Categories

## Problem on Inequality | ISI – MSQMS – B, 2018 | Problem 2a

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

## INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 2a

(a) Prove that if $x>0, y>0$ and $x+y=1,$ then $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq 9$

### Key Concepts

Algebra

Inequality

Numbers

But Try the Problem First…

Answer: $xy \leq \frac{1}{4}$

Source

ISI – MSQMS – B, 2018, Problem 2A

“INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA”

## Try with Hints

First hint

We have to show that ,

$(1+\frac{1}{x})(1+\frac{1}{y}) \geq 9$

i.e $1+ \frac{1}{x} + \frac{1}{y} +\frac{1}{xy} \geq 9$

Since $x+y =1$

Therefore the above equation becomes $\frac{2}{xy} \geq 8$

ie $xy \leq \frac{1}{4}$

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

Second hint

Applying AM $\geq$ GM on $x,y$

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

Final Step

Therefore, $\frac{x+y}{2} \geq (xy)^\frac{1}{2}$

$\Rightarrow \frac{1}{2} \geq (xy)^\frac{1}{2}$

Squaring both sides we get, $xy \leq \frac{1}{4}$

Hence the result follows.

Categories

## 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…

Source

ISI – MSQMS – B, 2015, Problem 7b

“INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA”

## Try with Hints

First hint

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.

Second hint

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………….

Final hint

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

Categories

## Inequality Problem From ISI – MSQMS – B, 2017 | Problem 3a

Try this problem from ISI-MSQMS 2017 which involves the concept of Inequality.

## INEQUALITY | ISI 2017| MSQMS | PART B | PROBLEM 3a

(a) Prove that $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a),$ for all positive distinct values of $a, b, c$

Algebra

Inequality

Numbers

## Try with Hints

First hint

We know if we have $n$ numbers say $a_1,a_2,…..,a_n$ then AM $\geq$ GM implies

$\frac{a_1+a_2+….+a_n}{n} \geq (a_1.a_2……..a_n)^\frac{1}{n}$

I assure you that the sum the can be done just by using this simple inequality,why don’t you just give it a try?

Second hint

Applying AM $\geq$ GM on $a^5,a^5,a^5,b^5,c^5$ we get

$3a^5+b^5+c^5 \geq 5a^3bc$

Similarly, $a^5+3b^5+c^5 \geq 5ab^3c$

$a^5+b^5+3c^5 \geq 5abc^3$

Adding the above three equations we get $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2)$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

Third hint

Now we have to show that $a^2+b^2+c^2 > ab+bc+ca$

Applying AM $\geq$ GM on $a^2,b^2$ we get,

$a^2+b^2 \geq 2ab$

Similarly,$b^2+c^2 \geq 2bc$

$a^2+c^2 \geq 2ca$

Adding the above three equations we get $a^2+b^2+c^2 \geq ab+bc+ca$

We are almost there ,so just try the last step yourself.

Final Step

Therefore, $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2) \geq abc(a^2+b^2+c^2)$

i.e, $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a)$

Categories

## Problem on Inequality | ISI – MSQMS – B, 2018 | Problem 4b

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

## INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 4b

Let $a>0$ and $n \in \mathbb{N} .$ Show that

$$\frac{a^{n}}{1+a+a^{2}+\ldots+a^{2 n}}<\frac{1}{2 n}$$

Algebra

Inequality

Numbers

## Try with Hints

First hint

We know if $\frac{a}{b} \geq c$

Then $\frac{b}{a} \leq c$

$\frac{a^n}{1+a+a^2+…….+a^{2n}}$ < $\frac{1}{2n}$

$\frac{1+a+a^2+…..+a^{2n}}{a^n}$ > $2n$

Now why don’t you just give it a try yourself,try to conclude something from the previous line,I know you can.

Second hint

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+………….+(a^n+\frac{1}{a^n})> 2n$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

Third hint

$a+\frac{1}{a} \geq 2$

$a^2+\frac{1}{a^2} \geq 2$

.

.

.

$a^n+\frac{1}{a^n} \geq 2$

Adding the above inequalities we get,$(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+……….+(a^n+\frac{1}{a^n}) \geq 2n$

We are almost there ,so just try the last step yourself.

