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

Check the Answer

Answer: is (B)

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 .

Outstanding Statistics Program with Applications

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

Check the Answer

Answer: is (C)

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 .

Outstanding Statistics Program with Applications

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

Check the Answer

Answer: is (B)

ISI MStat 2018 PSA Problem 10

Introduction to Real Analysis by Bertle Sherbert

Try with Hints

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

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 $

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.

Outstanding Statistics Program with Applications

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

Check the Answer

Answer: is (A)

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

Outstanding Statistics Program with Applications

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

Quadratic equation

Discriminant

Check the Answer

Answer: is (D)

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 $P_n$ has no real root so discriminant is $(a_n)^2-4<0$ so $|a_n|<2$ and $a_n$ 's are positive and decreasing so $0\leq a_n<2$ . 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.

Outstanding Statistics Program with Applications

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) – g(a)f(x))}{(x – a)} $ ?

5
3
- 3
-5

Key Concepts

Algebraic manipulation

Limit form of the Derivative

Check the Answer

Answer: is 5

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) – g(a)f(x))}{(x – a)} $ so that you can use the Limit form of the Derivative . Let's give a try .

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

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

Outstanding Statistics Program with Applications

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
(d) changes sign infinite number of times

Key Concepts

Calculus

Limit

Trigonometry

Check the Answer

Answer: (d)

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 $\pi, 2\pi, 3\pi, ….$.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.....

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 $n = 2, 3, ….$ Define $P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)$. Then $\lim\limits_{x \to \infty} {P_n}$?

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

Key Concepts

Calculus

Limit

Trigonometry

Check the Answer

Answer: (b)$e$

TOMATO, Problem 709

Challenges and Thrills in Pre College Mathematics

Try with Hints

Given that $P_n = (1 +1/a_1)(1 + 1/a_2)….(1 + 1/a_n)$

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.....

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) $y = e^x – e^-x + 2$

Key Concepts

Calculus

Graph

Functions

Check the Answer

Answer: (C)

TOMATO, Problem 699

Challenges and Thrills in Pre College Mathematics

Try with Hints

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

For $y = 2e^x$, this is not possibe because $y = 2e^x$ expression represent only positive side's of x axis
For $y = 2e^-x$,This expression also represents one side of x-axis
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
$y = e^x – e^-x + 2$, this expression also represents both sides of the x-axis

Can you now finish the problem ..........

Therefore ,the correct ans is (c)

Negative & Positive Roots | ISI-B.stat | Objective Problem 708

Try this beautiful problem on Negative & Positive Roots, useful for ISI B.Stat Entrance.

Negative & Positive Roots | ISI B.Stat Entrance | Problem 708

If n stands for the number of negative roots and p for the number of positive roots of the equation $e^x = x$, then

(a) $n = 1$, $p = 0$
(b) $n = 0$, $p = 1$
(c) $n = 0$, $p > 1$
(d) $n = 0$, $p = 0$

Key Concepts

Calculus

Limit

Trigonometry

Check the Answer

Answer: (d)

TOMATO, Problem 708

Challenges and Thrills in Pre College Mathematics

Try with Hints

Draw the figure of given equation $e^x = x$ and check all the points that are in the options.....

Can you now finish the problem ..........

The given equation is $e^x = x$

as $x $increases exponential function increases more rapidly than any polynomial function say $x$ . Hence from graph we can say they never intersect . So, there is no solution for the equation . Therefore $n=0$ and $p=0.$

Therefore option $(d)$ is correct.....

Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

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Limit of a Function - ISI MStat Year 2018 PSA Question 8

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Dirichlet Function - ISI MStat Year 2018 PSA Question 10

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Continuous Function - ISI MStat Year 2018 PSA Question 7

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Sequence of positive numbers - ISI MStat Year 2018 PSA Question 12

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Limit of a function

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Sign change | ISI B.Stat Entrance | Problem 709

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Limit Problem | ISI B.Stat Entrance | Problem 694

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

Key Concepts

Check the Answer

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Negative & Positive Roots | ISI B.Stat Entrance | Problem 708

Key Concepts

Check the Answer

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