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

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=~{{\mathbf{x}}^{\mathbf{3}}}~-~\mathbf{x}\]

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

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=~{{\mathbf{x}}^{\mathbf{3}}}~-~\mathbf{x}\]

Given f: R → R, defined by 

    \[f\left( x \right)\text{ }=~{{x}^{3}}~-~x\]

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 (R), such that f(x) = f(y).

f(x) = f(y)

    \[{{x}^{3~}}\text{ }x\text{ }=\text{ }{{y}^{3~}}-\text{ }y\]

Here, we cannot say x = y.

For example, 

    \[x=1\]

and 

    \[y=-1\]

    \[{{x}^{3~}}-\text{ }x~=\text{ }1\text{ }-\text{ }1\text{ }=~0\]

    \[{{y}^{3~}}\text{ }y\text{ }=\text{ }{{\left( -1 \right)}^{3}}-\text{ }\left( -1 \right)\text{ }\text{ }1\text{ }+\text{ }1\text{ }=\text{ }0\]

So, 

    \[1\]

 and

    \[-1\]

 have the same image 

    \[0\]

.

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

    \[{{x}^{3}}~-~x~=~y\]

By observation we can say that there exist some x in R, such that 

    \[{{x}^{3}}~-~x~=~y\]

.

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

i

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