Date : 11 - 5 - 2008

Day : Sunday

QUESTION OF THE DAY

The function  f  is defined such that  f ( x + yⁿ )   =   f ( x )  +  [ f ( y ) ]ⁿ  where n is a natural number and n > 1. Also  f '( 0 ) > 0.

Then the value of   f ( 5 )  +  f '( 5 )  is

( A )  1

( B )  5

( C )  2

( D )  6

put n=2 ...make it simpler ..

if not ...then i m trying.

wat about dis one ..

u got d ans ..

r u satisfied ..

atlest tell i\me

I KNOW THE ANSWER. I AM JUST TRYING TO PUT DOWN A MAXIMUM NUMBER OF GOOD QUESTIONS TO HELP YOU ALL TO SHARPEN YOUR BRAIN.



NOW TRY TO GET THE CORRECT ANSWER TO THIS QUESTION.

putting x and y both as 0 i get f(0) as 0........{f(0)=2f(0)}

so i have putting 0 in the origional eqn. that

f(yⁿ) =[f(y)]ⁿ ........

now this can happen in a special case of f(y)=y.............or say f(x)=x........

thus i have my f(5)=5............----------------1)

and as f(x)=x .........f'(x)=1.........--------------2)

so we have f(5) + f'(5) = 6 from 1) and 2)..............d)

this might not be the traditional mtd........this is the mtd i will use..............

Date : 12 - 5 - 2008

Day : Monday

QUESTION OF THE DAY

Let  f : N * N N  be a function such that 

• f ( 1 , 1 )   =   2
  
  


• f ( m + 1 , n )   =   f ( m , n )  +  m
  
  


• f ( m , n + 1 )   =   f ( m , n )  -  n
  
  

for all m, n    N. If  f ( p , q )  =  2001, then which of the following is(are) true?

( A )  ( p , q )  =  ( 2000 , 1999 )

( B )  ( p , q )  =  ( 1001 , 999 )

( C )  ( p , q )  =  ( 1001 , 1999 )

( D )  ( p , q )  =  ( 2000 , 999 )

-------------------------------soory wrong answer-----------------------------------------------------------

 

I think ans are A and B.The important thing to note is f(k,k)=f(k,k-1+1)=f(k,k-1)-(k-1)=f(k-1+1,k-1)+1-k=f(k-1,k-1).

So,f(2000,1999)=f(1999,1999)+1999=f(1998,1998)+1999=.......=f(1,1)+1999=2001

f(1001,999)=f(1000,999)+1000=f(999,999)+1999=f(1,1)+1999=2001

reply to the first qn .

Take y to be constant :

differentiating both sides wrt x , we get f'( x + y^n ) = f'(x ) for all y

which means f'( x ) = constant = c ( let )

so f( x ) = cx + d  ( d = constant )

Now putting x = y = 0 in the given reln we get that f(0) = 0

giving , d= 0

so f( x ) = cx

Now put x = 0  in the given relation , so we get

f( y^n ) = [f(y )]^n

so cy^n = c^n y^n for all y

giving , c= 1

so f( x ) = x  ; thus f(5 ) + f'(5 ) = 5 + 1 = 6