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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.

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Solution

Given: The function f : N → N defined by f(x) = x2 + x + 1 To show f is one-one: $\mathrm{Let}f\left({x}_{1}\right)=f\left({x}_{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}_{1}^{2}+{x}_{1}+1={x}_{2}^{2}+{x}_{2}+1\phantom{\rule{0ex}{0ex}}⇒{x}_{1}^{2}+{x}_{1}={x}_{2}^{2}+{x}_{2}\phantom{\rule{0ex}{0ex}}⇒{x}_{1}^{2}+{x}_{1}-{x}_{2}^{2}-{x}_{2}=0\phantom{\rule{0ex}{0ex}}⇒{x}_{1}^{2}-{x}_{2}^{2}+{x}_{1}-{x}_{2}=0\phantom{\rule{0ex}{0ex}}⇒\left({x}_{1}-{x}_{2}\right)\left({x}_{1}+{x}_{2}\right)+\left({x}_{1}-{x}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left({x}_{1}-{x}_{2}\right)\left({x}_{1}+{x}_{2}+1\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}_{1}-{x}_{2}=0\mathrm{or}{x}_{1}+{x}_{2}+1=0\phantom{\rule{0ex}{0ex}}⇒{x}_{1}={x}_{2}\mathrm{or}{x}_{1}=-\left({x}_{2}+1\right)\phantom{\rule{0ex}{0ex}}⇒{x}_{1}={x}_{2}\left(\because {x}_{1},{x}_{2}\in N\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},f\mathrm{is}\mathrm{one}-\mathrm{one}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{show}\mathit{}f\mathrm{is}\mathrm{not}\mathrm{onto}:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}f\left(x\right)={x}^{2}+x+1\phantom{\rule{0ex}{0ex}}\therefore f\left(1\right)=3\phantom{\rule{0ex}{0ex}}f\left(2\right)=7\phantom{\rule{0ex}{0ex}}f\left(3\right)=13\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{so}\mathrm{on}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Range}\mathrm{of}f=\left\{3,7,13,...\right\}\ne N\phantom{\rule{0ex}{0ex}}\mathrm{Hence},f\mathrm{is}\mathrm{not}\mathrm{onto}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{Let}f:N\to \mathrm{Range}\mathrm{of}f\phantom{\rule{0ex}{0ex}}y={x}^{2}+x+1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x+1-y=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x+\left(1-y\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{{1}^{2}-4\left(1\right)\left(1-y\right)}}{2\left(1\right)}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{1-4+4y}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{4y-3}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1+\sqrt{4y-3}}{2}\mathrm{or}x=\frac{-1-\sqrt{4y-3}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1+\sqrt{4y-3}}{2}\left(\because x\in N\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{f}^{-1}\left(x\right)=\frac{-1+\sqrt{4x-3}}{2}.$

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