{\partial u} [ \frac{\partial g}{\partial u} ]$

Hmm... I think you can easily do it from here ........

$\begin{aligned}\frac{\partial^{2} g}{\partial u^{2}}=\frac{\partial}{\partial u}\left(\frac{\partial u}{\partial u}\right) &=\frac{\partial}{\partial u}\left[f^{\prime}\left(v^{2}-v^{2}\right) \cdot 2 u\right] \\&=2 u \cdot f^{\prime \prime}\left(v^{2}-v^{2}\right) \cdot 2 u+f^{\prime}\left(v^{2}-v^{2}\right) \cdot 2 \\&=4 u^{2} f^{\prime \prime}\left(v^{2}-v^{2}\right)+2 f^{\prime}\left(v^{2}-v^{2}\right)\ldots\ldots(i)\end{aligned}$

Similarly,

$\begin{aligned}\frac{\partial^{2} g}{\partial v^{2}}=\frac{\partial}{\partial v}\left(\frac{\partial g}{\partial v}\right) &=\frac{\partial}{\partial v}\left[f^{\prime}\left(v^{2}-v^{2}\right) \cdot (-2 v)\right] \\&=(-2 v) \cdot f^{\prime \prime}\left(v^{2}-v^{2}\right) \cdot (-2 v)+f^{\prime}\left(v^{2}-v^{2}\right) \cdot (-2) \\&=(4 v^{2}) f^{\prime \prime}\left(v^{2}-v^{2}\right)-2 f^{\prime}\left(v^{2}-v^{2}\right) \ldots\ldots(ii) \end{aligned}$

Adding $(i)$ and (ii) we get,

$\frac{\partial^{2} g}{\partial u^{2}}+\frac{\partial^{2} g}{\partial x^{2}}$

$=4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)+2 f^{\prime}\left(u^{2}-v^{2}\right)-2 f^{\prime}\left(u^{2}-v^{2}\right)$

$=4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right) \textbf{[Ans]}$

Acute angles between surfaces: IIT JAM 2018 Qn 6

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Warm yourself up with an MCQ

[/et_pb_text][et_pb_code _builder_version="4.0.9"][/et_pb_code][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]In $latex \Bbb R^3$ the cosine of acute angle between the surfaces $latex x^2+y^2+z^2-9=0$ and $latex z-x^2-y^2+3=0$ at the point $latex (2,1,2)$ is

$latex \frac{8}{5\sqrt{21}}$
$latex \frac{10}{5\sqrt{21}}$
$latex \frac{8}{3\sqrt{21}}$
$latex \frac{10}{3\sqrt{21}}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" custom_padding="|0px||||"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0.9" hover_enabled="0"]IIT JAM 2018 Qn no 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0.9" hover_enabled="0" open="off"]Multivable calculus[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" hover_enabled="0" open="on"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" hover_enabled="0" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0.9" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]If we are asked to give the angle between two lines then it is very easy to calculate but our forehead will get skinned whenever we will be asked to find out the acute angle between two lines and even worse if we are asked to find the angle between two surfaces. Surprisingly it is not very hard if think stepwise. Observe when you are asked to find out the angle between two lines you calculate it in terms slope. So basically you are firing putting the gun on someone else's shoulder. Here the question is to find that shoulder when it comes in finding the angle between two curves. Observe from the conception of the intersection of two curves that the tangent line of those curves also intersects and we have their corresponding slopes. Bingo! why not calculating the acute angle of the tangent lines and call them the angle between two curves. Now can you think how to calculate the acute angle between to surfaces?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]The acute angle between two surfaces would be the acute angle between their tangent plane. You can stop here and try to do the problem by your own otherwise continue... The main idea of finding tangent planes revolves around finding gradient of the corresponding surfaces. (For more info see question no 5). Can you calculate the gradient of the surfaces $latex x^2+y^2+z^2-9$ and $latex z-x^2-y^2+3$ at $latex (2,1,2)$?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]The gradient of the surfaces $latex f=x^2+y^2+z^2-9$ and $latex g=z-x^2-y^2+3$ at $latex (2,1,2)$ are $latex n_1=f_xi +f_yj+f_zk$ and $latex n_2=g_xi +g_yj+g_zk$ at $latex (2,1,2)$ which is $latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$. Now given these two gradients, can you find out the angle between them?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]$latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$.

