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Home » 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.
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.
CBSE | Class 11 | Excercise 3A | Functions | Maths | RS Aggarwal
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.
  • Answer : 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.

i

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