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,

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

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

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

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

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

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

ISI 2018 Subjective Problem 8 , A Problem from Matrix

Problem

Key Concepts

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Try with Hints...

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ISI 2018 Objective Problem 8 | A Problem from Sequence

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ISI 2021 Objective Problem 23 I A Problem from Limit

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ISI 2021 Subjective Problem 5 I A Problem from Polynomials

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