Moderate -0.5 This is a straightforward proof by contradiction with the setup already provided. Students need only rearrange to get a quadratic inequality, recognize it has no real solutions (negative discriminant), reaching the contradiction. It's easier than average as the method is prescribed and the algebra is routine, though proof questions require more than pure recall.
Prove by contradiction the following proposition.
When \(x\) is real and positive,
$$4x + \frac{9}{x} \geq 12.$$
The first line of the proof is given below.
Assume that there is a positive and a real value of \(x\) such that
$$4x + \frac{9}{x} < 12.$$ [3]
Prove by contradiction the following proposition.
When $x$ is real and positive,
$$4x + \frac{9}{x} \geq 12.$$
The first line of the proof is given below.
Assume that there is a positive and a real value of $x$ such that
$$4x + \frac{9}{x} < 12.$$ [3]
\hfill \mbox{\textit{WJEC Unit 3 Q15 [3]}}