2022 UPDATE: PLEASE NOTE FOR 2022 APPLICATIONS THE ECAA IS NOT BEING USED. INSTEAD STUDENTS ARE BEING ASKED TO TAKE THE TMUA ENTRANCE TEST
This contains a worked guide to the ECAA 2020 questions.
For the workings for the ECAA 2020 maths questions, please see the document linked below:
Note that these answers are brief, do not provide full explanations and may be subject to errors. You are welcome to let me know if you spot an error.
For the corresponding questions, the link to the ECAA 2020 paper is here.
Below, for a selection of some of the more difficult questions, I have provided more detailed explanations below of the methods required.
Question 25: Wall around a Field
Firstly, one way to view this problem is as a constrained optimisation problem. See the article here for more information on this type of problem.
Let the length WZ be \(x\) and length WX be equal to \(y\).
The total length of the wall is then
\[5x+2y=260\]
The area of P is a quarter of the area of the rectangle, that is \(\frac{xy}{4}\).
Next, we must maximise the area subject to the constraint on the total wall length. This is a two-variable problem, so we transform this into a one-variable problem by substituting the constraint on the total wall length into the area:
\[y=\frac{260-5x}{2}\]
\[Area = \frac{x}{4} \frac{260-5x}{2} = \frac{x(260-5x)}{8}\]
Then to maximise the area, we differentiate with respect to \(x\) and set the derivative to zero:
\[\frac{d(Area)}{dx}=\frac{1}{8}(260-10x)=0\]
\[x=26\]
The length of WZ is 26 metres.
Question 34: Turning Points
Firstly, to find turning points, we can differentiate with respect to \(x\) and set the derivative equal to zero:
\[\frac{dy}{dx}=3x^2+6\sqrt{5}px+3p=0\]
\[x^2+2\sqrt{5}px+p=0\]
Now, for there to be two distinct turning points, there must be two distinct solutions to this quadratic equation. This requires the discriminant (from the quadratic formula: \(b^2-4ac\)) to be strictly positive, in other words:
\[20p^2-4p>0\]
\[4p(5p-1)>0\]
Then to solve the quadratic equation, note that \(p=0\) and \(p=\frac{1}{5}\) are the critical values.
There are a few ways to see that if \(p\) lies between \(0\) and \(\frac{1}{5}\), the inequality does not hold, but does hold for other values of \(p\). This could be by drawing a graph, by realising the quadratic function \(4p(5p-1)\) is a positive quadratic function in itself, or by observing the expression is negative if \(p\), when \(p\) is between \(0 \) and \(\frac{1}{5}\).
Finally, therefore the possible values of \(p\) are:
\[p<0, \enspace p>\frac{1}{5}\]