If the functional relationship f(xy) = 
+ 
holds for all real x and y greater than 0 and f(x) is a differentiable function for all x > 0 such that f(e) = 
, then find the maximum value of f(x).
Text Solution
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Sol. We have ,
f(xy) = 
+ 
for all x, y > 0
⇒ f(1) = f(1) + f(1) [Putting x = y = 1]
⇒ f(1) = 0
Now, f(x) is differentiable for all x > 0. Therefore,
f ′ (x) = 
= 
= 
= 
= – 
+ 
= – 
+ 
f ′ (x) = 
+ 
, where A = 
⇒ 
f(x) = – 
+ 
⇒ 
f(x) + 
= 
⇒ x 
f(x) + f(x) = 
⇒ 
[x f (x)] = 
⇒ x f (x) = A log e x + log C [On integration]
putting x = 1, we get
f(1) = A log e 1 + log C
⇒ 0 = log C [ f (1) = 0]
∴ xf (x) = A log e x
Putting x = e, we get
ef(e) = A log e e
⇒ A = 1 
∴ xf (x) = log e x
⇒ f(x) = 
⇒ f ′ (x) = 
For maximum or minimum, we have
f ′ (x) = 0 ⇒ 1 –log e x = 0 ⇒ log e x = 1 ⇒ x = e
Clearly, f ′′ (e) < 0
Hence, f(x) is maximum for x = e. The maximum value of f(x) is given by
f(e) = 
= 
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