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

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{4}\]

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

    \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{4}\]

Given f: R → R, defined by 

    \[f\left( x \right)\text{ }=\text{ }5{{x}^{3}}~+\text{ }4\]

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)

    \[5{{x}^{3~}}+\text{ }4~=~5{{y}^{3~}}+\text{ }4\]

    \[5{{x}^{3}}=~5{{y}^{3}}\]

    \[{{x}^{3~}}=~{{y}^{3}}\]

x = y

So, f is 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

    \[5{{x}^{3}}+\text{ }4~=~y\]

    \[{{x}^{3}}~=\text{ }\left( y\text{ }\text{ }4 \right)/5\in R\]

So, f is a surjection and f is a bijection.

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