MATHEMATICS BEHIND THE MAGIC

MIND READER

          

You might have come across mind reader, but ever wondered how it works. How it is able to give you the exact symbol each time. You can check out the mind reader from this site or download it from here.

What it does ?
Select a two digit number. Add the two digits and subtract it from the original number. You are not to disclose any number at any time. The symbol ahead of the result will be displayed. Check out this magic. It really works.

Let us see the Mathematics behind the Magic.
Select any two digit number. You can select any number from 10 to 99. Right ? This number can be expressed as 10x + y. Comfortable. ? For eg if you want to make 56 then x is 5 and y is 6, so that 10*5 + 6 = 56.


Now add both digits. i.e. 5 + 6 = 11
I shall add them x + y. Right !!!

Now subtract the result from the original number, means 56-(5+6) = 56-11 = 45. Right !!
I shall do the same now with algebra. 10x+y-(x+y) = 9x.

You see your symbol and press the button. The symbol that is displayed is same as you got.


Now from my answer which is 9x, it implies that for whatever value of x or y you take, you will always end up with a multiple of 9 as your final answer. And the trick is that all the multiples of 9 means the numbers 9,18,27,36,45,54,63,72,81,90 and 99, they all have the same symbol in the game. The symbol changes at all places in the next try but all these places hold a similar symbol again in the next game (though the symbol being a new one now).

So whatever number you select I can tell you what symbol you will arrive at.

YOU WILL FAINT TO SOLVE

A question can be really difficult one, but the answer need not be that difficult. Lets see a question which I hope shall use each inch of your brain and yet you shall be amazed to see the solution.

Problem:
If 2a + 3b + 6c =0, & a,b,c ε R, then prove that a.(x)^2 + b.x + c has atleast one root in
the interval (0,1).

The solution for the same can be found here....

FROM THE FUNCTIONS

Here is another simple but a really good concept building problem from the topic Functions and Graphs
The problem must be approached systematically to arrive at the result. After solving the question you will build some confidence to solve the problems related specially to the Greatest Integer Functions.


Problem :

The number of solutions of | [x] – 3x | = 6, where [x] is the greatest integer < or = x, is

1.   2
2.   4
3.   3
4.   No solution

 Solution:
          First of all we must know what is [x].

[x] is the greatest integer function i.e. if the value of x is 2.5 then the greatest integer less than 2.5 is 2. If the value of x is -2.5 then the greatest integer less than -2.5 is -3 and Not -2.

In this problem the nature of x is not specified, so we have to consider the whole range of real numbers.

First we will consider that x is an integer and later as a non-integer.

When x is an integer, which means
[x] = x,

hence the equation | [x] – 3x | = 6 reduces to | x – 3x | = 6 which can be easily solved as below:

x – 3x = 6    &   -(x – 3x) = 6

-2x = 6        &   -(-2x) = 6

x = -3          &      x = 3

When x is not an integer, we can write x as x = n+ k. Where n is the integer and k is the fraction part such that 0 < k < 1.

so the equation | [x] – 3x | = 6 will now reduce to | [n+k] – 3(n+k) | = 6 i.e.

| n – 3n - 3k | = 6

| – 2n - 3k | = 6

What is important to note is that for the result to be an integer as can bee seen on the R.H.S. the factor 3k on L.H.S. must be an integer which is possible only when 3k=1 i.e. when k = 1/3

With k = 1/3, the equation becomes | -2n - 1 | = 6 which can be solved very easily as

-2n - 1 = 6       &      -( -2n - 1 ) = 6

-2n = 7            &        2n = 5

n = -7/2          &        n = 5/2

Hence we see that four solutions are possible to the question. Hence the option (b)

This was a very simple problem and hardly takes two minutes.

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For support and queries mail to mathsprobe@gmail.com

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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

CO-ORDINATE WITH GEOMETRY

Geometry can be fun to learn and easy to score. But sometimes we take up the wrong approach, start solving unnecesssary equations and thus make the problem look all the more difficult and complex than it actually is. To get a better understanding of the topic, one must try solving the question by self rather than depending on the solution. The solution should help you get the right direction and not lessen your efforts and contribution.



