HSC 2015 Maths Marathon (archive) (1 Viewer)

InteGrand · May 18, 2015

Re: HSC 2015 2U Marathon

(Unless your emphasis was about 'areas' (the integral is exact, but this may not be the 'area').)

braintic · May 18, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
(Unless your emphasis was about 'areas' (the integral is exact, but this may not be the 'area').)

I'm not sure what content is in the current General course - I've never taught it. But Simpson's rule was in the old Maths in Society course (pre 2001).
But it wasn't connected with the number plane. Instead it was about approximating the areas of fields and paddocks, where you drew a 'baseline' (equivalent to the x-axis) and took regularly spaced measurements to the perimeter from each side of the baseline.
Many teachers taught that it gave the exact area if the boundary of the field was made up of parabolic arcs. This is nonsense, unless all the arcs are "concave outwards", which is of course impossible for a smooth boundary.

And I know many teachers in the 2 unit course refer to exact areas for parabolic arcs without making this clarification.
Yet most neglect to point out that Simpson's rule also happens to give exact values for integrals of cubics.

InteGrand · May 18, 2015

Re: HSC 2015 2U Marathon

braintic said:
Many teachers taught that it gave the exact area if the boundary of the field was made up of parabolic arcs. This is nonsense, unless all the arcs are "concave outwards".

Why isn't it possible for the area to be exact with some arcs concave up, and some concave down? I mean, it's unlikely to be true in practice, but theoretically we could have exact area. We could imagine the following: the x-axis is the base, we start at x = 0 and finish at x = 1. The measurements at x = 0, x = ½, and x = 1 could all be the same (say 1 metre), and the measurement at x = ¼ could be say 1.5 m, with a measurement of 0.5 m at x = ¾. Then we have one parabola being concave down, and the other concave up, and using Simpson's rule for each parabolic arc would give the exact area. (Probably an unrealistic fence structure, but still possible mathematically to get exact area.)

InteGrand · May 18, 2015

Re: HSC 2015 2U Marathon

And were the paddock questions similar to Q 3(d) from this HSC 2U paper?: http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-exam-mathematics.pdf (page 5)

braintic · May 18, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
Why isn't it possible for the area to be exact with some arcs concave up, and some concave down? I mean, it's unlikely to be true in practice, but theoretically we could have exact area. We could imagine the following: the x-axis is the base, we start at x = 0 and finish at x = 1. The measurements at x = 0, x = ½, and x = 1 could all be the same (say 1 metre), and the measurement at x = ¼ could be say 1.5 m, with a measurement of 0.5 m at x = ¾. Then we have one parabola being concave down, and the other concave up, and using Simpson's rule for each parabolic arc would give the exact area. (Probably an unrealistic fence structure, but still possible mathematically to get exact area.)

I'm not talking about concave up/down.
I'm talking about (for example) the area of the region between y=x^2 and the Y-axis. (ie. y = sqrt(x) and the x-axis)
That is, it gives the exact area on the convex side of the the parabola, but not on the concave side.

Edit: Oops ... I now see the confusion .... I'm saying it all wrong.
It's not about concave convex ... it's about whether the axis that bounds the region is parallel or perpendicular to the axis of the parabola.

InteGrand · May 18, 2015

Re: HSC 2015 2U Marathon

braintic said:
I'm not talking about concave up/down.
I'm talking about (for example) the area of the region between y=x^2 and the Y-axis. (ie. y = sqrt(x) and the x-axis)
That is, it gives the exact area on the convex side of the the parabola, but not on the concave side.

Edit: Oops ... I now see the confusion .... I'm saying it all wrong.
It's not about concave convex ... it's about whether the axis that bounds the region is parallel or perpendicular to the axis of the parabola.

Oh, I see.

Krossceeper1 · May 19, 2015

Re: HSC 2015 2U Marathon

Find the equations of the two circles (by solving simultaneously or otherwise) in the first quadrant which are tangent to the y axis, the x axis, and the line x + y = 2.

(Sorry I didn't answer a question, last 15 or so comments were just banter and I couldn't see an unanswered one.)

Flop21 · May 19, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
In that formula, n is the number of sub-intervals, you're right. Sub-intervals refers to how many points we are using. Basically, is our interval is [a, b], we can split it up evenly using the following x-values:

[x₀, x₁, x₂, ..., x_{n – 1}, x_n], where

x₀ = a (the starting point of our interval where we're integrating)

x_n = b (our ending point).

