Given f: R → R, defined by f(x) =\( \dfrac{x}{(x^2 + 1)}\)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
\( \dfrac{x}{(x^2 + 1)}= \dfrac{y}{(y^2 + 1)}\)
x y2+ x = x2y + y
xy2 − x2y + x − y = 0
−x y (−y + x) + 1 (x − y) = 0
(x − y) (1 – x y) = 0
x = y or x = \( \dfrac{1}{y}\)
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
\( \dfrac{x}{(x^2 + 1)}\)= y
y x2 – x + y = 0
x = \( (-1) ±\dfrac{(\sqrt{1-4y^2)}}{2y}\;if \;y ≠ 0\)
= \(1±\dfrac{(\sqrt{1-4y^2)}}{2y}\) which may not be in R
For example, if y=1, then
\( 1±\dfrac{(\sqrt{1-4y^2)}}{2y}\)
= \( \dfrac{ (1 ± i \sqrt{3})}{2}\) which is not in R
So, f is not surjection and f is not 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
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