Given f: Q − {3} → Q, defined by f (x) = \(\dfrac{ (2x +3)}{(x-3)}\)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).
f(x) = f(y)
\( \dfrac{ (2x +3)}{(x-3)}=\dfrac{ (2y +3)}{(y-3)}\)
(2x + 3) (y − 3) = (2y + 3) (x − 3)
2xy − 6x + 3y − 9 = 2xy − 6y + 3x − 9
9x = 9y
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).
f(x) = y
\( \dfrac{ (2x +3)}{(x-3)}=y\)
2x + 3 = x y − 3y
2x – x y = −3y − 3
x (2−y) = −3 (y + 1)
x = \( \dfrac{-3(y + 1)}{(2 – y)}\) which is not defined at y = 2.
So, f is not a surjection and f is not a bijection
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) = x3 + 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) = x3 −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) = 9x2 + 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?