Test of Mathematics Solution Subjective 90 - Graphing Inequality

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 90 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem:

:
Draw the region of points $ {\displaystyle{(x,y)}}$ in the plane, which satisfy $ {\displaystyle{|y| {\le} |x| {\le} 1}}$.


Solution:

$ {\displaystyle{|y| {\le} |x| {\le} 1}}$ region will bounded by lines $ {\displaystyle{x = y}}$, $ {\displaystyle{x = -y}}$, $ {\displaystyle{x = -1}}$ & $ {\displaystyle{x = 1}}$. Why is that?

First note that ( |x| \le 1 ) implies:

Inequality region

Similarly, if we demand ( |y| \le 1 ) (the double shaded zone).

Now if we want ( |y| \le |x| ) . This can be achieved by

Therefore the final region is the following shaded region:

Inequality region

Test of Mathematics Solution Subjective 84 - Comparing Equations

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 84 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

Show that there is exactly one value of \(x\) that satisfies the equation:

\(2 cos^2(x^3+x)=2^x+2^{-x}\)


Solution:

We know that \(cos \;x \leq 1\) for all \(x \in I\!R\)

\(=> cos(x^3 + x)\leq 1\)

\(=> cos^2(x^3 + x)\leq 1\)

\(=> 2cos^2(x^3 + x)\leq 2\)

Now consider \(2^x\) and \(2^{-x}\). By AM-GM inequality we have,

\(2^x+2^{-x}\geq 2\)

So \(2 cos^2(x^3+x)=2^x+2^{-x}\), only when \(2 cos^2(x^3+x)=2=2^x+2^{-x}\).

That means \(2^x+2^{-x} = 2, => x= 0\). So \(x=0\) being the only solution.

and \(x=0\) also satisfies \(2 cos^2(x^3+x)=2\)

Thus there is exactly one solution.

Hence Proved.

Test of Mathematics Solution Subjective 83 - Two numbers adding up to 1

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 83 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

If a and b are positive real numbers such that a + b = 1, prove that $ \displaystyle \left( a + \frac{1}{a} \right)^2 $ + \( \left( b + \frac{1}{b} \right)^2 \ge \frac{25}{2} \)


Solution:

We replace 1 by a+b and expand the squares.

$ \displaystyle \left( a + \frac{a+b}{a} \right)^2 + \left(b + \frac{a+b}{b} \right)^2 $
$ \displaystyle \left( a + \frac{b}{a} + 1 \right)^2 + \left(b + \frac{a}{b} + 1 \right)^2 $
$ \displaystyle \left( a^2 + \frac{b^2}{a^2} + 1 + 2a + 2b + 2\frac{b}{a} \right) + \left(b^2 + \frac{b^2}{a^2} + 1 + 2b + 2a + 2 \frac{a}{b} \right) $

Now we use A.M. - G.M. inequality to have

$ \displaystyle { \frac {\frac{b^2}{a^2} + \frac{a^2}{b^2}}{2}\ge \sqrt {\frac{b^2}{a^2} \times \frac{a^2}{b^2}} = 1 } $
$ \displaystyle { \frac {\frac{b}{a} + \frac{a}{b}}{2}\ge \sqrt {\frac{b}{a} \times \frac{a}{b}} = 1 } $

Also we apply the square mean inequality to have

$ \displaystyle {\frac{a^2 + b^2}{2} \ge \left(\frac{a+b}{2} \right)^2} $ Since a+b = 1, we have $ a^2 + b^2 \ge \frac{1}{2} $

Combining all of them we have $ \displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}} $


Discussion

Some theoretical background.

Test of Mathematics Solution Subjective 82 - Inequality on four positive real numbers

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 82 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Let a, b, c, d be positive real numbers such that abcd = 1. Show that $ (1+a)(1+b)(1+c)(1+d) \ge 16 $


Solution


Concept: Inequality (see this link for some background information).

Using A.M. - G.M. inequality we see that

$\frac{1+a}{2} \ge \sqrt {1 \times a} $
$ \frac{1+b}{2} \ge \sqrt {1 \times b} $
$ \frac{1+c}{2} \ge \sqrt {1 \times c} $
$ \frac{1+d}{2} \ge \sqrt {1 \times d} $

Hence $ \displaystyle {\frac{1+a}{2} \times \frac{1+b}{2} \times \frac{1+c}{2} \times \frac{1+d}{2}\ge \sqrt {abcd} = 1 }$

Therefore $ (1+a)(1+b)(1+c)(1+d) \ge 16 $

Test of Mathematics Solution Subjective 78 -Absolute Value Inequality

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 78 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

For real numbers $ {x}$, $ {y}$ and $ {\displaystyle{z}}$, show that
$ {\displaystyle{|x| + |y| + |z| {\le} |x + y - z| + |y + z - x| + |z + x - y|}}$.


