ISI MStat PSB 2012 Problem 2 | Dealing with Polynomials using Calculus

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 2 based on calculus . Let's give it a try !!

Problem- ISI MStat PSB 2012 Problem 2

Let \(f\) be a polynomial. Assume that \( f(0)=1, \lim _{x \rightarrow \infty} f''(x)=4\) and \( f(x) \geq f(1) \) for all \( x \in \mathbb{R} .\) Find \( f(2)\) .

Prerequisites

Limit

Derivative

Polynomials

Solution :

Here given \(f(x) \) is a polynomial and \( \lim _{x \rightarrow \infty} f''(x)=4\)

So, Case 1: If f(x) is a polynomial of degree 1 then f''(x)=0 hence limit can't be 4.

Case 2: If f(x) is a polynomial of degree 2 ,say \( f(x) = ax^2+bx+c \) then \( f''(x)= 2a \) .Hence taking limit we get \( 2a=4 \Rightarrow a=2 \)

Case 3: If f(x) is a polynomial of degree >2 then \( f''(x) = O(x) \) . So, it tends to infinity or - infinity as x tends to infinity .

Therefore the only case that satisfies the condition is Case 2 .

So , f(x) = \( 2x^2+bx+c \) ,say . Now given that \( f(0)=1 \Rightarrow c=1 \) .

Again , it is given that \( f(x) \geq f(1) \) for all \( x \in \mathbb{R} \) which implies that f(x) has minimum at x=1 .

That is f'(x)=0 at x=1 . Here we have \( f'(x)=4x+b=0 \Rightarrow x=\frac{-b}{4}=1 \Rightarrow b=-4 \)

Thus we get \( f(x)=2x^2-4x+1 \) . Putting x=2 , we get \( f(2)=1 \) .

Food For Thought

Assume f is differentiable on \( (a, b)\) and is continuous on \( [a, b]\) with \( f(a)=f(b)=0\). Prove that for every real \( \lambda\) there is some c in \( (a, b)\) such that \( f'(c)=\lambda f(c) \).

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Derivative Problem | TOMATO BStat Objective 764

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative.

Derivative Problem (B.Stat Objective Question )

If y=\(3^\frac{sinax}{cosbx}\), then \(\frac{dy}{dx}\) is

  • \(3^\frac{sinax}{cosbx}\frac{acosaxcosbx-bsinaxsinbx}{cos^{2}bx}log3\)
  • \(3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3\)
  • \(3^\frac{sinax}{cosbx}log3\)
  • \(3^\frac{sinax}{cosbx}\)

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer:\(3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3\)

B.Stat Objective Problem 764

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

\(y=3^\frac{sinax}{cosbx}\)

or, \(\frac{dy}{dx}\)=\(3^{\frac{sinax}{cosbx}}log_{e}{3}\frac{d}{dx}\frac{sinax}{cosbx}\)

or, \(\frac{dy}{dx}\)=\(3^{\frac{sinax}{cosbx}}log_{e}{3}{\frac{cosbxcosaxa-sinax(-sinbx)b}{cos^{2}bx}}\)

or, \(3^\frac{sinax}{cosbx}\frac{acosaxcosbx+bsinaxsinbx}{cos^{2}bx}log3\)

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Derivative of Function Problem | TOMATO BStat Objective 756

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function.

Derivative of Function Problem (B.Stat Objective Question )

Let f(x)=x[x] where [x] denotes the greatest integer smaller than or equal to x where x is not an integer, what is f'(x)?

  • 2x
  • [x]
  • 2[x]
  • it does not exist.

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer: [x].

B.Stat Objective Problem 756

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

\(f(x)=x[x]\) where f(x) discontinuous at integer points so non differentiable at those pts

so we find for non-integer points

if x is not an integer then [x] is a constant and not equal to x

x is non integer so here [x] is treated as constant here . Let [x] = a then f(x) =ax => f'(x)= a = [x] . Hence option [x] is correct.

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ISI MStat PSA 2019 Problem 17 | Limit of a function

This is a beautiful problem from ISI Mstat 2019 PSA problem 17 based on limit of a function . We provide sequential hints so that you can try this.

Limit of a function

If \( f(a)=2, f'(a)=1 , g(a)=-1\) and \(g'(a)=2\) , then what is

 \( \lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x))}{(x – a)} \) ?

