2

OPTIONS

A) $(-2,0)$

B) $(0,10)$

C) $(1,12)$

D) $(20,30)$

From the first part, it is clear that $$a+b+c=0$$ And $$49a+7b+c\in (50,60)$$ $$64a+8b+c\in (70,80)$$

Subtracting them $$15a+b\in (10,30)$$ While adding them gives $$113a+15b+2c\in (120,140)$$ $$111a+13b\in (120,140)$$

But we have to find the range of $$4a+2b+c$$ $$=3a+b$$

What should I do next?

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Aditya
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    Does "$a, b, c \in I$" mean that $a,b,c$ are integers? – mathlove Mar 17 '20 at 06:21
  • @mathlove yes, they are integers – Aditya Mar 17 '20 at 07:02
  • If the round brackets are inclusive, then since $48a+6b$ is an integer divisible by $6$, we get $48a+6b=54$ or $60$. Similarly, $63a+7b=70$ or $77$. Solving these gives $(a,b)=(2,-7),(1,2),(0,10),(1,1)$ implying $f(2)=-1,5,10,4$ respectively. – mathlove Mar 17 '20 at 07:33
  • @mathlove that doesn’t match any option. – Aditya Mar 17 '20 at 07:39
  • @mathlove As I've indicated in my answer comment, I've deleted my answer since I now believe the brackets were likely meant to be exclusive, so as your answer shows, $f(2) = -1$ is the only solution, with this matching option A). – John Omielan Mar 17 '20 at 07:55
  • @John Omielan : Oh, I realize that my deleted answer is correct. So, now I've just undeleted my answer. Thanks! – mathlove Mar 17 '20 at 09:20
  • @Aditya : If we assume that the brackets are exclusive, then the only correct option is A). See my answer. – mathlove Mar 17 '20 at 09:21

1 Answers1

3

This answer assumes that the round brackets are exclusive.

Since we have $$f(1)=0\iff c=-a-b$$ we get $$f(7)\in (50,60)\implies 48a+6b\in (50,60)$$ Since $48a+6b$ is an integer divisible by $6$, we get $$48a+6b=54\tag1$$ Similarly, we get $$f(8)\in (70,80)\implies 63a+7b\in (70,80)$$ implying $$63a+7b=77\tag2$$

It follows from $(1)(2)$ that $(a,b)=(2,-7)$.

Since we get $f(2)=-1$, the answer is option $\color{red}{A}$.

mathlove
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