This follows the cosine angles between two gradient is $latex cos \theta=|\frac{n_1.n_2}{|n_1||n_2|}|=|\frac{-16-4+4}{\sqrt{36 \times 21}}|=\frac{8}{3\sqrt{21}}$

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Explanation of hints with graph

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-61-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-62-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-63-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-64-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-65-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Knowledge Graph

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/drawit-diagram.png" _builder_version="4.0.9"][/et_pb_image][et_pb_code _builder_version="3.26.4"]https://apis.google.com/js/platform.js

[/et_pb_code][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="College Mathematics Program" url="https://cheenta.com/collegeprogram/" image="https://cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://cheenta.com/collegeprogram/" border_color_all="#e02b20"]

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Finding Tangent plane: IIT JAM 2018 problem 5

What are we learning?

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" text_font_size="18px" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]We will learn to find tangent plane by solving an IIT JAM 2018 Problem. This is the Question no. 5 of the IIT JAM 2018 Solved Paper Series. Go through this link for Question no. 6. Gradient is one of the key concepts of vector calculus. We will use this problem from IIT JAM 2018 to clear our concepts.

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="50px||50px||false|false" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The tangent plane to the surface $latex z= \sqrt{x^2+3y^2}$ at (1,1,2) is given by

$x-3y+z=0$
$x+3y-2z=0$
$2x+4y-3z=0$
$3x-7y+2z=0$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0.9"]IIT Jam 2018[/et_pb_accordion_item][et_pb_accordion_item title="Key competency" _builder_version="4.0.9" open="off"]Gradient[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Look at the knowledge graph...

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-2018-Problem-5.png" align="center" admin_label="knowledge graph" _builder_version="4.0.9"][/et_pb_image][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]Given a differentiable function $Z=f(x,y)$, Observe that when we are asked to find a tangent plane at $(x_0,y_0,z_0)$ then the picture that comes in our mind is a plane that touches the curve at a point.

When we are in dimension $2$ it is just a line, (easy to visualize), dim 3 a plane (still visible), dim 4,5,…. a surface which is hard to see, but we can plug in $x=x_0$ in the equation $z=f(x,y)$ to have $z=f(x_0,y)$ which is just a curve in 2D then we can visualize the tangent line at $y=y_0$ is a part of the tangent plane $z=f(x,y)$ isn’t it?? The same thing is true of about the tangent line at $x=x_0$ for the curve $z=f(x,y_0)$. These $f(x,y_0)$ and $f(x_0,y)$ are called sections of the curve $f(x,y)=z$ . Here $(x_0,y_0,z_0)=(1,2,3)$. So, quickly find out $f(1,y)$ and $f(x,1)$.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]You can see that $f(1,y)=\sqrt{1+3y^{2}}$ and $f(x,1)= \sqrt{x^{2}+3}$ Now observe that the tangent plane of the curve $z=f(x,y)$ is a plane right !! What will be the basic structure of a plane at $(x_0,y_0,z_0)$?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]

It is a $a(x-x_0)+ b(y-y_0)+ c(z-z_0)=0$ ----------------------(1) Now see that $(x_0,y_0,z_0)=(1,1,2)$ is already given in the question. Hence the unknown is $(a,b,c)$ . Equation (1) implies $z = z_0+ \frac{a}{c}(x-x_0)+ \frac{b}{c}(y-y_0)$ Differentiating the equation by $x$ we get, $z_x= \frac{a}{c}$ Differentiating the equation by $y$ we get, $z_y= \frac{b}{c}$ Hence the equation of the tangent plane is $z=z_0+z_x|_{(x_0,y_0)}(x-x_0)+ z_y|_{(x_0,y_0)}(y-y_0)$ So calculate $z_x$ and $z_y$ at $(x_0,y_0)$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]

