## Pages

### Tricks for solving problems of ISI Entrance exam

MATH-TRICK-6
This is on Functional Equations explained through some solved problems.
Result:- If f is a continuous function that satisfies f(x+y)=f(x)+f(y) for all x,y, then f(x)=xf(1).
Proof:- (Hints)
We know that if f is a function satisfying f(x+y)=f(x)+f(y) for all x,y then f(x)=kx ,where k is a constant.
[To show this, Check the differentiability first.Then
f'(x)=Lt{f(x+h)-f}/h ,where h->0.
f'(x)=Lt {f(h)/h} where h->0
= f'(0)=k(say).
Then integrate both sides, you will get  f(x)=kx+c ------(1)
Put x=y=0 in the given equation,then f(0)=0.
From (1), c=0, i.e. f(x)=kx. ]
Here f(1)=k ,so f(x)=kx=xf(1).

Problem:- [ISI Sample paper] Let g be a continuous function with g(1)=1 such that
g(x+y)=5g(x)g(y)
for all x,y. Find g(x).
Sol:- g(x+y)=5g(x)g(y)
5g(x+y)=5g(x)5g(y), taking log bothsides
=> log[5g(x+y)]=log[5g(x)] + log[5g(y)]
=> f(x+y)=f(x)+f(y), taking f(x+y)=log[5g(x+y)]
=> f(x)= xf(1) [By the above result]
So, log[5g(x)]=xlog[5g(1)]
=> 5g(x) = 5^x ,since g(1)=1 (given)
=> g(x) = 5^{x-1} .