Functional Equations

Note: While solving fuctional equations, Remember the following:

f(xy) = f(x).f(y) ⇒$\Rightarrow$ f(x) = xn$x^n$
f(x+y) = f(x) . f(y) ⇒$\Rightarrow$ f(x) = aKx$a^{Kx}$
f(xy) = f(x) + f(y) ⇒$\Rightarrow$ f(x) = K log x or f(x) = 0
f(x+y) = f(x) = f(y) ⇒$\Rightarrow$ f(x) = k
f(x).f(1x)$f\left({\displaystyle \frac{1}{x}}\right)$ = f(x) + f(1x)$f\left({\displaystyle \frac{1}{x}}\right)$ ⇒$\Rightarrow$ f(x) = ±xn+1$\pm x^n+1$

1.  Given f(x) is a function satisfying f(x+y) = f(x).f(y) for all real values of x and y.  If f(1) = 3 and f(1) + f(2) + f(3) + ...f(n) = 1092. Then find the vallue of n.
a.  5             b. 6          c. 7            d. 18
Sol: f(1) = 3
f(2) = f(1 + 1) = f(1) x f(1) = 32$3^2$
f(3) = f(1 + 2) = f(1) x f(2) = 33$3^3$
...
...
f(1) + f(2) +....f(n) = 1092
3+32+33....3n$3+3^2+3^3...{.3}^n$ =1092
ie., 3.3n−13−1=1092$3.{\displaystyle \frac{3^n-1}{3-1}=1092}$
Solving n = 6

2. Given g(x) is a function such that g(x+1) + g(x-1) = g(x).  For what minimum value of P does the relation g(x+p) = -g(x) necessarily hold true? 
a.  2         b. 3                c.  5           d. 6
Sol: given g(x+1) + g(x-1) = g(x) ......(1)
g(x + 2) + g(x) = g(x+1) .........(2)
adding
g(x+2) + g(x-1) = 0
g(x+3) + g(x) = 0
So g(x+3) = - g(x)
P =3

3. The odd function f(x) has period 6.  If f(5) = 4 then what is the value of f(1)-f(3)
a. -6   b. -4      c. 0   d. 4
Sol: f(-x) = -f(x)
f(x+6) = f(x)
f(-5) = -f(5) = -4
f(-5) = f(-5+6) = f(1) = -4
f(-3) = f(-3 +6) = f(3)
Also f(-3) = -f(3) So f(3) = 0

4.F(x) = 4x4x+2${\displaystyle {\displaystyle \frac{4^x}{4^x+2}}}$ Then F(11997)+F(21997)+......F(19961997)$F\left({\displaystyle {\displaystyle \frac{1}{1997}}}\right)+F\left({\displaystyle \frac{2}{1997}}\right)+......F\left({\displaystyle \frac{1996}{1997}}\right)$ is equal to
F(x)=4x4x+2$F(x)={\displaystyle \frac{4^x}{4^x+2}}$
F(x) + F(1-x) = 4x4x+2+41−x41−x+2=4x4x+2+22+4x=1${\displaystyle \frac{4^x}{4^x+2}+{\displaystyle \frac{4^{1-x}}{4^{1-x}+2}={\displaystyle \frac{4^x}{4^x+2}+{\displaystyle \frac{2}{2+4^x}=1}}}}$
⇒F(11997)+F(21997)+......F(19961997)$\Rightarrow F\left({\displaystyle \frac{1}{1997}}\right)+F\left({\displaystyle \frac{2}{1997}}\right)+......F\left({\displaystyle \frac{1996}{1997}}\right)$ = F(11997)+F(19961997)+F(21997)+F(19951997)+......=19962$F\left({\displaystyle \frac{1}{1997}}\right)+F\left({\displaystyle \frac{1996}{1997}}\right)+F\left({\displaystyle \frac{2}{1997}}\right)+F\left({\displaystyle \frac{1995}{1997}}\right)+......={\displaystyle \frac{1996}{2}}$ = 998

5. If f(x/y) = f(x) - f(y) (for y0) then which of these is equal in value to f(150/6)
a.  f(5) b.  2.f(5) c.  f(6) d.  f(20)
Sol: f(x/y) = f(x).f(y)
Put x = y⇒$\Rightarrow$ f(1) = 0
Put x = 1 ⇒$\Rightarrow$ y = 1/x
f(x) = f(1) - f(1/x) = 0 - f(1/x)
f(1/x) = -f(x)
Put x = x and y = 1/x
⇒$\Rightarrow$ f(x2)$f(x^2)$ =f(x) - f(1/x)
⇒$\Rightarrow$ f(x) - (-f(x) = 2f(x)
f(x2)$f(x^2)$ =2.f(x)
⇒$\Rightarrow$ f(25) = 2.f(5)

Alternatively:
When you see f(x/y) = f(x) - f(y) you should recall that it is of the form Log (a/b) = log a - log b
So the given function is a logarithmic function.
So log (150/6) = log (25) = log (52$5^2$) = 2log5