Test of Mathematics Solution Subjective 144 - Finding a Function's Upper Bound

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 144 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

 

Problem

Suppose $ f(x)$ is a real valued differentiable function defined on $ [1,\infty)$ with $ f(1)=1$. Suppose moreover $ f(x)$ satisfies

$ f'(x) = \frac {1}{x^2+f^2(x)}$

Show that $ f(x) \leq 1+\frac{\pi}{4}$ for every $ x \geq 1$

Solution

As the question doesn't require us to find an exact solution rather just an upper bound, we can easily find it by manipulating the given statement after establishing certain properties of $ f(x)$.

We see that $ f'(x)>0$ for all $ x$ which means $ f(x)$ is an increasing function.

As the domain is $ [1,\infty)$ we can say that $ f(x)\geq  f(1) $ for all $ x$.

$ => f^2(x)\geq f^2(1)$

$ => x^2+f^2(x)\geq x^2+f^2(1)$         (as $ x^2> 0$)

$ => \frac{1}{x^2+f^2(x)}\leq \frac{1}{x^2+f^2(1)}$

$ => f'(x)\leq \frac{1}{x^2+f^2(1)}$

Integrating both sides from 1 to $ x$

$ => \int_{1}^{x}f'(x)\leq \int_{1}^{x}\frac{1}{x^2+f^2(1)}\leq \int_{1}^{\infty}\frac{1}{x^2+f^2(1)}$

As $ f(1)=1$ we have,

$ => \int_{1}^{x}f'(x)\leq \int_{1}^{\infty}\frac{1}{x^2+1}$

$ => f(x) - f(1) \leq tan^{-1}\infty - tan^{-1}1$

$ => f(x) - f(1) \leq \frac{\pi}{2} - \frac{\pi}{4}$

Substituting $ f(x) = 1$

$ => f(x) \leq  1+\frac{\pi}{4}$

Hence Proved.

Important Resources:

REAL ANALYSIS PROBLEM | TIFR A 201O | PROBLEM 5

Try this problem of TIFR GS-2010 from Real analysis, Differentiantiation and Maxima and Minima.

REAL ANALYSIS | TIFR 201O| PART A | PROBLEM 5

The maximum value of $f(x)=x^n(1-x)^n$ for natural number $n\geq 1$ and $0\leq x\leq1$

  • $\frac{1}{2^n}$
  • $\frac{1}{3^n}$
  • $\frac{1}{5^n}$
  • $\frac{1}{4^n}$

Key Concepts

REAL ANALYSIS

MAXIMA AND MINIMA

DIFFERENTIATION

Check the Answer

Answer:$\frac{1}{4^n}$

TIFR 2010|PART A |PROBLEM 1

AN INTRODUCTION TO ANALYSIS DIFFERENTIAL CALCULUS PART-I RK GHOSH, KC MAITY

Try with Hints

Here first differentiate $f(x)$

Then equate the terms of $f'(x)$ containing $x$ to $0$ and find all possible values of $x$,since your answer is in terms of $n$ no need to perform any kind of operations on $n$

Now equating $x$ we get $x=0,\frac{1}{2},1$

Now put each of these values of $x$ in $f(x)$ and see for which value of $x$ you get the maximum value of $f(x)$

you will get the maximum value of $f(x)$ for $x=\frac{1}{2}$ that is $\frac{1}{4^n}$

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Relation Mapping (IIT JAM 2014)

Question 33 - Relation-Mapping (IIT JAM 2014)

A beautiful problem involving the concept of relation-mapping from IIT JAM 2014.

Let $f:(0,\infty)\to \mathbb R$ be a differentiable function such that $f'(x^2)=1-x^3$ for all $x>0$ and $f(1)=0$. Then $f(4)$ equals

  • $\frac{-47}{5}$
  • $\frac{-47}{10}$
  • $\frac{-16}{5}$
  • $\frac{-8}{5}$

Key Concepts

Relation/Mapping

Differentiation

Integration

Check the Answer

Answer: (A) $ \frac{-47}{5}$

Try with Hints

The above problem can be done in many ways we will try to solve this by the simplest method.

Now, as the function is given as $f'(x^2)=1-x^3$

So first try to change this $x^3$ into $x^2$. Try this. It's very easy !!!

To change $x^3$ into $x^2$ we can easily do

$f'(x^2)=1-(x^2)^{\frac32}$

Now we have to find the value of $f(4)$ so we have to change the second degree term, i.e., $x^2$ into some linear form. Can you cook this up ???

Let us assume $x^2=y$

i.e., $f'(y)=1-y^{\frac32}$

Now you know from previous knowledge that integration is also known as anti-derivative. So $f'(y)$ can be changed into $f(y)$ by integrating it with respect to $y$. Try to do this integration and we are half way done !!!

On integrating both side w.r.t $y$ we get :

$f(y)=y-\frac25y^{\frac52}+c$, (where $c$ is a integrating constant.)

Now we find the value to $c$

We know $f(1)=0$

$\Rightarrow c=-\frac35$

i.e., $f(y)=y-\frac25y^{\frac52}-\frac35$

Can you find the answer now ?

Now simply, putting $y=4$

we get $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$

Other useful links

Subscribe to Cheenta at Youtube

Finding Tangent plane: IIT JAM 2018 problem 5

[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"]

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
  1. \(x-3y+z=0\)
  2. \(x+3y-2z=0\)
  3. \(2x+4y-3z=0\)
  4. \(3x-7y+2z=0\)
[/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_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"]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
[/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"]

Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="12" show_meta="off" image_placement="left" _builder_version="4.0.9"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

TIFR 2013 Problem 38 Solution -Eigenvalue of differentiation

TIFR 2013 Problem 38 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 Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem Type:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).

Hint:

If 0 were an eigenvalue, what would be its eigenvector?

Discussion:

There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,...,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.

HELPDESK