Each 'sub-interval' refers to the following intervals: [x₀, x₁], [x₁, x₂], ..., [x_{n – 1}, x_n]. As you can see, there are n sub-intervals, because we have a total of n + 1 points being used.

Since we have equal spacing, the width of each sub-interval is h = (total width of intervals)/(number of intervals) = (b – a)/n.

Now for Simpson's rule, we actually apply each approximation by spanning over two sub-intervals. E.g. if we had our original interval of integration as [0, 1], we'd need to split it up into at least two sub-intervals, e.g. via the partition {0,½,1} (here, we're using 3 x-values, and the two sub-intervals are [0, ½], [½, 1]).

Then what Simpson's rule would do is it essentially finds the unique quadratic (or line if the function we're integrating is linear) that goes through the points (0, f(0)), (½, f(½)), and (1, f(1)) and computes the integral of this quadratic (or line), and gives us this value as an approximation to the integral of our original function. So you can see that for Simpson's rule to work, we'll need at least 3 x-values. If we want more x-values, we can add two more (so 5 x-values), but we can't add just one more, since Simpson's rule requires a multiple of 2 to be the number of sub-intervals. E.g. if we tried using the x-values {0, ⅓, ⅔, 1} (i.e. 4 x values, which is 3 sub-intervals), our Simpson's rule would first be applied over the interval [0, ⅔] (using the y-values at 0, ⅓, and ⅔), but for the next interval, we couldn't use Simpson's rule, since we don't have 3 remaining x values.

So in general, for Simpson's rule, n (the no. of sub-intervals) must be even, because we use two sub-intervals for each 'use' of Simpson's rule.

(Sorry if that's hard to understand without a picture. Here's a picture: http://www.mathwords.com/s/simpsons_rule.htm ; Simpson's rule would be used on the interval [x₀, x₂] (as 3 x-values are used), and then the next one would use the interval [x₂, x₄], so even number of sub-intervals required (a sub-interval is just between adjacent x values)

So if we had 5 x values to use {x₀ = a, x₁, x₂, x₃, x₄ = b}, we'd have n = 4, and Simpson's rule would be used on [x₀, x₂] and [x₂, x₄], and we'd sum up the results. (By using Simpson's rule on these intervals, I mean the rule for using it when there's just 3 values.) If you do this, you see that the 'odd' x's (i.e. the x's with odd-numbered subscripts) get counted four times, and the ones with even subscripts get counted twice (and the end points of the entire interval get counted once).

Thanks!

I really appreciate it. I'm going to read over this again later to fully understand it.

I think I got the right h in the test, yay. But see that's what my problem was, I couldn't remember how many function values you needed for Simpson's rule, and therefore didn't know n.

Ambility · May 22, 2015

Re: HSC 2015 2U Marathon

Krossceeper1 said:
Find the equations of the two circles (by solving simultaneously or otherwise) in the first quadrant which are tangent to the y axis, the x axis, and the line x + y = 2.

(Sorry I didn't answer a question, last 15 or so comments were just banter and I couldn't see an unanswered one.)

What? I can only think of one possible circle which is tangent to all three lines and in the first quadrant...

InteGrand · May 22, 2015

Re: HSC 2015 2U Marathon

Ambility said:
What? I can only think of one possible circle which is tangent to all three lines and in the first quadrant...

There can be one inside the triangle formed by the line and the axes, and also one outside.

InteGrand · May 22, 2015

Re: HSC 2015 2U Marathon

Krossceeper1 said:
Find the equations of the two circles (by solving simultaneously or otherwise) in the first quadrant which are tangent to the y axis, the x axis, and the line x + y = 2.

(Sorry I didn't answer a question, last 15 or so comments were just banter and I couldn't see an unanswered one.)

