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Question Number 28124 by ajfour last updated on 20/Jan/18

f (R^{+} \to R) i s a d i f f e r e n t i a b l e

f u n c t i o n o b e y i n g

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

f o r a l l x, y \in R^{+} a n d

f (1) = 0, f^{'} (1) = 1 .

F i n d f (x) . M o r e q u e s t i o n s m a y

f o l l o w . .

Commented by prakash jain last updated on 20/Jan/18

y = x

f (x^{2}) = 2 f (x)

f (x) = c \ln x

c h e c k

2 c \ln x = c \ln x + c \ln y + c \ln x - c \ln y

f^{'} (x) = \frac{c}{x}

f^{'} (1) = 1 \Rightarrow c = 1

f (x) = \ln x

Commented by ajfour last updated on 20/Jan/18

A r e t h e f o l l o w i n g t r u e :

(i) N o . o f s o l u t i o n s o f f^{- 1} (x) = x^{5}

i s 3

(i i) t h e r e i s a t l e a s t o n e p o i n t α \in

(0, \infty) f o r w h i c h f^{- 1} (x) - x^{5} > 0

f o r x \in (α, \infty) ?

Commented by ajfour last updated on 20/Jan/18

t h a n k y o u S i r .

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