Let-f-R-R-be-a-function-satisfying-the-functional-relation-f-x-y-f-y-x-2f-xy-for-all-x-y-R-and-it-is-given-that-f-1-1-2-Answer-the-following-questions-i-f-

Question Number 120325 by Ar Brandon last updated on 30/Oct/20

Let f : R \to R be a function satisfying the

functional relation

{(f (x))}^{y} + {(f (y))}^{x} = 2 f (xy)

for all x, y \in R and it is given that f (1) = 1 / 2 . Answer

the following questions .

(i) f (x + y) =

(A) f (x) + f (y) (B) f (x) f (y)

(C) f (x^{y} y^{x}) (D) \frac{f (x)}{f (y)}

(ii) f (xy) =

(A) f (x) f (y) (B) f (x) + f (y)

(C) {(f (x))}^{y} (D) {(f (xy))}^{xy}

(iii) \underset{k = 0}{\sum^{\infty}} f (k) =

(A) 5 / 2 (B) 3 / 2 (C) 3 (D) 2

Commented by Dwaipayan Shikari last updated on 30/Oct/20

f (x) = {(C e)}^{x}

f (x + y) = {(C e)}^{x + y} = C^{x} e^{x} . C^{y} e^{y} = f (x) f (y)

f (x y) = C^{x y} e^{x y} = {(f (x))}^{y}

f (1) = C e = \frac{1}{2} \Rightarrow C = \frac{1}{2 e}

\underset{k = 0}{\sum^{\infty}} f (k) = 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots = \frac{1}{1 - \frac{1}{2}} = 2

Commented by Ar Brandon last updated on 30/Oct/20

Wow! Thanks bro . Are there any

calculations to get f (x) = {(Ce)}^{x} ?

Commented by Dwaipayan Shikari last updated on 30/Oct/20

I j u s t a s s u m e d i t

Commented by mindispower last updated on 30/Oct/20

f {(x)}^{y} + f {(y)}^{x} = 2 f (x y), x = 1

\Rightarrow f {(1)}^{y} + f (y) = 2 f (y)

f (y) = f {(1)}^{y} = \frac{1}{2^{y}}

Commented by Dwaipayan Shikari last updated on 30/Oct/20

f {(1)}^{y} + f {(y)}^{1} = 2 (f (y))

{(\frac{1}{2})}^{y} = f (y) S o {(\frac{1}{2})}^{x y} = f (x y) = {({(\frac{1}{2})}^{x})}^{y} = f {(x)}^{y}

{(\frac{1}{2})}^{x} = f (x) S o \underset{k = 0}{\sum^{\infty}} f (k) = {(\frac{1}{2})}^{0} + {(\frac{1}{2})}^{1} + . . = 2

{(\frac{1}{2})}^{x + y} = f (x + y)

f (x) f (y) = f (x + y)

Commented by Ar Brandon last updated on 30/Oct/20

Thanks��