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tmua.me Extreme Set B Paper 2 · Q2

Paper 2Number & countingAlgebra

The quadratic polynomial

q(x)=ax2+bx+cq(x)=ax^2+bx+c

has integer coefficients, with a>0a>0. It satisfies q(n)>0q(n)>0 for every integer nn, but q(x)<0q(x)<0 for at least one real number xx.

A student writes

q(x)=a(x−t)2+h,q(x)=a(x-t)^2+h,

where h<0h<0, and chooses an integer nn nearest to tt, so that ∣n−t∣≤1/2|n-t|\leq1/2.

Which of the following arguments validly rules out every integer aa with 1≤a≤41\leq a\leq4?

ASince q(n)>0q(n)>0, we have h=q(n)−a(n−t)2>−a/4h=q(n)-a(n-t)^2>-a/4. If a≤4a\leq4, this implies h≥0h\geq0, a contradiction.
BBecause the coefficients are integers, hh is an integer. Thus h≤−1h\leq-1, so if a≤4a\leq4 then q(n)≤a/4−1≤0q(n)\leq a/4-1\leq0, a contradiction.
CFor an integer nearest to tt, we always have ∣n−t∣=1/2|n-t|=1/2. Thus 1≤q(n)=a/4+h<a/41\leq q(n)=a/4+h<a/4, so a>4a>4.
DThe number q(n)q(n) is a positive integer, so q(n)≥1q(n)\geq1. Hence h=q(n)−a(n−t)2≥1−a/4h=q(n)-a(n-t)^2\geq1-a/4. If a≤4a\leq4, then h≥0h\geq0, a contradiction.
EThe inequalities q(n)≥1q(n)\geq1 and (n−t)2≤1/4(n-t)^2\leq1/4 give h≤1−a/4h\leq1-a/4. Since h<0h<0, this forces 1−a/4<01-a/4<0, so a>4a>4.
FSince (n−t)2≤1(n-t)^2\leq1 and q(n)≥1q(n)\geq1, we have h≥1−ah\geq1-a. If a≤4a\leq4, then h≥−3h\geq-3, contradicting h<0h<0.

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