Solution

Applying Ravi transformation
$ {x = a + b}$, $ {y = b + c}$ and $ {z = c + a}$.
Our inequality reduces to $ {\displaystyle{|a + b| + |b + c| + |c + a| {\le} |2b| + |2c| + |2a|}}$.
$ {\Leftrightarrow}$ $ {\displaystyle{|a + b| + |b + c| + |c + a| {\le} 2(|a| + |b| + |c|)}}$.
Now we know, $ {\displaystyle{|m + n| {\le} |m| + |n|}}$.
Applying this we get
L.H.S = $ {\displaystyle{|a + b| + |b + c| + |c + a| {\le} |a| + |b| + |b| + |c| + |c| + |a|}}$
= $ {\displaystyle{2(|a| + |b| + |c|)}}$
= R.H.S (proved)

Test of Mathematics Solution Subjective 77

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 77 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

For $ {x > 0}$, show that $ {\displaystyle{\frac{x^n - 1}{x - 1}}{\ge}{n{x^{\frac{n - 1}{2}}}}}$, where $ {n}$ is a positive integer.


Solution

$ {\displaystyle{\frac{x^n - 1}{x - 1}}{\ge}{n{x^{\frac{n - 1}{2}}}}}$

$ {\Leftrightarrow}$ $ {\displaystyle{\frac{(x - 1)(x^{n - 1} + x^{n - 2} + ......... + x + 1)}{x - 1}}}$ $ {> n x^{\frac{n - 1}{2}}}$

$ {\Leftrightarrow}$ $ {\displaystyle{\frac{x^{n - 1} + x^{n - 2} + ......... + x^1 + x^0}{n}}}$ $ {> x^{\frac{n - 1}{2}}} (\dagger) $

Now to prove \( ( \dagger) \) we observe:

But  $ \displaystyle{\frac{x^{n - 1} + x^{n - 2} + ......... + x^1 + x^0}{n} \\ > \{x^{n-1}\cdot x^{n-2} \cdots x^0 \}^{\frac{1}{n}} \\ = \{x^{(n-1) + \cdots 0} \}^{\frac{1}{n}} \\ =\{ x^{\frac{n(n-1)}{2}}\}^\frac{1}{n}} \\ = x^{\frac{(n-1)}{2}}$

Now this follows directly from AM-GM inequality.

Watch Video Solution

Useful Resources:

What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

Here is a video solution for a Problem based on finding the area of a quadrilateral. This question is from American Mathematics Competition, AMC 12, 2018. Watch and Learn!

Here goes the question…

Connect the centroids of the four triangles in a square. Can you find the area of the quadrilateral?

We will recommend you to try the problem yourself.

Done?

Let’s see the solution in the video below:

Some Useful Links:

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Solving Weird Equations using Inequality | TOMATO Problem 78

Here is a video solution for ISI Entrance Number Theory Problems based on solving weird equations using Inequality. Watch and Learn!

Here goes the question…

Solve: 2 \cos ^{2}\left(x^{3}+x\right)=2^{x}+2^{-x}

We will recommend you to try the problem yourself.

Done?

Let’s see the proof in the video below:

Some Useful Links:

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Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

Positive Integer - AIME I, 1996


For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that \(n \lt 1000\) and that \([log_{2}n]\) is a positive even integer.

  • is 107
  • is 340
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequality

Greatest integer

Integers

Check the Answer


Answer: is 340.

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

Try with Hints


here Let \([log_{2}n]\)=2k for k is an integer

\(\Rightarrow 2k \leq log_{2}n \lt 2k+1\)

\(\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}\) and \(n \lt 1000\)

\(\Rightarrow 4 \leq n \lt 8\)

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

\(\Rightarrow 4+16+64+256\)=340.

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ISI MStat 2015 PSA Problem 17 | Basic Inequality

This is a problem from ISI MStat 2015 PSA Problem 17. First, try the problem yourself, then go through the sequential hints we provide.

Basic Inequality - ISI MStat Year 2015 PSA Question 17


Let \( X=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}\). Then,

  • x<1
  • \( x>\frac{3}{2} \)
  • \( 1<x< \frac{3}{2} \)
  • None of these

Key Concepts


Basic Inequality

Check the Answer


Answer: is \( 1<x< \frac{3}{2} \)

ISI MStat 2015 PSA Problem 17

Precollege Mathematics

Try with Hints


Take it easy. Group things up. Use \(\frac{1}{n+k} < \frac{1}{n}\) for all natural (k)

\( \frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000} \)
\( \frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}\)

\( \Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2} \)

Again see that we can write , x=\( (\frac{1}{1001}+\frac{1}{3001}) + (\frac{1}{1002}+\frac{1}{3000})+ \dots + (\frac{1}{2000}+\frac{1}{2002}) + \frac{1}{2001} \) > \( \frac{2}{2001} +\frac{2}{2001} + \dots + \frac{2}{2001} +\frac{1}{2001} > \frac{2001}{2001}=1 \)

Hence , \( 1<x< \frac{3}{2} \).

Similar Problems and Solutions



ISI MStat 2015 PSA Problem 17
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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