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Question Number 33649 by rahul 19 last updated on 21/Apr/18

C o n s i d e r f : R^{+} \to R s u c h t h a t

f (3) = 1 f o r a \in R^{+} a n d

f (x) . f (y) + f (\frac{3}{x}) . f (\frac{3}{y}) = 2 f (x y)

\forall x, y \in R^{+} . T h e n f i n d f (x) ?

Commented by rahul 19 last updated on 21/Apr/18

p r o f A b d o i m a d, M J S s i r p l s h e l p .

Commented by math khazana by abdo last updated on 21/Apr/18

\Rightarrow {(f (1))}^{2} + {(f (3))}^{2} = 2 f (1) \Rightarrow {(f (1))}^{2} - 2 f (1) + 1 = 0

\Rightarrow {(f (1) - 1)}^{2} = 0 \Rightarrow f (1) = 1 f o r y = \frac{1}{x} \Rightarrow

f (x) f (\frac{1}{x}) + f (\frac{3}{x}) . f (3 x) = 2 l e t x \to + \infty

f (+ \infty) f (0) + f (+ \infty) f (0) = 2 \Rightarrow f (0) f (+ \infty) = 1

t h i s c o n d i t i o n p r o v e t h a t f (x) = c (c o n s t a n t)

\forall x \in R^{+} f (x) f (\frac{1}{x}) + f (\frac{3}{x}) f (3 x) = 2 \Rightarrow

c^{2} + c^{2} = 2 \Rightarrow 2 c^{2} = 2 \Rightarrow c = \bar{+} 1 b u t f (1) = f (3) = 1

\Rightarrow f (x) = 1 \forall x \in R^{+}

Commented by rahul 19 last updated on 22/Apr/18

t h a n k u s i r!

Answered by MJS last updated on 21/Apr/18

x = 1; y = 1

f (1) f (1) + f (3) f (3) = 2 f (1)

f {(1)}^{2} - 2 f (1) + 1 = 0

{(f (1) - 1)}^{2} = 0

f (1) = 1

x = 3; y = 3

f (3) f (3) + f (1) f (1) = 2 f (9)

1 + 1 = 2 f (9)

f (9) = 1

so at least one solution is

f (x) = 1

we could also start with

x = 1; y = 3

f (1) f (3) + f (\frac{3}{1}) f (\frac{3}{3}) = 2 f (1 \times 3)

f (1) + f (1) = 2 \Rightarrow f (1) = 1

Commented by rahul 19 last updated on 21/Apr/18

s i r, a c t u a l l y i a l s o g o t o n e o f t h e s o l .

a s f (x) = 1 .

B u t i a m f i n d i n g h a r d t o p r o v e t h a t

f o r a n y v a l u e o f x, y \in R^{+} .

f (x) = 1 o n l y!

i m e a n o n l y o n e s o l u t i o n .

Commented by MJS last updated on 21/Apr/18

if the function is y = 1 then the given equation

is true for all x \in R because

f (x) = f (y) = f (\frac{3}{x}) = f (\frac{3}{y}) = f (x y) = 1 and

1 \times 1 + 1 \times 1 = 2 \times 1

if x \neq y was demanded {we}^{'} d need a different

function, but I could not find one

Commented by rahul 19 last updated on 22/Apr/18

t h a n k u s i r .