$It is clear that the circles' centres will need to lie on the line $y=x$, so we can write the circles as $(x-r)^2 + (y-r)^2 = r^2$ (since the axes are tangent) and seek two positive values for $r$.$

$Since the line $y=-x+2$ is tangent to the circle, we require a discriminant of 0 when we substitute the equation of the line into that of the circle. Doing this substitution, we have at the point where the line is tangent to the circle: $(x-r)^2 + ((-x+2)-r)^2 = r^2$. Expansion and simplification yields $2x^2-4x+(r-2)^2=0$. Setting the discriminant to 0 yields $(-4)^2 - 4(2)(r-2)^2 = 0\Rightarrow (r-2)^2 =2 \Rightarrow r-2=\pm\sqrt{2}\Rightarrow r=2\pm\sqrt{2}$.$

leehuan · May 29, 2015

Re: HSC 2015 2U Marathon

Radian measure was used because I couldn't find the degree symbol.

$(a) P, Q and R are collinear. What is the value of $\angle $PQR? \\ (b) Prove that the angle sum of any triangle is always equal to $ \pi. \\ $(c) Deduce that the angle sum of any quadrilateral is always equal to 2$\pi. \\$ (d) Hence, explain why the angle sum of any polygon is $\pi$($n$-2)$

For the sake of making (b) easier:

Use parallel lines

astroman · May 30, 2015

Re: HSC 2015 2U Marathon

help
$\int_{\frac{1}{2}}^{1} cos\frac{\pi }{2}x dx$

InteGrand · May 30, 2015

Re: HSC 2015 2U Marathon

astroman said:
help
$\int_{\frac{1}{2}}^{1} cos\frac{\pi }{2}x dx$

$We use the rule $\int \cos ax \text{ d}x=\frac{1}{a}\sin ax +C$ (alternatively, we could substitute $u=\frac{\pi}{2}x$, but 2U students aren't expected to know $u$-substitution).$

$\int _\frac{1}{2} ^1 \cos \frac{\pi}{2}x\text{ d}x = \left[ \frac{1}{\frac{\pi}{2}}\sin \frac{\pi}{2}x\right] ^1 _{\frac{1}{2}}$

$=\frac{2}{\pi}\left(\sin \frac{\pi}{2} - \sin \frac{\pi}{4}\right)$

$=\frac{2}{\pi}\left(1 - \frac{\sqrt{2}}{2}\right)$

$$$=\frac{2-\sqrt{2}}{\pi}$$.$

keepLooking · Jun 2, 2015

Re: HSC 2015 2U Marathon

I'm not sure if I'm retarded but this is Q2 from Baulko 2011 2U Trials.

How exactly do I find the answer of vi?

In the solutions, it says

To find the point from M, move <- 6 units, ^ 3units. P(-3, 4)

How does he come to this conclusion? .__.

Sy123 · Jul 25, 2015

Re: HSC 2015 2U Marathon

$\\ $Consider the function$ \ f(x) = \tan x - x, \ 0 \leq x < \frac{\pi}{2} \\ \\ $i) Show that$ \ f(x) \ $has exactly one stationary point, and that it is a minimum$ \\ \\ $ii) Hence explain why$ \ f(x) \geq 0 \\ \\ $iii) Hence show that$ \ \frac{1}{\cos x} > \frac{x}{\sin x} \ $for$ \ 0 < x < \frac{\pi}{2}$

$\\ $iv) Explain with the use of a diagram or otherwise, that$ \ x > \sin x \ $for$ \ x > 0$

$\\ $v) Hence finally show that$ \ 1 < \frac{x}{\sin x} < \frac{1}{\cos x} \ $for$ \ 0 < x < \frac{\pi}{2}$

Sy123 · Jul 25, 2015

Re: HSC 2015 2U Marathon

Flop21 · Aug 22, 2015

Re: HSC 2015 2U Marathon

Find the range of the function y= 2 [(square root) 25-x^2]

rand_althor · Aug 22, 2015

Re: HSC 2015 2U Marathon

$y=\sqrt{25-x^2}$ is in the form $y=\sqrt{r^2-x^2}$, which is the equation of a semicircle with radius $r$ units and centre $(0,0)$. So $y=\sqrt{25-x^2}$ is a semicircle with radius 5 units. For this function the range is: $0 \leq y \leq 5$. For $y=2\sqrt{25-x^2}$, we double this range to get $0 \leq y \leq 10$.$

Drsoccerball · Aug 22, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
Find the range of the function y= 2 [(square root) 25-x^2]

One way to do this is by noticing what values of x give you the max and min value for y.
This case if x=0 that means it is at a max since you're subtracting x^2.
Therefore max value = 2 times 5 =10
Similarly with min when x=5 y=0
Therefore range $0 \leq y \leq 10$

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