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Classify the following function as injection, surjection or bijection: f: Z → Z, defined by

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=~{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}\]

CBSE | Class 12 | Exercise 2.1 | Function | Maths | RD Sharma
Classify the following function as injection, surjection or bijection: f: Z → Z, defined by

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=~{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}\]

Given f: Z → Z, defined by 

    \[f\left( x \right)\text{ }=~{{x}^{2}}~+~x\]

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 (Z), such that f(x) = f(y).

f(x) = f(y)

    \[{{x}^{2}}+~x~=~{{y}^{2~}}+~y\]

Here, we cannot say that x = y.

For example, 

    \[x~=~2~\]

and 

    \[y~=~~3\]

Then,

    \[{{x}^{2~}}+\text{ }x\text{ }=\text{ }{{2}^{2~}}+\text{ }2\text{ }=~6\]

    \[{{y}^{2~}}+\text{ }y\text{ }=\text{ }{{\left( -3 \right)}^{2~}}\text{ }3\text{ }=~6\]

So, we have two numbers 

    \[2\]

 and 

    \[-3\]

 in the domain Z whose image is same as 6.

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (Z),

such that f(x) = y for some element x in Z (domain).

f(x) = y

    \[{{x}^{2}}~+~x~=~y\]

Here, we cannot say x ∈ Z.

For example, 

    \[y~=\text{ }\text{ }4\]

.

    \[{{x}^{2}}~+~x~=~-~4\]

    \[{{x}^{2~}}+~x~+~4~=~0\]

    \[x~=\text{ }\left( -1\text{ }\pm \text{ }\sqrt{-5} \right)/2\text{ }=\text{ }\left( -1\text{ }\pm \text{ }i\text{ }\sqrt{5} \right)/2~\]

which is not in Z.

So, f is not a surjection and f is not a bijection.

i

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