  • 5
  • 3
  • - 3
  • -5

Key Concepts

Algebraic manipulation

Limit form of the Derivative

Check the Answer

Answer: is 5

ISI MStat 2019 PSA Problem 17

Introduction to real analysis Robert G. Bartle, Donald R., Sherbert.

Try with Hints

Try to manipulate \( \frac{(g(x)f(a) – g(a)f(x))}{(x – a)} \) so that you can use the Limit form of the Derivative . Let's give a try .

\( \frac{(g(x)f(a) – g(a)f(x))}{(x – a)} \) =

\( \frac{(g(x)f(a) –g(a)f(a) +g(a)f(a) - g(a)f(x))}{(x – a)} \) =

\( f(a)\frac{g(x)-g(a)}{(x-a)} - g(a)\frac{f(x)-f(a)}{(x-a)} \) .

Now calculate the limit using Limit form of the Derivative.

So, we have \( \lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x)}{(x – a)} \) =

\( \lim\limits_{x\to a} f(a)\frac{g(x)-g(a)}{(x-a)} - \lim\limits_{x\to a} g(a)\frac{f(x)-f(a)}{(x-a)} \) =

\( f(a) g'(a) - g(a)f'(a)= 2.(2)-1.(-1)=5 \).

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Derivative of Function Problem | TOMATO BStat Objective 757

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function.

Derivative of Function Problem (B.Stat Objective Question )

If f(x)=(sinx)(sin2x).....(sinnx), then f'(x) is

  • \(\sum_{k=1}^{n}(kcos{kx})f(x)\)
  • \(\sum_{k=1}^{n}(kcot{kx})f(x)\)
  • \((cosx)(2cos2x)(3cos3x).....(ncosnx)\)
  • \(\sum_{k=1}^{n}(kcos{kx})(sin{kx})\)

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer:\(\sum_{k=1}^{n}(kcot{kx})(f(x)\).

B.Stat Objective Problem 757

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

f(x)=(sinx)(sin2x).....(sinnx)

or, \(f'(x)=cosx(sin2x).......(sinnx)\)

+\(2sinxcos2x....(sinnx)+.....+n(sinx)(sin2x)....(cosnx)\)

=\(\sum_{k=1}^{n}k\frac{coskx}{sinkx}f(x)\)

=\(\sum_{k=1}^{n}kcot{kx}f(x)\).

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Derivative of Function Problem | TOMATO BStat Objective 759

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function.

Derivative of Function Problem (B.Stat Objective Question )

Consider the function \(f(x)=|sin(x)|+|cos(x)|\) defined for x in the interval \((0,{2\pi})\)

  • f(x) is differentiable everywhere
  • f(x) is not differentiable at x=\(\frac{\pi}{2}\), \({\pi}\) and \(\frac{3\pi}{2}\) and differentiable everywhere else.
  • f(x) is not differentiable at x=\(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) and differentiable everywhere else
  • none of the foregoing statements is true.

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer:f(x) is not differentiable at x=\(\frac{\pi}{2}\), \({\pi}\) and \(\frac{3\pi}{2}\) and differentiable everywhere else.

B.Stat Objective Problem 759

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

\(f(x)=|sin(x)|+|cos(x)|\)

or, \(f'(x)=\frac{sinxcosx}{|sinx|} - \frac{cosxsinx}{|cosx|}\)

f(x) is not differentiable at x=\(\frac{\pi}{2}\), \({\pi}\) and \(\frac{3\pi}{2}\) and differentiable everywhere else.

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Derivative of Function | TOMATO BStat Objective 767

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function and absolute value.

Derivative of Function and absolute value (B.Stat Objective Question )

Let \(f(x)=|sin^{3}x|\), \(g(x)=sin^{3}x\), both being defined for x in the interval \((\frac{-\pi}{2}, \frac{\pi}{2})\). Then

  • \(f'(x)=g'(x) for all x\)
  • \(g'(x)=|f'(x)| for all x\)
  • \(f'(x)=|g'(x)| for all x\)
  • \(f'(x)=-g'(x) for all x\)

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer:\(g'(x)=|f'(x)| for all x\)

B.Stat Objective Problem 767

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

Let \(f(x) =|sin^{3}x|=sin^{2}x|sinx|\)

\(f'(x)=2cos(x)sin(x)|sin(x)|+\frac{cos(x)(sin^{3}x)}{|sin(x)|}\)

=\(3sin(x)cos(x)|sin(x)|\)

or, \(g(x)=sin^{3}x\)

or, \(g'(x)=3 cosx sin^{2}x\)

or, |f'(x)| for all \(x \in (\frac{-\pi}{2},\frac{\pi}{2})\) = \(3cos x|sin x||sin x|=g'(x)\)

or, |f'(x)|=g'(x) for all \(x \in (\frac{-\pi}{2},\frac{\pi}{2})\)

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Derivative of Function | TOMATO BStat Objective 763

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Derivative of Function.

