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Calculus I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

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

Check the Answer


Answer: is (A)

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 .

ISI MStat
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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

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 .

ISI MStat 2018 PSA Problem 11
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Calculus I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

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 .

ISI MStat 2018 PSA Problem 8
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Calculus I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

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


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.

ISI MStat 2018 PSA Problem 10
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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

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

ISI MStat 2018 PSA Problem 7
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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.

ISI MStat 2018 PSA Problem 12
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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 \( \lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x)}{(x – a)} \) =

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

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

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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
  • (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 ……….

Sign change - graph

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…..

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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…..

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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

graph
  • (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


Griaph in calculus

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. \(y = e^x – e^-x + 2\), this expression also represents both sides of the x-axis

Can you now finish the problem ……….

Therefore ,the correct ans is (c)

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