# ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

This is a very simple but beautiful sample problem from ISI MStat PSB 2009 Problem 3. It is based on recognizing density function and then using CLT. Try it !

## Problem- ISI MStat PSB 2009 Problem 3

Using and appropriate probability distribution or otherwise show that,

$$\lim\limits_{x\to\infty}\int^n_0 \frac{exp(-x)x^{n-1}}{(n-1)!}\,dx =\frac{1}{2}$$.

### Prerequisites

Gamma Distribution

Central Limit Theorem

Normal Distribution

## Solution :

Here all we need is to recognize the structure of the integrand. Look, that here, the integrand is integrated over the non-negative real numbers. Now, event though here it is not mentioned explicitly that $$x$$ is a random variable, we can assume $$x$$ to be some value taken by a random variable $$X$$. After all we can find randomness anywhere and everywhere !!

Now observe that the integrand has a structure which is very identical to the density function of gamma random variable with parameters $$1$$ ande $$n$$. So, if we assume that $$X$$ is a $$Gamma(1, n)$$, then our limiting integral transforms to,

$$\lim\limits_{x\to\infty}P(X \le n)$$.

Now, we know that if $$X \sim Gamma(1,n)$$, then its mean and variance both are $$n$$.

So, as $$n \uparrow \infty$$, $$\frac{X-n}{\sqrt{n}} \to N(0,1)$$, by Central Limit Theorem.

Hence, $$\lim\limits_{x\to\infty}P(X \le n)=\lim\limits_{x\to\infty}P(\frac{X-n}{\sqrt{n}} \le 0)=\lim\limits_{x\to\infty}\Phi (0)=\frac{1}{2}$$. [ here $$\Phi(z)$$ is the cdf of Normal at $$z$$.]

Hence proved !!

## Food For Thought

Can, you do the proof under the "Otherwise" condition !!

Give it a try !!

# ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

## Problem- ISI MStat PSB 2006 Problem 2

Maximize $$x+y$$ subject to the condition that $$2x^2+3y^2 \le 1$$.

### Prerequisites

Cauchy-Schwarz Inequality

Tangent-Normal

Conic section

## Solution :

This is a beautiful problem, but only if one notices the trick, or else things gets ugly.

Now we need to find the maximum of $$x+y$$ when it is given that $$2x^2+3y^2 \le 1$$. Seeing the given condition we always think of using Lagrange Multipliers, but I find that thing very nasty, and always find ways to avoid it.

So let's recall the famous Cauchy-Schwarz Inequality, $$(ab+cd)^2 \le (a^2+c^2)(b^2+d^2)$$.

Now, lets take $$a=\sqrt{2}x ; b=\frac{1}{\sqrt{2}} ; c= \sqrt{3}y ; d= \frac{1}{\sqrt{3}}$$, and observe our inequality reduces to,

$$(x+y)^2 \le (2x^2+3y^2)(\frac{1}{2}+\frac{1}{3}) \le (\frac{1}{2}+\frac{1}{3})=\frac{5}{6} \Rightarrow x+y \le \sqrt{\frac{5}{6}}$$. Hence the maximum of $$x+y$$ with respect to the given condition $$2x^2+3y^2 \le 1$$ is $$\frac{5}{6}$$. Hence we got what we want without even doing any nasty calculations.

Another nice approach for doing this problem is looking through the pictures. Given the condition $$2x^2+3y^2 \le 1$$ represents a disc whose shape is elliptical, and $$x+y=k$$ is a family of straight parallel lines passing passing through that disc.

Hence the line with the maximum intercept among all the lines passing through the given disc represents the maximized value of $$x+y$$. So, basically if a line of form $$x+y=k_o$$ (say), is a tangent to the disc, then it will basically represent the line with maximum intercept from the mentioned family of line. So, we just need to find the point on the boundary of the disc, where the line of form $$x+y=k_o$$ touches as a tangent. Can you finish the rest and verify weather the maximum intercept .i.e. $$k_o= \sqrt{\frac{5}{6}}$$ or not.

## Food For Thought

Can you show another alternate solution to this problem ? No, Lagrange Multiplier Please !! How would you like to find out the point of tangency if the disc was circular ? Show us the solution we will post them in the comment.

Keep thinking !!

# Problem on Integral Inequality | ISI - MSQMS - B, 2015

Try this problem from ISI-MSQMS 2015 which involves the concept of Integral Inequality.

## INTEGRAL INEQUALITY | ISI 2015 | MSQMS | PART B | PROBLEM 7b

Show that $1<\int_{0}^{1} e^{x^{2}} d x<e$

### Key Concepts

Real Analysis

Inequality

Numbers

But Try the Problem First...

ISI - MSQMS - B, 2015, Problem 7b

"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"

## Try with Hints

We have to show that ,

$1<\int_{0}^{1} e^{x^{2}} d x<e$

$0< x <1$

It implies, $0 < x^2 <1$

Now with this reduced form of the equation why don't you give it a try yourself, I am sure you can do it.

Thus, $e^0 < e^{x^2} <e^1$

i.e $1 < e^{x^2} <e$

So you are just one step away from solving your problem, go on.............