Derivative of Function (B.Stat Objective Question )

\(y=sin^{-1}(3x-4x^{3})\), then \(\frac{dy}{dx}\) equals

  • 3x
  • \(\frac{3}{\sqrt{1-x^{2}}}\)
  • 3
  • none of the foregoing expressions

Key Concepts

Equation

Derivative

Algebra

Check the Answer

Answer:\(\frac{3}{\sqrt{1-x^{2}}}\)

B.Stat Objective Problem 763

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

Let x=\(sin {\theta}\)

or, \(y=sin^{-1}(3sin {\theta}-4sin^{3} {\theta})\)

or, \(y=sin^{-1}(sin3{\theta})\)=\(3{\theta}\)=\(3sin^{-1}{x}\)

or, \(\frac{dy}{dx}=\frac{3}{\sqrt{1-x^{2}}}\).

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

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Directional Derivative : IIT JAM 2018 Problem 42

[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_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Please give a warm-up quiz

[/et_pb_text][et_pb_code _builder_version="4.1"][/et_pb_code][et_pb_text _builder_version="4.1" 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="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $latex \phi (x,y,z) = 3y^2+3yz $ for $latex (x,y,z) \in \mathbb{R}^{3}$ Then the absolute value of the directional derivative of $latex \phi $ in the direction of the line $latex \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} $ , at the point $latex (1,-2,1)$ is _____

[/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_code _builder_version="3.26.4"]
[/et_pb_code][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" 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="4.1"]IIT JAM 2018[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Directional Derivative [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]

Generalized Directional Derivatives  By Pastor Karel

 [/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="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][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="4.1" hover_enabled="0"] The rate of change of $latex f(x,y,z) $ in the direction of the unit vector $latex \vec u = <a,b,c>$ is called the directional derivative and is denoted by $latex D_{\vec u}f(x,y,z) $. The defination of directional derivative is  $latex D_{\vec u}f(x,y,z) = \lim h \to 0 \frac{f(x+ah,y+bh,z+ch)-f(x,y)}{h} $. Here if you consider $latex u$ to be $latex (1,0,0), (0,1,0)$ and $latex (0,0,1)$ then we will get $latex D_{\vec u}f(x,y,z)$ to be $latex f_x$, $latex f_y$ and $latex f_z$ respectively. If you observe closely, this is nothing but differentiating the function $latex f$ w.r.t $latex x,y$ and $latex z$ keeping all the other thing constant.

Now $latex \nabla (f)=(f_x,f_y,f_z)$ by the above definition can you try to solve the problem ???????

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]

 The points on the straight line $latex \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} = k $ $latex \implies x=1+2k , y=2-k , z=-2k $ So any point on the straight line would be of the form $latex (x,y,z)= (1,2,0) + (2,-1,-2)k $ Here the direction of the straight line would be $latex (2,-1,-2) $

Now we have to consider the unit vector in that direction and so the unit vector would be $latex u= \frac{(2,-1,-2)}{\sqrt{4+1+4}} = (\frac{-2}{3},\frac{-1}{3},\frac{-2}{3}) $Here the absolute value of the directional derivative would be $latex \nabla \phi . u $ . Would you like to calculate this dot product?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

$latex \nabla \phi = (\phi_{x},\phi_{y},\phi_{z}) $  i.e the co-ordinates involving partial derivative of $latex \phi $ . So, $latex \nabla \phi = (0, 6y+3z , 3y)|_{(1,-2,1)} = (0,-9,-6)$

Then $latex \nabla \phi . u= \frac{9}{3} + \frac{6 \times 2}{3} = 3+4 = 7 $ (Ans) .

[/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"]

A graphical view point

[/et_pb_text][et_pb_video src_webm="https://cheenta.com/wp-content/uploads/2020/01/2020-01-21-04-29-14.mp4" _builder_version="4.1" custom_margin="77px||77px||true|"][/et_pb_video][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"]

Take A Look At This Knowledge Graph 

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/Untitled-Diagram-6.png" align="center" _builder_version="4.1"][/et_pb_image][et_pb_video _builder_version="4.1"][/et_pb_video][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

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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

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