AM - GM Inequality and Minimum Value | ISI BStat BMath Entrance 2015 Objective 5

In this video we discuss the Arithmetic Mean-Geometric Mean (AM-GM) inequality and its application to find minimum value of an algebraic expression. We use a problem from the ISI BStat BMath Entrance exam 2015 which involves finding the minimum value of an expression, demonstrating algebraic manipulation without AM-GM inequality.

We look at the relationship between AM and GM and use the fact that the arithmetic mean is always greater than or equal to the geometric mean. We also explore a geometric interpretation of the AM-GM inequality using the example of lengths forming a circle's diameter.

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Math Kangaroo 2021 Problem 17 | Kadett

Try this beautiful Problem based on simple Arithmetic appeared in Math Kangaroo (Kadett) 2021 Problem 17.

Math Kangaroo 2021 Problem 17 | Kadett

There are 20 questions in a quiz. Each correct answer scores 7 points, each wrong answer scores $-4$ points, and each question left blank scores 0 points. Eric took the quiz and scored 100 points. How many questions did he leave blank?

Key Concepts

Arithmetic

Divisibility

Logic Question

Suggested Book | Source | Answer

Algebra by Gelfand

Math Kangaroo (Kadett), 2021

$1$

Try with Hints

Here given that,

Total number of questions $=20$
Score of correct answer $=7$ points
Score of wrong answer $=-4$
Eric scored $=100$ points

So that mean, the score for all of the correct answers is greater than $100$.

Now we know $7\times 16 = 112$.

Now we have to count how many wrong answers are there if the score is $100$.

Remaining question $=20-(16+3)$
So, $1$ is the correct answer.

Math Kangaroo 2021 Problem 21 | Ecolier

Try this beautiful Problem based on simple Divisibility Rules appeared in Math Kangaroo (Ecolier) 2021 Problem 21.

Math Kangaroo 2021 Problem No 21 | Ecolier

A box has fewer than 50 cookies in. The cookies can be divided evenly between 2,3 , or 4 children. However, they cannot be divided evenly between 7 children because 6 more cookies would be needed. How many cookies are there in the box?

Key Concepts

Arithmetic

Divisibility

Algebra

Suggested Book | Source | Answer

Algebra by Gelfand

Math Kangaroo (Ecolier), 2021

$36$

Try with Hints

Here

The box contain less than 50 cookies.

Cookies are distributed among 2, 3 or 4 children.

So, all of the options are correct.

Now apply the second condition in the question.

They cannot be divided evenly between 7 children because 6 more cookies would be needed.

Apply the condition and check which option is satisfying the condition.

So, $36$ is the correct answer.

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 2015 Subjective Problem 8 | A Problem from Sequence

Try this beautiful Subjective Sequence Problem appeared in ISI Entrance - 2015.

Problem

(b) For any integer $k>0$, give an example of a sequence of $k$ positive integers whose reciprocals are in arithmetic progression.

Key Concepts

Sequence

Arithmetic Progression

Suggested Book | Source | Answer

IIT Mathemathematics by Asit Dasgupta

ISI UG Entrance - 2015 , Subjective problem number - 8

Try to prove using the following hints.

Try with Hints

Assume , $d$ is the common difference for the given AP.

Therefore,

$ d = \frac{1}{m_2} - \frac{1}{m_1} \leq \frac{1}{m_1+1} - \frac{1}{m_1} = d' (say)$

[Equality holds when , $m_2 = m_1 + 1 $ ]

Hence proceed.

Now , $ \frac{1}{m_k} = \frac{1}{m_1} + d(k-1) \leq \frac{1}{m_1} + d'(k-1)$

Notice that till now we haven't used $m_1 < m_2 < \ldots < m_k$ are positive integers.

So , $\frac{1}{m_k} > 0 $.

Therefore , $\frac{1}{m_1} + d'(k-1) > 0 $.

