Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian



Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

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