Given A = {1, 2, 3}

Number of elements in A = 3

Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y

Express R and R^{-1} as sets of ordered pairs. Determine also

**(i) **the domain of R^{‑1}

**(ii)** The Range of R.

A function f: R → R is defined as f(x) = x^{3} + 4. Is it a bijection or not? In case it is a bijection, find f^{−1} (3).

If f: R → R be defined by f(x) = x^{3} −3, then prove that f^{−1} exists and find a formula for f^{−1}. Hence, find f^{−1 }(24) and f^{−1} (5).

Consider f: R^{+} → [−5, ∞) given by f(x) = 9x^{2} + 6x − 5. Show that f is invertible with f^{-1}(x) = \(
\dfrac{\sqrt{(x+6)-1}}{3}\)

If f(x) = \(\dfrac{ (4x + 3)}{(6x – 4)}\), x ≠ (\( \dfrac{2}{3}\)) show that fof(x) = x, for all x ≠ (\( \dfrac{2}{3}\)). What is the inverse of f?