Secret Specialist Business: Exam 2 Discussion

This is our post for teachers and students to discuss Specialist Exam 2 (not online). There are also posts for Methods Exam 1, Methods Exam 2 and Specialist Exam 1.

28 Replies to “Secret Specialist Business: Exam 2 Discussion”

  1. It was good to see a few *new* multiple choice questions that didn’t reward blatant button-pushing. Of particular note, I thought Q4, 5 and 8 were pretty nice (come on – tell me why I’m wrong…)

    Q9 though… I think I know what the intended answer was (B) but with the absence of any reference to a differential equation, I am a bit lost as to exactly what “curve” the question is actually referring to.

    It seemed a bit… odd. But I do not know enough about differential equations to know if this is an error or not (and I get the feeling some regular readers will soon be able to assist)

    1. I agree, RF. And I also liked Section A Q2. But Q8 was trivial with button pushing: Just let x = 2 and y = 1, say, calculate the powers and then choose the consistent option.

      Re: Q9. A simple diagram facilitates getting the DE \displaystyle \frac{dy}{dx} = \frac{y}{x-y}. Then you can either button push to get the slope field and choose the compatible option (trivial with Mathematica using StreamPlot), or you can use your brain and eliminate all the wrong options in the following way:

      \displaystyle y = 0 \implies \frac{dy}{dx} = 0 therefore only options B, C and E are viable.

      \displaystyle x = 0 and \displaystyle y > 0 \implies \frac{dy}{dx} = -1 therefore only option B remains.

      A couple of superficial observations (I haven’t had a chance to sit down and do the exam in detail):

      I disliked Q13 (“The vectors … will be linearly dependent when the value of \lambda is”) because well-taught students can just set the determinant of an appropriate 3×3 matrix equal to zero – absolutely no understanding is required.

      I thought Section B was *mostly harmless*. It’s been a long time between drinks when students last had to state the geometrical interpretation (Q2 (c)) of |z - u| = |z - v| as ‘the line that is the perpendicular bisector of the line segment joining z = u and z = v.’

      I thought Q3 was very interesting – parts (d) and (e) are clearly intended to ‘force’ teachers to include a generalisation in their SACs …

      1. I found Q8 to be pretty simple without choosing specific values, but yes, in this case it would be an easy way to get a correct answer.

        The geometry of the Argand diagram has been so much about circles and rays lately that seeing a perpendicular bisector was nice, although all my students will know this ad nauseum from Year 11.

        Overall, I felt Part B was a lot easier than previous years. Part of me wonders why (given what happened in Methods)

        Still not convinced MCQ9 actually works as a question, mainly because I’m not convinced that “a curve” and a solution to a DE have to be the same thing just because a slope field is given. Don’t have the mathematical knowledge to know where to begin to understand if the question is valid or not – hence I come to Marty’s blog!

      2. I should add that Q11 is an improper integral (I wonder if the exam setting panel even understand what these are since they keep appearing) but no calculation is required so I suppose it can go through to the wicket keeper.

        1. And a lucky thing no calculation was required. The integral doesn’t converge! Maybe that was deliberate so that students couldn’t compute the value with a CAS and then compare with the value from the options … (Still, it’s a cheap trick). Here is the question:

          Question 11
          With a suitable substitution \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2{x}}{\sec^2{x} - 3 \tan{x} + 1} dx can be expressed as

      3. I should add that Q11 is an improper integral (I wonder if the exam setting panel even understand what these are since they keep appearing) but no calculation is required so I suppose it can go through to the wicket keeper …

    2. Re: Q4

      If we’re going by the maximal capability of what any of the allowed CAS calculators can do (i.e Mathematica) then unfortunately, Mathematica can completely steamroll the question. The code below does just that:
      FullSimplify[f[g[x]], 0 < x < \[Pi]/2] // TrigReduce
      FunctionRange[{1/2 Sin[2 x], 0 < x < \[Pi]/2}, x, y]
      If we're talking about the TI-nspire, it will also simplify the function (albeit with a mod sign) and all that's left is to determine the range of the function, which is more or less trivial. Nothing really tested if any student just threw the expression into their CAS.

      Re: Q5
      I don't really have any issues here, it's a decent way of testing properties of complex numbers.

      Re: Q8
      Just a question here, would it be frowned upon if instead of manipulating the expression into one that relates to the original expression, a student chose to plug in arbitrary values of x and y?

      Re: Q9
      What JF said, although personally using StreamPlot requires some good eyesight… In case the derivation was not so obvious, you would do it like so:

          \[y-y_0 = \frac{dy}{dx} (x-x_0)\]

      Substitute y=0 then rearrange for x as follows:

          \[\frac{-y_0}{\frac{dy}{dx}} +x_0 = x\]

      Then finally equate that for the value at that point:

          \[\frac{-y_0}{\frac{dy}{dx}} +x_0 = y_0\]

      Finally, rearrange for \frac{dy}{dx} and you're ready to squint your eyes at 5 different options. I'm not really certain on the validity of the question myself however….

