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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