Math, asked by fussygamer92, 7 months ago

15. Let R^+ be the set of all positive real numbers.
Let f:R^+→ R: f(x) = logex.
Find (i) range (f) (ii) {x : XE R* and f(x) = -2).
(iii) Find out whether f(xy) = f(x) + f(y) for all x, y eR*.​

Answers

Answered by saounksh
2

ᴀɴsᴡᴇʀ

1)

Here,

 \:\:\:\:\:\:\: f(x) = log_e(x) = ln(x)

or  \: e^{f(x)} = x ... (1)

 \to f'(x) = \frac{1}{x}

 \to f'(x) > 0

f is a strictly increasing function. So its maxima and minima will occurs at end points of its domain  (0, \infty).

Now,

 \:\:\:\:\:\: \lim \limits_{x \to 0^+}e^{f(x)} = \lim \limits_{x \to 0^+} x

 \to \lim \limits_{x \to 0^+}e^{f(x)} = 0^+

This is possible only if

 \to \lim \limits_{x \to 0^+} f(x) = - \infty

Also,

 \to \lim \limits_{x \to \infty} f(x) =  \infty

Hence, Range of f(x) is  (-\infty, \infty) .

2)

Here,  f(x) = - 2

\implies ln(x) = - 2

\implies x= e^{-2}

{  x: x \in \R, f(x) =-2} = {  e^{-2} }

3)

Now,

 \:\:\:\: f(xy) = ln(xy), f(x) = ln(x)

and  \:\:\:\:\:f(y) = ln(y)

 \to e^{f(xy)} = xy, e^{f(x)} = x, e^{f(y)} = y

 \to e^{f(xy)} = e^{f(x)}e^{f(y)}

 \to e^{f(xy)} = e^{f(x) + f(y)}

Tanking log on both side, we get

 \to f(xy) = f(x) + f(y)

Hence Proved

Similar questions