      1. Hey Sai,

        Q4 for MCQ would’ve been a nice composite functions question if it wasn’t trivialised by Mathematica/CAS. I solved the question in reading time (it caught my interest!) whereas my friends just brute forced it on the TI Inspire, LOL!

        I was surprised that the geometry of circles didn’t come up this time for Q5. I felt that this exam was very physics themed and would be nicer with more creative geometry.

    3. Please allow me to reiterate my objection to multiple choice questions. The purpose of an examination is to assess where students are up to in learning about the subject. Suppose that my answer to a multiple choice question is (C). What does that tell you about my learning?

      If (C) is the correct answer, can you infer that I understand what is being tested? Not really – I might have guessed, or I might have got to (C) because I eliminated the alternatives but have no idea why (C) was correct, or I might have meant (B) but accidentally wrote (C). In any case, my score is 1.

      If (C) is not the correct answer, can you infer that I don’t understand what is being tested? Not really – perhaps I made a mistake in a calculation, or I knew the correct answer and accidentally wrote (C). In any case, my score is zero.

      Now the response of experts in assessment is that you don’t assess a student’s learning about a topic by a single MC question. You should ask several MC questions on the same topic to get an understanding of the student’s learning on the topic. This process is called triangulation. But in an examination, one does not ask several questions on the same topic.

      What does (C) tell us? Thus endeth the lesson.

      1. TM, as far as the VCAA exams are concerned, I am in complete agreement with you. I hate them on the VCAA exam – they are purely a stupid cost saving measure. I’d like to see them wiped from the face of the Earth and get either a shorter Exam 2 (90 minutes) or two extra extended-response questions.

      1. The complex numbers question where the argument of u is pie on 4 and it asked for the equation of the line. I had y=x+1 and stated the correct domain but it wasn’t in the full like function format

  2. I’m curious how Section B Question 4d (asking if the drone makes contact with the aeroplane, giving reasons) will be marked – in particular, what will count as sufficient evidence.

    I imagine many students will just use CAS to solve when the horizontal components are equal and then to solve when the vertical components are equal; CAS gives approximate solutions only, but there is no common solution, so no collision. But since CAS only gives approximate solutions, will students be required to give evidence that there aren’t OTHER solutions that perhaps the CAS is not finding? For instance, perhaps by sketching appropriate graphs and showing the locations of zeroes.

    1. Good question. Who knows what pinheaded reasons will be required. However if common sense prevails (a huge assumption), then

      “use CAS to solve when the horizontal components are equal and then to solve when the vertical components are equal; CAS gives approximate solutions only, but there is no common solution, so no collision”

      will be sufficient to get the 3 marks:
      1 mark for both equations,
      1 mark for the approximate solutions to each equation,
      1 mark for a conclusion based on those solutions.
      And I suspect explicit inclusion of the domain 0 leq t leq 40 somewhere will be required.

      A CAS calculator does not warn that other solutions might be possible. Mathematica gives all of them (this is certainly expected from Mathematica and assesors won’t know who’s using Mathematica …)

      The plot of y = 450 – 150 Sin[pi*t/6] and y = 30*t for 0 leq t leq 40 is interesting – there is *nearly* a second solution which is very close to one of the solutions to 400 – 200 Cos[pi*t/6] = -t^2 + 40*t ….

      1. JF what about parts of question 5 where they ask for the time and distance in terms of k?? Are answers which are in terms of g acceptable or do we have to further simplify it??

  3. Question 2 (11 marks)

    Two complex numbers, \displaystyle u and \displaystyle v, are defined as u = -2 - i and v = -4 - 3i.

    b. Plot the points that represent \displaystyle u and \displaystyle v … on the Argand diagram below. (2 marks)

    d. i. Sketch the ray given by \displaystyle \text{Arg}(z - u) = \frac{\pi}{4} on the Argand diagram in part b. (1 mark)

    Hands up those who see the conundrum here. (If \displaystyle u was a cat it would surely be called Schrodinger).

    Incidentally, surely the preamble would have been much better as “Let two complex numbers u = -2 - i and v = -4 - 3i.” (But VCAA loves its bloated sentences).

    1. JF what made that part d. i. so annoying was that the ray plays absolutely no role in the rest of the question – it could just as well have been Arg(z – i) = pi/4 without any loss.

      I read part d. ii. – which strongly suggests that students need to write down the domain of the line – as a tacit admission that it’ll be difficult to tell from responses to d. i. if students have drawn the graph correctly, so hopefully this means students won’t be penalised for an ambiguous graph at the endpoint.

    2. JF, apart from part (d), I really liked this question because it was refreshingly different to the way loci of the complex plane have been examined in recent years (what with the areas of sectors and over-reliance on rays and circles).

      Sure, it isn’t anything a middling Specialist Unit 2 student couldn’t handle (but that could be said for any of the non-calculus parts to a certain extent). I just felt this was actually a decent question…

      …until part (d).

  4. So what do you reckon I’d get if I score 37/40 in exam 1 and 74-76/80 in exam 2.
    This is based on me being ranked 1 in a cohort with a sac average of 93%.

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