Rolle's Theorem | IIT JAM 2017 | Problem 10

Try this problem from IIT JAM 2017 exam (Problem 10).This problem needs the concept of Rolle's Theorem.

Rolle's Theorem | IIT JAM 2017 | Problem 10

$$f(x)=\left\{\begin{array}{ll}1+x & \text { if } x<0 \\ (1-x)(p x+q) & \text { if } x \geq 0\end{array}\right.$$

satisfies the assumptions of Rolle's theorem in the interval $[-1,1],$ then the ordered pair $(p, q)$ is

$(2,-1)$
$(-2,-1)$
$(-2,1)$
$(2,1)$

Key Concepts

Real Analysis

Continuity / Differentiability

Mean-value theorem of differential calculus

Check the Answer

Answer: $(2,1)$

IIT JAM 2017 , Problem 10

Real Analysis : Robert G. Bartle

Try with Hints

Rolle's Theorem :

Let a function $f:[a, b] \rightarrow R$ be such that

$f$ is continuous on $[a, b]$
$f$ is differentiable at every point of $(a, b)$
$f(a)=f(b)$

Then there exists at least one point $c \in(a, b)$ such that $f^{\prime}(c)=0$

We can easily see that $3^{rd}$ assumption of Rolle's theorem is satisfied for $f(x)$ irrespective of the values of $p,q$.

Since $f(-1)=0=f(1)\quad \forall p,q$

Since $f(x)$ satisfies $1^{st}$ assumption, then

$\begin{aligned}& \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\\&\text { ie, } \lim _{x \rightarrow 0^{-}}(1+x)=\lim _{x \rightarrow 0^{+}}(1-x)(px+q)=q\\&\Rightarrow 1=q\end{aligned}$

$L f^{\prime}(0)=R f^{\prime}(0) \cdots \cdots(*)$

$\begin{aligned} \text{Now, } L f^{\prime}(0) &=\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h} \\&=\lim _{h \rightarrow 0^{-}} \frac{(1+h)-q}{h} \\ &=\lim _{h \rightarrow 0^{-}} \frac{1+h-1}{h}[\text{because } q=1] \\&=\lim_{h \to 0^{-}} \frac hh\\&=1\end{aligned}$

$\begin{aligned}\text{and, } R f^{\prime}(0)&=\displaystyle\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)\left(ph+q\right)-q}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)(ph +1)-1}{h}\quad[\text{because } q=1]\\&=\lim _{h \rightarrow 0^{+}}\frac{ph+1- ph^{2}-h-1}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{h(p-ph-1)}{h}\\&=\lim_{h \ to 0^{+}} (p-ph-1)\\&=p-1\end{aligned}$

Then by $(*) \text{we have}, \quad P-1=1 \Rightarrow P=2$

Then order pair $(p,q)\equiv (2,1)$ [ANS]

Limit and differentiability of a function-I.S.I B.Stat. Entrance 2017, UGA Problem 5

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What are we learning ?

This problem from I.S.I. B.Stat. Entrance 2017 , it is based on simple manipulations, differentiation from first principles and squeeze theorem.

[/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"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/ISI-BSTAT-entrance-2017-UGA-problem-5-1.png" align="center" force_fullwidth="on" _builder_version="4.1" hover_enabled="0"][/et_pb_image][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"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let f : R → R be a continuous function such that for any two real
numbers x and y, $ |f(x)-f(y)| \le {|x-y|}^{201} $.
Then,[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][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="4.1" 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"]I.S.I. B.Stat. Entrance 2017, UGA Problem 5[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" open="off"]Differentiation from first principles and squeeze (or sandwich) theorem [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

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Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="HINT 0" _builder_version="4.1"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.1"]Given that $ |f(x)-f(y)| \le {|x-y|}^{201} $ for any two real number x and y . So we can replace x by y+h for any real number y and h .Then we have , $ |f(y+h)-f(y)| \le {|(y+h)-y|}^{201} \implies |f(y+h)-f(y)| \le {|h|}^{201} $ . Now, what can we use ?[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.1"]

See that we can write $\frac{ |f(y+h)-f(y)|}{h} \le {|h|}^{200} \implies -{h}^{200} \le \frac {|f(y+h)-f(y)|}{h} \le h^{200}$ . Now taking limit h tends to 0 on both sides gives what ?[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1"]

See that $\lim_{h \to 0} -{h}^{200}=\lim_{h \to 0} {h}^{200}=0 $ and differentiation from first principles say's $\lim_{h \to 0} \frac{f(y+h)-f(y)}{h} =f'(y) $ .

