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## ISI MStat 2019 PSA Problem 16 | Area bounded by the curve

This is a problem from ISI MStat 2019 PSA Problem 16 based on calculating area bounded by the curve.

## Area bounded by the curve – ISI MStat 2018 PSA Problem 16

The functions $f, g:[0,1] \rightarrow[0,1]$ are given by $f(x)=\frac{1}{2} x(x+1)$ and $g(x)=\frac{1}{2} x^{2}(x+1) .$ What is the area enclosed between the graphs of $f^{-1}$ and $g^{-1} ?$

• (A) 1/8
• (B) 1/4
• (C)5/12
• (D) 7/24

### Key Concepts

Integration

Graph of a function

ISI MStat 2019 PSA Problem 16

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Inverse of a function is basically reflection about y=x line .

So , we can get $f^{-1}$ and $g^{-1}$ from $f(x)$ and $g(x)$ respectively by replacing x by y and y by x .

Let’s draw the curves .

This is graph of inverses of f and g when they are defined in $R \to R$ . But in our problem we should consider one positive x axis and y axis .

Therefore area of the curve bounded by the graphs of $f^{-1}$ and $g^{-1}$ is $\int^1_{0}\frac{y^2+y-y^3-y^2}{2} \,dx$ = 1/8 .

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## ISI MStat 2018 PSA Problem 11 | Sequence & it’s subsequence

This is a problem from ISI MStat 2018 PSA Problem 11 based on Sequence and subsequence.

## Sequence & it’s subsequence – ISI MStat Year 2018 PSA Problem 11

Let ${a_{n}}_{n \geq 1}$ be a sequence such that $a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots$
Suppose the subsequence ${a_{2 n}}_{n \geq 1}$ is bounded. Then

• (A) $\{a_{2 n}\}_{n \geq 1}$ is always convergent but $\{a_{2 n+1}\}_{n \geq 1}$ need not be convergent.
• (B) both $\{a_{2 n}\}_{n \geq 1}$ and $\{a_{2 n+1}\}_{n \geq 1}$ are always convergent and have the same limit.
• (C) $\{a_{3 n}\}_{n \geq 1}$ is not necessarily convergent.
• (D) both $\{a_{2 n}\}_{n \geq 1}$ and $\{a_{2 n+1}\}_{n \geq 1}$ are always convergent but may have different limits.

### Key Concepts

Sequence

Subsequence

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Given that $a_{2n}$ is bounded . Again we have $a_{1} \leq a_{2} \leq \cdots \leq a_{2n} \leq a_{2n+1} \leq a_{2n+2} \leq \cdots$ which shows that if $a_{2n}$ is bounded then $a_{2n+1}$ is also bounded .

Again both $a_{2n}$ and $a_{2n+1}$ both are monotonic sequence . Hence both converges .

Now we have to see whether they converges to same limit or not ?

As both $a_{2n}$ and $a_{2n+1}$ are bounded hence $a_{n}$ is bounded and it’s already given that it is monotonic . Hence $a_{n}$ converges . So, it’s subsequences must converges to same limit . Hence option (B) is correct .

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## ISI MStat 2018 PSA Problem 8 | Limit of a Function

This is a beautiful problem from ISI MStat 2018 PSA Problem 8 based on limit of a function. Try yourself and use hints if required.

## Limit of a Function – ISI MStat Year 2018 PSA Question 8

The value of $\lim _{x \rightarrow \infty}(\log x)^{1 / x}$

• (A) is e
• (B) is 0
• (C) is 1
• (D) does not exist

### Key Concepts

Limit

L’hospital Rule

ISI MStat 2018 PSA Problem 8

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

What is the form of the limit? Can you convert it to some known limit?

$\infty^0$
$\frac{\infty}{\infty}$
L’Hospital Rule

Let , $\lim _{x \rightarrow \infty}(\log x)^{1 / x}$ =l (say) then taking log on both sides we get , $\lim _{x \rightarrow \infty} \frac{ log( logx) }{x}$ =log (l) . Now we will apply L’hospital rule .

Applying L’hospital rule we get , log (l)= $\lim _{x \rightarrow \infty} \frac{1}{logx x}$= 0 $\Rightarrow l=e^0 \Rightarrow l=1$

Hence , option (C) is correct .

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## ISI MStat 2018 PSA Problem 10 | Dirichlet Function

This is a problem from ISI MStat 2018 PSA Problem 10 based on Dirichlet Function.

## Dirichlet Function – ISI MStat Year 2018 PSA Question 10

Let $x$ be a real number. Then $\lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right)$

• (A) does not exist for any x
• (B) exists for all x
• (C) exists if and only if x is irrational
• (D) exists if and only if x is rational

### Key Concepts

Limit

Sandwich Theorem

ISI MStat 2018 PSA Problem 10

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

First hint

Check two cases separately one when x is rational and other is when x is irrational.

