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Prove that the function $f: N \rightarrow N$ , defined by $f(x) = x^2 + x + 1$ is one-one but not onto.

Find inverse of $f: N \rightarrow S$ , where S is range of f.

Answers (1)

Given $f: N \rightarrow N$ natural numbers

$f(x) = x^2 + x + 1$

$\\f'(x) = 2x + 1 \\ x = \frac{-1}{2}$

For one-one:

Let $x_1,x_2\in N \qquad f(x_1) = f(x_2)$

$\\x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1 \\ x_1^2 - x_2^2 +x_1 - x_2 = 0 \\ (x_1 - x_2)[x_1 -x_2 +1] = 0 \\ x_1 - x_2 = 0 \quad [\therefore x_1 - x_2 + 1 \neq 0] \\\Rightarrow x_1 = x_2$

It is one-one

For Onto: Let $y = 1 \in N\ (Codomain)$

$\\x^2 + x + 1 = 1 \\ x^2 + x = 0 \\ x(x+1) = 0 \therefore x = 0,-1\notin N$

So, f is not onto. Hence proved.

Now: $f:N \rightarrow S$ where S (range of f given) $f:N \rightarrow S$ is onto. Hence f is inversible.

Let $y = x^2 + x +1$

$y = x^2 + 2x\cdot\frac{1}{2} + \left (\frac{1}{2}\right)^2 - \left (\frac{1}{2}\right)^2 + 1$

$y = \left(x + \frac{1}{2} \right )^2 + \frac{3}{4}$

$y- \frac{3}{4} = \left(x + \frac{1}{2} \right )^2$

$\frac{4y -3}{4} = \left(x + \frac{1}{2} \right )^2$

$\Rightarrow \pm \frac{\sqrt{4y -3}}{2} = x + \frac{1}{2}$

$\therefore \pm \frac{\sqrt{4y -3}}{2} = \frac{2x + 1}{2}$

$\therefore x = \frac{-1 \pm\sqrt{4y -3}}{2}$

$\Rightarrow x = \frac{-1 \pm\sqrt{4y -3}}{2} \qquad x\in N$

Posted by

Ravindra Pindel

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