Classify the function as injection, surjection or bijectionf: f: R → R, defined by f(x) = $$\dfrac{x}{(x^2 + 1)}$$

Asked by Aaryan | 1 year ago |  65

##### Solution :-

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

xy− 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 x– 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 ago

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