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Question Number 12632 by @ANTARES_VY last updated on 27/Apr/17

$$\mathbf{y}=\frac{{\mathbf{x}}^2-4\mathbf{x}+9}{{\mathbf{x}}^2-4\mathbf{x}+5}\mathbf{the}\mathbf{values}\mathbf{of}\mathbf{the}$$$$\mathbf{feture}\mathbf{set}\mathbf{of}\mathbf{prime}\mathbf{numbers}?$$

Answered by abcd last updated on 27/Apr/17

Commented by prakash jain last updated on 28/Apr/17

$$\mathrm{There}\mathrm{is}\mathrm{a}\mathrm{mistake}\mathrm{in}\mathrm{calculation}$$$$\mathrm{of}\mathrm{determinant}\mathrm{above}.$$$$\mathrm{Discriminant}$$$$\mathrm{D}=b^2-4ac$$$$=4^2{(y-1)}^2-4(y-1)(5y-9)$$$$=4(y-1)(-y+5)$$$$=-4(y-1)(y-5)$$$$D⩾0\Rightarrow 1⩽y⩽5$$$$y=1\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{original}$$$$\mathrm{function}\mathrm{except}\mathrm{in}\mathrm{limiting}\mathrm{case}$$$$\mathrm{of}x\to \mathrm{\infty }$$$$1<y⩽5$$

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