[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1"]Now we know the the squeeze theorem, stated as follows Let I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have

g(x)\leq f(x)\leq h(x)

and also suppose that

\lim _{x\to a}g(x)=\lim _{x\to a}h(x)=L.

Then

\lim_{x \to a} f(x) = L.

So, what we have ?

[/et_pb_tab][et_pb_tab title="HINT 5" _builder_version="4.1"]f'(y)=0 for all y belongs to real . Thus integrating both sides we have f(y)=c ( constant ) for all y belongs to real . Therefore , f(101)=f(202)=f(200).[/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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Differentiability at origin | I.S.I. B.Stat, B.Math Subjective 2017

Try this problem from ISI B.Stat, B.Math Subjective Entrance Exam, 2017 Problem no. 3 based on Differentiability at origin.

Problem: Differentiability at origin

Suppose $ f : \mathbb{R} \to \mathbb{R} $ is a function given by $$
f(x) = \left\{\def\arraystretch{1.2}%
\begin{array}{@{}c@{\quad}l@{}}
1 & \text{if x=1}\\ e^{(x^{10} -1)} + (x-1)^2 \sin \left (\frac {1}{x-1} \right ) & \text{if} x \neq 1\ \end{array}\right.
$$

Find f'(1)
Evaluate $ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

Discussion:

a)First of all we need to check whether $f'(1)$ exists or not.

We will proceed with the first principle.

Let us check the Right hand derivative(RHD) and Left hand derivative(LHD) of $f$ at $x=1$.

RHD at $x=1$ is

$\lim_{h\to0}\frac{f(1+h)-f(1)}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}+h^2\sin( \frac{1}{h})-1}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{h}+\lim_{h\to0}h\sin( \frac{1}{h})\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{(1+h)^{10}-1}\frac{(1+h)^{10}-1}{h}+0=10$

LHD at $x=1$ is

$\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}+h^2\sin( \frac{1}{-h})-1}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{-h}+\lim_{h\to0}(-h)\sin( \frac{1}{-h})\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{(1-h)^{10}-1}\frac{(1-h)^{10}-1}{-h}+0=10$

Thus,LHD=RHD.

Hence $f'(1)$ exists and it is equal to $10$.

(b)

$ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

As u becomes infinitely large k/u becomes arbitrarily small for finite value of k (clearly k is finite as we are interested in k=1 to 100).

Hence $ f \left ( 1 + \frac {k}{u} \right )$ is nothing but f of (1 plus an infinitesimal positive quantity). This tells us $ f \left ( 1 + \frac {k}{u} \right )$ is almost waiting to become the derivative of f at x=1. And we already know that such a derivative exists from part (a).

With this motivation, divide and multiply by $ \frac{k}{u} $.

$ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] \\ =\lim_ {u \to \infty} \left [ 100 u -\sum_{k=1}^{100} k \frac{f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\=\lim_ {u \to \infty} \left [ \sum_{k=1}^{100} k \frac{1-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\ =\sum_{k=1}^{100} k \lim_ {u \to \infty}\frac{f(1)-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \\ =\sum_{k=1}^{100} k \times(- f'(1)) \\ = -10\times \left(\frac{100\times101}{2} \right )\\ =-50500 }$

Back to the question paper

Some Useful Links:

One-One function and differentiability

Let's understand one-one function and differentiability with the help of a problem. Try it yourself before reading the solution.

Let f be real valued, differentiable on (a, b) and $ f'(x) \ne 0 $ for all $ x \in (a, b) $. Then f is 1-1.

True

Discussion:

Suppose f is not 1-1. Then there exists $x_1 , x_2 \in (a, b) $ such that $ f(x_1 ) = f(x_2) $. Since f(x) is differentiable it must be continuous as well. Applying Rolle's Theorem in the interval $ (x_1 , x_2 ) $ we conclude that there exists a number c in this interval such that f'(c) = 0. But this contradicts the given conditions. Hence f must be 1-1

Some Useful Links:

Chinese Remainder Theorem

Math Olympiad in India

Our Math Olympiad Program

A Math Game in Symmetry - Video

Rolle's Theorem | IIT JAM 2017 | Problem 10