ISI MStat PSB 2013 Problem 2 | Application of sandwich Theorem

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 2 based on use of Sandwich Theorem . Let's give it a try !!

Problem- ISI MStat PSB 2013 Problem 2

Let f be a real valued function satisfying \(|f(x)-f(a)| \leq C|x-a|^{\gamma}\) for some \(\gamma>0\) and \(C>0\)
(a) If \(\gamma=1,\) show that f is continuous at a
(b) If \(\gamma>1,\) show that f is differentiable at a

Prerequisites

Differentiability

Continuity

Limit

Sandwich Theorem

Solution :

(a) We are given that \(|f(x)-f(a)| \leq C|x-a|\) for some \(C>0\).

We have to show that f is continuous at x=a . For this it's enough to show that \(\lim_{x\to a} f(x)=f(a)\).

\(|f(x)-f(a)| \leq C|x-a| \Rightarrow f(a)-C|x-a| \le f(x) \le f(a) + C|x-a| \)

Now taking limit \( x \to a\) we have , \( \lim_{x\to a} f(a)-C|x-a| \le \lim_{x\to a} f(x) \le \lim_{x\to a} f(a) + C|x-a| \)

Using Sandwich theorem we can say that \( \lim_{x\to a} f(x) = f(a) \) . Since \(\lim_{x\to a} -C|x-a| = \lim_{x\to a} C|x-a|=0 \)

Hence f is continuous at x=a proved .

(b) Here we have to show that f is differentiable at x=a for this it's enough to show that the \(\lim_{x\to a} \frac{f(x)-f(a)}{x-a} \) exists .

We are given that , \(|f(x)-f(a)| \leq C|x-a|^{\gamma}\) for some \(\gamma>1\) and \(C>0\) ,

which implies \( |\frac{f(x)-f(a)}{x-a} | \le C|x-a|^{\gamma -1} \)

\(\Rightarrow -C|x-a|^{\gamma -1} \le \frac{f(x)-f(a)}{x-a} \le C|x-a|^{\gamma -1} \)

Now taking \(\lim_{x\to a} \) we get by Sandwich theorem \(\lim_{x\to a}\frac{f(x)-f(a)}{x-a} =0 \) i.e f'(a)=0 .

Since , \( \lim_{x\to a} C|x-a|^{\gamma -1} = \lim_{x\to a} -C|x-a|^{\gamma -1} = 0 \) , for \( \gamma >1 \).

Hence f is differentiable at x=a proved .

Food For Thought

\( f : R \to R \) be such that \( |f(x)-f(a)| \le k|x-y| \) for some \( k \in (0,1) \) and all \( x,y \in R \) . Show that f must have a unique fixed point .

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ISI MStat PSB 2014 Problem 2 | Properties of a Function

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let's give it a try !!

Problem- ISI MStat PSB 2014 Problem 2

Let \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) be real numbers such
that \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R} \)
Show that \(m=n\) and \(a_{j}=b_{j}\) for \(1 \leq j \leq n\)

Prerequisites

Differentiability

Mod function

continuity

Solution :

Let , \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R} \)

Then , \( f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right| \) is not differentiable at \( x=a_1,a_2, \cdots , a_m \) ---(1)

As we know the function \(|x-a_i|\) is not differentiable at \(x=a_i\) .

Again we have , \( f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right| \) it also not differentiable at \( x= b_1,b_2, \cdots , b_n \) ----(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) .

So , we have \(a_{j}=b_{j}\) for \(1 \leq j \leq n\) .

Food For Thought

\(a<b \in \mathbb{R} .\) Let \(f:[a, b] \rightarrow[a, b]\) be a continuous and differentiable on (a,b) . Suppose that \(\left|f^{\prime}(x)\right| \leq \alpha<1\) for all \(x \in(a, b)\) for some \(\alpha .\) Then prove that there exists unique \(x \in[a, b]\) such that \(f(x)=x\)

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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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ISI MStat PSB 2012 Problem 5 | Application of Central Limit Theorem

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 5 based on central limit theorem . Let's give it a try !!

Problem- ISI MStat PSB 2012 Problem 5

Let \( X_{1}, X_{2}, \ldots, X_{j} \ldots \) be i.i.d. \(N(0,1)\) random variables. Show that for any \(a>0\)

\( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) = 0 \)

Prerequisites

Limit

Central Limit Theorem

Normal Distribution

Chi- Square Distribution

Solution :

\( X_{1}, X_{2}, \ldots, X_{j} \ldots \) are i.i.d. \(N(0,1)\) random variables .

Let \( S_n = \sum_{i=1}^{n} {X_i}^2 \) , then \( S_n \sim \chi^{2}(n) \) , where \( \chi^{2}(n) \) is Chi-Square distribution with n degrees of freedom .

Therefore , \( E(S_n)= n \) and \( Var(S_n)=2n \) .

In this type of problems obvious thing that would come to our mind is to apply Central Limit Theorem right ! Let's try to apply it .

Now by Lindeberg Levy Central Limit Theorem we can say \( \frac{S_n-E(S_n)}{\sqrt{Var(S_n)}} \) = \( \frac{S_n-n}{\sqrt{2n}} {\to }^{d} N(0,1) \) , as n approaches infinity.

So, \( \lim {n \rightarrow \infty} P(\sum_{i=1}^{n} {X_i}^2 \leq a) \)

= \( \lim {n \rightarrow \infty} P( \frac{S_n-n}{\sqrt{2n}} \le \frac{a-n}{\sqrt{2n}} ) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a-n}{\sqrt{2n}}) \)

= \( \lim {n \rightarrow \infty} \Phi(\frac{a}{\sqrt{2n}}- \sqrt{\frac{n}{2}}) = \Phi(0- \infty) \) (Since \( \Phi(x) \) is right continuous ) \( = 0 \) .

Hence Proved .

Food For Thought

Let \( \{X_{1}: i \geq 1 \}\) be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5.Then \( (\frac{1}{n} \sum_{i=1}^{n} x_{i})^{2} \) converges in probability to ?

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Limit of a function | IIT JAM 2017 | Problem 8

Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8

Let $$ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $$
Write $L=\displaystyle\lim_{x \to 0^{-}} f(x)$ and $R=\displaystyle\lim_{x \to 0^{+}} f(x) .$

Then which one of the following is true?

  • $L$ exists but $R$ does not exist
  • $L$ does not exist but $R$ exists
  • Both $L$ and $R$ exist
  • Neither $L$ nor $R$ exists

Key Concepts

Real Analysis

Function

Limit

Check the Answer

Answer: $L$ exists but $R$ does not exists

IIT JAM 2017 , Problem 8

Try with Hints

Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.

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