In this post I review Part 2 of the examination paper and provide worked solutions. To view my post on Part 1 of the examination, click here. This portion of the test consisted of 9 questions, worth a total of 70 marks, to be solved within a 70 minute time limit, and a calculator was permitted. That meant students had to average 7 minutes 46 seconds per question, or 1 minute per mark in order to finish.
This problem requires a bit of ingenuity, but doesn't require the application of any advanced skills. To find the solution the student had to remember the definitions of altitude and median when referring to the geometry of triangles, as well as how to define the equation of a line and do some algebraic manipulation to find a point of intersection. Easy if the student could see how to solve it upon reading the question, but impossible if he or she forgot a relevant detail like what median meant.
|Solution to Problem 1|
This was pretty easy problem involving factoring and expanding a second-order polynomial. Most students should have no issues solving this.
|Solution to Problem 2|
Part (a) of the problem is very easy, especially considering that calculators were permitted on the exam. But it's also only worth 1 mark. The meat of the problem is in part (b), where the student has to determine if the unfortunate animals could escape the well. This too is actually quite easy solve though because recurrence relations modelling the problem are provided and students had a calculator. Even using "brute force" to iteratively calculate each animal's progress over time, a student should only need a few minutes to find that the frog cannot escape and the toad could escape on the 12th day. However, it would have saved time and demonstrated a higher degree of mastery if the student recognized that, due to how the problem is defined, there is a maximum height each animal could theoretically climb to. That maximum height occurs when the distance climbed during the day exactly equals the distance slid overnight. Using that approach, the student would find that the frog cannot climb past 48 feet and the toad cannot climb past 52 feet.
|Solution to Problem 3|
This problem requires the student to solve a definite integral. The functions involved here are easy to integrate so this question shouldn't have causeed too many hangups. I don't know what sort of calculator was permitted on this exam, but many scientific calculators can solve definite integrals, which would have made this problem even easier. There are two areas here that need to be calculated (one on either side of the line of symmetry). But because of symmetry, there's no need to do two calculations because both areas are equal. However, a student who didn't recognize that simplification could still plough ahead and solve both integrals. It would waste a few minutes but could probably still be done within the 1 minute per mark allotment.
|Solution to Problem 4|
I didn't find this question difficult either, though I found it to be time consuming and went well over the targeted 1 minute per mark on it. I have a feeling that maybe I missed something that would've made this problem easier to solve. At any rate, the student here would have to recognize distance between the centres of C1 and C2 has to equal the sum of the radii R1 and R2 in order to solve part (a). To solve part (b), the student would have to recognize that the R3 would be equal to R1 + R2 and then find where center of C3 would have to lie in order to just touch both C1 and C2.
|Solution to Problem 5|
Not a particularly difficult problem, though the student did need to recall that the angles in an equilateral triangle are 60 degrees. Not having that detail would have cost a lot of marks. Otherwise, the problem involves some simple vector algebra.
|Solution to Problem 6|
The formula for the integral of the cosine function was provided in the formulae list that was included with the test, so part (a) wasn't difficult at all. Part (b) required use of trigonometric identities to prove the two expressions were equivalent. Double angle identities were provided with the formulae list included with the test and using those alone part (b) could have been solved, though some students may have encountered some difficulty here. The problem is easier to solve if the student also remembered the Pythagorean trigonometric identity. Part (c) is very easy if the student demonstrated a bit of ingenuity, using the information in Part (b) to show that the integral in (c) is really -1/2 of the integral in (a). Failing that, the student could still have solve this integral, though it's only worth 2 marks and solving this one the hard way would likely have been a big waste of time.
|Solution to Problem 7|
Part (a) of the problem was very easy. Part (b) required the student to find a local minimum, which means determining where the first derivative of the function is zero. It was not necessary to verify that the local extreme value was a local minimum here because the problem statement tells you it's a minimum. The derivative here wasn't very difficult, but it did require an application of the chain rule so it wasn't a fluff question either. Again, I'm not sure what sort of calculator students were permitted to use, but many scientific calculators can perform root solving. Finding the value of x that makes the derivative equal to zero could have been simplified with a calculator if that functionality was available. However, it was still easily possible to solve this problem within the 1 minute per mark targeted time allotment without the aid of a root solver.
|Solution to Problem 8|
I thought this was probably one of the most difficult questions on the exam. The student had to use an angle sum identity to help get started on a solution to the values of k and a. The student also had to remember that a calculator only gives one solution when using the inverse trigonometric functions, though there are infinite possible solutions. The first solution represents the point where the tip of the blade is moving up towards its maximum height. The second solution represents the point where the tip of the blade is on the way back down.
|Solution to Problem 9|