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Home » Prove that the function f: N → N, defined by f(x) =     , is one-one but not onto
Prove that the function f: N → N, defined by f(x) =

    \[{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}~+\text{ }\mathbf{1}\]

, is one-one but not onto
CBSE | Class 12 | Exercise 2.1 | Function | Maths | RD Sharma
Prove that the function f: N → N, defined by f(x) =

    \[{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}~+\text{ }\mathbf{1}\]

, is one-one but not onto

Given f: N → N, defined by f(x) = 

    \[{{\mathbf{x}}^{\mathbf{2}}}~+~\mathbf{x}~+\text{ }\mathbf{1}\]

Now we have to prove that given function is one-one

Injectivity:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

⇒

    \[{{x}^{2}}~+\text{ }x\text{ }+\text{ }1\text{ }=\text{ }{{y}^{2}}~+\text{ }y\text{ }+\text{ }1\]

⇒

    \[({{x}^{2}}~\text{ }{{y}^{2}})\text{ }+\text{ }\left( x\text{ }\text{ }y \right)\text{ }=\text{ }0\]

`

⇒

    \[\left( x\text{ }+\text{ }y \right)\text{ }\left( x-\text{ }y\text{ } \right)\text{ }+\text{ }\left( x\text{ }\text{ }y\text{ } \right)\text{ }=\text{ }0\]

⇒

    \[\left( x\text{ }\text{ }y \right)\text{ }\left( x\text{ }+\text{ }y\text{ }+\text{ }1 \right)\text{ }=\text{ }0\]

⇒

    \[x\text{ }\text{ }y\text{ }=\text{ }0\]

[

    \[x\text{ }+\text{ }y\text{ }+\text{ }1~\]

cannot be zero because x and y are natural numbers

⇒

    \[x\text{ }=\text{ }y\]

So, f is one-one.

Surjectivity:

When

    \[x\text{ }=\text{ }1\]

    \[{{x}^{2}}~+\text{ }x\text{ }+\text{ }1\text{ }=\text{ }1\text{ }+\text{ }1\text{ }+\text{ }1\text{ }=\text{ }3\]

⇒

    \[{{x}^{2}}~+\text{ }x\text{ }+1\text{ }\ge \text{ }3\]

, for every x in N.

⇒ f(x) will not assume the values

    \[1\]

and

    \[2\]

.

So, f is not onto.

i

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