Given f: R^{+}→ R and f(x) = log_{e }x.

**(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

log_{e }x = –2

∴ x = e^{-2} [since, log_{b }a = c ⇒ a = b^{c}]

∴ {x: f(x) = –2} = {e^{–2}}

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

We have f (x) = log_{e }x ⇒ f (y) = log_{e }y

Now, let us consider f (xy)

F (xy) = log_{e }(xy)

f (xy) = log_{e }(x × y) [since, log_{b }(a×c) = log_{b }a + log_{b }c]

f (xy) = log_{e }x + log_{e }y

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

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

Answered by Aaryan | 1 year agoLet 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.

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Let R = {(x, x^{2}) : x is a prime number less than 10}.

**(i) **Write R in roster form.

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

If A = {5} and B = {5, 6}, write down all possible subsets of A × B.