Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = logx. Determine

(i) the image set of the domain of f

(ii) {x: f (x) = –2}

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

Asked by Sakshi | 1 year ago |  83

1 Answer

Solution :-

Given f: R+→ R and f(x) = logx.

(i) the image set of the domain of f

Domain of f = R+ (set of positive real numbers)

We know the value of logarithm to the base e (natural logarithm) can take all possible real values.

∴ The image set of f = R

(ii) {x: f(x) = –2}

Given f(x) = –2

logx = –2

∴ x = e-2 [since, loga = c ⇒ a = bc]

∴ {x: f(x) = –2} = {e–2}

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

We have f (x) = logx ⇒ f (y) = logy

Now, let us consider f (xy)

F (xy) = log(xy)

f (xy) = log(x × y) [since, log(a×c) = loga + logc]

f (xy) = logx + logy

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

∴ the equation f (xy) = f (x) + f (y) holds.

Answered by Aaryan | 1 year ago

Related Questions

Let R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.

Class 11 Maths Relations and Functions View Answer

Let R = {(x, x2) : x is a prime number less than 10}.

(i) Write R in roster form.

(ii) Find dom (R) and range (R).

Class 11 Maths Relations and Functions View Answer