Problem :
Determine the values of β if (0, β) lies inside the Δ formed by 2x + y + 2 = 0 , 3y -2x -5 = 0 and 4y +x = 14.


Try this problem yourself before checking out the solution. Click here to Download the solution 



Go to Next Chapter on Geometry


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Your suggestions and querries are welcome.
Mail to mathsprobe@gmail.com
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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

PROBABILITY OF SOLVING THIS PROBLEM

Probability can be a real fun to solve and scoring topic too but only when right approach is used. A wrong approach can take you far from the answer. Its Probability is much more a case to be thought practically rather than jumping here and there with equations. Probability of solving questions on this topic is inversely proportional to the probability of your losing concentration. So check your concentration.....



Problem :
From a pack of n cards no 1,2,3,4,5.......... n. A draws a card and then puts it back. Now B draws a card. Find the probability that B has drawn the same card as A.
Solution :
A can draw card in n ways.
Similarly, B can also draw card in n ways.
So the total no ways in which A and B can select a card is = n x n.
This becomes the total no of possible cases.

Now A and B can draw the same card nos like in these cases(1,1) or (2,2) or (3,3) or (4,4)....... or (n,n). Hence the total no of favourable cases are n.

So P = No of favourable cases / Total no of cases = n/(nxn) = 1/n



That was quite a simple problem. But the important thing to consider is that how you start to think. If you start with a negative approach of  " its too difficult ..very hard problem... bla bla bla....", it oughts to become the same.  So be positive and explore the ways to achieve the desired results.

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Your suggestions and querries are welcome.
Mail to mathsprobe@gmail.com
*************************************************************

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

TEST YOUR SKILLS

If you think you are best prepared for the toughest one, test yourself for the lighter one. A ten minute test for all. It will help you understand better. The questions here asked require a little more concentration. Do not worry its not a mountain load. If you are able to solve it easily and quickly, you need not worry. The rest better pick up their books and attend schools :)


Ideally you should be able to solve atleast 3 or 4.

Go ON THE PIC TO ENLARGE

You may download the .pdf for your use from this download-link 

Click Here to Download the Solutions.

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For queries and support mail to :  mathsprobe@gmail.com

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IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

INTRODUCTION

Welcome to all readers !
This is a special effort to help all the students with the subject which is mis-taught as difficult. You might have experienced difficulties every now and then while working on it. But the real problem lies in the fact that the approach to solve problem is not same in every question. It is not the amount or the quantity of exercises you solve to have a grasp on the subject but it is how deep understanding you have for the problem to be solved. Today when the competition is more tougher than ever before, you can no longer sit and try the older hit and trial error or expect for any shortcut to work at all times. To save time you must know widen your experience with different types and level of problems. My advice to all my readers is never rush to get and answer to a problem and throw away, rather learn how you reach to the solution.


Let me show a simple case !


If g(x) = f(x) + f(1-x)
& f ''(x)<0  V x € (0,1),
   then discuss the monotonicity of g(x).


I know this looks like a bit twisted question, but it can be so simple if solved step by step if you know what you really want to reach at !


Lets solve it


g(x) = f(x) + f(1-x)
g '(x) = f '(x) + f '(1-x)


Now as given f ''(x)<0, this implies f '(x) is a decreasing function


If g '(x) > 0
=>  f '(x) - f '(1-x)>0
      f '(x) > f '(1-x)
      x < 1-x
      x < 1/2


Similarly if you go for   g '(x) < 0 , you will get x >1/2


Hence, g(x) is increasing for x < 1/2 & decreasing for x > 1/2.


Right !!!! See How easy it was !!!






XXXX Wrong  XXXXX
Though we solved the question but we presented wrong answer and I hope many of you might have reached the same result as above if at all.


The correct solution for the above problem should be
g(x) is increasing for x € (0, 1/2 ) & decreasing for x € (1/2 , 1).
This is surely different from the earlier one which has no limiting boundaries.





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Your suggestions and querries are welcome.
Mail to mathsprobe@gmail.com
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