The given relation f is defined as:

It is seen that, for 0 ≤ x < 3,

f(x) = x^{2 }and for 3 < x ≤ 10,

f(x) = 3x

Also, at x = 3

f(x) = 3^{2} = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9 [Single image]

Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Therefore, the given relation is a function.

Now,

In the given relation g is defined as

It is seen that, for x = 2

g(x) = 2^{2} = 4 and g(x) = 3 × 2 = 6

Thus, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6.

Therefore, this relation is not a function.

Answered by Pragya Singh | 11 months 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).

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