$z_x = \frac{d}{dx}f(x,1)= \frac{2x}{2\sqrt{x^{2}+3}}|_{(1,1)} = \frac{2}{4}= \frac{1}{2}$ $z_y=\frac{d}{dy}f(1,y)=\frac{6y}{2\sqrt{1+3y^2}}|_{(1,1)}=\frac{6}{2 \times 2}=\frac{3}{2}$ So the equation of the tangent line is $z= 2+\frac{1}{2}(x-1)+\frac{3}{2}(y-1)$ $\Rightarrow 2z= 4+x-1+3y-3$

$x+3y-2z=0$ (Ans)

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="20px||20px||false|false" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Try to answer this question

[/et_pb_text][et_pb_code admin_label="quiz" _builder_version="4.0.9"][/et_pb_code][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Play with graph

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/JAM-5-1-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-51-1-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-52-scaled.jpg" _builder_version="4.1" hover_enabled="0"][/et_pb_image][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/IIT-JAM-53-scaled.jpg" _builder_version="4.0.9"][/et_pb_image][et_pb_divider _builder_version="4.0.9"][/et_pb_divider][et_pb_code _builder_version="3.26.4"]https://apis.google.com/js/platform.js

Connected Program at Cheenta

TIFR 2014 Problem 22 Solution -An application of Intermediate Value Theorem

TIFR 2015 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a continuous map such that $f(x)=0$ for only finitely many values of (x). Which of the following is true?

A. either $f(x) \le 0$ for all (x) or $f(x) \ge 0$ for all (x).

B. the map $f$ is onto

C. the map $f$ is one-to-one

D. none of the above

Discussion:

Let $f$ be the map $f(x_1,x_2)=x_1^2 +x_2^2 $. Then $f$ is zero at only $(0,0)$. $f$ is continuous because $x=(x_1,x_2) \to x_1 \to x_1^2 $ is continuous from $\mathbb{R}^2 \to \mathbb{R} \to \mathbb{R}$. i.e, each arrow is continuous. The first arrow is the projection map, and such maps are always continuous, and the second arrow is just squaring, which is continuous. And composition of continuous functions are continuous, so $x \to x_1^2 $ is continuous function from $\mathbb{R}^2 \to \mathbb{R}$. Where here and henceforth $x=(x_1,x_2)\in \mathbb{R}^2 $.

Similar reasoning will show that $x \to x_2^2 $ is continuous function from $\mathbb{R}^2 \to \mathbb{R}$.

Sum of continuous functions is continuous, so the map $x \to x_1^2+ x_2^2 $ is continuous function from $\mathbb{R}^2 \to \mathbb{R}$.

This function $f$ is not one-one since $f(1,0)=f(0,1)=1$ and it is not onto since it only takes values in $[0,\infty ) $.

So we now are sure that B,C are false options.

We will prove A now.

Let $x=(x_1,x_2)$ and $y=(y_1,y_2)$ be two points in $\mathbb{R}^2$ such that $f(x)>0$ and $f(y)<0$.

We will prove that this will imply infinitely many zeros in between $x$ and $y$. But wait a second... what does between mean in this context? For that we consider the paths between $x$ and $y$. Note that there are infinitely many paths between any two points in $\mathbb{R}^2$. Further, we can in fact have infinitely many paths completely disjoint except for the initial and final points. We show that corresponding to each path $\alpha: [0,1] \to \mathbb{R}^2$ which connects $x$ and $y$ we have a zero in the path. Since there are infinitely many disjoint paths, we get infinitely many distinct zeros for $f$.

Now, $\alpha: [0,1] \to \mathbb{R}^2$ is a continuous function , $\alpha(0)=x$, $\alpha(1)=y$.

Consider the composition $g=f o \alpha : [0,1] \to \mathbb{R}$. $g$ is continuous. $g(0)=f(x)>0$ and $g(1)=f(y)<0$.

Therefore by the intermediate value theorem, $g(c)=0$ for some $c\in [0,1]$.

That means, $f(\alpha(c))=0$. And using the discussion above we get a contradiction.

This proved the option A.

Chatuspathi

What is this topic: Real Analysis
What are some of the associated concept: Continuity,Intermediate Value Theorem
Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

Partial Differentiation | IIT JAM 2017 | Problem 5