ISI 2018 Subjective Problem 8 , A Problem from Matrix

Try this beautiful Subjective Matrix Problem appeared in ISI Entrance - 2018.

Problem

$a_{i j} \in\{1,-1\}$ for all $1 \leq i,j \leq n$.

Suppose that

\[a_{k 1}=1 \text { for all } 1 \leq k \leq n \]

and \sum k =1 n a ki a kj =0 for all i \neq j

Show that $n$ is a multiple of $4 $.

Key Concepts

Algebra

Matrix

Suggested Book | Source | Answer

Suggested Book: IIT mathematics by Asit Das Gupta

Source of the Problem: ISI UG Entrance - 2018 , Subjective problem number - 8

Try with Hints...

Hint 1:
We have $a_{k 1}=1 \forall k=1,2, \ldots, n$.
Now,
\[\sum_{k=1}^{n} a_{k 1} a_{k 2}=0 \]

\[\Rightarrow \sum_{k=1}^{n} a_{k 2}=0\]

Similarly,
\[ \sum_{k=1}^{n} a_{k 3}=0\]
Hence proceed.

Hint 2:

As $a_{i j}=+1$ or $-1$
so number of $+1^{\prime} s$ and $-1^{\prime} s$ are same in every column.
Therefore $n$ must be even . $(n=2 m$ say $)$
Proceed to work with the following:

\[\sum_{k=1}^{n} a_{k 2} a_{k 3}=0\]

Hint 3
Observe in column 2,
\[\prod_{k=1}^{n} a_{k 2}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots \ldots(1)\]

Similarly in column 3,
\[\prod_{k=1}^{n} a_{k 3}=(-1)^{m} \ldots \ldots(2)\]

And we also have

\[\sum_{k=1}^{n} a_{k 2} a_{k 3}=0 \ldots \ldots\]
So proceed.

Hint 4:
Obviously,
Hence, $m$ of the $a_{k 2} a_{k 3}$ are $+1^{\prime}$ s and $m$ of them are $-1$ 's.
Now
\[\prod_{k=1}^{n} a_{k 2} a_{k 3}=(1)^{m}(-1)^{m}=(-1)^{m} \ldots(4)\]

Hint 5:
But,
\[\prod_{k=1}^{n} a_{k 2} a_{k 3}=\prod_{k=1}^{n} a_{k 2} \Pi_{k=1}^{n} a_{k 3}\]
we get by (1) and (2)
\[=(-1)^{m}(-1)^{m}=1 \ldots \ldots(5)\]

Hint 6:
Comparing (4) and (5),
we get $(-1)^{m}=1$
Hence, $m$ is even .
Therefore, $n$ is obviously a multiple of 4 .

ISI 2018 Objective Problem 8 | A Problem from Sequence

Try this beautiful Objective Sequence Problem appeared in ISI Entrance - 2018.

Problem

Consider the real valued function $h:\{0,1,2, \ldots, 100\} \longrightarrow \mathbb{R}$ such that $h(0)=5, h(100)=20$ and satisfying $h(i)=\frac{1}{2}(h(i+1)+h(i-1))$, for every $i=1,2, \ldots, 99$. Then the value of $h(1)$ is:

(A) $5.15$
(B) $5.5$
(C) $6$
(D) $6.15.$

Key Concepts

Sequence

Arithmetic Progression

Suggested Book | Source | Answer

IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2018 , Objective problem number - 8

(B) $5.15$

Try with Hints

Observe the following,

$2 h(i)=h(i-1)+h(i+1)$

$2h(1) = h(0) + h(2) $

$2h(2) = h(1) + h(3) $

and so on.

Therefore we have the following,

$h(i+1)-h(i)=h(i)-h(i-1).$

Means

$h(0) , h(1) , h(2) , \ldots \ldots ,h(100) $ are in Arithmetic Progression.

$h(0) $ and $ h(100)$ are the first and last terms of the AP.

Common difference

\[=\frac{h(100) - h(0)}{100}\]

\[=\frac{20 - 5}{100}= 0.15\]

Therefore ,

$h(1) = h(0) + 0.15 = 5.15$

ISI 2021 Objective Problem 23 I A Problem from Limit

Try this beautiful Objective Limit Problem appeared in ISI Entrance - 2021.

Problem

Let us denote the fractional part of

a real number $x$ by $\{x\}.$

(Note ${x}=x-[x]$ where $[x]$

is the integer part of $x$.)

Then
\[
\lim _{n \rightarrow \infty}{(3+2 \sqrt{2})^{n}}.
\]

(A) equals 0
(B) equals 1
(C) equals $\frac{1}{2}$
(D) does not exist

Key Concepts

Fractional part of a real number

Greatest Integer function

Suggested Book | Source | Answer

IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2021 , Objective problem number - 23

(B) equals 1

Try with Hints

Try to find the fractional part of $(3+2 \sqrt{2})^{n} = N $(Let) .

