Consider the statement: x( - x) < y( - y) for all x, y with 0 < x < y < 1 . The statement is true

Let f(t) = t( - t) = t - t^2 . The statement says f(x) 0 on (0,1) , so f is strictly increasing and the statement holds. - If < 2 : pick t close to 1 ; then f'(t) = - 2t < 0 , so f decreases somewhere in (0,1) and the statement fails. Hence the statement is true if and only if 2 . Answer: A.

Inequalities Difficulty 4

TMUA practice: Inequalities, problem 3

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

The question

Consider the statement:

x(\alpha - x) < y(\alpha - y)

for all

x, y

with

0 < x < y < 1

. The statement is true

\text{iff } \alpha \ge 2

\text{iff } \alpha > 2

\text{iff } \alpha < -1

\text{for no value of } \alpha

The correct answer is highlighted. A

Worked solution

Let $f(t) = t(\alpha - t) = \alpha t - t^2$ . The statement says $f(x) < f(y)$ whenever $0 < x < y < 1$ — that is, $f$ is strictly increasing on $(0, 1)$ .

$f'(t) = \alpha - 2t.$

$f$ is increasing on $(0,1)$ exactly when $f'(t) \ge 0$ throughout, and since $t$ ranges up to (but not including) $1$ , this needs

$\alpha \ge 2t \ \text{ for all } t \in (0,1) \iff \alpha \ge 2.$

If $\alpha \ge 2$ : $f'(t) = \alpha - 2t \ge 2 - 2t > 0$ on $(0,1)$ , so $f$ is strictly increasing and the statement holds.
If $\alpha < 2$ : pick $t$ close to $1$ ; then $f'(t) = \alpha - 2t < 0$ , so $f$ decreases somewhere in $(0,1)$ and the statement fails.

Hence the statement is true if and only if $\alpha \ge 2$ .

Answer: A.