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Problem Solving Strategy-ON Functional Equation

Math Trick-7
Functional Equations is a chapter which is essential for Olympiad,ISI,CMI,IOMA entrances.
An equation involving an unknown function is called a functional equation.

Ex. Considering Cauchy's functional equation:-
f(x+y)=f(x) + f(y) ;

Generally while you are solving a problem of functional equation with two variables x & y then in most of the situations to solve a given problem you are to put these values
Put x= -x
Put y= -x
Put x=f(y)
Put x=y
Put x=f(x)
etc . . .
The choices differ & it completely depends on the given problem which requires these selections. So, just use your brain & start solving problem. I am giving you some examples.

Ex.1. [ISI B.Math sample paper] If f(x+y)=f(x)+f(y) for all y in R. Then f(x) is a function which is
A. Odd   B.Even   C. None
Sol:-(A)  Putting x= -x ,then
 f(x)+f(-x)=f(0) -----(1)
Again put x=0 in (1), so f(0)=0.
So, from (1), f(-x)= -f(x) which is an odd function.

Ex.2. [RMO paper] If the function f satisfies the relation f(x+y)+f(x-y)=2f(x).f(y) for all x,y in R.
And f(0)≠0. Prove that f(x) is even function.
Sol:- f(x+y)+f(x-y)=2f(x).f(y) ---(1)
Replace x by y & y by x
Then f(y+x)+f(y-x)=2f(y)f(x) ---(2)
Solving (1) & (2), we get
   f(y-x)=f(x-y)
Put y=2x then f(x)=f(-x).
Hence f(x) is an even function.

Ex.3.[RMO paper] Determine all functions f:R->R such that
 f(x-f(y))=f(f(y))+xf(y)+f(x)-1,  x,y€R.
Sol:- Put x=f(y)=0
Then f(0)=1
Again put x=f(y)=k,then
   f(0)=2f(k)-1+(k^2)
=> f(k)= 1- (k^2)/2 .
Hence, f(x)=1- (x^2)/2 is the unique equation.

Ex.4.[ISI MMA SAMPLE PAPER] If f(x) is a real valued fiction such that
                  2f(x) + 3f(-x) = 15 - 4x , for all x.
Then f(2) is
A.-15     B.22    C.11     D.0
Sol:- (C) Given 2f(x) + 3f(-x) = 15 - 4x -----(1)
 put  x = -x in the given functional equation (1),
we have 2f(-x) + 3f(x) = 15 + 4x ------------(2)
Solving (1) & (2) we have f(x) = 3 + 4x .
f(2)=11.