Second Hint

If $m!x$  is an integer, then $cos ^{2 n}(m ! \pi x) =1$

If x is rational $\frac{p}{q}$, then, eventually, for large enough m, m! will be divisible by q , so that $m!x$ will be an integer, and we have $\lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) =1$

Final Step

If x is irrational, $m!x$  will never be an integer, and $|cos(m! {\pi } x)|<1$ , so that $\lim {m \rightarrow \infty}\left(\lim {n \rightarrow \infty} \cos ^{2 n}(m ! \pi x)\right) =0$  for all m>0 by Sandwich Theorem.

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## ISI MStat 2018 PSA Problem 7 | Continuous Function

This is a problem from ISI MStat 2018 PSA Problem 7 based on Continuity.

## Continuous Function – ISI MStat Year 2018 PSA Question 7

Let $f$ be a function defined from $(0, \infty)$ to $\mathbb{R}$ such that
$\lim _{x \rightarrow \infty} f(x)=1$ and $f(x+1)=f(x)$ for all x
Then $f$ is

• (A) continuous and bounded.
• (B) continuous but not necessarily bounded.
• (C) bounded but not necessarily continuous.
• (D) neither necessarily continuous nor necessarily bounded.

### Key Concepts

Epsilon-Delta definition of limit

Continuity

Bounded function

ISI MStat 2018 PSA Problem 7

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Try to use the epsilon-delta definition of limit and the property that $f(x+1)=f(x)$ for all x.

$\lim_{x \to \infty} f(x) = 1$ $\Rightarrow \exists M > 0$ such that $x > M \Rightarrow$ $|f(x) – 1| < \epsilon$. Now $f(x) = f(x+1)$ for all $x \in \mathbb{R} \Rightarrow f(x) = f(x+n)$ for every $n \in \mathbb{Z}$.

Let’s try to use this .

Given any $y \in \mathbb{R}$ we can select a suitable $n \in \mathbb{Z}$ such that $y+n > M$. Then $|f(y+n) – 1| < \epsilon$. But $f(y+n) = f(y)$. Hence , $|f(y) – 1| < \epsilon$. Hence , for all $y \in \mathbb{R},$ we have $|f(y) – 1| < \epsilon$. Since , $\epsilon > 0$ is arbitrary , we must have $f(y) = 1$ for all $y \in \mathbb{R}.$

So $f$ is continuous and bounded

Categories

## ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

This is a problem from ISI MStat 2018 PSA Problem 12 based on Sequence of positive numbers

## Sequence of positive numbers – ISI MStat Year 2018 PSA Question 12

Let $a_n$ ,$n \ge 1$ be a sequence of positive numbers such that $a_{n+1} \leq a_{n}$ for all n, and $\lim {n \rightarrow \infty} a{n}=a .$ Let $p_{n}(x)$ be the polynomial $p_{n}(x)=x^{2}+a_{n} x+1$ and suppose $p_{n}(x)$ has no real roots for every n . Let $\alpha$ and $\beta$ be the roots of the polynomial $p(x)=x^{2}+a x+1 .$ What can you say about $(\alpha, \beta)$?

• (A) $\alpha=\beta, \alpha$ and $\beta$ are not real
• (B) $\alpha=\beta, \alpha$ and $\beta$ are real.
• (C) $\alpha \neq \beta, \alpha$ and $\beta$ are real.
• (D) $\alpha \neq \beta, \alpha$ and $\beta$ are not real

### Key Concepts

Sequence

Discriminant

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Write the discriminant. Use the properties of the sequence $a_n$ .

Note that as  has no real root so discriminant is  so  and ‘s are positive and decreasing so  . So , what can we say about a ?

Therefore we can say that $0 \le a < 2$ hence discriminant of P  , $a^2-4$ must be strictly negative so option D.

Categories

## 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

$\lim\limits_{x\to a}\frac{(g(x)f(a) â€“ g(a)f(x))}{(x â€“ a)}$ ?

• 5
• 3
• – 3
• -5

### Key Concepts

Algebraic manipulation

Limit form of the Derivative

ISI MStat 2019 PSA Problem 17

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

## Try with Hints

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

$\frac{(g(x)f(a) â€“ g(a)f(x))}{(x â€“ a)}$ =

$\frac{(g(x)f(a) â€“g(a)f(a) +g(a)f(a) – g(a)f(x))}{(x â€“ a)}$ =

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

Now calculate the limit using Limit form of the Derivative.

So, we have $\lim\limits_{x\to a}\frac{(g(x)f(a) â€“ g(a)f(x)}{(x â€“ a)}$ =

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

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

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## Sign change | ISI-B.stat | Objective Problem 709

Try this beautiful problem on Sign change, useful for ISI B.Stat Entrance.