Final Step

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+……….+(a^n+\frac{1}{a^n}) \geq 2n+1$ > $2n$

Thus, $\frac{a^n}{1+a+a^2+……+a^n}$<$\frac{1}{2n}$

Categories

## Inequality Problem | ISI – MSQMS 2018 | Part B | Problem 4

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality and Combinatorics.

## INEQUALITY | ISI 2018 | MSQMS | PART B | PROBLEM 4

Show that $\sqrt{C_{1}}+\sqrt{C_{2}}+\sqrt{C_{3}}+\ldots+\sqrt{C_{n}} \leq 2^{n-1}+\frac{n-1}{2}$ where
$C_k={n\choose k}$

INEQUALITIES

COMBINATORICS

## Try with Hints

First hint

Use Cauchy Schwarz Inequality $\left(\displaystyle\sum_{i} a_{i} b_{i}\right)^{2} \leq\left(\displaystyle\sum_{i} a_{i}^{2}\right)\left(\displaystyle\sum_{i} b_{i}^{2}\right)$

Second hint

Apply Cauchy Schwarz Inquality in two sets of real numbers ($\sqrt C_1$,$\sqrt C_2$,…..,$\sqrt C_n$)and ($1$,$1$,$1$,……$1$)

($C_1+C_2+$……..$+C_n$)($1+1+$……$+1$) $\geq$ ($\sqrt C_1+\sqrt C_2+………+\sqrt C_n$)

($2^n-1$)$n \geq$ ($\sqrt C_1+\sqrt C_2+$……….$+\sqrt C_n$)$^2$

$\sqrt C_1+\sqrt C_2+$……….$+\sqrt C_n \leq \sqrt n\sqrt (2^n-1)$

The proof is still not done,why don’t you try the remaining part yourself?

Third hint

We know AM $\geq$ GM

i.e

For $n$ positive quantities $a_{1}, a_{2}, \dots, a_{n}$
$$\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq \sqrt[n]{a_{1} a_{2} \cdot \cdot a_{n}}$$
with equality if and only if $a_{1}=a_{2}=\ldots=a_{n}$

Now you have all the ingredients,why don’t you cook it yourself? I firmly believe that you can cook a food tastier than mine.

Final Step

$\frac{n+2^n-1}{2} \geq \sqrt n\sqrt {2^n-1}$

Thus,$\sqrt C_1+\sqrt C_2+$……..$+\sqrt C_n \leq \frac {n+2^n-1}{2}$

Categories

## 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

But try the problem first…

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

Source

ISI 2018|MSQMS |QMA|PROBLEM 22

INTRODUCTION TO REAL ANALYSIS :BARTLE SHERBERT

## Try with Hints

First hint

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

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

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

Second hint

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

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

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

$J$<$2$

Third hint

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;

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

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

Final Step

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

Categories

## Problem on Natural Numbers | TIFR B 2010 | Problem 4

Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on natural numbers.

## Problem on Natural Numbers | TIFR 201O| PART B | PROBLEM 4

Which of the following statements is false?

• There exists a natural number which when divided by $3$ leaves remainder $1$ and when divided by $4$ leaves remainder $0$
• There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$
• There exists a natural number which when divided by $7$ leaves remainder $1$ and when divided by $11$ leaves remainder $3$
• There exists a natural number which when divided by $12$ leaves remainder $7$ and when divided by $8$ leaves remainder $3$

### Key Concepts

NUMBER THEORY

CONGRUENCE

CHINESE REMAINDER THEOREM

But try the problem first…

Answer:There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$

Source

TIFR 2010|PART B |PROBLEM 12

ELEMENTARY NUMBER THEORY DAVID M.BURTON

## Try with Hints

First hint

Let us take the equations $x\equiv1(mod 3)$ and $x\equiv0(mod 4)$

Now we will apply Chinese remainder theorem to get the value of $x$

Second Hint

Since $3$,$4$ are relatively prime,gcd($3$,$4$)$=1$. Let $m=3\times4=12$

Then $M_1=4$,$M_2=3$.

