![\[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{x}/({{\mathbf{x}}^{\mathbf{2}~}}+\text{ }\mathbf{1})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc12c6c1e62db31a8f0a4982bc01aedb_l3.png "Rendered by QuickLaTeX.com")
) Given f: R → R, defined by
![\[f\left( x \right)\text{ }=\text{ }x/({{x}^{2~}}+\text{ }1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-46fb8400c883bc2e672646431eba74d5_l3.png "Rendered by QuickLaTeX.com")
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)
![\[x\text{ }/({{x}^{2~}}+\text{ }1)\text{ }=\text{ }y\text{ }/({{y}^{2}}~+\text{ }1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-56b35dcf23735a4d1179ac74ac829b11_l3.png "Rendered by QuickLaTeX.com")
![\[x\text{ }{{y}^{2}}+~x~=~{{x}^{2}}y~+~y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-25a8e88170d1d659faab65e5f8989b7e_l3.png "Rendered by QuickLaTeX.com")
![\[x{{y}^{2~}}-\text{ }{{x}^{2}}y~+~x~-\text{ }y~=~0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-255691b18f80d42d4f6ea579dc0fa3e9_l3.png "Rendered by QuickLaTeX.com")
![\[-x\text{ }y~\left( -y\text{ }+\text{ }x \right)\text{ }+~1~\left( x\text{ }-\text{ }y \right)~=~0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-768970ae09215c97621b79e89b2514ea_l3.png "Rendered by QuickLaTeX.com")
![\[\left( x\text{ }-\text{ }y \right)~\left( 1\text{ }\text{ }x\text{ }y \right)~=~0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2c4e1ee3c705205215de439fb54aed1f_l3.png "Rendered by QuickLaTeX.com")
![\[x~=~y~or~x~=\text{ }1/y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cdb671543154db4b6cc350a814eb0215_l3.png "Rendered by QuickLaTeX.com")
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
![\[x\text{ }/({{x}^{2}}~+\text{ }1)\text{ }=\text{ }y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6189cb532b0d24135c075d04777238a6_l3.png "Rendered by QuickLaTeX.com")
![\[y\text{ }{{x}^{2~}}\text{ }x\text{ }+\text{ }y\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf9c2159be3efe3e8c33067deedf21a1_l3.png "Rendered by QuickLaTeX.com")
![\[x\text{ }=\text{ }(-\left( -1 \right)\text{ }\pm \text{ }\sqrt{(1-4{{y}^{2}})})/\left( 2y \right)~if~y~\ne ~0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a7e4fd494939f55f29696299baacb0cf_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }(1\text{ }\pm \text{ }\sqrt{(1-4{{y}^{2}})})/\text{ }\left( 2y \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eda492c5cc0ec591bb003b7a8a35130f_l3.png "Rendered by QuickLaTeX.com")
, which may not be in R
For example, if
![\[y=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-430688ed68b48bacead95da52f3ffd4a_l3.png "Rendered by QuickLaTeX.com")
, then
![\[\left( 1\text{ }\pm \text{ }\sqrt{\left( 1-4 \right)} \right)\text{ }/\text{ }\left( 2y \right)\text{ }=\text{ }\left( 1\text{ }\pm \text{ }i\text{ }\sqrt{3} \right)/2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cd466aea4d68710950ee93138def509_l3.png "Rendered by QuickLaTeX.com")
, which is not in R
So, f is not surjection and f is not bijection.