Now use $ d' = \frac{1}{m_1+1} - \frac{1}{m_1} $.

\[\frac{1}{m_1} + d'(k-1) > 0 \]

\[\Rightarrow \frac{1}{m_1} + ( \frac{1}{m_1+1} - \frac{1}{m_1})(k-1) > 0 \]

Proceed with the above inequality and get $m_1 + 2 > k .$

As $m_1 < m_2 < \ldots < m_k$ ,

so $\frac{1}{m_1}, \frac{1}{m_2}, \ldots, \frac{1}{m_k}$

is a decreasing $AP.$

Think about the LCM of $m_1 < m_2 < \ldots < m_k$ .

Then you proceed.

Suppose , $L = LCM(m_1 < m_2 < \ldots < m_k).$

Now multiply $L$ with all the reciprocals

i.e. with $\frac{1}{m_1}, \frac{1}{m_2}, \ldots, \frac{1}{m_k}.$

Then observe the pattern and try get such a sequence.

E.g. if $k=5$

Take $5,4,3,2,1$ and $LCM(5,4,3,2,1)=60$.

So the AP : $\frac{5}{60},\frac{4}{60} , \frac{3}{60} , \frac{2}{60}, \frac{1}{60}$

with common difference $\frac{1}{60}.$

ISI 2019 Subjective Problem 2 | Removable Discontinuity

Try this beautiful Subjective Calculus Problem appeared in ISI Entrance - 2019.

Problem

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be defined by
\[
f(x)=\lim _{n \rightarrow \infty} \cos ^{n}\left(\frac{1}{n^{x}}\right)
\]
(a) Show that $f$ has exactly one point of discontinuity.
(b) Evaluate $f$ at its point of discontinuity.

Key Concepts

Calculus

Limit , Continuity

Suggested Book | Source | Answer

Suggested Reading: IIT mathematics by Asit Das Gupta

Source of the Problem: ISI UG Entrance - 2019 , Subjective problem number - 2

Try with Hints

Hint 1:

Observe that,

we have $\frac{1}{n^x} \rightarrow 0 $

as x∈(0,∞) and n→∞.

So the indeterminate form of the given limit is $1^{\infty}.$

Hint 2:

\[f(x)=\lim _{n \rightarrow \infty} \cos ^{n}\left(\frac{1}{n^{x}}\right)\]

\[=e^{\lim _{n \rightarrow \infty} n \log \cos \left(\frac{1}{n^{x}}\right)}\]

= e lim n \to\infty n log [1-2 sin 2 (1 2 n x)]

\[=e^{\lim _{n \rightarrow \infty}\frac{ n \log [1-2 \sin ^{2}(\frac{1}{2 n^{x}})]}{-2 \sin ^{2}(\frac{1}{2 n^{x}})}(-2 \sin ^{2}(\frac{1}{2 n^{x}}))}\]

Hint 3:

As we have $\frac{1}{n^x} \rightarrow 0 $ ,

\[lim_{n\rightarrow \infty}-2 \sin ^{2}\left(\frac{1}{2 n^{x}}\right)=0.\]

And we have the standard result :

\[lim_{x \rightarrow 0}\frac{log(1+x)}{x} = 1 \]

Hint 4:

Therefore ,

\[f(x)=e^{\lim _{n \rightarrow \infty} n [-2 \sin ^{2}(\frac{1}{2 n^{x}})]}\]

\[f(x)=e^{\lim _{n \rightarrow \infty} -2n [\frac{ \sin ^{2}(\frac{1}{2 n^x})}{(\frac{1}{2 n^{x}})^2}](\frac{1}{2 n^{x}})^2}\]

And here again,

$lim_{n \rightarrow \infty}\frac{1}{2n^x} \rightarrow 0 .$

So $f(x)$ reduces to ,

\[ f(x) = e^{\lim _{n \rightarrow \infty}-2n (\frac{1}{2 n^{x}})^2}\]

\[f(x)=e^{-\frac{1}{2} \lim _{n \rightarrow \infty} n^{1-2 x}}\]

Hint 5:

Therefore ,

\[ f(x) = \Bigg{\{}{\begin{matrix}
0 & , & x < \frac{1}{2}\\
e^{-\frac{1}{2}} & ,& x= \frac{1}{2} \\1 & , & x> \frac{1}{2}
\end{matrix}}\]

Discontinuity at $x= \frac{1}{2}$

Math Kangaroo (Benjamin) 2014 Problem No 24

Try this beautiful Problem based on simple Algebra appeared in Math Kangaroo (Benjamin) 2014 Problem 24.