$\bullet $ Observe $(3+2 \sqrt{2})^{n} + (3-2 \sqrt{2})^{n}$ is an integer.

$\bullet $ And use $ 0 < (3-2 \sqrt{2}) < 1.$

$\bullet $ Obviuosly $0<(3-2 \sqrt{2})^{n}<1.$

$\bullet $ Also assume , $p=(3-2 \sqrt{2})^{n}.$

$\bullet $ As $N+p = integer = [N] + \{N\} + p ,$

so $ \{N\} + p = integer - [N]= integer.$

$\bullet $ Hence proceed.

$\bullet $ It is very easy to find that $ \{N\} + p =1.$

∙ Therefore ,

$\lim _{n \rightarrow \infty}(3+2 \sqrt{2})^{n}$

= lim n \to\infty {N} = lim n \to\infty (1 - p) =1

[As, lim n \to\infty p = lim n \to\infty (3 -2 2 - \sqrt) n =0]

ISI 2021 Subjective Problem 5 I A Problem from Polynomials

Try this beautiful Subjective Problem from Polynomials appeared in ISI Entrance - 2021.

Problem

Let $a_{0}, a_{1}, \ldots, a_{19} \in \mathbb{R}$ and
\[
P(x)=x^{20}+\sum_{i=0}^{19} a_{i} x^{i} x \in \mathbb{R}
\]
If $P(x)=P(-x)$ for all $x \in \mathbb{R}$ and

$P(k)=k^{2}$, for all $k=0,1,2, \ldots, 9$

then find
\[
\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}.\]

Key Concepts

Monic Polynomial

Even Polynomial

Degree of a Polynomial

Suggested Book | Source | Answer

An Excursion in Mathematics (Chapter - 2.1)

ISI Entrance - 2021 , Subjective problem number - 5

$1-(9 !)^{2}$

Try with Hints

$\bullet $ Recall the Fundamental Theorem of Algebra i.e. every polynomial $P(z)$ of degree $n$ has $n$ values $z_{i}$ (some of them possibly degenerate) for which $P\left(z_{i}\right)=0$.

$\bullet $ And apply it to construct the polynomial.

$\bullet $ Observe $P(0) =0$ i.e. $0$ is a root of $P(x) .$

Use $P(x)=P(-x)$ to observe $P(\pm k)=k^{2}$ for every $k=1,2, \ldots, 9$.
Define a new polynomial $Q(x)$
\[
Q(x)=P(x)-x^{2}
\]
And see constructing $Q(x)$ is easier . Hence find $Q(x)$ and eventually $P(x)$.

To construct $Q(x)$ use followings:

$Q(x)$ has 19 roots and those are 0 and $\pm 1, \pm 2, \ldots, \pm 9$.

As $P(x)$ is monic and of degree 20 , so $Q(x)$ is also. Hence all factors of $Q(x)$ are like $(x+a)$.

Therefore,
\[Q(x)=x(x-1)(x+1)(x-2) \]

\[\ldots (x-9)(x+9) \times(x+c)
\]
(Observe extra $(x+c)$ is multiplied to make the degree of $Q(x)$ to be 20 .)

$\bullet$ As

\[Q(x)=x(x-1)(x+1)(x-2) \]

\[\ldots (x-9)(x+9) \times(x+c)\]

$\bullet $

\[
P(x)=x(x-1)(x+1)(x-2) \]

\[\ldots (x-9)(x+9) \times(x+c)+x^{2} .\]

\[\Rightarrow P(x)=x^{2}(x^{2}-1)(x^{2}-4) \]

\[\ldots (x^{2}-81)+x^{2}\]

$\bullet $ Have you noticed the coefficients of all odd exponents of $x$ in $P(x) $ are $0?$

$\bullet $ Recall we are given that $P(x)=P(-x)$ for all $x \in R$ means that $P(x)$ is an even function and so all odd degree coefficients are 0 . That is, $a_{i}=0$ for $i=1,3,5, \ldots, 17,19$.

$\bullet $ Therefore,

\[P(x)=x^{2}(x^{2}-1)(x^{2}-4) \]

\[\ldots(x^{2}-81)+x^{2} .\]

\[\Rightarrow \frac{P(x)}{x^{2}}=(x^{2}-1)(x^{2}-4)\]

\[ \ldots(x^{2}-81)+1 .\]

\[\Rightarrow \lim _{x \rightarrow 0} \frac{P(x)}{x^{2}}\]

\[=(-1)(-4) \ldots(-81)+1\]

Roots and coefficients of equations | PRMO 2017 | Question 4

Try this beautiful problem from the PRMO, 2017 based on Roots and coefficients of equations.

