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Question Number 21234 by Tinkutara last updated on 17/Sep/17

$$\mathrm{Let}f\left(x\right)={ax}^2+bx+c,\mathrm{where}a,b,c$$$$\mathrm{are}\mathrm{real}\mathrm{numbers}.\mathrm{If}\mathrm{the}\mathrm{numbers}2a,$$$$a+b,\mathrm{and}c\mathrm{are}\mathrm{all}\mathrm{integers},\mathrm{then}\mathrm{the}$$$$\mathrm{number}\mathrm{of}\mathrm{integral}\mathrm{values}\mathrm{between}1$$$$\mathrm{and}5\mathrm{that}f\left(x\right)\mathrm{can}\mathrm{take}\mathrm{is}$$

Answered by Tinkutara last updated on 17/Sep/17

$$f\left(0\right)=c\in Z$$$$f\left(1\right)=a+b+c$$$$f(-1)=a-b+c$$$$Since2a,a+b,careintegers,so$$$$2a-(a+b)+c=a-b+c\in Z\Rightarrow f(-1)\in Z$$$$Similarly(a+b)+c=f\left(1\right)\in Z.$$$$f\left(1\right)-f(-1)=2b\in Z$$$$f\left(2\right)=4a+2b+c\in Z$$$$f\left(3\right)=9a+3b+c=a+b+c+8a+2b\in Z$$$$f\left(4\right)=16a+4b+c\in Z$$$$f\left(5\right)=25a+5b+c=a+b+c+24a+4b\in Z$$$$\therefore f\left(x\right)has5integralvaluesin[1,5].$$

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