## Sign change | ISI B.Stat Entrance | Problem 709

In the interval $(-2\pi, 0)$ the function $f(x) = sin(1/x^3)$

• (a) never changes sign
• (b) changes sign only once
• (c) changes sign more than once, but a finite number of times

### Key Concepts

Calculus

Limit

Trigonometry

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

## Try with Hints

As x becomes $\leq 1$ and tends to zero then it crosses $\pi, 2\pi, 3\pi, â€¦.$.can you draw the graph?

Can you now finish the problem ……….

If we draw the graph then we can see that the function $f(x) = sin(1/x^3)$ crosses many times. Therefore number of sign changes is infinite.

Therefore option $(d)$ is correct…..

Categories

## Limit Problem | ISI-B.stat | Objective Problem 694

Try this beautiful problem on Limit, useful for ISI B.Stat Entrance.

## Limit Problem | ISI B.Stat Entrance | Problem 694

Let $a_1 = 1$ and $a_n = n(a_{n-1} + 1)$ for $n = 2, 3, â€¦.$ Define $P_n = (1 +1/a_1)(1 + 1/a_2)â€¦.(1 + 1/a_n)$. Then $\lim\limits_{x \to \infty} {P_n}$?

• (a) $1+e$
• (b) $e$
• (c) $1$
• (d) $\infty$

### Key Concepts

Calculus

Limit

Trigonometry

Answer: (b)$e$

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Given that $P_n = (1 +1/a_1)(1 + 1/a_2)â€¦.(1 + 1/a_n)$

Therefore $P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}$

Now $a_n = n(a_{n-1} + 1)$

Put $n=2$, we will get $a_1+1=\frac{a_2}{2}$

$a_2+1=\frac{a_3}{3}$……………….

………………………..

…………………………

$a_n+1=\frac{a_n}{n}$

Therefore $P_n=\frac{a_1 +1}{a_1}.\frac{a_2 +1}{a_2}.\frac{a_3 +1}{a_3}…..\frac{a_n +1}{a_n}$

$\Rightarrow {P_n}= \frac{a_2}{2a_1}.\frac{a_3}{3a_2}.\frac{a_4}{4a_3}……..\frac{a_{n+1}}{(n+1).{a_n}}$

$\Rightarrow {P_n}=\frac{a_{n+1}}{{a_1}\{2.3.4………..(n+1)\}}$

$\Rightarrow {P_n}=\frac{a_{n+1}}{\{1.2.3.4………..(n+1)\}}$ (as $a_1=1$)

$\Rightarrow {P_n}=\frac{a_{n+1}}{(n+1)!}$

$\Rightarrow {P_n}=\frac{(n+1)(a_n +1)}{(n+1)!}$

$\Rightarrow {P_n}=\frac{(a_n +1)}{n!}$

$\Rightarrow {P_n}=\frac{a_n}{n!} +\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{n(a_{n-1}+1)}{n!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_{n-1}+1}{(n-1)!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{a_2}{2!}+\frac{1}{2!}+\frac{1}{3!}+…….+\frac{1}{n!}$

$\Rightarrow {P_n}=\frac{2(1+1)}{2!}+\frac{1}{2!}+….+\frac{1}{n!}$

$\Rightarrow {P_n}=1+\frac{1}{1!}+…..+\frac{1}{n!}$

Can you now finish the problem ……….

Now we have to find out $\lim\limits_{x \to \infty} {P_n}$

we know that $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+……….+\infty$

So,$e^1=1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty$

$\lim\limits_{x \to \infty} {P_n}$=$1+\frac{1}{1!}+\frac{1^2}{2!}+……….+\infty$

$\lim\limits_{x \to \infty} {P_n}$=$e$

Therefore option (b) is correct…..

Categories

## Graph in Calculus | ISI-B.stat | Objective Problem 699

Try this beautiful problem on Graph in Calculus, useful for ISI B.Stat Entrance.

## Graph in Calculus | ISI B.Stat Entrance | Problem 699

The adjoining figure is the graph of

• (a) $y = 2e^x$
• (b) $y = 2e^-x$
• (c) $y = e^x + e^-x$
• (d) $y = e^x â€“ e^-x + 2$

### Key Concepts

Calculus

Graph

Functions

TOMATO, Problem 699

Challenges and Thrills in Pre College Mathematics

## Try with Hints

At first we have to check all the options…..

1. For $y = 2e^x$, this is not possibe because $y = 2e^x$ expression represent only positive side’s of x axis
2. For $y = 2e^-x$,This expression also represents one side of x-axis
3. For $y = e^x + e^-x$.This expression represents both sides of the x-axise.as we put $x=0$ then $y=2$ i.e the curve not starts from the origin
4. $y = e^x â€“ e^-x + 2$, this expression also represents both sides of the x-axis

Can you now finish the problem ……….

Therefore ,the correct ans is (c)