Math Kangaroo (Benjamin) 2014 | Problem No 24

Grandma gives 180 marbles to her ten grandchildren. No two children get the same amount of marbles. Anna gets the most. What is the minimum number of marbles that Anna could get?

Key Concepts

Arithmetic

Equation solving

Algebra

Suggested Book | Source | Answer

Algebra by Gelfand

Math Kangaroo (Benjamin), 2014

$23$

Try with Hints

Here

Number of children is $10$.
Number of marbles is $180$.
And Anna gets the most and no 2 children gets the same number of marbles.

Let us assume that Anna could get $x$ marbles and also the other 9 children receiving 1 less each step. Apply the condition to construct the equation

The equation will be-

$x$+$x$-$1$+$x$-$2$+$x$-$3$+$x$-$4$+$x$-$5$+$x$-$6$+$x$-$7$+$x$-$8$+$x$-$9$=$180$

Now solve for $x$.

So, Anna can have $23$ marbles.

AMC 10A 2021 Problem 9 | Factorizing Problem

Try this beautiful Problem based on Factorizing Problem from AMC 2021 Problem 9.

Factorizing Problem: AMC 10A 2021 Problem 9

What is the least possible value of $(x y-1)^{2}+(x+y)^{2}$ for real numbers $x$ and $y$ ?

0
$\frac{1}{2}$
$\frac{1}{4}$
1
2

Key Concepts

Expansion of Polynomial

Factorization

Suggested Book | Source | Answer

AMC 10A 2021 Problem 9

Try with Hints

Expand the expression.

So, we get that the expression is $x^{2}+2 x y+y^{2}+x^{2} y^{2}-2 x y+1$ or $x^{2}+y^{2}+x^{2} y^{2}+1$.

Then the minimum value for this is 1 , which can be achieved at $x=y=0$. .

AMC 10A 2021 Problem 22 | System of Equations

Try this beautiful Problem based on System of Equations from AMC 10A, 2021 Problem 22.

System of Equations | AMC 10A 2021, Problem 22

Hiram's algebra notes are 50 pages long and are printed on 25 sheets of paper; the first sheet contains pages 1 and 2 , the second sheet contains pages 3 and 4 , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly 19 . How many sheets were borrowed?

Key Concepts

Arithmetic Sequence

System of Equations

Algebra

Suggested Book | Source | Answer

Problem-Solving Strategies by Arthur Engel

AMC 10A 2021 Problem 22

Try with Hints

Let us assume that the roommate took sheets $a$ through $b$.
So, try to think what will be the changes in the page number?

So, page numbers $2 a-1$ through $2 b$. Because there are $(2 b-2 a+2)$ numbers.

Now apply the condition given there.

So we get, $\frac{(2 a-1+2 b)(2 b-2 a+2)}{2}$+$19(50-(2 b-2 a+2))$=$\frac{50 \cdot 51}{2}$

Now simplify this expression.

So , $2 a+2 b-39=25, b-a+1=13$

Now solve for $a, b$.

Find the number of pages using the values.

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 ,

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

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Math Kangaroo 2021 Problem 17 | Kadett

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Math Kangaroo 2021 Problem No 21 | Ecolier

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Problem

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Problem

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Problem

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Math Kangaroo (Benjamin) 2014 | Problem No 24

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Factorizing Problem: AMC 10A 2021 Problem 9

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System of Equations | AMC 10A 2021, Problem 22

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Problem

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