FUNCTIONS

EVEN AND ODD FUNCTIONS


If f(x)=f(-x), then f(x) is even function
and if f(x) =-f(-x) then f(x) is odd function
For example Sin(x) is odd function while cos(x) is even function. This means that sin(x)=-sin(-x) and cos(x)=cos(-x)


With this basic understanding, let us see this question.


Problem : If function f satisfies the relation f(x+1)+f(1-x)=2f(x) for all real values of x and f(0)< or >0, then prove that f(x) is even.
Solution:
f(x+1)+f(1-x)=2f(x)
Replacing x with -x
f(1-x)+f(1+x)=2f(-x)
Comparing both we get, 2 f(x) = 2 f(-x)
Hence f(x) = f(-x)
Hence proved that, f(x) is an even function.


COMPOSITE FUNCTIONS






Let f : A à B, and  g : B à C, be two functions, then gof : A à C
This is known as the product function or composite of f and g, given by gof(x)=g{f(x)} for all real values of x.
The symbols  f : A à B is read as Function F maps A to B.


It is  to be noted that fog = f{g (x)} and (f ± g) x = f(x) ± g(x) and (f/g)x=f(x)/g(x)

Example:

If F : RàR and G : RàR, be two mappings such that f(x) = sin x and g(x)=x²

Then show that fog ≠ gof

Solution : Let x € R,

So (fog) x = f {g(x)} = sin(x²)         -----(1)

While (gof) x = g {f(x)} = g {sinx} = (sinx)² = sin²x    -----(2)

From (1) & (2), (fog) x ≠ (gof) x

PERIODIC FUNCTIONS
   


A function f(x) is said to be periodic function of x, if there exists a positive real number T such that f(x+T) = f(x). Then the smallest value of T is called the Period of the function.
For example Sin x is periodic because

Sin(2π+x)= Sin(x) and also sin(4π+x)=Sin(x) = Sin(6π+x) and so on…. But 2π being the smallest value is the period of the function.

Example: Show tha cos(√x) is non-periodic.

Solution: Let Cos(√x) be periodic with T as the period.

So, Cos(√x) = Cos{ √ (T+x) }

à √ (T+x)  = 2nπ ± √x

Putting x = 0, we get,  √T= 2nπ    -----(1)

Putting x=T, we get  √2T = 2nπ ± √T     ----(2)

From (1) and (2) we get, √2T = √T ± √T

Or √T x √2 = √T (1 ± 1)

Or √2 = 1 ± 1, which is not possible, Hence Cos√x is not periodic function.

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

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.

__________________________________

For support and queries 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.

********************************************

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

********************************************

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.





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