Question

Show that the function f : N → N defined by f(x) = x² + x + 1 is one-one but not onto. Find the inverse of f : N → S, where S is range of f.

Answer 1

Given: The function f : N → N defined by f(x) = x² + x + 1

To show f is one-one:

$$\begin{gathered} \mathrm{Let}f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow x_1^2+x_1+1=x_2^2+x_2+1 \\ \Rightarrow x_1^2+x_1=x_2^2+x_2 \\ \Rightarrow x_1^2+x_1-x_2^2-x_2=0 \\ \Rightarrow x_1^2-x_2^2+x_1-x_2=0 \\ \Rightarrow \left(x_1-x_2\right)\left(x_1+x_2\right)+\left(x_1-x_2\right)=0 \\ \Rightarrow \left(x_1-x_2\right)\left(x_1+x_2+1\right)=0 \\ \Rightarrow x_1-x_2=0\mathrm{or}{x}_1+x_2+1=0 \\ \Rightarrow x_1=x_2\mathrm{or}{x}_1=-\left(x_2+1\right) \\ \Rightarrow x_1=x_2\left(\because x_1,x_2\in N\right) \\ \mathrm{Hence},f\mathrm{is}\mathrm{one}-\mathrm{one}. \\ \mathrm{To}\mathrm{show}\mathit{}f\mathrm{is}\mathrm{not}\mathrm{onto}: \\ \mathrm{Since}f\left(x\right)=x^2+x+1 \\ \therefore f\left(1\right)=3 \\ f\left(2\right)=7 \\ f\left(3\right)=13 \\ \mathrm{and}\mathrm{so}\mathrm{on} \\ \mathrm{Thus},\mathrm{Range}\mathrm{of}f=\left\{3,7,13,...\right\}\ne N \\ \mathrm{Hence},f\mathrm{is}\mathrm{not}\mathrm{onto}. \\ \mathrm{Now},\mathrm{Let}f:N\to \mathrm{Range}\mathrm{of}f \\ y=x^2+x+1 \\ \Rightarrow x^2+x+1-y=0 \\ \Rightarrow x^2+x+\left(1-y\right)=0 \\ \Rightarrow x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(1-y\right)}}{2\left(1\right)} \\ \Rightarrow x=\frac{-1\pm \sqrt{1-4+4y}}{2} \\ \Rightarrow x=\frac{-1\pm \sqrt{4y-3}}{2} \\ \Rightarrow x=\frac{-1+\sqrt{4y-3}}{2}\mathrm{or}x=\frac{-1-\sqrt{4y-3}}{2} \\ \Rightarrow x=\frac{-1+\sqrt{4y-3}}{2}\left(\because x\in N\right) \\ \mathrm{Hence},f^{-1}\left(x\right)=\frac{-1+\sqrt{4x-3}}{2}. \end{gathered}$$