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Let-f-x-ax-2-bx-c-where-a-b-c-are-real-numbers-If-the-numbers-2a-a-b-and-c-are-all-integers-then-the-number-of-integral-values-between-1-and-5-that-f-x-can-take-is-

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Question Number 21234 by Tinkutara last updated on 17/Sep/17
Let f(x) = ax^2 + bx + c, where a, b, c are real numbers. If the numbers 2a, a + b, and c are all integers, then the number of integral values between 1 and 5 that f(x) can take is
Letf(x)=ax2+bx+c,wherea,b,carerealnumbers.Ifthenumbers2a,a+b,andcareallintegers,thenthenumberofintegralvaluesbetween1and5thatf(x)cantakeis
Answered by Tinkutara last updated on 17/Sep/17
f(0)=c∈Z f(1)=a+b+c f(−1)=a−b+c Since 2a,a+b,c are integers, so 2a−(a+b)+c=a−b+c∈Z⇒f(−1)∈Z Similarly (a+b)+c=f(1)∈Z. f(1)−f(−1)=2b∈Z f(2)=4a+2b+c∈Z f(3)=9a+3b+c=a+b+c+8a+2b∈Z f(4)=16a+4b+c∈Z f(5)=25a+5b+c=a+b+c+24a+4b∈Z ∴ f(x) has 5 integral values in [1,5].
f(0)=c∈Zf(1)=a+b+cf(−1)=a−b+cSince2a,a+b,careintegers,so2a−(a+b)+c=a−b+c∈Z⇒f(−1)∈ZSimilarly(a+b)+c=f(1)∈Z.f(1)−f(−1)=2b∈Zf(2)=4a+2b+c∈Zf(3)=9a+3b+c=a+b+c+8a+2b∈Zf(4)=16a+4b+c∈Zf(5)=25a+5b+c=a+b+c+24a+4b∈Z∴f(x)has5integralvaluesin[1,5].

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