 ## Book: RS Aggarwal - Mathematics

### Chapter: 3. Functions

#### Subject: Maths - Class 11th

##### Q. No. 15 of Exercise 3A

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15
##### Let R+ be the set of all positive real numbers. Let f : R+→ R : f(x) = logex. Find(i) range (f)(ii) {x : x ϵ R+ and f(x) = -2}.(iii) Find out whether f(x + y) = f(x). f(y) for all x, y ϵ R.

Given that f: R+ R such that f(x) = logex

To find: (i) Range of f

Here, f(x) = logex

We know that the range of a function is the set of images of elements in the domain.

the image set of the domain of f = R

Hence, the range of f is the set of all real numbers.

To find: (ii) {x : x ϵ R+ and f(x) = -2}

We have, f(x) = -2 …(a)

and f(x) = logex …(b)

From eq. (a) and (b), we get

logex = -2

Taking exponential both the sides, we get  x = e-2

{x : x ϵ R+ and f(x) = -2} = {e-2}

To find: (iii) f(xy) = f(x) + f(y) for all x, y ϵ R

We have,

f(xy) = loge(xy)

= loge(x) + loge(y)

[Product Rule for Logarithms]

= f(x) + f(y) [f(x) = logex]

f(xy) = f(x) + f(y) holds.

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