Dear student Given e-xyf(xy)=e-xf(x)+e-yf(y) (1)Putting x=y=1 in (1), we get f(1)=0Now, f'(x)=limh→0f(x+h)-f(x)h=limh→0fx1+hx-f(x 1)h=limh→0ex+he-xf(x)+e-1-hxf1+hx-2xe-xf(x)+e-1f(1)h=limh→0ehf(x)+ex+h-1-hxf1+hx-f(x)-ex-1f(1)h=f(x)limh→0eh-1h+ex-1limh→0eh-hxf1+hxx×hx As f(1)=0=f(x) 1+ex-1f'(1)x=f(x)+ex-1 ex As f'(1)=eSo, f'(x)=f(x)+exx⇒e-xf'(x)=e-xf(x)=1x⇒ddxe-xf(x)=1xOn integrating, we havee-xf(x)=logx+C at x=1C=0SO, f(x)=exlogx Regards

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Solve this: 48 [f V x, y e R', and deterrnine f(x).

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Given e^{- xy} f (xy) = e^{- x} f (x) + e^{- y} f (y) . . . . (1) Putting x = y = 1 in (1), we get f (1) = 0 Now, f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f (x (1 + \frac{h}{x})) - f (x . 1)}{h} = \lim_{h \to 0} \frac{e^{x + h} \{e^{- x} f (x) + e^{- 1 - \frac{h}{x}} f (1 + \frac{h}{x})\} - 2^{x} (e^{- x} f (x) + e^{- 1} f (1))}{h} = \lim_{h \to 0} \frac{e^{h} f (x) + e^{x + h - 1 - \frac{h}{x}} f (1 + \frac{h}{x}) - f (x) - e^{x - 1} f (1)}{h} = f (x) \lim_{h \to 0} (\frac{e^{h} - 1}{h}) + e^{x - 1} \lim_{h \to 0} \frac{e^{h - \frac{h}{x}} f (1 + \frac{h}{x})}{x \times \frac{h}{x}} (As f (1) = 0) = f (x) . 1 + e^{x - 1} \frac{f' (1)}{x} = f (x) + \frac{e^{x - 1} . e}{x} (As f' (1) = e) So, f' (x) = f (x) + \frac{e^{x}}{x} \Rightarrow e^{- x} f' (x) = e^{- x} f (x) = \frac{1}{x} \Rightarrow \frac{d}{dx} (e^{- x} f (x)) = \frac{1}{x} On integrating, we have e^{- x} f (x) = logx + C at x = 1 C = 0 SO, f (x) = e^{x} logx

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Solve this: Solve this: 48 [f V x, y e R', and deterrnine f(x).

Solve this:

Solve this: 48 [f V x, y e R', and deterrnine f(x).