Inequalities Mathematical Reasoning (Paper 2) Quadratics Difficulty 4

TMUA practice: Necessary Sufficient, problem 5

A TMUA-calibre necessary sufficient problem at difficulty 4 of 5, with the full worked solution.

The question

The quadratic

f(x) = x^2 - 2px + q

has two distinct real roots, and the difference between the roots is greater than

2

and less than

4

— call this statement

(\ast)

. Which one of the following is the **necessary and sufficient** condition for

(\ast)

q < p^2 < q + 3

q + 1 < p^2 < q + 4\ \text{and}\ p > 0

q + 1 \le p^2 \le q + 4

q < p^2 - 1 < q + 3

q < p^2 - 1\ \text{and}\ p^2 < q + 4

The correct answer is highlighted. D

Worked solution

Roots: $p \pm \sqrt{p^2 - q}$ . Difference: $2\sqrt{p^2 - q}$ .

Real and distinct: $p^2 - q > 0$ . Difference between 2 and 4: $1 < \sqrt{p^2 - q} < 2$ , i.e. $1 < p^2 - q < 4$ , i.e. $q + 1 < p^2 < q + 4$ , i.e. $q < p^2 - 1$ AND $p^2 - 1 < q + 3$ .

So $q < p^2 - 1 < q + 3$ . Answer: D.