Roots and coefficients of equations - PRMO 2017

Let a,b be integers such that all the roots of the equation $(x^{2}+ax+20)(x^{2}+17x+b)$=0 are negetive integers, find the smallest possible values of a+b.

is 107
is 25
is 840
cannot be determined from the given information

Key Concepts

Polynomials

Roots

Coefficients

Check the Answer

Answer: is 25.

PRMO, 2017, Question 4

Polynomials by Barbeau

Try with Hints

$(x^{2}+ax+20)(x^{2}+17x+b)$

where sum of roots $ \lt $ 0 and product $ \gt 0$ for each quadratic equation $x^{2}$+ax+20=0 and

$(x^{2}+17x+b)=0$

$a \gt 0$, $b \gt 0$

now using vieta's formula on each quadratic equation $x^{2}$+ax+20=0 and $(x^{2}+17x+b)=0$, to get possible roots of $x^{2}$+ax+20=0 from product of roots equation $20=(1 \times 20), (2 \times 10), (4 \times 5)$

min a=4+5=9 from all sum of roots possible

again using vieta's formula, to get possible roots of $(x^{2}$+17x+b)=0 from sum of roots equation $17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),$

$(-8,-9)$

min b=(-1)(-16)=16 from all products of roots possible

$(a+b)_{min}=a_{min}+b_{min}$=9+16=25.

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 2016 Problem 1 | Area bounded by the curves | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 1

In the diagram below, $L(x)$ is a straight line that intersects the graph of a polynomial $P(x)$ of degree 2 at the points $A=(-1,0)$ and $B=(5,12) .$ The area of the shaded region is 36 square units. Obtain the expression for $P(x)$.

Prerequisites

Area bounded by the curve
Polynomials of degree 2
Area of a triangle

Solution

Let $P(x)=ax^2 +bx + c $ as given $P(x)$ is of degree 2 .

Now from the figure we can see that $L(x)$ intersect $P(x)$ at points $A=(-1,0)$ and $B=(5,12) .$

Hence we have $P(-1)=0$ and $P(5)=12$ , which gives ,

$ a-b+c=0 $ ---(1) and $ 25a+5b +c =12 $ ----(2)

Then ,

ISI MStat 2016 Problem 1 graph — **Fig-1**

See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )

= $ \int^{L}_{-1} P(x)\,dx - \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx $

=$\int^{L}_{-1} P(x)\,dx - \int^{5}_{L} P(x)\,dx $ -36

=$ \int^{5}_{-1} P(x)\,dx $ -36

Again it is given that area of the shaded region is 36 square units.

So, $ \int^{5}_{-1} P(x)\,dx $ -36 =36 $ \Rightarrow $ $ \int^{5}_{-1} P(x)\,dx $ =$ 2 \times 36 $

$ \int^{5}_{-1} (ax^2+bx+c) \,dx = 2 \times 36 $ . After integration we get ,

$ 7a + 2b +c =12 $ ---(3)

Now we have three equations and three unknows

$ a-b+c=0 $

$ 25a+5b +c =12 $

$ 7a + 2b +c =12 $

Solving this three equations by elimination and substitution we get ,

$ a=-1 , b=6 , c=7 $

Therefore , the expression for $P(x)$ is $ P(x)= -x^2+6x+7 $ .

Previous MStat Posts:

Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin - AIME I, 2009 Question 3

A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac{1}{25}$ the probability of five heads and three tails. Let p=$\frac{m}{n}$ where m and n are relatively prime positive integers. Find m+n.

Key Concepts

Probability

Theory of equations

Polynomials

Check the Answer

Answer: 11.

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints

here $\frac{8!}{3!5!}p^{3}(1-p)^{5}$=$\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}$

then $(1-p)^{2}$=$\frac{1}{25}p^{2}$ then 1-p=$\frac{1}{5}p$

then p=$\frac{5}{6}$ then m+n=11

Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers - AIME, 2009

There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

Key Concepts

Complex Numbers

Theory of equations

Polynomials

Check the Answer

Answer: 697.

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints

Taking z=a+bi

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

Key Concepts

Combinations

Theory of equations

Polynomials

Check the Answer

Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

Problem

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Problem

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Problem

Key Concepts

Suggested Book | Source | Answer

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Roots and coefficients of equations - PRMO 2017

Key Concepts

Check the Answer

Try with Hints

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

Key Concepts

Check the Answer

Try with Hints

Similar Problems and Solutions

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Outstanding Statistics Program with Applications

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Problem- ISI MStat 2016 Problem 1

Prerequisites

Solution

Previous MStat Posts:

Probability of tossing a coin - AIME I, 2009 Question 3

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Complex Numbers - AIME, 2009

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Combinations- AIME, 2009

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube