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PROBLEM SOLVING TRICK FOR ISI ENTRANCE

MATHS-TRICK-3

This is how to find value of functional equation with this kind of form:-
Ex.1. Let f(x)= 9^x /(9^x + 3) .Show that f(x)+f(1-x)=1. Hence evaluate f(1/1996)+f(2/1996)+f(3/1996)+ . . . +f(1995/1996).

Sol:- f(1-x)=3/(3+9^x) .
So,we have f(x)+f(1-x)=1 -------------------(1)
Putting x=1/1996,2/1996,3/1996, . . . ,998/1996 in (1),we get
f(1/1996)+f(1995/1996)=1
f(2/1996)+f(1994/1996)=1
.
.
.
and lastly f(997/1996)+f(999/1996)=1
So,(1/1996)+f(2/1996)+f(3/1996)+ . . . +f(1995/1996)
=(1+1+1+ . . . . upto 997 terms)+f(998/1996)
=997+1/2=997.5, since f(1/2)=1/2 from(1).

Ex.2. [ISI SAMPLE PAPER]
Let  f(x)= e^(2x-1) /{1+e^(2x-1)}.
Then f(1/1234)+f(3/1234)+. . . . . . +f(1231/1234)+f(1233/1234) =?

Sol:- see f(1/2)=1/2 .
Now also see that f(x)+f(1-x)=1 ------------(1)
Putting x=1/1234,3/1234, . . . ,615/1234 in (1).
So, f(1/1234)+f(1233/1234)=1
       f(3/1234)+f(1231/1234)=1
                .
                 .
 upto 308 terms
              f(615/1234)+f(619/1234)=1.           
we have f(617/1234)=f(1/2)=1/2.
now,  f(1/1234)+f(3/1234)+. . . . . . +f(1231/1234)+f(1233/1234)
=(1+1+1+ . . . +upto 308 terms)+1/2
=308+1/2=308.5