Then gcd($M_1$,$3$)$=1$,gcd($M_2$,$4$)$=1$

Since gcd($4$,$3$)$=1$,therefore the linear congruence equation $4x\equiv1(mod 3)$ has a unique solution and $x\equiv1(mod 3)$ is the solution.

Since gcd($3$,$4$)$=1$,therefore the linear congruence equation $3x\equiv0(mod 4)$ has a solution and $x\equiv4(mod 4)$ is the solution.

Therefore,$x=1\times4\times1 +0\times3\times4=4$ is a solution.

The solution of the given system is $x\equiv4(mod 12)$

Final Step

So we have used the Chinese Remainder Theorem to check the statements, you may use it to check for other options.

Categories

## Positive Integers Problem | TIFR B 201O | Problem 12

Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on Positive Integers.

## Positive Integers Problem | TIFR 201O | PART B | PROBLEM 12

If $n$ and $m$ are positive integers and $n^{19}=19m+r$ then the possible values for $r$ modulo $19$ are:-

• only $0$
• only $0$,$\pm1$
• only $\pm1$
• none of the above

### Key Concepts

NUMBER THEORY

CONGRUENCE

FERMAT’S LITTLE THEOREM

But try the problem first…

Answer:only $0$,$\pm1$

Source

TIFR 2010|PART B |PROBLEM 12

ELEMENTARY NUMBER THEORY DAVID M.BURTON

## Try with Hints

First hint

$r(mod 19)=(n^9-19m)(mod 19)=n^9(mod 19)-19m(mod 19)=n^9(mod 19)$ [because $19\mid19m$]

Second Hint

Now there can be two cases

casei) $19\mid n$

caseii) n is not divisible by 19

Final Step

If $19\mid n$ then $19\mid n^9$ resulting in $n^9\equiv 0(mod 19)$

But if $n$ is not divisible by $19$ then according to Fermat’s little theorem,

$n^{19-1}\equiv1(mod 19) =n^{18}\equiv 1(mod 19) = 19\mid n^{18}-1 =19\mid[n^9+1][n^9-1]$

i.e If $19\mid n^9+1$ then $n^9\equiv(-1)(mod 19) =r\equiv(-1)(mod 19)$

If $19\mid n^9-1$ then $n^9\equiv1(mod 19) = r\equiv1(mod 19)$

Thus $r(mod 19)=0$,$\pm1$

Categories

## CYCLIC GROUP Problem | TIFR 201O | PART A | PROBLEM 1

Try this problem from TIFR GS-2010 which involves the concept of Cyclic Group.

## CYCLIC GROUP | TIFR 201O | PART A | PROBLEM 1

A cyclic group of order $60$ has

• $12$ generators
• $15$ generators
• $16$ generators
• $20$ generators

### Key Concepts

CYCLIC GROUP

ORDER OF AN ELEMENT

EULER’S PHI FUNCTION

But try the problem first…

Answer:$16$ generators

Source

TIFR 2010|PART A |PROBLEM 1

CONTEMPORARY ABSTRACT ALGEBRA: JOSEPH GALLIAN

## Try with Hints

First hint

If G be a cyclic group of order n ,then the number of generators of G is $\phi(n)$

Second Hint

Let us define Euler’s phi function:-

$\phi(n)$=the number of positive integers less than $n$ and prime to $n$

i)If $n$ is prime then $\phi(n)=n-1$

ii)If $m$,$n$ be two integers which are relatively prime $\phi{mn}=\phi(m)\phi(n)$

iii)If $n$ be a prime and $k$ be any positive integer,$\phi(p^k)=p^k(1-\frac{1}{p})$

Final Step

Now $60$ can be written as product of $2^2$,$3$,$5$

Therefore $\phi(60)=\phi(2^2).\phi(3).\phi(5)$

Now $\phi(2^2)=2$…….i

Now $\phi(3)=(3-1)=2$ ……..ii

Now $\phi(5)=(5-1)=4$ ……..iii

Now multiply i,ii &iii and get the result

Categories

## 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

But try the problem first…

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

Source

TIFR 2010|PART A |PROBLEM 1

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

## Try with Hints

First hint

Here first differentiate $f(x)$

Second Hint

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 $n$

Final Step

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