The document discusses finding the local maxima of a function g(x), given that g(x) is defined in terms of another function f(x). It is shown that: 1) The points where g(x) has local maxima are the same as the points where f'(x) is equal to 0, as g'(x) only changes sign at these points. 2) f(x) is given as a 4th degree polynomial, so it has 4 real roots where the derivative is 0. 3) By analyzing the signs of f'(x) at these roots, it is determined that g(x) has only one local maximum, at x = 2009.

1. Physics Helpline
L K Satapathy
Maxima and Minima 3

2. Physics Helpline
L K Satapathy
Application of Derivative 7
Question : Let f be a function defined on R ( the set of all real numbers )
such that for all x  R .
If g is a function defined on R with values in the interval ( 0 ,  ) , such that
for all x  R , then the number of points in R at which g has a
local maximum is
2 3 4
( ) 2010( 2009)( 2010) ( 2011) ( 2012)f x x x x x     
( ) ln[ ( )]f x g x
( ) 1 ( ) 2 ( ) 3 ( ) 4a b c d
Answer :
( )
( ) ln[ ( )] ( ) f x
f x g x g x e  The given function :
We are required to find the points where g(x) has local maximum
For these points , g (x) = 0 and changes sign from + ve to –ve across the points
Given that g(x)  (0,) ( )
( ) 0 0f x
g x e   

3. Physics Helpline
L K Satapathy
Application of Derivative 7
( )
( ) . ( )f x
g x e f x Differentiating we get
 g(x) is +ve , -ve or zero whenever f  (x) is +ve , -ve or zero
 Points of maximum for g(x) are the same as those for f (x)
( )
, ( ) 0 . ( ) 0f x
Now g x e f x   
( )
0 ( ) 0 ( ) 0f x
As e g x f x     
Given that 2 3 4
( ) 2010 ( 2009) ( 2010) ( 2011) ( 2012)f x x x x x     
( ) 0 2009 , 2010 , 2011 2012f x x or   
 We test these points for maximum / minimum

4. Physics Helpline
L K Satapathy
Application of Derivative 7
2 3 4
( ) 2010 ( 2009) ( 2010) ( 2011) ( 2012)f x x x x x     
Sign of derivative
Correct option = (a)
 g(x) is maximum only at x = 2009 [ f (x) changes from +ve to – ve ]
+   + +
2 3 4
2009 ( ) ( )( ) ( ) ( )x f x ve       
2 3 4
2009 2010 ( ) ( )( ) ( ) ( )x f x ve        
2 3 4
2010 2011 ( ) ( )( ) ( ) ( )x f x ve        
2 3 4
2011 2012 ( ) ( )( ) ( ) ( )x f x ve        
2 3 4
2012 ( ) ( )( ) ( ) ( )x f x ve       
(Max) (None) (Min) (None)
 Only 1 point where g(x) is maximum
4 factors on right
3 factors on right
2 factors on right
1 factor on right
No factor on right
Arranging factors in the increasing order , factors on right are –ve

5. Physics Helpline
L K Satapathy
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