Therefore, Integrating the inequality with limits $0$ to $1$ we get, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$

# Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

## Geometric Sequence Problem - AIME 2009

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

• is 500
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Series

Real Analysis

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

## Try with Hints

First hint

3-digit sequence a, ar, $$ar^{2}$$. The largest geometric number must have a<=9.

Second Hint

ar $$ar^{2}$$ less than 9 r fraction less than 1 For a=9 is $$\frac{2}{3}$$ then number 964.

Final Step

a>=1 ar and $$ar^{2}$$ greater than 1 r is 2 and number is 124. Then difference 964-124=840.

# Definite Integral Problem | ISI 2018 | MSQMS- A | Problem 22

Try this problem from ISI-MSQMS 2018 which involves the concept of Real numbers, sequence and series and Definite integral.

## DEFINITE INTEGRAL | ISI 2018| MSQMS | PART A | PROBLEM 22

Let $I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x$ and $J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x,$ then which of the following is
true?

• (a) $I<\frac{2}{3}$ and $J>2$
• (b) $I>\frac{2}{3}$ and $J<2$ (c) $I>\frac{2}{3}$ and $J>2$
• (d) $I<\frac{2}{3}$ and $J<2$

### Key Concepts

REAL NUMBERS

REIMANN INTEGRATION

SEQUENCE AND SERIES

Answer:(d) $I<\frac{2}{3}$ and $J<2$

ISI 2018|MSQMS |QMA|PROBLEM 22

INTRODUCTION TO REAL ANALYSIS :BARTLE SHERBERT

## Try with Hints

We know when $f(x)$>$g(x)$

$\int \limits_a^bf(x)$>$\int \limits_a^bg(x)$

We know for $0<x<1$, $\cos x <1$

$\frac{\cos x}{\sqrt x}$< $\frac{1}{\sqrt x}$ implies $\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx$

$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx = 2$

$\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$2$

$J$<$2$

Again we claim $x-s\sin x$>$0$ for $0 \leq x\leq 1$

Let $f(x)=x-\sin x$

$f'(x)=1-\cos x\geq 0$

hence $f(x)$ is monotonic increasing.

Therefore $x-\sin x$> $0$, $x\epsilon [0,1]$

So,$x$>$sinx$

$\sqrt x$ > $\frac{\sin x}{\sqrt x}$ $x\epsilon [0,1]$

integrating both sides with limits $0$ to $1$ we get;

$\int \limits_0^1\frac{\sin x}{\sqrt x} \mathrm dx$<$\frac{2}{3}$

$I$<$\frac{2}{3}$

Therefore,$I<\frac{2}{3}$ and $J<2$

# 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

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

# 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

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,

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

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

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

# 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

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

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

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]

# Radius of Convergence of a Power series | IIT JAM 2016

Try this problem from IIT JAM 2017 exam (Problem 48) and know how to determine radius of convergence of a power series.

## Radius of Convergence of a Power Series | IIT JAM 2016 | Problem 48

Find the radius of convergence of the power series
$$\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}$$

### Key Concepts

Real Analysis

Series of Functions

Power Series

Answer: $\frac12$

IIT JAM 2016 , Problem 48

Real Analysis : Robert G. Bartle

## Try with Hints

Given, the power series is $\sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(n+1)}(x+2)^{2 n}$.

Let us put $2n=m$ to get the standard form of a power series.

We get,

$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(x+2)^{ m}$.

Now let us make the transformation $z=x+2$ to get a power series about 0 :

We have,

$\sum_{m=2}^{\infty} \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}(z)^{ m}$

Compairing with $\sum_{m=2}^{\infty} a_m (z)^m$

we get,

$a_m= \frac{(-4)^{\frac m2}}{\frac m2(\frac m2+1)}$

Now we have to test the convergence of the series.

Can you apply Ratio Test to check the convergence of the series.

Ratio Test : Let $\sum_{n=0}^{\infty} a_{n} x^{n}$ be a power series and let $\lim \left|\frac{a_{n+1}}{a_{n}}\right|=\mu .$ Then

1. if $\mu=0$ the series is everywhere convergent;
2. if $0<\mu<\infty$ the series is absolutely convergent for all $x$ satisfyir $|x|<\frac{1}{\mu}$ and the series is divergent for all $x$ satisfying $|x|>\frac{1}{\mu}$
3. if $\mu=\infty,$ the series is nowhere converegnt.

\begin{aligned}\left|\frac{a_{m+1}}{a_m}\right| &=\left| \frac{4^{\frac m2}\cdot 2\cdot4}{(m+1)(m+3)} \times \frac{m(m+2)}{4^{\frac m2} \cdot 4}\right| \\&=\left| \quad \frac{2\left(1+\frac{2}{m}\right)}{\left(1+\frac{1}{m}\right)(1+\frac 3m)}\right|\end{aligned}

Now

$\lim \left|\frac{a_{m+1}}{a_{m}}\right|=2 \in (0,\infty)$

Then, The given power series is absolutely convergent i.e., convergent $\forall x$ such that $|x+2|<\frac 12$

Then the answer is $\frac 12$

# 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

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.